p
fβ Y β ,
Fp (ϕ) =
|β|6p
X
fβ ϕβ .
|β|6p
Then f = Ep + Fp , f (ϕ) = Ep (ϕ) + Fp (ϕ), and Ep (ϕ) ∈ M p+1 and hence, ˜ p )(ϕ) and ∂Ep (ϕ) belong d(Ep (ϕ)) ∈ M p . Consequently, the elements d(E ∂yi p to M , for every i = 1, . . . , s. Since Fp is a polynomial in R[y1 , . . . , ys ], we have (see Section 1.2): ˜ p )(ϕ) − Ps d(Fp (ϕ)) − d(F i=1
∂Fp ∂yi (ϕ)d(ϕi )
= 0.
14
Part I. Preliminary concepts
Therefore, the element A, defined by (1.2), is equal to ˜ p )(ϕ) − Ps d(Ep (ϕ)) − d(E i=1
∂Ep ∂yi (ϕ)d(ϕi ),
and so, it is an element of M p . Corollary 1.5.4. Let d be a k-derivation of k[[T ]], let f ∈ k[[y1 , . . . , ys ]] and let ϕ = (ϕ1 , . . . , ϕs ) ∈ k[[T ]]s . Assume that the series ϕ1 , . . . , ϕs have no constant terms. Then f (ϕ) ∈ k[[T ]] and s X ∂f d(f (ϕ)) = (ϕ)d(ϕi ). ∂yi i=1
Assume now that {Bα }α∈Ω is a family of k[[T ]]. Then there Pof elements α exists an element of k[[T ]] of the form α∈Ω Bα TP. This element is the α limit of the Cauchy sequence (Cp ), where Cp = |α|6p Bα T . If d is a derivation of k[[T ]], then {d(Bα )}α∈Ω isPa family of elements in k[[T ]] and hence, we have the the form α∈Ω d(Bα )T α . Moreover, we have Pelement of α also the element α Bα d(T ). P P P Proposition 1.5.5. d( α Bα T α ) = α d(Bα )T α + α Bα d(T α ). Proof. Use the same argument as in the proof of Proposition 1.5.3.
1.6
Systems of differential equations
In this section we prove two theorems concerning formal solutions of systems of ordinary differential equations which will be useful in the next chapters of this paper. Throughout this section k is a ring containing Q, k[X] = k[x1 , . . . , xn ] is the polynomial ring and k[[t]], k[X][[t]] are power series rings over k and k[X], respectively. The following theorem is a power series version of the well known Cauchy and Picard theorem from the theory of ordinary differential equations (see for example [90] or [106]). Theorem 1.6.1. Let f1 , . . . , fn ∈ k[X][[t]] and a1 , . . . , an ∈ k. Then there exist unique series ϕ1 , . . . , ϕn ∈ k[[t]] such that: i (1) ∂ϕ ∂t = fi (ϕ1 , . . . , ϕn ), for i = 1, . . . , n, and (2) the constant terms of ϕ1 , . . . , ϕn are equal to a1 , . . . , an , respectively.
Chapter 1. Definitions, notations and basic facts
15
For the proof of this theorem we need some notations and lemmas. Assume B0 , B1 , . . . is a sequence of series belonging to k[[t]] and put Pp that i Cp = i=1 Bi t for any p ∈ N0 . Since Cp+1 − Cp ∈ (t)p , thesePseries form a p Cauchy sequence in k[[t]] and hence, there exists the series ∞ p=0 Bp t . It is easy to check the following P∞ P∞ p = i Lemma 1.6.2. If Bp P = p=0 Bp t i=0 bpi t with bpj ∈ k, then P ∞ s s=0 as t , where as = i+j=s bij for any s ∈ N0 . Let f be a series inP k[X][[t]] and let ϕ1 , . . . , ϕn ∈ k[[t]]. Set ϕ = p (ϕ1 , . . . , ϕn ) and let f = ∞ p=0 fp t , where each fp is a polynomial in k[X]. Then we have the sequencePf0 (ϕ), f1 (ϕ), . . . of series in k[[t]] and so, we p can form the series f (ϕ) = ∞ p=0 fp (ϕ)t . When the ring k has no zero divisors then the following lemma is easy to be proved. We present a proof of it in the general case. Lemma 1.6.3. If g is a nonzero polynomial in k[X], then there exists a point a ∈ k n such that g(a) 6= 0. Proof. Let us recall that k[X] = k[x1 , . . . , xn ] and Q ⊆ k. P First assume that n = 1. Set x = x1 and let g = pj=0 gj xj with gj ∈ k and gp 6= 0. If p = 0 then our lemma is evident. Let p > 0 and suppose that g(a) = 0 for all a ∈ k 1 . Put h(x) = 2p g(x) − g(2x). Then h(a) = 0 (for all a ∈ k 1 ) and deg h < deg g. So, by induction, h = 0. This implies that g0 = · · · = gp−1 = 0 and we have a contradiction: 0 = g(1) = gp 6= 0. This completes the proof for n = 1. P Assume now that n > 1 and let g = pj=0 gj xjn with gj ∈ k[x1 , . . . , xn−1 ] and gp 6= 0. Then, by induction, there exists b ∈ k n−1 such that gp (b) 6= 0. Consider the polynomial g(b, xn ). This is a nonzero polynomial in a one variable. Therefore, by the first part of this proof, g(b, an ) 6= 0, for some an ∈ k. As a consequence of the above lemma we get Lemma 1.6.4. Let f, g ∈ k[X][[t]]. If f (a) = g(a) for any a ∈ k n , then f = g. P∞ i If ψ = i=0 ai t ∈ k[[t]] and p ∈ N0 , then denote by ψ[p] the p-th coefficient of ψ (that is, ψ[p] = ap ). Moreover, denote by sp (ψ) the sum Pp i i=0 ai t .
16
Part I. Preliminary concepts
Lemma 1.6.5. Let f ∈ k[X][[t]], p ∈ N0 and ϕ1 , . . . , ϕn ∈ k[[t]]. Then f (ϕ1 , . . . , ϕn )[p] = f (sp (ϕ1 ), . . . , sp (ϕn ))[p] . Proof. Observe that if ϕ and ψ are two series in k[[t]], then (ϕ+ψ)[p] = (sp (ϕ) + sp (ψ))[p] and (ϕ · ψ)[p] = (sp (ϕ) · cp (ψ))[p] . So Lemma holds for f ∈ k[X]. The rest follows from Lemma 1.6.2. Proof of Theorem 1.6.1. We construct the coefficients (ϕ1 )[p] , . . . , (ϕn )[p] using induction on p. If p = 0 then put (ϕ1 )[0] = a1 , . . . , (ϕn )[0] = an .
(1.3)
Let p > 0 and assume that, for all j 6 p, the coefficients (ϕ1 )[j] , . . . , (ϕn )[j] are already constructed. If i ∈ {1, . . . , n}, then we define: p
(ϕi )[p+1] =
X 1 fi (g1 , . . . , gn )[p] , where gi = (ϕi )[j] tj . p+1
(1.4)
j=0
P∞Thus, wep have constructed the series ϕ1 , . . . , ϕn ∈ k[[t]] (where ϕi = p=0 (ϕi )[p] t for i = 1, . . . , n) satisfying (2). It is easy to check (using Lemma 1.6.5) that they also satisfy (1). Assume now that ϕ = (ϕ1 , . . . , ϕn ) is an arbitrary sequence of series i in k[[t]] satisfying (1) and (2). Then ( ∂ϕ ∂t )[p] = fi (ϕ1 , . . . , ϕn )[p] , for any p ∈ N0 and i = 1, . . . , n. Hence (by Lemma 1.6.5), ϕ satisfies the equalities (1.4). Since ϕ also satisfies (1.3), we see (by a simple induction) that ϕ is unique. If f = (f1 , . . . , fn ) is a sequence of series in k[X][[t]] and a = (a1 , . . . , an ) ∈ k n , then denote by ϕ(t, a) = (ϕ1 (t, a), . . . , ϕn (t, a)) the unique sequence (ϕ1 , . . . , ϕn ) of series from k[[t]] satisfying the conditions (1) and (2) of Theorem 1.6.1. The sequence ϕ(t, a) is called the formal solution of the differential system ∂X = f (X), X[0] = a. ∂t
(1.5)
Chapter 1. Definitions, notations and basic facts
17
Now we prove the following polynomial property of the formal solutions. Theorem 1.6.6. If f ∈ k[X][[t]]n , then there exist uniquely determined polynomials ωij ∈ k[X] (for i ∈ {1, . . . , n} and j ∈ N0 ) such that ϕi (t, a) =
∞ X
ωij (a)tn ,
(1.6)
j=0
for all a ∈ k n and i ∈ {1, . . . , n}, where ϕ(t, a) is the formal solution of (1.5). Proof. Similarly as in the proof of Theorem 1.6.1 we construct (using an induction on p ∈ N0 ) the polynomials ω1p , ω2p , . . . , ωnp . If p = 0 then put ω01 = x1 , . . . , ω0n = xn . Let p > 0 and assume that, for all j 6 p, the polynomials ω1j , . . . , ωnj are already defined. Then define: ωi(p+1) =
1 fi (G1 , . . . , Gn )[p] , p+1
P where Gi = pj=0 ωij tj for any i = 1, . . . , n. Now, using Lemma 1.6.5, one can easily deduce that the polynomials ωij satisfy (1.6). The uniqueness follows from Lemma 1.6.3 and Theorem 1.6.1. If a = (a1 , . . . , an ) ∈ k n , then we denote by πa the surjective homomorphism from k[X][[t]] to k[[t]] defined by πa (f ) = f (a). Corollary 1.6.7. If f ∈ k[X][[t]]n , then there exist unique series W1 , . . . , Wn ∈ k[X][[t]] such that ϕi (t, a) = πa (Wi ) for all a ∈ k n and i ∈ {1, . . . , n}.
2
Useful facts and preliminary results
This chapter is devoted to the general properties of derivations in polynomial rings. We present some concepts and facts which will be often used in the next chapters. We have here six sections concerning different, to some extent independent, matters. We assume that n is a fixed natural number and we use the abbreviated denotations: k[X] = k[x1 , . . . , xn ] and k(X) = k(x1 , . . . , xn ) for the ring of polynomials and the field of rational functions, respectively. By a direction we mean a nonzero sequence γ = (γ1 , . . . , γn ) of integers. In Section 2.1 we first characterize all the polynomials and rational functions which are homogeneous with respect to a direction, and then, we give some information on homogeneous derivations and, in particular, on monomial derivations, that is, on derivations d of k[X] such that d(x1 ), . . . , d(xn ) are monomials. Section 2.2 contains the basic facts concerning Darboux polynomials for k-derivations of k[X]. We prove here, among other things , a useful proposition (see Proposition 2.2.4) which states that every homogeneous k-derivation of k[x, y] has a Darboux polynomial. If d is a k-derivation of k[X], then we denote by d? the divergence of Pn ? d, that is, d = i=1 ∂d(xi )/∂xi . A derivation d of k[X] is called special if d? = 0. In Section 2.3 we present some initial properties of the divergence and special derivations. More such properties and their applications will be given in the next sections. Section 2.4 deals with the automorphism Ed = exp(td) of the power series ring k[X][[t]]. First we recall some well known facts concerning this automorphism and its applications to the autonomous systems of differential equations, and then, we study the jacobian of Ed . In Section 2.5 we concentrate on the bases of the free R-module Derk (R), where R = k[X] or k[[X]]. We present descriptions of all the bases and all the commutative bases of Derk (R). The descriptions come from the author’s paper [79]. Moreover, we prove (Theorem 2.5.5) that every component of a commutative basis of Derk (k[X]) is a special derivation. The whole of Section 2.6 is a copy of paper [47], by K. Kishimoto and the author, devoted to the images of derivations in k(X). Assuming that k is a field of characteristic zero, we prove that if d is a k-derivation of k(X) (in particular, of k[X]), then d((X)) 6= k(X). We obtain it as a consequence 18
Chapter 2. Useful facts and preliminary results
19
of a more general theorem which states that if d is a k-derivation of a field L containing k with tr.degk L < ∞, then d(L) 6= L. Note that our theorem is important for the considerations contained in Chapter 3. Using this theorem it is easy to construct k-derivations d of k[X] such that k(X)d = k.
2.1
Homogeneous derivations
Let γ = (γ1 , . . . , γn ) be a direction and let s ∈ Z. If α = (α1 , . . . , αn ) ∈ Ω = Nn0 , then γα denotes the sum γ1 α1 + · · · + γn αn and X α denotes the monomial xα1 1 · · · xαnn . A nonzero polynomial f ∈ k[X] is said to be a γ-form of degree P s (or a γhomogeneous polynomial of degree s) if f is of the form: f = γα=s aα X α , where aα ∈ k. We assume that the zero polynomial is a γ-form of any degree. For example, if n = 2 and k[X] = k[x, y], then y 5 + xy 3 − 6x2 y is a (2, 1)-form of degree 5, x4 y + x7 y 3 is a (−2, 3)-form of degree −5 and x8 y 4 + x6 y 3 + 7 is a (1, −2)-form of degree 0. Proposition 2.1.1. Assume that k is a domain of characteristic zero and k0 is its field of fractions, If f is a nonzero polynomial in k[X], then the following conditions are equivalent: (1) f is a γ-form of degree s. (2) f (tγ1 x1 , . . . , tγn xn ) = ts f (x1 , . . . , xn ) (in the ring k0 (t)[X]). ∂f ∂f (3) γ1 x1 ∂x = sf . + · · · + γn xn ∂x n 1 (3) ⇒ (2). Set R = k0 (t)[X] = k0 (t)[x1 , . . . , xn ], u = ∂ , and let ϕ : k0 [X] −→ R be the g = t−s f (u), δ = ∂t homomorphism of k0 -algebras defined by ϕ(xi ) = tγi xi , for i = 1, . . . , n. Applying ϕ for (3) we get Proof.
(tγ1 x1 , . . . , tγn xn ),
∂f ∂f γ1 tγ1 x1 ∂x (u) + · · · + γn tγn xn ∂x (u) − sf (u) = 0. n 1
Next, using this equality and a simple calculation, we deduce that g ∈ Rδ . This implies (by Proposition 1.3.4) that g is a polynomial in k0 [X]. Thus, we have the equality: f (tγ1 x1 , . . . , tγn xn ) = f (u) = ts g(x1 , . . . , xn ). Substituting now t = 1, we get: g(x1 , . . . , xn ) = f (x1 , . . . , xn ) and hence, we have (2). The implications (2) ⇒ (1) and (1) ⇒ (3) are well known and easy to be proved.
20
Part I. Preliminary concepts
The implication (1) ⇒ (3) (for γ = (1, . . . , 1) and k = R) is the well known Euler theorem on homogeneous functions ([24], [25]). Equality (3) is called the Euler formula. (s)
(s)
Denote by Aγ the group of all γ-forms of degree s. Each Aγ is a L (s) (s) (t) (s+t) k-submodule of k[X] and k[X] = s∈Z Aγ . Moreover, Aγ Aγ ⊆ Aγ for all s, t ∈ Z. Thus, k[X] is a graded ring. Such a gradation on k[X] is said to be a γ-gradation. P Every polynomial f ∈ k[X] has the γ-decomposition f = fs into γ-components fs of degree s. If f 6= 0, then γ-deg(f ) denotes the γ-degree of f , that is, the maximal s such that fs 6= 0. We assume also that γ-deg(0) = −∞. It is easy to check the following Lemma 2.1.2. Let k be a domain and let f, g be nonzero polynomials in k[X]. If f g is a γ-form, then f and g are γ-forms too. Assume now that k is a field (of characteristic zero). An element ϕ of k(X) is said to be γ-homogeneous of degree s if, in the field k(t, x1 , . . . , xn ), the following equality holds: ϕ(tγ1 x1 , . . . , tγn xn ) = ts ϕ(x1 , . . . , xn ). Proposition 2.1.3. Let f, g be nonzero coprime polynomials in k[X] and let ϕ = f /g. The following conditions are equivalent: (1) ϕ is γ-homogeneous of degree s. (2) Polynomials f and g are γ-forms of degrees p and q, respectively, where s = p − q. ∂ϕ ∂ϕ (3) γ1 x1 ∂x + · · · + γn xn ∂x = sϕ. n 1 Proof. (1) ⇒ (2). Consider the γ-decompositions f = fp1 +· · ·+fp , g = gq1 + · · · + gq , where p1 < · · · < p, q1 < · · · < q, and put X = (x1 , . . . , xn ), tγ X = (tγ1 x1 , . . . , tγn xn ). By the equality ϕ(tγ X) = ts ϕ(X) we obtain the following equality of polynomials in k(t)[X]: f (tγ X)g(X) = ts f (X)g(tγ X). Hence, by Proposition 2.1.1, (tp1 fp1 (X) + · · · + tp fp (X))g(X) = ts f (X)(tq1 gq1 (X) + · · · + tq gq (X)).
Chapter 2. Useful facts and preliminary results
21
Comparing now powers of t we see that p = s + q and fp (X)g(X) = gq (X)f (X). Since the polynomials f and g are coprime, fp = f h and gq = gh for some h ∈ k[X] and hence, by Lemma 2.1.2, f and g are γ-forms. (2) ⇒ (3) It is a consequence of Proposition 2.1.1(3). (3) ⇒ (2) Use the same arguments as in the proof of the implication (3) ⇒ (2) of Proposition 2.1.1. The implication (2) ⇒ (1) is obvious. A k-derivation d of k[X] is called γ-homogeneous of degree s if (s+p) d(A(p) , γ ) ⊆ Aγ
for any p ∈ Z. For example, if k[X] = k[x, y] and d(x) = x + y 2 , d(y) = y, then d is a (2, 1)-homogeneous k-derivation of degree 0. The zero derivation is γ-homogeneous of every degree. The sum of γhomogeneous derivations of the same degree s is γ-homogeneous of degree s. If d1 , d2 are γ-homogeneous derivations of degree s1 and s2 , respectively, then the derivation [d1 , d2 ] = d1 d2 −d2 d1 is γ-homogeneous of degree s1 +s2 . Proposition 2.1.4. The following conditions are equivalent: (1) d is γ-homogeneous of degree s. (s+γ ) (2) d(xi ) ∈ Aγ i for i = 1, . . . , n. (γ )
Proof. (1) ⇒ (2). It is clear, because each xi belongs to Aγ i . (p) (2) ⇒ (1). Every polynomial in Aγ is a sum of monomials of the form aα X α , where γα = p and aα ∈ k. Hence, if α ∈ Ω is such an (s+p) element that γα = p, then it suffices to show that d(X α ) ∈ Aγ . But P d(X α ) = ni=1 αi xα1 1 · · · xαi i −1 · · · xαnn d(xi ), α1 γ1 + · · · + (αi − 1)γi + · · · + αn γn = αγ − γi = p − γi , (s+γi )
and d(xi ) ∈ Aγ
(p−γi )
. Therefore, d(X α ) ∈ Aγ
Corollary 2.1.5. The derivation
∂ ∂xi
(s+γi )
Aγ
(s+p)
⊆ Aγ
.
is γ-homogeneous of degree −γi .
A k-derivation d of k[X] is called monomial if d(x1 ), . . . , d(xn ) are monomials. Proposition 2.1.6. If d is a monomial k-derivation of k[X], then there exists a direction γ such that d is γ-homogeneous.
22
Part I. Preliminary concepts
Proof. Assume that d(xi ) = ai X αi , for i = 1, . . . , n, where ai ∈ k and αi = (αi1 , . . . , αin ) ∈ Ω. We must find a nonzero sequence (γ1 , . . . , γn ) of integers and an integer s such that γ1 αi1 + · · · + γn αin = γi + s, for all i = 1, . . . , n. For this purpose consider the following system of n+1 linear equations over Q: (α11 − 1)γ1 + · · · + α1n γn + (−1)s = 0 α21 γ1 + · · · + α2n γn + (−1)s = 0 ··· ··· αn1 γ1 + · · · + (αnn − 1)γn + (−1)s = 0 0γ1 + · · · + 0γn + 0s = 0 Since the determinant of the main matrix of this system is equal to 0, there exists a nonzero solution (γ1 , . . . , γn , s) belonging to Zn+1 and it is clear that (γ1 , . . . , γn ) 6= 0. Note also the following proposition which is easy to prove. Proposition 2.1.7. Let d be a γ-homogeneous k-derivation of k[X] and let f ∈ k[X]. If f ∈ k[X]d , then each γ-homogeneous component of f belongs also to k[X]d . Corollary 2.1.8. If d is a γ-homogeneous k-derivation of k[X], then k[X]d , the ring of constants with respect to d, is generated over k by γ-forms. In the present section we introduced several terms such as: γhomogeneous, γ-form, γ-degree, . . . . If γ = (1, . . . , 1), then we have the ordinary direction and we omit the sign γ. So, in this case, we say: homogeneous, form, degree, . . . . Moreover, in this case, we write A(s) instead of (s) Aγ . From Proposition 2.1.4 we get Corollary 2.1.9. The k-derivation d is homogeneous of degree s if and only if the polynomials d(x1 ), . . . , d(xn ) are forms belonging to A(s+1) . In particular, every Pn linear k-derivation, that is, a k-derivation d of k[X] such that d(xi ) = j=1 aij xj for i = 1, . . . , n, is homogeneous of degree 0.
2.2
Darboux polynomials
Let us introduce (as in [63], [64]) a new notion that dates back to Darboux’s memoir [15]. Let d be a k-derivation of k[X]. We say that a
Chapter 2. Useful facts and preliminary results
23
polynomial f ∈ k[X] is a Darboux polynomial of d if f 6= 0 and d(f ) = hf , for some h ∈ k[X]. If f is a Darboux polynomial of d, then every h ∈ k[X], such that d(f ) = hf , is said to be a polynomial eigenvalue of f . If k is a domain then such an h is unique. Darboux polynomials with nonzero eigenvalues (for k = R or C) are well known in the theory of polynomial differential equations. They coincide with the so-called partial first integrals (see, for example, [63] and [117]) of the system of polynomial differential equations determined by d. Every element belonging to the ring of constants with respect to d is of course a Darboux polynomial. In the vocabulary of differential algebra, Darboux polynomials coincide with generators of principal differential ideals, that is, f ∈ k[X] is a Darboux polynomial iff f 6= 0 and the ideal (f ) is differential (i. e., d(f ) ∈ (f )). We know ([98], [95], [76]) that every differential ideal (in a noetherian case) has a differential primary decomposition. As a consequence of this fact we have the following proposition which, in our case, is easy to be proved. Proposition 2.2.1. Assume that k is a UFD. If f ∈ k[X] is a Darboux polynomial of d, then all factors of f are also Darboux polynomials of d. Thus, looking for Darboux polynomials of a given k-derivation d (where k is a field) reduces to looking for irreducible ones. Note now some simple, but useful, propositions. Proposition 2.2.2. Let d be a k-derivation of k(X) such that d(k[X]) ⊆ k[X] (where k is a field). Let f and g be nonzero coprime polynomials in k[X]. Then f /g ∈ k(X)d iff f and g are Darboux polynomials with the same eigenvalue. Proposition 2.2.3. Let γ = (γ1 , . . . , γn ) be a direction and let d be a γhomogeneous k-derivation of k[X] (where k is a domain). If f ∈ k[X] is a Darboux polynomial of d, then the eigenvalue h of f is a γ-form and all γcomponents of f are also Darboux polynomials with the common eigenvalue equal to h. Proof. Compare γ-degrees of the equality d(f ) = hf . See [64] for a detail. Note that, even for the ordinary direction, Darboux polynomials of a homogeneous derivation are not necessarily homogeneous. Indeed, let
24
Part I. Preliminary concepts
n = 2, d(x1 ) = x1 , d(x2 ) = 2x2 , and let f = x21 + x2 . Then d is homogeneous, f is a Darboux polynomial of d (because d(f ) = 2f ), and f is not homogeneous. If n = 2 then homogeneous k-derivation of k[X] (with respect to the ordinary direction) have the following special property Proposition 2.2.4. If k is a UFD then every homogeneous k-derivation of k[x1 , x2 ] has a Darboux polynomial. Proof. Let d be a homogeneous k-derivation of degree s. Put f1 = d(x1 ), f2 = d(x2 ) and let f = x1 f2 − x2 f1 . It follows from the Euler formula that d(f ) = hf , where h = ∂f1 /∂x1 + ∂f2 /∂x2 . So, if f 6= 0 then f is a Darboux polynomial of d. If f = 0 then it is easy to see (since k is UFD), that x1 − x2 is a Darboux polynomial of d. If n > 2 then the above property does not hold, in general. Example 2.2.5 ([40]). Let d be the C-derivation of C[x, y, z] defined by: d(x) = y 2 , d(y) = z 2 and d(z) = x2 . Then d is homogeneous and d has no Darboux polynomial (see T heorem 11.0.2).
2.3
The divergence and special derivations
Let k be a ring containing Q, and let d be a k-derivation of the polynomial ring k[X] = k[x1 , . . . , xn ]. Denote by d? the divergence of d, i. e., d? =
∂d(x1 ) ∂d(xn ) + ··· + . ∂x1 ∂xn
The derivation d is said to be special if d? = 0. In this section we present some initial properties of the divergence and special derivations. Farther such properties and their applications will be given in the next sections. We start from the following easy Proposition 2.3.1. If d, δ ∈ Derk (k[X]) and r ∈ k[X], then: (1) (d + δ)? = d? + δ ? . (2) (rd)? = rd? + d(r). (3) [d, δ]? = d(δ ? ) − δ(d? ).
Chapter 2. Useful facts and preliminary results
25
The partial derivatives are special derivations. The above proposition implies that the set of all special derivations of k[X] is closed under the sum and the Lie product. Denote by [h1 , . . . , hn ] the jacobian of h1 , . . . , hn ∈ k[X], that is, ∂hi [h1 , . . . , hn ] = det . ∂xj Proposition 2.3.2. Let d : k[X] −→ k[X] be a k-derivation and let h1 , . . . , hn ∈ k[X]. Then ?
d([h1 , . . . , hn ]) = −[h1 , . . . , hn ]d +
n X
[h1 , . . . , d(hp ), . . . , hn ].
p=1 ∂fi ∂hi Proof. Put fi = d(xi ), fij = ∂x , hij = ∂x , for all i, j ∈ {1, . . . , n}, j j and let Sn denote the group of all permutations of {1, . . . , n}. Observe that n
d(hσ(p)p ) =
X ∂ d(hσ(p) ) − hσ(p)q fqp , ∂xp
(2.1)
q=1
for all σ ∈ Sn and p ∈ {1, . . . , n}, and P |σ| σ∈Sn (−1) hσ(1)1 · · · hσ(p−1)(p−1) hσ(p)q hσ(p+1)(p+1) · · · hσ(n)n
(2.2)
= [h1 , . . . , hn ]δpq , for all p, q ∈ {1, . . . , n}, where |σ| is the sign of σ, and δpq is the Kronecker delta. The above determines that n X X d([h1 , . . . , hn ]) = (−1)|σ| hσ(1)1 · · · d(hσ(p)p ) · · · hσ(n)n p=1 σ∈Sn (2.1)
=
n X
n
X
p=1 σ∈Sn (2.2)
=
n X
(−1)|σ| hσ(1)1 · · · (
X ∂ d(hσ(p) ) − hσ(p)q fpq ) · · · hσ(n)n ∂xp
[h1 , . . . , d(hp ), . . . , hn ] −
p=1
=
n X p=1
=
n X p=1
q=1
[h1 , . . . , d(hp ), . . . , hn ] −
X p=1
n X
n
n X
fpq [h1 , . . . , hn ]δpq
q=1
fpp [h1 , . . . , hn ]
p=1
[h1 , . . . , d(hp ), . . . , hn ] − [h1 , . . . , hn ]d? .
26
Part I. Preliminary concepts
Corollary 2.3.3. If d is a special k-derivation of k[X] and h1 , . . . , hn ∈ k[X], then d([h1 , . . . , hn ]) =
n X
[h1 , . . . , d(hp ), . . . , hn ].
p=1
2.4
Automorphism Ed
Assume that k is a ring containing Q. Let A be a k-algebra, A[[t]] the power series ring over A in one variable, and d a k-derivation of A. Denote by d˜ the k[[t]]-derivation of A[[t]] defined by ∞ ∞ X X ˜ d( ap tp ) = d(ap )tp , p=0
and set Ed (ϕ) =
p=0 ∞ X 1 ˜p d (ϕ)tp p! p=0
for all ϕ ∈ A[[t]]. It is well known that Ed is a k[[t]]-automorphism of A[[t]], which is very useful in the differential algebra (see, for example: [98], [109], [75], [23]). We recall here the known facts concerning the automorphism Ed and its applications to autonomous systems of differential equations. In the second part of this section we concentrate on the jacobian of Ed and we prove some facts which will be useful in the next sections. Let us start from the following consequence of Proposition 1.6.2 P Proposition 2.4.1. If ψ = ∞ p=0 P where bp = i+j=p hi, jidi (aj ).
1 p p! ap t ,
then Ed (ψ) =
P∞
1 p p=0 p! bp t ,
The next proposition contains some of the main properties of the automorphism Ed . Proposition 2.4.2. (1) Ed−1 = E−d . (2) Ed ◦ d˜ = d˜ ◦ Ed . (3) If d and δ are k-derivations of A, such that dδ = δd, then Ed ◦ EP δ = Ed+δ . P∞ p p (4) Ed ( ∞ p=0 ap t ) = p=0 Ed (ap )t .
Chapter 2. Useful facts and preliminary results
27
The proof is straightforward. In Section 1.6 we proved that every differential system of the form ∂X = f (X), X[0] = a, ∂t
(2.3)
(where f = (f1 , . . . , fn ) and a ∈ k n ) admits a unique formal solution. Elements f1 , . . . , fn of this system were series from k[X][[t]], where k[X] = k[x1 , . . . , xn ]. Now we concentrate on a special case. We assume that f1 , . . . , fn do not depend on t, that is, f1 , . . . , fn are polynomials in k[X]. Such a differential system (2.3) is called autonomous (or stationary) and its formal solution is called a formal flow. Let us recall (see Section 1.6) that if a ∈ k n , then we denote by πa the surjective homomorphism from k[X][[t]] to k[[t]] defined by πa (ϕ) = ϕ(a). The following theorem gives a formula describing formal flows. Theorem 2.4.3 ([116], [12], [22]). Let f = (f1 , . . . , fn ) ∈ k[X]n , a ∈ k n , and let ϕ(t, a) = (ϕ1 (t, a), . . . , ϕn (t, a)) be the formal flow of the autonomous differential system Then ϕi (t, a) = πa Ed (xi ), for any Pn (2.3). ∂ i = 1, . . . , n, where d = i=1 fi ∂xi . Proof. It is suffices to show that if i ∈ {1, . . . , n}, then ∂ϕi = fi (ϕ1 , . . . , ϕn ), ∂t where each ϕi is equal to Ed (xi ). Using the properties of Ed we get: ∂ϕi ∂t
P∞
1 p p p=0 p! d (xi )t
=
∂ ∂t
=
P∞
1 p p p=0 p! d (d(xi ))t
=
P∞
1 p p−1 p=1 p! pd (xi )t
= Ed (d(xi )) = Ed (fi )
= fi (Ed (x1 ), . . . , Ed (xn )) = fi (ϕ1 , . . . , ϕn ).
Note also the following two lemmas which will be used in Section 5.3. Lemma 2.4.4. ∂ ˜ to ∂t + d.
∂ ∂t Ed
= Ed ∆, where ∆ is the k-derivation of k[X][[t]] equal
28
Part I. Preliminary concepts
∂ Proof. It is easy to check that d˜∂t = we have ∂ ∂t Ed (F )
∂ ˜ ∂t d.
Thus, if F ∈ k[X][[t]] then
P∞
1 ˜p p p=0 p! d (F )t
=
∂ ∂t
=
P∞
=
P∞
=
P∞
=
P∞
1 ∂ ˜p p p=0 ( p! ∂t d (F )t p=0
∂ p + p!1 d˜p (F ) ∂t (t )) P ∞ 1 ˜p+1 1 ˜p ∂ p (F )tp p=0 p! d p! d ( ∂t F )t +
1 ˜p ∂ p=0 p! d ( ∂t F
˜ ))tp + d(F
1 ˜p p p=0 p! d (∆(F ))t
= Ed ∆(F ). Lemma 2.4.5. If F ∈ k[X][[t]] and a ∈ k n , then F (ϕ(t, a)) = πa Ed (F ). P p Proof. Set F = ∞ p=0 Fp t with Fp ∈ k[X]. Since ϕi (t, a) = πa Ed (xi ) for any i = 1, . . . , n (Theorem 2.4.3), we have: F (ϕ(t, a)) =
P∞
=
P∞
=
P∞
p=0 Fp (ϕ1 (t, a), . . . , ϕn (t, a))t
p
p=0 Fp (πa Ed (x1 ), . . . , πa Ed (xn ))t p=0 πa Ed (Fp (x1 , . . . , xn ))t
p
p
P p = πa E d ( ∞ p=0 Fp t ) = πa Ed (F ).
Now we present some facts concerning the jacobian of Ed . Let k[X] = k[x1 , . . . , xn ] be the polynomial ring and let d be a kderivation of k[X]. Denote by J the jacobian of Ed (x1 ), . . . , Ed (xn ), that is, ∂Ed (xi ) J = [Ed (x1 ), . . . , Ed (xn )] = det . ∂xj P 1 p The jacobian J is an element of k[X][[t]]. Set J = ∞ p=0 p! Bp t , where each Bp is in k[X]. ∂f Assume that n = 2. Put x = x1 , y = x2 and replace ∂f ∂x , ∂y by fx , fy ,
Chapter 2. Useful facts and preliminary results
29
respectively, for any f ∈ k[x, y]. Then we have: J
= Ed (x)x Ed (y)y − Ed (x)y Ed (y)x P P ∞ 1 p ∞ 1 p p p = d (x) t d (y) t x y p=0 p! p=0 p! P P ∞ 1 p ∞ 1 p p p − d (x) t d (y) t y x p=0 p! p=0 p! P∞ P 1 i (x) dj (y) − di (x) dj (y) ) tp (d = x y y x p=0 i+j=p i!j! P∞ P 1 i j p = p=0 i+j=p i!j! [d (x), d (y)] t
Thus, we see that if n = 2 then X Bp = hi, ji[di (x), dj (y)], i+j=p
for every p > 0. If n is arbitrary then, repeating the same argument, we get Proposition 2.4.6. For every p > 0, X Bp = hi1 , . . . , in i[di1 (x1 ), . . . , din (xn )]. i1 +···+in =p
Let us continue our calculations for n = 2. From Propositions 2.4.6 and 2.3.2 we get X d(Bp ) = hi, jid([di (x), dj (y)]) i+j=p
=
X
hi, ji([di+1 (x), dj (y)] + [di (x), dj+1 (y)]
i+j=p
−[di (x), dj (y)]d? ) X = −Bp d? + hi, ji([di+1 (x), dj (y)] + [di (x), dj+1 (y)]), i+j=p
where d? is the divergence of d (see Section 2.3).
30
Part I. Preliminary concepts By standard formulas and Proposition 2.4.6,
d(Bp ) + Bp d? = [dp+1 (x), y] + [x, dp+1 (y)] p−1 X + hi, p − ii[di+1 (x), dp−i (y)] i=0 p X
+
hi, p − ii[di (x), dp+1−i (y)]
i=1
= [d
p+1
(x), y] + [x, dp+1 (y)]
p X + (hi − 1, p + 1 − ii + hi, p − ii)[di (x), dp+1−i (y)] i=1
=
X
hi, ji[di (x), dj (y)]
i+j=p+1
= Bp+1 . We used the well known equality: hi−1, ji+hi, ji = hi, j +1i (where i > 1). There exists also the following its generalization hi1 , . . . , in i +
n−1 X
hi1 , . . . , ij − 1, . . . , in−1 , in i = hi1 , . . . , in−1 , in + 1i, (2.4)
j=1
for i1 , . . . , in−1 > 1. Now, using (2.4), Propositions 2.4.6, 2.3.2 and repeating the same arguments, one can easily deduce the following Theorem 2.4.7. Let d be a k-derivation of k[X] = k[x1 , . . . , xn ] and let [Ed (x1 ), . . . , Ed (xn )] =
∞ X 1 Bp tp , p! p=0
where Bp ∈ k[X]. Then B0 = 1, B1 = d? and Bp+1 = Bp d? + d(Bp ) for all p > 0. Corollary 2.4.8. Let d be a k-derivation of k[X]. If d? ∈ k then [Ed (x1 ), . . . , Ed (xn )] =
∞ X 1 p p b t = ebt , p! p=0
where b = d? .
Chapter 2. Useful facts and preliminary results
31
Corollary 2.4.9. If d is a special k-derivation of k[X], then [Ed (x1 ), . . . , Ed (xn )] = 1.
2.5
Bases of derivations in polynomial and power series rings
Let k be a ring containing Q and let R be either the ring k[X] or the ring k[[X]] = k[[x1 , . . . , xn ]] of formal power series over k. Consider the R-module Derk (R) of all k-derivations of R. In view of Theorems 1.2.1 and 1.5.2 we know that Derk (R) is a free R-module and the set {∂/∂xi , . . . , ∂/∂xn } is one of its bases. We will present a description of all the bases of Derk (R). The results of this section come from the author’s paper [79]. Proposition 2.5.1. Let d1 , . . . , dn be k-derivations of R. The set {d1 , . . . , dn } is a basis of Derk (R) if and only if the matrix [di (xj )] is invertible. Proof. If d1 , . . . , dn form a basis P of Derk (R),∂then there exist elements bij ∈ R (i, j = 1, . . . , n) such that np=1 bip dp = ∂x for i = 1, . . . , n. Hence, i for each i, j, we have P ∂ δij = ∂x (xj ) = np=1 bip dp (xj ), i where δij is the Kronecker delta. Hence the matrix [di (xj )] is invertible. Conversely, if the matrix [di (xj )] is invertible, thenP there exists an invertible matrix [bij ] of elements of R such that δij = np=1 bip dp (xj ), for each i, j = 1, . . . , n. Pn Since Di Denote by Di (for i = 1, . . . , n) the map p=1 bip dp . ∂ is a k-derivation of R and Di (xj ) = δij , we have Di = ∂xi . Hence Pn ∂ p=1 bip dp (xj ) = ∂xi , for any i = 1, . . . , n, and so the derivations d1 , . . . , dn form a basis of Derk (R). We say that a basis {d1 , . . . , dn } of Derk (R) is commutative if di dj = dj di for any i, j = 1, . . . , n. We have the following characterization of commutative bases of Derk (R). Theorem 2.5.2. Let d1 , . . . , dn be k-derivations of R (where R = k[X] or k[[X]]). The following conditions are equivalent (1) The set {d1 , . . . , dn } is a commutative basis of Derk (R). (2) There exist elements F1 , . . . , Fn ∈ R such that di (Fj ) = δij , for any i, j = 1, . . . , n, where δij is the Kronecker delta.
32
Part I. Preliminary concepts For the proof of this theorem we need two lemmas.
Lemma 2.5.3. If f1 , . . . , fn ∈ R then the following conditions are equivalent: ∂ (a) There exists F ∈ R such that ∂x (F ) = fi for i = 1, . . . , n. i ∂ ∂ (b) ∂xi (fj ) = ∂xj (fi ) for any i, j = 1, . . . , n. ∂ ∂ Proof. (a) ⇒ (b) follows from the equality ∂x = i ∂xj i, j = 1, . . . , n. (b) ⇒ (a). Let X [α1 , . . . , αn ]i xα1 1 · · · xαnn , fi =
∂ ∂ ∂xj ∂xi ,
for any
α1 ,...,αn
for i = 1, . . . , n, where the coefficients of the form [α1 , . . . , αn ]i are elements ∂ of k. Since ∂x (fj ) = ∂x∂ j (fi ), for any αi , αj > 1 we have i (1/αi )[α1 , . . . , αi−1 , αi − 1, αi+1 , . . . , αn ]i = (1/αj )[α1 , . . . , αj−1 , αj − 1, αj+1 , . . . , αn ]j . Put X
F =
aα1 ···αn xα1 1 · · · xαnn ,
α1 ,...,αn
where a0···0 = 0 and, if αi > 1 for some i, then aα1 ···αn = (1/αi )[α1 , . . . , αi−1 , αi − 1, αi+1 , . . . , αn ]i . It is easy to check, that
∂ ∂xi (F )
= fi for i = 1, . . . , n.
Lemma 2.5.4. Pn Let {d1 , . . . , dn } be a commutative basis of Der(R). Assume ∂ = that ∂x j=1 bij dj for all i = 1, . . . , n, where bij (i, j = 1, . . . , n) are i elements of R. Then ∂ ∂ ∂xp (bqj ) = ∂xq (bpj ), for any p, q, j = 1, . . . , n. P Proof. It suffices to prove that nj=1 ( ∂x∂ p (bqj ) − ∂x∂ q (bpj ))dp = 0. From the commutativity of the partial derivatives we have 0 =
Therefore
∂ ∂ ∂xp ∂xq
=
∂ ∂xp
=
Pn
Pn
−
∂ ∂ ∂xq ∂xp
j=1 bqj dj
−
∂ j=1 ( ∂xp (bpj )dj
∂ ∂xq
Pn
j=1 bpj dj
+ bqj ∂x∂ p dj −
∂ ∂xq (bpj )dj
− bpj ∂x∂ q dj ).
Chapter 2. Useful facts and preliminary results
Pn
∂ j=1 ( ∂xp (bqj )
−
∂ ∂xq (bpj ))dj
=
Pn
=
Pn
∂ j=1 (bpj ∂xq dj
33
− bqj ∂x∂ p dj )
Pn
j=1 bpj (
i=1 bqi di dj ) Pn − j=1 bqj ( i=1 bpi di dj ) Pn Pn = i=1 bpj bqi (di dj − dj di ) = 0. j=1
Pn
This completes the proof. Proof of Theorem 2.5.2. (1) ⇒ (2). Assume that the set {d1 , . . . , dn } is a commutative basis of Derk (R). Then, by Proposition 2.5.1, the matrix −1 = [b ], where b A = [di (xj )] is invertible. ij ij ∈ R for i, j = Pn Put A ∂ 1, . . . , n. Then ∂xi = b d , for any i = 1, . . . , n (see the proof of j=1 ij j Proposition 2.5.1). Now fix j ∈ {1, . . . , n} and consider the elements b1j , . . . , bnj . Since ∂ ∂ ∂xq (bpj ) = ∂xp (bqj ) for any p, q = 1, . . . , n (Lemma 2.5.4), there exists an element Fj ∈ R such that ∂x∂ p (Fj ) = bpj for p = 1, . . . , n (by Lemma 2.5.3). Moreover, we have P P di (Fj ) = np=1 ∂x∂ p (Fj )di (xp ) = np=1 bpj di (xp ) = (A · A−1 )ij = δij . Pn ∂ (2) ⇒ (1). Since di (Fj ) = δij , we have r=1 ∂xr (Fj )di (xr ) = δij . Thus the matrix [di (xj )] is invertible, and hence, by Proposition 2.5.1, the derivations d1 , . . . , dn form a basis of Derk (R). Since dp dq − dq dp is an element of Derk (R), there exist elements a1 , . . . , an of R such that dp dq − dq dp = a1 d1 + · · · + an dn . But we have ai = (a1 d1 + · · · + an dn )(Fi ) = (dp dq − dq dp )(Fi ) = 0 for i = 1, . . . , n. Therefore dp dq = dq dp . This completes the proof of Theorem 2.5.2. The next theorem shows that every component of a commutative basis of Derk (k[X]) is a special derivation (see Section 2.3). Theorem 2.5.5. Let k be a reduced ring containing Q and let k[X] = k[x1 , . . . , xn ] be the polynomial ring over k. If {d1 , . . . , dn } is a commutative basis of Derk (k[X]), then d?i = 0, for all i = 1, . . . , n.
34
Part I. Preliminary concepts
Proof. Denote by A the matrix [di (xj )] and let w = det(A). Then w is an invertible element of k[X] (Proposition 2.5.1) and so, since k is reduced, ∂ w is an invertible element of k. Let [bij ] be the matrix such that ∂x = i Pn Pn −1 j=1 bij dj , for i = 1, . . . , n. Then j=1 bij dj (xi ) = δij . Thus, [bij ] = A i+j −1 and bij = (−1) Aji w , where Apq is the determinant of the (n − 1) × (n − 1) matrix obtained from A by removing in A the p-th row and the q-th column. Set d1 (r) d1 D = ... and D(r) = ... , for r ∈ R. dn
dn (r)
Then A = [D(x1 ), . . . , D(xn )] and ∂ = w−1 det[D(x1 ), . . . , D(xi−1 ), D, D(xi+1 ), . . . , D(xn )], ∂xi for all i = 1, . . . , n. Therefore, if p ∈ {1, . . . , n} then 0 = w−1 dp (w) = w−1 dp (det[D(x1 ), . . . , D(xn )]) P = w−1 ni=1 det[D(x1 ), . . . , dp D(xi ), . . . , D(xn )] P = w−1 ni=1 det[D(x1 ), . . . , D(dp (xi )), . . . , D(xn )] Pn ∂ = i=1 ∂xi (dp (xi )) = d∗p . This completes the proof. The following example shows that the divergence of a noncommutative basis is not a constant, in general. Example 2.5.6. Let d1 = ∂/∂x1 + x22 ∂/∂x2 , d2 = ∂/∂x2 , . . . , dn = ∂/∂xn . Then {d1 , . . . , dn } is a basis of Derk (K[x]) (since det[di (xj )] = 1), and d∗1 = 2x2 6∈ k.
2.6
The image of derivations
Let k be a field of characteristic zero and let d be a k-derivation of k[X]. Let us extend the derivation d to the field k(X).
Chapter 2. Useful facts and preliminary results
35
This section is based on the result of K. Kishimoto and the author [47]. We will show that d(k(X)), the image of d, is always different from k(X). We will prove the following theorem, which is a more general one. Theorem 2.6.1. Let L be a field containing k and let d be a k-derivation of L. If transcendence degree of L over k is finite, then d(L) 6= L. First we note the following evident Lemma 2.6.2. Let L be a field, d : L −→ L a derivation, a ∈ L r {0} and δ = ad. Then d(L) = L if and only if δ(L) = L. The next fact is a special case of our theorem. Proposition 2.6.3. Let L be a finite field extension of k(X). If d is a k-derivation of L, then d(L) 6= L. Proof. Let M = k(X), s+1 = dimM L and denote by R the polynomial ring k[X]. Since the extension M ⊂ L is finite separable, there exists an element γ ∈ L such that L = M [γ] is a simple algebraic field extension. Let f = ts+1 + as ts + · · · + a1 t + a0 be the minimal polynomial of γ over M and let q be a common multiple of all denominators of the fractions a0 , . . . , as . Multiplying by q s+1 the equality γ s+1 + as γ s + · · · + a1 γ + a0 = 0 we obtain (qγ)s+1 + bs (qγ)s + · · · + b1 (qγ) + b0 = 0, for some b0 , . . . , bs ∈ R. Moreover, M [qγ] = M [γ]. Therefore we may assume that all the coefficients a0 , . . . , as of the polynomial f belong to R. By this assumption we see that the set S = {b0 + b1 γ + · · · + bs γ s ; b0 , . . . bs ∈ R} is a subring of L. Suppose now that d(L) = L. If d(x1 ) = · · · = d(xn ) = 0, then d is an M -derivation of L, and so d = 0 because the extension M ⊂ L is algebraic (Theorem 1.4.1). In this case we have a contradiction L = d(L) = 0. Therefore, there exists i ∈ {1, . . . , n} such that d(xi ) 6= 0. By a permutation of variables we may assume that d(x1 ) 6= 0.
36
Part I. Preliminary concepts
Consider the k-derivation d1 = d(x1 )−1 d. Then d1 (x1 ) = 1 and, by Lemma 2.6.2, d1 (L) = L. Moreover (since L = M [γ]): d1 (xi ) =
s X
Aij γ j , f or i = 2, . . . , n,
j=0
d1 (γ) =
s X
Bj γ j ,
j=0
where all elements Aij and Bj belong to M . Denote by A a common multiple of denominators of all elements Aij and Bj . Put δ = Ad1 . Then δ is a k-derivation of L such that δ(x1 ) = A, s X δ(xi ) = Pij γ j , f or i = 2, . . . , n, j=0
δ(γ) =
s X
Qj γ j ,
j=0
where A 6= 0, Pij , Qj are elements of R. Moreover, by Lemma 2.6.2, δ(L) = L and observe that δ(S) ⊂ S. Since k is infinite, there exists an element a ∈ k such that (x1 − a) - A. Denote y = x1 − a. Now we will show that 1/y 6∈ δ(L). Suppose that there exists u ∈ L such that 1/y = δ(u). Put u = w/b, w = w0 + w1 γ + · · · + ws γ s , where b, w0 , . . . , ws ∈ R and gcd(b, w0 , . . . , ws ) = 1. Then 1/y = δ(w/b) = (δ(w)b − δ(b)w)/b2 , so b2 = y(δ(w)b − δ(b)w), and so y|b2 , because the element (δ(w)b−δ(b)w) belongs to S. Let b = y m b1 , where b1 ∈ R, y - b1 and m > 1. Then y 2m b21 = b2 = y(δ(w)b − δ(b)w) = y(δ(w)y m b1 − y m δ(b1 )w − my m−1 Ab1 w) = y m+1 (δ(w)b1 − δ(b1 )w) − my m Ab1 w,
Chapter 2. Useful facts and preliminary results
37
and we see that yv = mAb1 w, for some v ∈ S. Let v = v0 + v1 γ + · · · + vs γ s , where v0 , . . . , vs ∈ R. Then (yv0 )+(yv1 )γ +· · ·+(yvs )γ s = (mAb1 w0 )+(mAb1 w1 )γ +· · ·+(mAb1 ws )γ s , that is, y|Ab1 w0 , y|Ab1 w1 , . . . , y|Ab1 ws . Since y - A, y - b1 and y = (x1 − a) is an irreducible polynomial in R we have: y|w0 , . . . , y|ws . Moreover y|b. But this contradicts to the assumption gcd(b, w0 , . . . , ws ) = 1. Therefore we showed that 1/y 6∈ δ(L) which completes the proof. Lemma 2.6.4 ([107]). Let F ⊂ L be an algebraic field extension and let d : L −→ L be a derivation such that d(F ) ⊂ F . If δ(F ) 6= F , then d(L) 6= L, where δ = d | F . Proof. Let u ∈ F r δ(F ). We will show that u 6∈ d(L). Suppose that u = d(γ) for some γ ∈ L. Since γ is algebraic over F , γ s + as−1 γ s−1 + · · · + a1 γ + a0 = 0, for some a0 , . . . as−1 ∈ F . Assume that s is minimal. Then 0 = d(0) = (sγ s−1 + (s − 1)as−1 γ s−2 + · · · + a1 )u + δ(as−1 )γ s−1 + · · · + δ(a1 )γ + δ(a0 ), so, by minimality of s, su + δ(as−1 ) = 0. Hence u = δ(−as−1 /s) ∈ δ(F ), but this is a contradiction. Now we are ready to prove our theorem. Proof of Theorem 2.6.1. Let {x1 , . . . , xn } be a transcendental basis of L over k and denote M = k(X), F = M (d(x1 ), . . . , d(xn )). Then k ⊂ M ⊂ F ⊂ L, the extension M ⊂ F is finite and the extension F ⊂ L is algebraic. Observe that d(M ) ⊂ F . In fact, if w ∈ M then d(w) =
∂w ∂x1 d(x1 )
+ ··· +
∂w ∂xn d(xn )
∈ F.
Since M ⊂ F is finite, there exists γ ∈ F such that F = M [γ] is a simple algebraic field extension. Let f = ts + as−1 ts−1 + · · · + a1 t + a0 ∈ M [t] be the minimal polynomial of γ over M. Then we have 0 = d(f (γ)) = f d (γ) + f 0 (γ)d(γ),
38
Part I. Preliminary concepts
where f d = d(1)ts +d(as−1 )ts−1 +· · ·+d(a1 )t+d(a0 ), and f 0 is the derivative of f . It is clear that f 0 (γ) 6= 0, f 0 (γ) ∈ F and f d (γ) ∈ F (since d(M ) ⊂ F ). Therefore d(γ) = −f d (γ)/f 0 (γ) is an element of F and hence d(F ) ⊂ F . Denote δ = d | F . Then, by Proposition 2.6.3, δ(F ) 6= F and, by Lemma 2.6.4, d(L) 6= L. This completes the proof of Theorem 2.6.1. As a consequence of Theorem 2.6.1 (or Proposition 2.6.3) we obtain the following Corollary 2.6.5. If d is a k-derivation of k(X), then d(k(X)) 6= k(X). We end this section with the following example which shows that if L = k(x1 , x2 , . . .) is a field of rational functions in an infinite set of variables, then the above corollary is not true in general. Example 2.6.6. Let L = Q(x1 , x2 , . . .). Then L = {a1 , a2 , . . .} is a countable set. Consider the derivation d of L such that d(xn ) = an , for n = 1, 2, . . .. Then d(L) = L.
Part II
Characterization of subalgebras of the form Ad Let k be a field of characteristic zero and let A be a finitely generated k-domain. We present a description of all k-subalgebras of A which are the rings of constants with respect to a k-derivation (or a family of kderivations) of A. The following theorem (which we prove here) is one of the main results of this paper (see Theorem 4.1.4): Theorem A. If B is a k-subalgebra of A, then the following conditions are equivalent: (1) There exists a k-derivation d of A such that B = Ad ; (2) The ring B is integrally closed in A and B0 ∩ A = B, where B0 is the field of fractions of B. Using this theorem we show (see Theorem 4.1.5) that if D is a family of k-derivations of A, then there exists a k-derivation d of A such that AD = Ad . Assume that G ⊆ GLn (k) is a connected algebraic group which acts on the polynomial ring k[X] = k[x1 , . . . , xn ]. As a consequence of Theorem A we obtain Theorem 4.2.4 which states that there exists a k-derivation d of k[X] such that the invariant ring k[X]G is equal to k[X]d . The present part of the paper is based on the author’s paper [83]. We have divided this part into two chapters (Chapter 3 and Chapter 4). Chapter 4 contains the proofs of the above theorems. We use here several facts from the previous Chapter 3 devoted to the characterization of all the subfields of a field L containing k which are of the form Ld , for some kderivation d of L. It is well known (see [107], [16]) that if k ⊆ L is a field extension of finite transcendence degree, then an intermediate field M is of the form Ld if and only if M is algebraically closed in L. This result is very important in the proof of Theorem A. It is a simple consequence of the fact that for every n ∈ N there exists a k-derivation dn of k[x1 , . . . , xn ] with the trivial field of constants, that is, k(x1 , . . . , xn )dn = k. In Section 3.2 we have collected several known examples of derivations with the trivial field of constants 39
40
Part I. Preliminary concepts
and we describe (see the proof of Theorem 3.2.2) an effective method for a construction of such derivations. In Chapter 3 we also show (Theorem 3.3.2) that the result on the intermediate fields is valid for an arbitrary field extension (without any assumption on the transcendence degree). For this aim we prove (see Theorem 3.2.7) that if the extension k ⊂ L is purely transcendental, then there exists a derivation d of L such that Ld = k.
3
Characterization of subfields
3.1
Initial observations
Let A be a k-domain and let A0 be its field of fractions. Assume that D is a family of k-derivations of A and denote by D the set {d; d ∈ D}, where d is the natural extension of d to A0 . In this situation we have two subfields of A0 : (AD )0 = the field of fractions of AD , (A0 )D = the field of constants of A0 with respect to D. The following example shows that these subfields could be different. Example 3.1.1. Let d be a k-derivation of A = k[x, y] such that d(x) = x and d(y) = y. Then (Ad )0 6= (A0 )d . Proof. It is clear that Ad = k. Thus, (Ad )0 = k. However (A0 )d 6= k, since x/y ∈ (A0 )d . Indeed: d(x/y) = (d(x)y − xd(y))/y 2 = (xy − xy)/y 2 = 0/y 2 = 0. Proposition 3.1.2. If D is a family of k-derivations of a k-domain A, then AD ⊆ (AD )0 ⊂ (A0 )D ⊆ A0 and (AD )0 ∩ A = (A0 )D ∩ A = AD . The proof is straightforward. Note the following two propositions. Proposition 3.1.3. If D is a family of k-derivations of a k-domain A, then the ring AD is integrally closed in A. Proof. Let x ∈ A be an integral element over AD and let xn + c1 xn−1 + · · · + cn−1 x + cn = 0, where c1 , . . . , cn ∈ AD and n is minimal. If d ∈ D then 0 = d(0) = ud(x), where u = nxn−1 + (n − 1)c1 xn−2 + · · · + cn−1 . Since u 6= 0 (because n is minimal and char(k)=0), d(x) = 0 and hence, x ∈ Ad for any d ∈ D, i. e., x ∈ AD . Proposition 3.1.4. If D is a family of k-derivations of a field L containing k, then the field LD is algebraically closed in L. The proof is similar to the proof of Proposition 3.1.3. 41
Part II. Characterization of subalgebras of the form Ad
42
3.2
Derivations of purely transcendental field extensions
Let S be a set of algebraically independent elements over k. Denote by |S| the cardinality of S and consider the field k(S), the pure transcendental extension of k. In this section we present k-derivations d of k(S) such that k(S)d = k. Let us start from the following Lemma 3.2.1 ([107] Lemma 4). Let k ⊂ k(t) ⊆ L be fields of characteristic zero, where t ∈ L is a transcendental element over k. Let d be a derivation of k and let u be an element from k r d(k). Assume that δ is the unique derivation of k(t) such that δ|k = d and δ(t) = u. Then k(t)δ = k d . Proof. Let g be a polynomial in k[t]. Then δ(g) ∈ k[t] (because u ∈ k) and deg δ(g) 6 deg g. If g is monic, then deg δ(g) < deg g. Observe that if δ(g) = 0, then g ∈ k. Indeed: set g = an tn + · · · + a0 , where an , . . . , a0 ∈ k with an 6= 0, and suppose that n > 0. Then 0 = δ(g) = d(an )tn + (d(an−1 ) + nan u)tn−1 + · · · , and hence, d(an ) = 0 and d(an−1 ) + nan u = 0. This implies that n−1 u = − na1n d(an−1 ) = d(− ana ) ∈ d(k), n
which is a contradiction. Assume now that α ∈ k(t)δ . Put α = f /g, g 6= 0, where f, g ∈ k[t] and gcd(f, g) = 1. We may assume that g is monic. Then 0 = δ(α) = δ(f /g) = g −2 (δ(f )g − δ(g)f ) so, δ(f )g = δ(g)f . Hence, g | δ(g) (because f and g are coprime) and deg δ(g) < deg g (because g is monic). Therefore, δ(g) = 0 and hence, g ∈ k and consequently, f ∈ k, that is, α ∈ k. This means that k(t)δ ⊆ k d . The opposite inclusion is obvious. Using this lemma and Theorem 2.6.1 we may prove the following Theorem 3.2.2. If L = k(x1 , . . . , xn ) is the field of rational functions over a field k of characteristic zero, then there exists a derivation d of L such that Ld = k.
Chapter 3. Characterization of subfields
43
Proof. If n = 1 then k(x1 )d = k, where d is the derivation ∂/∂x1 . Let n > 1. Put M = k(x1 , . . . , xn−1 ) and assume that δ is a derivation of M with M δ = k. In view of Theorem 2.6.1 there exists an element u in M which is not in δ(M ). Let d be the derivation of L = M (xn ) such that d|M = δ and d(xn ) = u. Then (by Lemma 3.2.1) Ld = M δ = k. The above proof gives an effective way to construct k-derivations of k(x1 , . . . , xn ) with trivial field of constants. Look at the known examples of such derivations. Proposition 3.2.3. Let d1 , d2 , d3 , d4 be k-derivations of k(x1 , . . . , xn ) defined as follows d1 =
∂ ∂x1
+ (x1 x2 + 1) ∂x∂ 2 + (x2 x3 + 1) ∂x∂ 3 + . . . + (xn−1 xn + 1) ∂x∂n ,
d2 =
∂ ∂x1
+
1 ∂ x1 ∂x2
+
1 ∂ x2 ∂x3
d3 =
∂ ∂x1
+
1 ∂ x1 ∂x2
+
1 ∂ x1 x2 ∂x3
d4 =
∂ ∂x1
+ x2 ∂x∂ 2 + x2 x3 ∂x∂ 3 + . . . + x2 . . . xn ∂x∂n .
+ ··· +
∂ 1 xn−1 ∂xn ,
+ ... +
1 ∂ x1 ...xn−1 ∂xn ,
Then k(x1 , . . . , xn )di = k, for i = 1, 2, 3, 4. Proof. If i = 1 then it is a consequence of Shamsuddin’s result [101] mentioned in [38] (we will prove it in Section 13.4, see Example 13.4.3). For i = 2 and i = 3 see Suzuki [107]. If i = 4 then see Derksen [16]. It follows also from the property of d3 , because d4 = x1 . . . xn−1 d3 (under a permutation of variables). Another examples of k-derivations of k(x1 , . . . , xn ) with trivial field of constants we may find in Chapters 13, 10 and 12. Applying the same argument as Suzuki [107] in his proof for the derivation d2 we obtain (for countable S): Proposition 3.2.4. Let d be the k-derivation of L = k(x1 , x2 , . . .) defined 1 by d(xi ) = xi−1 , for i = 1, 2, . . ., where x0 = 1. Then Ld = k. There exists a simpler example of k-derivation in k(x1 , x2 , . . .) with the trivial field of constants. It is not difficult to prove the following Proposition 3.2.5. Let d be the k-derivation of L = k(x1 , x2 , . . .) defined by d(xi ) = xi+1 , for i = 1, 2, . . .. Then Ld = k.
Part II. Characterization of subalgebras of the form Ad
44
A similar derivation, as in Proposition 3.2.5, may be constructed for any infinite cardinality of S. Proposition 3.2.6. Let S be an infinite set of algebraically independent elements over k and let L = k(S). Then there exists a k-derivation d of L such that Ld = k. Proof. Since S is infinite, there exists a well-order 6 on S without maximal element. If s ∈ S then denote by s? the next element of s, that is, s? = min{t ∈ S; s < t}. Now let d be the k-derivation of L defined by d(s) = s? , for any s ∈ S. It is clear that Ld = k. From the above propositions we get the following Theorem 3.2.7. If k ⊂ L is a purely transcendental field extension, then there exists a derivation d of L such that Ld = k.
3.3
Algebraically closed subfields
Suzuki, in [107], and Derksen, in [16], showed that if k ⊂ L is an extension of fields (of characteristic zero) of finite transcendence degree, then every intermediate field, which is algebraically closed in L, is the field of constants for a k-derivation of L. In the proofs they used the derivations d2 , d3 and d4 from Proposition 3.2.3 and, moreover, they used the following evident Lemma 3.3.1. Let k ⊂ L be an algebraic field extension, d : k −→ k a derivation, and δ : L −→ L the derivation which is the unique extension of d to L. If the field k d is algebraically closed in L, then Lδ = k d . Thanks to Lemma 3.3.1 and Theorem 3.2.7 we see that the proofs of Suzuki and Derksen are valid for arbitrary field extension (without any assumption on the transcendence degree). Theorem 3.3.2. Let K ⊆ L be fields of characteristic zero. The following conditions are equivalent: (1) There exists a derivation d of L such that Ld = K; (2) K is algebraically closed in L.
Chapter 3. Characterization of subfields
45
Proof. (1) ⇒ (2) follows from Proposition 3.1.4. (2) ⇒ (1) Let S be a transcendence basis of L over K. Then the extension K(S) ⊆ L is algebraic. Let d0 : K(S) −→ K(S) be a derivation 0 such that K(S)d = K (Theorem 3.2.7), and let d : L −→ L be the unique 0 extension of d0 to L. Then, by Lemma 3.3.1, Ld = K(S)d = K. Applying this theorem and Proposition 3.1.4 one can prove, for instance, the following Theorem 3.3.3. Let D be a family of derivations of a field L. Then there exists a derivation d of L such that LD = Ld .
4
Characterization of subalgebras
4.1
Integrally closed subrings
It follows from Theorem 3.3.2 that the converse of Proposition 3.1.4 is also true. Now let us return to Proposition 3.1.3. Let A be a k-domain and let B be a k-subalgebra of A which is integrally closed in A. We may ask the following Question 4.1.1. Is B a ring of constants with respect to k-derivations of A? This question has a negative answer in general Example 4.1.2. Let A = k[x1 , . . . , xn ] (n > 2) be the polynomial ring over k and let B be the integral closure of the ring k[x1 , x1 x2 ] in A. Then of course B is integrally closed in A and x2 6∈ B (see the example of Gustafson in [110] p. 489). Therefore k[x1 , x1 x2 ] ⊆ B ⊂ k[x1 , x2 ] ⊆ A. Suppose that D is a family of k-derivations of A such that AD = B. Let d ∈ D. Then d(x1 ) = 0 and 0 = d(x1 x2 ) = x1 d(x2 ). Hence x2 ∈ B and we have a contradiction. Observe that if B and A are as in Example 4.1.2, and B0 is the field of fractions of B, then B0 = k(x1 , x2 ), so B0 ∩ A = k[x1 , x2 ] 6= B. Rings of constants have the following additional property: Proposition 4.1.3. Let D be a family of k-derivations of a k-domain A and let B = AD . Then B0 ∩ A = B. Proof. Denote by D0 the set {d0 ; d ∈ D}, where d0 is the k-derivation of A0 defined by d0 ( ab ) = (d(a)b − ad(b))b−2 , for all a, b ∈ A and b 6= 0. Let 0 M be the field AD 0 . Then it is clear that B0 ⊆ M and M ∩ A = B. Hence B ⊆ B0 ∩ A ⊆ M ∩ A = B, that is, B0 ∩ A = B. Now we are able to prove the following description of all k-subalgebras of a finitely generated k-domain, which are rings of constants with respect to k-derivations. 46
Chapter 4. Characterization of subalgebras
47
Theorem 4.1.4. Let A be a finitely generated k-domain, where k is a field of characteristic zero. Let B be a k-subalgebra of A. The following conditions are equivalent: (1) There exists a k-derivation d of A such that B = Ad ; (2) The ring B is integrally closed in A and B0 ∩ A = B. Proof. (1) ⇒ (2) follows from Propositions 3.1.3 and 4.1.3. (2) ⇒ (1). Let M be the algebraic closure of the field B0 in the field A0 . By Theorem 3.3.2 there exists a k-derivation δ : A0 −→ A0 such that M = Aδ0 . Since A is finitely generated over k, A = k[f1 , . . . , fs ] and A0 = k(f1 , . . . , fs ), for some f1 , . . . , fs ∈ A. Let w be a nonzero element of A such that the elements wδ(f1 ), . . . , wδ(fs ) belong to A, and let δ 0 = wδ. 0 Then Aδ0 = Aδ0 = M and δ 0 (A) ⊆ A. Consider the k-derivation d of A which is the restriction of δ 0 to A. We will show that Ad = B. For this purpose observe, at first, that d A = M ∩ A. In fact; If x ∈ M ∩ A then x ∈ A and δ 0 (x) = 0, hence d(x) = δ 0 (x) = 0, i. e, x ∈ Ad . If x ∈ Ad then x ∈ A and δ 0 (x) = d(x) = 0, so x ∈ M ∩ A. Now we will prove that M ∩ A = B. The inclusion B ⊆ M ∩ A is clear. Assume that x ∈ M ∩ A. Then x ∈ A and x is algebraic over B0 . So, there exists a natural number n such that pn n pn−1 n−1 p0 p1 x + =0 x + . . . + x1 + qn qn1 q1 q0 where p0 , . . . , pn ; q0 , . . . , qn ∈ B, pn 6= 0 and q0 q1 . . . qn 6= 0. Multiplying the two sides of the above equality by q0 q1 . . . qn we have cn xn + cn−1 xn−1 + . . . + c1 x + c0 , where c0 , . . . , cn ∈ B and cn 6= 0. Denote y = cn x. Then y ∈ A and y n + cn−1 y n−1 + cn−2 cn y n−2 + . . . + c1 cnn−2 y 1 + c0 cnn−1 = 0. This means that y is an element of A which is integral over B. So y ∈ B, because B is integrally closed in A. Hence x = yc−1 n ∈ B0 and hence x ∈ B0 ∩ A = B. Therefore M ∩ A = B and we have Ad = M ∩ A = B. The following is an immediate consequence of Theorem 4.1.4, Proposition 3.1.3 and Proposition 4.1.3. Theorem 4.1.5. Let A be a finitely generated k-domain and let D be a family of k-derivations of A. Then there exists a k-derivation d of A such that AD = Ad .
48
4.2
Part II. Characterization of subalgebras of the form Ad
Rings of invariants
Let A be a k-domain and G a subgroup of Autk (A), the group of all k-automorphisms of A. Denote AG = {a ∈ A; σ(a) = a, f or any σ ∈ G}. The set AG is a k-subalgebra of A. We may ask the following Question 4.2.1. Is AG of the form Ad , for some k-derivation d of A ? It is evident that if B = AG then B0 ∩ A = B. Therefore, by Theorem 4.1.4, our question reduces to Question 4.2.2. Is AG integrally closed in A ? If G is finite then it is well known that A is integral over AG (see, for instance, [5] Exercises in Section 5). This means that our questions have negative answers in general. Proposition 4.2.3. Let A be a k-domain and G ⊆ Autk (A) a group. Assume that G has no proper subgroups of finite index. Then AG is integrally closed in A. Proof. Let us denote B = AG . Assume that a ∈ A is an integral element over B and f ∈ B[t] is a monic polynomial such that f (a) = 0. Then f (σ(a)) = σ(f (a)) = σ(0) = 0, for any σ ∈ G, hence S = {σ(a); σ ∈ G} is a set of roots of f . Since B is a k-domain, the polynomial f has only a finite set of roots. Let {r1 = a, r2 , . . . , rs } be the set of all roots of f belonging to S and let Gi = {σ ∈ G; σ(a) = ri }, for i = 1, . . . , s. Then G = G1 ∪ . . . ∪ Gs and Gi ∩ Gj = ∅, for i 6= j, and we see that G1 is a subgroup of G and its index is equal to s < ∞. So s = 1 and hence σ(a) = a, for any σ ∈ G. Therefore a ∈ B = AG . Assume now that G ⊆ Gln (k) is an algebraic group which acts on k[x1 , . . . , xn ], the polynomial ring over k. If G is connected then G has no closed proper subgroup of finite index (see for instance [36] p. 53). Repeating the argument of the proof of Proposition 4.2.3, we see that (if G is connected) the ring k[x1 , . . . , xn ]G is integrally closed in k[x1 , . . . , xn ]. Therefore, by Theorem 4.1.4, we obtain
Chapter 4. Characterization of subalgebras
49
Theorem 4.2.4. Let k be a field of characteristic zero, and G ⊆ Gln (k) a connected algebraic group. Then there exists a k-derivation d of k[x1 , . . . , xn ] such that k[x1 , . . . , xn ]G = k[x1 , . . . , xn ]d . Let d be a k-derivation of the polynomial ring k[X] = k[x1 , . . . , xn ]. Quite naturally the question arises whether k[X]d is a finitely generated kalgebra. This question has an obvious connection with the fourteenth problem of Hilbert. In his recent paper [16] Derksen showed that the Nagata’s counterexample [67] can be put in the form k[X]d for some k-derivation d with n = 32. Thus, he proved the following Theorem 4.2.5 ([16]). Let A = k[x1 , . . . , xn ], where n = 2r2 , r = 4, 5, . . .. There exists a k-derivation d of A such that the ring Ad is not finitely generated over k. The above theorem is a simple consequence of Theorem 4.2.4. It is also a consequence of Theorem 4.1.5 because it is clear that the ring in the Nagata’s counterexample is of the form k[X]D , where D is a family of locally nilpotent k-derivations of k[X]. In 1990, P. Roberts in [94], constructed a new counterexample to the fourteenth problem of Hilbert. From this result and from Theorem 4.2.4 one can deduce that if n = 7 then there exists a k-derivation d of k[X] such that k[X]d is not finitely generated. The last fact implies that if n > 7 then there always exists such a k-derivation d of k[X] that k[X]d is not finitely generated.
Part III
Finiteness and properties of Ad Let d be a k-derivation of k[X] = k[x1 , . . . , xn ]. As we mentioned at the end of the previous chapter, it is known that the ring k[X]d need not be finitely generated over k. Recently Deveney and Finston, in [18], gave an explicit example of a k-derivation of k[X] (for n = 7) with the nonfinitely generated ring of constants (see Section 6.3). The derivation d in this example is locally nilpotent. In general, it is not known when the ring k[X]d is finitely generated. However, in several special cases the finiteness of k[X]d is known. In 1932, Weitzenb¨ock in [111] proved that every linear locally nilpotent k-derivation of k[X] has a finitely generated ring of constants. Nagata [66] has generalized this fact lately. He has proved that it is also true for every linear k-derivation of k[X]. In 1988, Nagata and the author, in [85], observed that using a result of Zariski [113] one can prove that if n 6 3, then k[X]d is always finitely generated (see Section 7.1). It is well known ([73], [91], [21], [23]) that if d is locally nilpotent and there exists an element b ∈ k[X] such that d(b) = 1, then k[X]d is finitely generated. In this case one can give an explicit formula for a generating set of k[X]d (see Section 6.4). The present part of the paper is devoted to the above problems and to some properties of the rings of constants of polynomial derivations. This part consists of the following three chapters: Chapter 5. General properties of the rings of constants for polynomial derivations, Chapter 6. Constants for locally nilpotent derivations, Chapter 7. Rings of constants for small n.
50
5
General properties of the rings of constants for polynomial derivations
This chapter has three independent sections relating to the ring of constants of a k-derivation d of k[X] = k[x1 , . . . , xn ]. In Section 5.1 we show that some problems concerning the ring k[X]d can be reduced to the case when the field k is algebraically closed or even, to the case when k is the field C of complex numbers. In the short Section 5.2 we prove that generators of k[X]d have a special property; they are the so-called closed polynomials. We will study closed polynomials in the next chapter. The aim of Section 5.3 is to exhibit the basic relation between the ring k[X]d and the set of all the first integrals of a polynomial system of ordinary differential equations. A useful formula describing all the formal first integrals is also given in this section (see Corollary 5.3.6).
5.1
Extension of scalars
Let k ⊂ k 0 be a field extension and let d be a k-derivation of a kalgebra A. Consider the k 0 -algebra A ⊗k k 0 and the k 0 -linear mapping d ⊗ 1 : A ⊗k k 0 −→ A ⊗k k 0 . It is obvious that d ⊗ 1 is a k 0 -derivation of A ⊗ k 0 . Note the following two propositions Proposition 5.1.1. The k 0 -algebras Ad ⊗k k 0 and (A ⊗k k 0 )d⊗1 are isomorphic. i
d
Proof. The exact sequence 0 −→ Ad −→ A −→ A of vector spaces over k (where i(x) = x, for x ∈ Ad ) induces the exact sequence i⊗1
d⊗1
0 −→ Ad ⊗k k 0 −→ A ⊗k k 0 −→ A ⊗k k 0 of vector spaces over k 0 . Thus, (A ⊗k k 0 )d⊗1 = Ker(d ⊗ 1) = Im(i ⊗ 1) ≈ Ad ⊗k k 0 . Proposition 5.1.2. A is a finitely generated algebra over k if and only if A ⊗k k 0 is a finitely generated algebra over k 0 . Proof. If {x1 , . . . , xn } is a finite generating set of A over k then {x1 ⊗ 1, . . . , xn ⊗ 1} is a finite generating set of A ⊗k k 0 over k 0 . So, if A is finitely generated then A ⊗k k 0 is too. 51
Part III. Finiteness and properties of Ad
52
Assume now that A ⊗k k 0 is finitely generated over k 0 . Since {a ⊗ 1; a ∈ A} is a generating set of A ⊗k k 0 , there exists a finite subset X of A such that {x ⊗ 1; x ∈ X} is a finite generating set of A ⊗k k 0 . We will show that k[X] = A. For this aim let us consider the exact sequence i
p
0 −→ k[X] −→ A −→ A/k[X] −→ 0 of vector spaces over k, where i(u) = u for u ∈ k[X] and p is the natural map. This sequence induces the exact sequence over k 0 : i⊗1
p⊗1
0 −→ k[X] ⊗k k 0 −→ A ⊗k k 0 −→ A/k[X] ⊗k k 0 −→ 0. Hence, A/k[X] ⊗k k 0 ≈ (A ⊗k k 0 )/Im(i ⊗ 1) ≈ (A ⊗k k 0 )/(k[X] ⊗k k 0 ) = 0, and hence k[X] = A. Corollary 5.1.3. If k ⊂ k 0 is a field extension and d is a k-derivation of a k-algebra A, then (1) Ad = k if and only if (A ⊗k k 0 )d⊗1 = k 0 . (2) Ad is a finitely generated k-algebra if and only if (A ⊗k k 0 )d⊗1 is a finitely generated k 0 -algebra. The above corollary we will use mostly in a case when A is a polynomial ring in a finite set of variables. In such a case from Corollary 5.1.3 we get Theorem 5.1.4. Let d be a k-derivation of the polynomial ring k[X] = k[x1 , . . . , xn ], where k is a field. Let k 0 be an overfield of k and denote by d0 the k 0 -derivation of k 0 [X] such that d0 (xi ) = d(xi ) for i = 1, . . . , n. Then 0 the rings of constants k[X]d and k 0 [X]d have the following properties 0 (1) k[X]d = k if and only if k 0 [X]d = k 0 . 0 (2) k[X]d is a finitely generated k-algebra if and only if k 0 [X]d is a finitely generated k 0 -algebra. Thus when proving the nonexistence (or the finiteness) of the nontrivial polynomial solutions of the equation d(F ) = 0, where d is a k-derivation of k[x1 , . . . , xn ] and F ∈ k[x1 , . . . , xn ], without any restriction of generality one can suppose that k is an algebraically closed field. The next proposition shows that the field k(X)d has a similar property. Proposition 5.1.5. Let d be a k-derivation of k[X], and let k 0 be an overfield of k. Denote by d0 the k 0 -derivation of k 0 [X] such that d0 (xi ) = d(xi ) 0 for i = 1, . . . , n. Then k(X)d = k if and only if k 0 (X)d = k 0 .
Chapter 5. General properties of the rings of constants. . .
53
For the proof of this proposition we need three lemmas. Lemma 5.1.6. Assume that L(t) is the field of rational functions in one variable t over a field L containing k. Let d be a k-derivation of L such that Ld = k, and let δ be the k-derivation of L(t) such that δ | L = d and δ(t) = 0. Then L(t)δ = k(t). Proof. It is obvious that k(t) ⊆ L(t)δ . Let w = f /g ∈ L(t)δ with f, g ∈ L[t], and put p = deg(g). If p = 0 then it is clear that w ∈ k[t] ⊂ k(t). Let p > 1. We may assume that g is monic and that p is minimal. Set g = tp + b1 tp−1 + · · · + bp , where b1 , . . . , bp ∈ L. Then we have: δ(g) = d(b1 )tp−1 + · · · + d(bp ). Observe that δ(g) = 0. Indeed, if δ(g) 6= 0 then w = δ(f )/δ(g) and we have a contradiction with the minimality of p. Thus, d(b1 ) = · · · = d(bp ) = 0 and hence, g ∈ k[t] (because Ld = k). Consequently, δ(f ) = δ(g)w = 0 · w = 0, that is, f ∈ k[t]. Therefore, w = f /g ∈ k(t). Lemma 5.1.7. Let L be a field containing k and let S be a set of algebraically independent elements over L. Let d be a k-derivation of L such that Ld = k. If δ is the k-derivation of L(S) such that δ | L = d and δ(s) = 0, for every s ∈ S, then L(S)δ = k(S). Proof. It is obvious that k(S) ⊆ L(S)δ . Let w ∈ L(S)δ . Since w ∈ L(S), there exists a finite subset S 0 of S such that w ∈ L(S 0 ). Denote by δ 0 0 the restriction of the derivation δ to L(S 0 ). Then w ∈ L(S 0 )δ and hence, by Lemma 5.1.6 and by induction, w ∈ k(S 0 ) ⊆ k(S). Lemma 5.1.8. Let F be a field containing k and let F ⊆ k 0 be an algebraic field extension. Assume that δ is an F -derivation of F [X] and let d0 be the k 0 -derivation of k 0 [X] such that d0 (xi ) = δ(xi ) for i = 1, . . . , n. If 0 F (X)δ = F then k 0 (X)d = k 0 . 0
0
Proof. Suppose that k 0 (X)d 6= k 0 and let w ∈ k 0 (X)d r k 0 . Since 0 k (X) is algebraic over F (X), there exist a1 , . . . , ap ∈ F (X) such that wp + a1 wp−1 + · · · + ap = 0. Let us assume that p is minimal. Then δ(a1 )wp−1 + · · · + δ(ap ) = d0 (wp + a1 wp−1 + · · · + ap ) = d0 (0) = 0 and hence, by the minimality of p, δ(a1 ) = · · · = δ(ap ) = 0, that is, a1 , . . . , ap ∈ F (because F (X)δ = F ). Thus, w is algebraic over F . This implies that w
Part III. Finiteness and properties of Ad
54
is algebraic over k 0 . But k 0 is algebraically closed in k 0 (X). Hence, w ∈ k 0 and we have a contradiction. 0
Proof of Proposition 5.1.5. If k 0 (X)d = k 0 then it is clear that k(X)d = k. Assume now that k(X)d = k. Let S be a subset of k 0 such that the extension k ⊂ k(S) is purely transcendental and the extension k(S) ⊂ k 0 is algebraic. Put L = k(X), F = k(S). Then, by Lemma 5.1.7, F (X)δ = L(S)δ = k(S) = F , where δ is the F -derivation of F [X] such that δ(xi ) = d(xi ) for all i = 1, . . . , n. Now Lemma 5.1.8 implies that 0 k 0 (X)d = k 0 . Now we will show that in some cases we may suppose that k is the field C of complex numbers. Proposition 5.1.9. Let f1 , . . . , fn be polynomials in Z[x1 , . . . , xn ] and consider the equation ∂F ∂F f1 ∂x = PF. (5.1) + · · · + fn ∂x n 1 If this equation does not admit a solution F, P ∈ C[x1 , . . . , xn ] with F 6∈ C, then this equation has no solutions F, P ∈ R[x1 , . . . , xn ] with F 6∈ R, where R is a domain of characteristic zero. Proof. Let R be a domain of characteristic zero and assume that polynomials F, P ∈ R[x1 , . . . , xn ] with F 6∈ R form a solution of (5.1). Consider the finite set S ⊂ R of all coefficients of polynomials F, P and the field k = Q(S), the smallest subfield of R0 containing S, where R0 is the field of fractions of R. It is clear that the polynomials F and P form the solution of (5.1) in k[x1 , . . . , xn ] and F 6∈ k. Since S is finite, there exists a field embedding of k in C such that F 6∈ C. So, we may assume that F, P ∈ C[x1 , . . . , xn ] with F 6∈ C. Therefore, we obtain a contradiction, because the equation (5.1) does not admit any solution in C[x1 , . . . , xn ] with F 6∈ C.
5.2
Closed polynomials
Consider the following family L of subrings in k[X] = k[x1 , . . . , xn ]: L = {k[f ]; f ∈ k[X] r k}. If k[f ] ( k[g], for some f, g ∈ k[X] r k, then deg(f ) > deg(g) and hence, we see that in the family L there exist maximal elements. We will say that a polynomial f ∈ k[X] r k is closed if the ring k[f ] is a maximal element in L.
Chapter 5. General properties of the rings of constants. . .
55
Proposition 5.2.1. Let f ∈ k[X] r k. Then f is closed if and only if the ring k[f ] is integrally closed in k[X]. In the proof of this proposition we will use the following result of Zaks ([112], see also [19] or [96]). Theorem 5.2.2 ([112]). If R is a Dedekind subring of k[X] containing k, then there exists a polynomial f ∈ k[X] such that R = k[f ]. Proof of Proposition 5.2.1. Assume that k[f ] is a maximal element in L and denote by E the integral closure of k[f ] in k[X]. Then E is a Dedekind subring of k[X] containing k so, by Theorem 5.2.2, E = k[h], for some h ∈ k[X]. Now, by the maximality of k[f ] in L, k[f ] = k[h] = E and so, the ring k[f ] is integrally closed in k[X]. Assume now that k[f ] is integrally closed in k[X] and let k[f ] ⊆ k[g] for some g ∈ k[X]. Then f ∈ k[g], that is, f = as g s + · · · + a1 g + a0 , for some a0 , . . . , as ∈ k with as 6= 0. Hence s−1 −1 g s + a−1 + · · · + a−1 s as−1 g s a1 g + (as a0 − f ) = 0
and hence g is integral over k[f ]. Since k[f ] is integrally closed in k[X], k[f ] = k[g] and we see that k[f ] is maximal in L.
Theorem 5.2.3. Let d be a k-derivations of k[X]. Then there exists a family S of closed polynomials in k[X] such that k[X]d = k[S]. Proof. If k[X]d = k then we put S = ∅. Assume that k[X]d 6= k and let f ∈ k[X]d rk. Let hf be a polynomial in k[X]rk such that k[f ] ⊆ k[hf ] and k[hf ] is a maximal element in L. Then f = u(hf ) for some polynomial u(t) ∈ k[t]. We may assume that u has the minimal degree. Now, using the same argument as in the proof of Proposition 3.1.3, we see that hf ∈ k[X]d . Put S = {hf ; f ∈ k[X]d r k}. Then S is a family of closed polynomials in k[X] and it is clear that k[X]d = k[S].
5.3
First integrals and the ring of constants
The aim of this section is to exhibit the basic relation between the ring of constants for a k-derivation of k[X] = k[x1 , . . . , xn ] and the set of all the first integrals of a polynomial system of differential equations.
56
Part III. Finiteness and properties of Ad
Let us recall the definition and some facts concerning first integrals in the theory of ordinary differential equations. Consider a system of differential equations in Rn of the form dxi = fi (t, x1 , . . . , xn ), i = 1, . . . , n, dt
(5.2)
where t is an independent variable and x1 , . . . , xn are unknown functions of t. Let F (t, x1 , . . . , xn ) be a function defined and continuous together with its ∂F partial derivatives ∂F ∂t , ∂xi , i = 1, . . . , n, in a domain D. It is said to be the first integral of system (5.2) if, for an arbitrary solution x1 (t), . . . , xn (t) of system (5.2) this function takes a constant value, that is, the function F (t, x1 , . . . , xn ) depends only on the choice of a solution x1 (t), . . . , xn (t), not on the variable t. Example 5.3.1. Consider the following differential system in R2 : dx dy = y, = −x. dt dt
(5.3)
It is not difficult to check (using for example Theorem 2.4.3) that if x(t), y(t) is a solution of (5.3) then x(t) = a cos t + b sin t,
y(t) = −a sin t + b cos t
(5.4)
for some a, b ∈ R. Let F (t, x, y) = x2 + y 2 . Substituting (5.4) we get F (t, x(t), y(t)) = a2 + b2 . Hence, function F (t, x(t), y(t)) is constant (not dependent on t). Therefore F , is a first integral of system (5.3). Example 5.3.2. Consider the following differential system in R3 : dy1 dy2 dy3 = y2 + y3 , = y 1 + y3 , = y1 + y2 . dt dt dt
(5.5)
Substracting termwise the second equation from the first we get d(y1 − y2 ) = −(y1 − y2 ), that is, y1 − y2 = ae−t , dt where a ∈ R. This means that the function F1 (t, y1 , y2 , y3 ) = (y1 −y2 )et is a first integral of system (5.5). In a similar way we obtain next first integrals of (5.5): F2 = (y2 − y3 )et , F3 = (y1 + y2 + y3 )e−2t . Multiplying F22 by F3 we get a polynomial first integral F22 F3 = (y1 − y2 )2 (y1 + y2 + y3 ).
Chapter 5. General properties of the rings of constants. . .
57
It is well known that first integrals play an important role in the problem of finding solutions (in a given class of functions) of system (5.2). If n independent first integrals of system (5.2) are obtained, then its integration is complete; if, however, m independent first integrals are obtained, where m < n, then system (5.2) is reduced to a system with a smaller number of unknown functions. For a function F (t, x1 , . . . , xn ) to be a first integral of system (5.2) it is necessary and sufficient that the condition n
∂F X ∂F =0 + fi (t, x1 , . . . , xn ) ∂t ∂xi i=1
should hold in the domain D. Now we will use the notations as in Sections 1.6 and 2.4. Assume that f = (f1 , . . . , fn ) is a sequence of polynomials in k[X] = k[x1 , . . . , xn ] and consider the autonomous system of differential equations ∂X = f (X) ∂t
(5.6)
and its flow ϕ(t, a) = (ϕ1 (t, a), . . . , ϕn (t, a)). Introduce the following Definition 5.3.3. A series F ∈ k[X][[t]] r k is said to be a formal first integral of autonomous system (5.6) if ∀a∈kn F (ϕ(t, a)) ∈ k. Observe that all the first integrals from Examples 5.3.1 and 5.3.2 are formal first integrals. P ∂ Denote by d (as in Section 2.4) the derivation ni=1 fi ∂x and let d˜ be the i P P ∞ ∞ p p ˜ derivation of k[X][[t]] defined by d( p=0 d(Fp )t . Moreover, p=0 Fp t ) = ˜ let ∆ be the k-derivation of k[X][[t]] equal to ∂ + d. ∂t
P∞
Theorem 5.3.4. Let F = p=0 Fp tp be a series from k[X][[t]] r k. The following conditions are equivalent: (1) F is a formal first integral of system (5.6). (2) πa Ed (F ) ∈ k, for any a ∈ k n . (3) Ed (F ) = F0 . (4) F = E−d (F0 ). (5) ∆(F ) = 0. Proof. (1) ⇔ (2) follows P from Lemma 2.4.5. p (2) ⇔ (3). Let Ed (F ) = ∞ p=0 Gp t , where Gp ∈ k[X] for all p ∈ N0 . Then G0 = F0 (see Proposition 2.4.1). Condition (2) is equivalent (by
Part III. Finiteness and properties of Ad
58
Lemma 1.6.3) to the condition: Gp = 0 for all p > 1, that is, Ed (F ) = G 0 = F0 . (3) ⇔ (4) follows from the equality E−d = Ed−1 . ∂ (2) ⇔ (5). In view of Lemma 2.4.4 we know that ∂t Ed = Ed ∆. Therefore: ∀a∈kn πa Ed (F ) ∈ k ⇔ ∀a∈kn
∂ ∂t πa Ed (F )
=0
⇔ ∀a∈kn πa Ed (∆(F )) = 0 ⇔ Ed (∆(F )) = 0 (Lemma 1.6.3) ⇔ ∆(F ) = 0. This completes the proof. Corollary 5.3.5. If G is a polynomial belonging to k[X] r k then E−d (G) is a formal first integral of system (5.6). Proof. Put F = E−d (G) and let F0 be the constant term of F with respect to t. Then F0 = G (by Proposition 2.4.1) and hence, F ∈ k[X][[t]rk and Ed (F ) = Ed E−d (G) = G = F0 . This means (by Theorem 5.3.4) that F is a formal first integral of (5.6). Corollary 5.3.6. The series E−d (x1 ), . . . , E−d (xn ) are formal first integrals of system (5.6). They are algebraically independent over k. Proof. It follows from Corollary 5.3.5 that each E−d (xi ) is a formal first integral of (5.6). Suppose that H(E−d (x1 ), . . . , E−d (xn )) = 0 for some H ∈ k[X]. Then E−d (H) = 0 and hence, H = 0 because E−d is an automorphism. Corollary 5.3.7. The ring of constants of the derivation ∆ is finitely generated over k. More precisely, k[X][[t]∆ = k[G1 , . . . , Gn ] = E−d (k[X]), where Gi = E−d (xi ) for i = 1, . . . , n. Proof. We already know, in view of Theorem 5.3.4 and Corollary 5.3.5, that ∆(Gi ) = 0 for all i = 1, . . . , n, i. e., k[G1 , . . . , Gn ] ⊆ k[X][[t]]∆ . Assume now that F ∈ k[X][[t]]∆ . Then (by Theorem 5.3.4) F = Ed (F0 ), where F0 ∈ k[X] is the constant term of F with respect to t. Thus, F ∈ E−d (k[X]) = k[G1 , . . . , Gn ]. We conclude this section with the following characterization of the ring of constants for k-derivations of polynomial rings.
Chapter 5. General properties of the rings of constants. . .
59
Corollary 5.3.8. Let F be a polynomial belonging to k[X] r k. Then F is a formal first integral of system (5.6) if and only if d(F ) = 0. Proof. It is a consequence of Theorem 5.3.4 because ∆(F ) = d(F ).
6
Constants for locally nilpotent derivations
Let k be a ring containing the field Q of rational numbers and let d be a k-derivation of a k-algebra A. Let us denote by Nil(d) the set of all the elements r ∈ A such that dn (r) = 0 for some n ∈ N0 . This set is a k-subalgebra of A containing Ad . The derivation d is called locally nilpotent if Nil(d) = A. Locally nilpotent derivations play an important role in the commutative algebra and algebraic geometry. If d is locally nilpotent, then the mapping t 7→ etd (see Section 6.1) determines an action of the additive group (k, +) on have a useful homomorphism of k-algebras A −→ A[t], a 7→ P∞A and we p p p=0 1/p!d (a)t . Several known problems may be formulated using locally nilpotent derivations. In particular, one of the equivalent formulations of the Jacobian Conjecture states that some special k-derivations of k[X] = k[x1 , . . . , xn ] are locally nilpotent (it will be explained in Chapter 9, see Section 9.8). Another equivalent formulation of this conjecture says that if {d1 , . . . , dn } is a commutative basis of Derk (k[X]), then each di is locally nilpotent (see also Section 9.8). Using such derivations Rentschler [91] gave a simple proof of the well known fact that every k-automorphism of k[x, y] is tame, and M. Smith [104] proved that some known k-automorphisms of k[X] are stably tame. Applications and various properties of the locally nilpotent derivations can be found in many papers (see, for example, [91], [109], [61], [74], [23]). The majority of facts presented in this chapter is well known. We recall here the results of Weitzenb¨ock, Deveney, Finston, van den Essen and others. Some of these results are supplied with simple proofs. The fundamental role in our proofs is played by the function, denoted by degd , which has properties similar to the ordinary degree function (Section 6.1). Using degd we see that some classical results are valid in a more general situation. Other applications of this function will also be given in Chapter 8. Let d be a locally nilpotent k-derivation of k[X]. An element b ∈ k[X] is said to be principal if d(b) = 1. If d has a principal element, then the ring k[X]d is finitely generated over k and it is easy to find its finite generating set (Theorem 6.4.4). If d has no principal elements, then the problem of the finiteness of k[X]d is difficult and it is known (see Section 6.3) that there exists a locally nilpotent k-derivation of k[X] with the non finitely generated ring of constants. In Section 6.7 we present the algorithm of A. van den Essen [21], based on the theory of Gr¨obner bases, which seems to be useful for a solution of this problem. By this algorithm we may 60
Chapter 6. Rings of constants for locally nilpotent derivations
61
find, in some cases, a finite generating set for k[X]d . With the help of this algorithm we give, in Sections 6.8 and 6.9, several examples related to the finite generating sets of the Weitzenb¨ock derivations.
6.1
The function degd and the automorphism ed
Assume that d is a locally nilpotent k-derivation of a k-algebra A. If a is an element of A then (as in [28]) we define degd (a) as follows: degd (a) = s degd (a) = −∞
if a 6= 0, ds (a) 6= 0 and ds+1 = 0; if a = 0.
Using Leibniz’s formula we get Proposition 6.1.1. Let a, b ∈ A. Then (1) degd (ab) 6 degd (a) + degd (b), and the equality is valid if A is a domain. (2) degd (a + b) 6 max{degd (a), degd (b)}, and the equality is valid if d(a) 6= d(b). If a is a nonzero element of A then degd (a) = 0 ⇔ d(a) = 0. Therefore, by Proposition 6.1.1(1), we obtain Proposition 6.1.2. Assume that A is a domain and d is locally nilpotent. Let a, b ∈ A r {0}. If ab ∈ Ad then a ∈ Ad and b ∈ Ad . As an immediate consequence of this proposition we get Proposition 6.1.3 ([65], [91], [71]). Let d be a locally nilpotent kderivation of a k-domain A. If A is UFD, then Ad is too. P∞ 1 p If d is a locally nilpotent k-derivation of A, then ed = p=0 p! d = I + d + 2!1 d2 + 3!1 d3 + · · · (where I is the identity function) is a well defined mapping from A into A. It is easy to check the following Proposition 6.1.4. (1) ed is a k-algebra automorphism of A. (2) (ed )−1 = e−d . (3) ed d = ded . d d (4) Ad = Ae , where Ae = {a ∈ A; ed (a) = a}. (5) If d1 and d2 are commutative locally nilpotent k-derivations of A, then d1 + d2 is locally nilpotent ed1 +d2 = ed1 ed2 . P∞ andp+1 (6) d = log(I + τ ) = p=1 (−1) p−1 τ p , where τ = ed − I.
Part III. Finiteness and properties of Ad
62
It follows from the above proposition that if d is locally nilpotent, then the mapping α 7→ eαd is a group homomorphism of the additive group (k, +) into the group Autk (A) of all the k-algebra automorphisms of A. Thus, every locally nilpotent k-derivation of A determines an action of the additive group (k, +) on A.
6.2
The theorem of Weitzenb¨ ock
Now let k be a field of characteristic zero and let k[X] = k[x1 , . . . , xn ] be the polynomial ring over k. The well known result of Weitzenb¨ock [111] states that every linear action of the additive group (k, +) on An has a finitely generated ring of invariants. A modern proof of this result is due to C. S. Seshadri [100]; (cf. [66] pp. 36 – 40). In the vocabulary of derivations this is equivalent to the following Theorem 6.2.1 ([111]). Let d be a k-derivation of k[X] such that d(xi ) =
n X
aij xj for i = 1, . . . , n, with aij ∈ k.
(6.1)
j=1
If the matrix [aij ] is nilpotent then the ring of constants k[X]d is finitely generated over k. In other words, the above theorem says that the ring of constants with respect to a locally nilpotent linear k-derivation of k[X] is finitely generated. Note the following generalization of this fact: Theorem 6.2.2. Every k-derivation of k[X] of the form (6.1) has a finitely generated (over k) ring of constants. Proof. See [66].
6.3
The example of Deveney and Finston
As we mentioned in Chapter 4, in 1990, P. Roberts in [94], gave a new counterexample to the fourteenth problem of Hilbert. He constructed a non finitely generated subring of a polynomial ring in seven variables. As a consequence of this fact and the result of the author (Theorem 4.2.4) one can deduce that there exists a k-derivation d of A = k[x1 , . . . , x7 ] such that Ad is not finitely generated. Recently, J. K. Deveney and D. R. Finston [18] realized Roberts’ subring as a ring of constants with respect to a locally nilpotent derivation.
Chapter 6. Rings of constants for locally nilpotent derivations
63
Example 6.3.1 ([18]). Let δ be the C-derivation of A = C[x1 , . . . , x7 ] given by δ(x1 ) = 0,
δ(x2 ) = 0,
δ(x3 ) = 0,
δ(x4 ) = x31 ,
δ(x5 ) = x32 ,
δ(x6 ) = x33 ,
δ(x7 ) = (x1 x2 x3 )2 . Then Aδ is not finitely generated over C.
6.4
Principal elements
Let us assume that d is a locally nilpotent k-derivation of a k-algebra A. An element b ∈ A is called principal if d(b) = 1. Proposition 6.4.1. If b ∈ A is a principal element, then b is not a unit in A and b is not a zero divisor in A. Proof. Let c denote either 0 or 1. Suppose that ab = c for some nonzero a ∈ A. Let n = degd (a). Then n > 0 and 0 = dn+1 (c) = hn, 1idn (a)d(b) = (n + 1)dn (a). Thus dn (a) = 0 which is a contradiction. Corollary 6.4.2. Let k ⊆ L be fields of characteristic zero and let d be a k-derivation of L. If d is locally nilpotent, then d = 0. Proof. Suppose that d(x) 6= 0, for some x ∈ L. Let n = degd (x). Then n > 1 and d(b) = 1, where b = dn−1 (x)/dn (x). It is a contradiction with Proposition 6.4.1. Let A[[t]] be the power series ring over A in the P variable t and let us 1 p p consider the mapping ε = εd : A −→ A[[t]], a 7−→ ∞ p=0 p! d (a)t . This mapping is the restriction of the automorphism Ed : A[[t]] −→ A[[t]] to A (see Section 2.4). It is easy to check that ε is an injective homomorphism ∂ of k-algebras, ε ◦ d = ∂t ◦ ε and Ad = {a ∈ A; ε(a) = a}. Since d is locally nilpotent, all the values of this homomorphism belong to A[t], the polynomial ring over A in the variable t. If b is a nonzero element of A then, for any x ∈ A, we denote by σb (x) the element ε(x)(−b), i. e., σb (x) =
∞ X 1 (−b)p dp (x). p! p=0
(6.2)
Part III. Finiteness and properties of Ad
64
Then we have a map σb : A −→ A which is a k-endomorphism of k-algebra A, and it is easy to check that dσb (x) = (1 − d(b))σb (d(x)), for any x ∈ A. This implies that if b is principal, then σb (A) ⊆ Ad and hence, we have the map σb : A −→ Ad which is a homomorphism of k-algebras. Moreover, σb is surjective (because σb (x) = x for x ∈ Ad ) and Kerσb = Ab. Thus, we have the following Theorem 6.4.3 ([73], [91]). If b is a principal element of A, then the homomorphism σb : A −→ Ad induces an isomorphism of the k-algebras Ad and A/Ab. As a consequence of Theorem 6.4.3 we get Theorem 6.4.4 ([91], [23]). Let A = k[a1 , . . . , an ] be a finitely generated k-algebra and let d be a locally nilpotent k-derivation of A. Assume that there exists a principal element b ∈ A. Then Ad is a finitely generated k-algebra. More precisely: Ad = k[σb (a1 ), . . . , σb (an )]. Proof. Ad = σb (A) = σb (k[a1 , . . . , an ]) = k[σb (a1 ), . . . , σb (an )]. Note also the following Lemma 6.4.5 ([21]). Assume that b is a principal element of A. Let ε = εdP . Then every element u in ε(A) can be written uniquely in the form m i u = i=0 ui (t + b) where m is a nonnegative integer and the elements u0 , . . . , um belong to Ad . The elements u0 , . . . , um , from the above lemma, are the coefficients of Taylor’s presentation of the polynomial u at the point t0 = −b, i. e., ui = i!−1 ∆i (u)(−b), where ∆ = ∂/∂t. In particular, if x ∈ A, then all the elements of the form ∆i (ε(x))(−b), where i = 0, 1, . . . , belong to d . Observe that ∆0 (ε(x))(−b) = ε(x)(−b) = σ (x) and x = ε(x)(0) = A b P∞ −1 ∆i (ε(x))(−b)bi . Thus, we have proved the following theorem i! i=1 which we may find in many papers (see for example [73], [91], [17], [23]; compare [114], see also [50]). Theorem 6.4.6. If b is a principal element of A, then A is a polynomial ring in b over Ad , that is, A = Ad [b].
Chapter 6. Rings of constants for locally nilpotent derivations
6.5
65
Derivations without principal elements
Throughout this section A = k[a1 , . . . , an ] is a finitely generated kdomain, and d is a nonzero locally nilpotent k-derivation of A. We do not assume that there exists a principal element in A. Our assumptions imply that there exists a nonzero element c ∈ A such that d(c) 6= 0 and d2 (c) = 0. Put h = d(c) and consider the k-algebra A˜ = S −1 A, where S is the multiplicative set {1, h1 , h2 , . . . }. Let d˜ : A˜ −→ A˜ be the natural extension of d. Then d˜ is a locally r ) = d(x)/hr , for any x ∈ A, r > 0, and ˜ ˜ d(x/h nilpotent k-derivation of A, it is easy to see that ˜ ˜ A˜d = S −1 Ad = Ad [1/h] and A˜d ∩ A = Ad .
˜ Put b = c/h. Then d(b) = d(c)/h = h/h = 1, that is, b is a principal ˜ element of A. The derivation d˜ satisfies the assumption of Section 6.4. ˜ Therefore, by Theorem 6.4.4, A˜d = k[σb (a1 ), . . . , σb (an )]. Let us recall that si si X c p 1 X 1 p 1 ˜p d (ai )(− ) = s d (ai )(−c)p hsi −p σb (ai ) = i p! h h p! p=0
p=0
for i = 1, . . . , n, where si = degd (ai ). If i ∈ {1, . . . , n} then we denote by a ˜i the element hsi σb (ai ), i. e., a ˜i =
si X 1 p d (ai )(−c)p hsi −p . p!
(6.3)
p=0
It is clear that the elements h and a ˜1 , . . . , a ˜n belong to Ad . Thus, we have proved the following Proposition 6.5.1 ([108], [21], [22]). Let A = k[a1 , . . . , an ] be a finitely generated k-domain, and let d be a nonzero locally nilpotent k-derivation of A. Then k[h, a ˜1 , . . . , a ˜n ] ⊆ Ad ⊆ k[h, a ˜1 , . . . , a ˜n ][1/h], k[h, a ˜1 , . . . , a ˜n ][1/h] ∩ A = Ad , where h = d(c) 6= 0 for some c ∈ A with d2 (c) = 0, and the elements a ˜1 , . . . , a ˜n are defined by (6.3).
Part III. Finiteness and properties of Ad
66
6.6
Results of van den Essen
In this section we present some results of A. van den Essen ([21] and [22]) concerning the ring of constants of locally nilpotent derivations without principal elements. These results will be useful in the next section. Let us start from the following Lemma 6.6.1 ([21]). Let R ⊆ S be algebras over a field k and let r0 be an element of R which is not a zero divisor in S. Let P be the k-subalgebra of S generated by the set {s ∈ S; r0 s ∈ R}. If R is finitely generated over k, then P is too. Proof. Set R = k[u1 , . . . , um ] and let I be the set of all the polynomials F in the polynomial ring k[Y ] = k[y1 , . . . , ym ] with F (u1 , . . . , um ) ∈ Sr0 . One easily checks that I is an ideal of k[Y ]. Let F1 , . . . , Fq be its generators. Then there exist elements v1 , . . . , vq of P such that vi r0 = Fi (u1 , . . . , um ), for i = 1, . . . , q. We will show that k[u1 , . . . , um , v1 , . . . , vq ] = P . The inclusion ⊆ is trivial. It remains to show that every generator of P belongs to k[u1 , . . . , um , v1 , . . . , vq ]. Therefore let x be such a generator. Then xr0 ∈ R = k[u1 , . . . , um ] and thus there exists a polynomial G ∈ k[Y ] such that xr0 = G(u1 , . . . , um ). It follows that G ∈ I. So, G = H1 F1 +· · ·+Hq Fq for some H1 , . . . , Hq ∈ k[Y ]. Put u = (u1 , . . . , um ). Then we have xr0 = G(u) = H1 (u)F1 (u)+· · ·+Hq (u)Fq (u) = (H1 (u)v1 +· · ·+Hq (u)vq )r0 . Since r0 is not a zero divisor of S, we get x = H1 (u)v1 + · · · + Hq (u)vq , i. e., x ∈ R[v1 , . . . , vq ] = k[u1 , . . . , um , v1 , . . . , vq ]. Assume now that k is a field, A = k[a1 , . . . , an ] is a finitely generated k-domain, and let d be a nonzero locally nilpotent k-derivation of A. Let c be an element of A such that d(c) 6= 0, d2 (c) = 0, and put h = d(c). Denote by M the family of all k-subalgebras B of A such that h ∈ B ⊆ Ad ⊆ B[1/h]. In view of Proposition 6.5.1 we know that there exists a finitely generated k-algebra B belonging to M. If B ∈ M, then we denote by B the k-subalgebra of A generated by all the elements a ∈ A such that ah ∈ B. It is clear that if B ∈ M, then B ⊆ B and B ∈ M. For every integer m > 0 we define (as in [21]) the following sequence (Bm ) of k-subalgebras of A: ( B0 = k[h, a ˜1 , . . . , a ˜n ], (6.4) Bm = Bm−1 , for m > 1.
Chapter 6. Rings of constants for locally nilpotent derivations
67
It follows from Lemma 6.6.1 that every Bm is a finitely generated k-algebra S d B belonging to M. Moreover, B0 ⊆ B1 ⊆ · · · ⊆ Ad and ∞ m=0 m = A . Therefore we get Proposition 6.6.2 ([21], [22]). If Ad is finitely generated over k, then Ad = Bm for some m > 0. By a similar way one can prove the following two propositions Proposition 6.6.3. Let A = k[a1 , . . . , an ] be a finitely generated k-domain, and let d be a nonzero locally nilpotent k-derivation of A with the elements c and h as above. Let (Bm ) be the sequence of k-algebras defined by (6.4). The following conditions are equivalent: (1) Ad is finitely generated over k. (2) Ad = Bm , for some m > 0. (3) The sequence B0 ⊆ B1 ⊆ . . . is stationary. (4) There exists m > 0 such that Bm = Bm+1 . (5) There exists a finitely generated k-algebra B ∈ M with B = B. Proposition 6.6.4. Assume that there exists a finitely generated k-algebra B ∈ M with B = B. Then Ad is finitely generated over k and Ad = B.
6.7
On van den Essen’s algorithm
In 1989 L. Tan in [108] gave a method of computing generators of the ring of constants for some linear locally nilpotent k-derivations of k[x1 , . . . , xn ]. It is known (Theorem 6.2.1) that in this case the ring of constants is finitely generated. Later, in 1992, A. van den Essen ([21], [22]) showed that this method works in a more general case. He gave an elegant algorithm computing a finite set of generators of the ring of constants for any locally nilpotent derivation (of a finitely generated k-domain), in the case when the finiteness of the ring of constants is known. Assume that A = k[a1 , . . . , an ] is a finitely generated k-domain and d is a locally nilpotent k-derivation of A. By A. van den Essen’s algorithm we may compute a finite set of generators of the algebras Bm defined in Section 6.4. Let us recall that every Bm is a finitely generated over k. If we know that Ad is finitely generated, then we know, by Lemma 6.6.2, that Ad = Bm , for some m. The ascending chain B0 ⊆ B1 ⊆ . . . becomes stationary after a finite number of steps where they are equal to Ad . Using the theory of Gr¨obner basis we may compute Bm if Bm−1 is known (for all
68
Part III. Finiteness and properties of Ad
m > 1) and also we may decide if Bm = Bm−1 . Note that if Bm = Bm−1 , then (see Proposition 6.6.3) Ad = Bm . Let us recall how we may compute B from B and decide if B = B (where B is a finitely generated algebra from M). We assume that the reader is familiar with the basic notions of Gr¨obner bases (see for example [9], [89]). For our aim we may assume that A = k[X] = k[x1 , . . . , xn ] is a polynomial ring. Let B = k[u1 , . . . , um ], h ∈ B, where h, u1 , . . . , um are polynomials in k[x1 , . . . , xn ]. Put k[Y ] = k[y1 , . . . , ym ]. Let us look at the proof of Proposition 6.6.1 (where R = B, S = A, P = B and r0 = h). First we must find generators of the ideal I = {F ∈ k[Y ]; F (u1 , . . . , um ) ∈ k[X]h}. Observe that I is the kernel of the ring k-homomorphism ϕ : k[Y ] −→ k[X]/(h) defined by yi 7→ ui + (h), where i = 1, . . . , m. Consider in the ring k[X, Y ] the ideal J = (u1 − y1 , . . . , um − ym , h). It is easy to see that I = J ∩ k[Y ]. Let us choose an admissible ordering on k[X, Y ] such that xi > yj , for any i, j. A suitable lexicographic ordering is an example of such an ordering. If G is a Gr¨obner basis of J, with respect to this ordering, then it is known that the finite set G0 = G ∩ k[Y ] is a Gr¨obner basis of I. Assume that G0 = {F1 , . . . , Fs }. Then h is a divisor of all the polynomials F1 (u), . . . , Fs (u), where u is the sequence (u1 , . . . , um ). It follows from the proof of Proposition 6.6.1 that the polynomials u1 , . . . , um and F1 (u)/h, . . . Fm (u)/h form a finite set of generators of the k-algebra B. Now we may use the known algebra membership algorithm given in [102] and decide if B = B. As an illustration we will prove the following result of Bass ([6] Theorem (b)). Example 6.7.1. Let d be the k-derivation of k[x1 , x2 , x3 ] defined by d(x1 ) = −2x2 d(x2 ) = x3 d(x3 ) = 0. Then k[x1 , x2 , x3 ]d = k[x3 , x1 x3 + x22 ]. Proof. Put u1 = x3 , u2 = x1 x3 + x22 . Let B = k[u1 , u2 ] and h = u1 = x3 . Then, by Proposition 6.5.1, B ⊆ k[x1 , x2 , x3 ]d ⊆ B[1/h], that is, B is a finitely generated k-algebra belonging to M. Now consider the ideal
Chapter 6. Rings of constants for locally nilpotent derivations
69
I = (u1 − y1 , u2 − y2 , h) in the polynomial ring A = k[x1 , x2 , x3 , y1 , y2 ]. Observe that I = (y1 , x22 − y2 , x3 ). Let us fix on A the lexicographic ordering x1 > x2 > x3 > y1 > y2 and let G be the reduced Gr¨obner basis of I with respect to this ordering. Then it is easy to see that G = {y1 , x22 − y2 , x3 } and hence, G ∩ k[y1 , y2 ] = {y1 }. This means that the algebra B is generated over k by the polynomials u1 , u2 (the generators of B) and by the element f (u1 , u2 ), where f (y1 , y2 ) is the polynomial in k[y1 , y2 ] defined by f (y1 , y2 ) = y1 . But f (u1 , u2 ) = u1 so, {u1 , u2 } is a generating set of B. Therefore, B = B and hence, by Proposition 6.6.3, k[x1 , x2 , x3 ]d = B. It is remarkable that the assumption of finiteness of Ad in van den Essen’s algorithm is unessential. By this assumption we only know that our calculations must be stopped in a finite number of steps, but we do not know when. We may use the same procedure for arbitrary locally nilpotent derivation of A. If we are successful, i. e., the calculations are stopped, then, by Proposition 6.6.3, we have a proof that Ad is finitely generated, and we have a finite set of its generators. The author was trying to use the above algorithm to several derivations of polynomial rings. There exist some difficulties. If we start from the algebra B0 than, in many cases, we must (for the next steps) introduce new polynomial rings in much bigger number of variables. This number is often too big for a computer. There exists however another way. We may start not from B0 but from an algebra B which is in family M described in Section 6.6 and, in the next steps, we construct the algebras of the form B. In this way the author described generators for some locally nilpotent derivations studied in the next two sections.
6.8
Generating sets for some Weitzenb¨ ock derivations
The theorem of Weitzenb¨ock [111] (mentioned in Section 6.2) states that every linear locally nilpotent k-derivation of k[x1 , . . . , xn ] has a finitely generated ring of constants. All the known proofs of this fact are not constructive. Given a linear k-derivation it is not easy to describe its ring of constants even if we assume that its matrix is nilpotent. In this section we present several examples of the generating sets for such derivations. We obtained these examples with the help of van den Essen’s algorithm (described in the previous section) and using the theory of Gr¨obner bases and the computer programme CoCoA [3]. All the examples (except of Example 6.8.4) are given without proofs.
Part III. Finiteness and properties of Ad
70
A k-derivation d of k[x1 , . . . , xn ] is called Weitzenb¨ ock if d is linear and the matrix of d is a Jordan’s matrix with the zeros on the main diagonal. It is obvious that when studying the ring of constants of a linear locally nilpotent k-derivation d, without loss of generality, we may suppose that d is a Weitzenb¨ock derivation. Let us make some notations. Assume that Rn = k[x0 , x1 , . . . , xn−1 ] is the polynomial ring in n variables over k, where n > 2. and let dn be the Weitzenb¨ock derivation of Rn corresponding to the one n × n Jordan cellJ, that is, dn (x0 ) = 0, dn (x1 ) = x0 , dn (x2 ) = x1 , . . . , dn (xn−1 ) = xn−2 .
(6.5)
Every Weitzenb¨ock derivation of the form dn is called basic. We would like to find (for any n) a finite set of generators over k of Rndn , the ring of constants of dn . If n = 2 then it is easy. Since d2 = x0 ∂/∂x1 , R2d2 = k[x0 ]. For arbitrary n the problem seems to be difficult. If d = dn is a basic Weitzenb¨ock derivation, then dp (xi ) = xi−p , for p = 0, 1, . . . , i, and degd (xi ) = i, for i = 0, 1, . . . , n − 1. The variables x0 and x1 play the same role as elements h and c from Section 6.5, that is, d(x1 ) = x0 6= 0, d2 (x1 ) = 0. Thus, by (6.3) (putting ai = xi , si = i, c = x1 , h = x0 ), we get x ˜0 = x0 , x ˜1 = 0 and, for i > 2, x ˜i =
i X
(−1)p
p=0
1 i−p p x x xi−p p! 0 1 i−2
X 1 1 (−1)p xi−p xp xi−p = (−1)i x00 xi1 x0 + (−1)i−1 x10 x1i−1 x1 + i! p! 0 1 p=0
= (−1)i−1
i−1 i! x0 · ( xi1 + i! i−1
i−2 X p=0
1 (−1)i−1−p x0i−1−p xp1 xi−p ). i!
Denote by t the variable x0 . i! Multiplying every element x ˜i (for i > 2) by (−1)i−1 i−1 and dividing by t (recall that d(t) = 0) we obtain the polynomial zi (belonging to Rndn ) defined by i−2 i! X 1 (−1)i−1−p ti−1−p xp1 xi−p . (6.6) zi = xi1 + i−1 i! p=0
Chapter 6. Rings of constants for locally nilpotent derivations In particular, z2 = x21 − 2tx2 , z3 = x3 − 3tx1 x2 + 3t2 x3 , 1 z4 = x41 − 4tx21 x2 + 8t2 x1 x3 − 8t3 x4 , z5 = x51 − 5tx31 x2 + 15t2 x21 x3 − 30t3 x1 x4 + 30t4 x5 .
71
(6.7)
The above polynomials one may find also in [111] p. 242. It is not difficult to see that they are algebraically independent over k(t, x1 ) and hence, they are algebraically independent over k. Moreover, by Proposition 6.5.1, we have k[t, z2 , . . . , zn−1 ] ⊆ Rndn ⊆ k[t, z2 , . . . , zn−1 ][1/t], (6.8) k[t, z2 , . . . , zn−1 ][1/t] ∩ Rn = Rndn . This implies that the k-algebra k[t, z2 , . . . , zn−1 ] is an element of the family M defined in Section 6.6. The derivation dn is homogeneous; if F is a homogeneous polynomial in Rn , then dn (F ) is homogeneous and deg dn (F ) = deg F . There exists also an another homogeneity of dn . Consider the direction α = (0, 1, 2, . . . , n − 1). Recall (see Section 2.1) that a polynomial F ∈ Rn is called an α-form in−1 of degree s if every monomial of F is of the form axi00 xi11 . . . , xn−1 , with a ∈ k, where 0i0 + 1i1 + 2i2 + · · · + (n − 1)in−1 = s. It is easy to see that if F ∈ Rn is an α-form of degree s then dn (F ) is an α-form of degree s − 1. Thus, The ring Rndn is generated over k by homogeneous polynomials which are α-forms. Note that every polynomial zi is homogeneous of degree i and it is also an α-form of degree i. Using the same method as in the proof of Example 6.7.1 we obtain the following two examples Example 6.8.1. R3d3 = k[t, z2 ]. Set u = x21 x22 − 2x31 x3 + 6tx1 x2 x3 − 38 tx32 − 3t2 x23 . Example 6.8.2. R4d4 = k[t, z2 , z3 , u]. The case of five variables seems to be more complicated. Generators of the ring R5d5 are homogeneous in the usual sense and also they are homogeneous with respect to the direction α = (0, 1, 2, 3, 4). It is not difficult to find such homogeneous polynomials which are of small degrees. Here are four of them:
Part III. Finiteness and properties of Ad
72 Table 6.8.3. z2 p2 z3 p3
x21 − 2tx2 2x1 x3 − x22 − 2tx4 x31 − 3tx1 x2 + 3t2 x3 6x21 x4 − 6x1 x2 x3 + 2x32 − 12tx2 x4 + 9tx23
2 2 3 3
2 4 3 6
In the first column we write the name of a polynomial and in the last two columns its degree and α-degree, respectively. We will show that {t, z2 , z3 , p2 , p3 } is a set of generators of the k-algebra R5d5 . For this aim set B = k[t, z2 , z3 , p2 , p3 ]. Observe at first that the polynomial z4 (see (6.7)) belongs to B. In fact; z4 = z22 + 4t2 p2 . Thus, we have: B ⊆ R5d5 ⊆ B[1/t], that is, B is an element of M. Note that the polynomial u from Proposition 6.8.2 also belongs to B (since u = −z2 p2 − 31 tp3 ). Example 6.8.4. R5d5 = B = k[t, z2 , z3 , p2 , p3 ]. Proof. Consider the ring A = k[t, x1 , x2 , x3 , x4 , y1 , y2 , y3 , y4 , y5 ] and its ideal I = (t − y1 , z2 − y2 , p2 − y3 , z3 − y4 , p3 − y5 , t) = (t, y1 , x21 − y2 , 2x1 x3 − x22 − y3 , x31 − y4 , 6x21 x4 − 6x1 x2 x3 + 2x32 − y5 ). Using the CoCoA programme [3] we compute that the elements b1 , . . . , b27 below form the set GB(I), the reduced Gr¨obner basis of I (with respect to the lexicographic ordering x1 > · · · > x4 > t > y2 > · · · > y5 > y1 ). b1 b2 b3 b4 b5 b6 b7 b8 b9 b10
= = = = = = = = = =
y1 , x21 − y2 , x1 x3 − 1/2x22 − 1/2y3 , x1 y2 − y4 , x32 + 3x2 y3 − 6x4 y2 + 2y5 , t, x22 y2 − 2x3 y4 + y2 y3 , x1 y4 − y22 , x1 x22 + x1 y3 − 2x3 y2 , y23 − y42 ,
Chapter 6. Rings of constants for locally nilpotent derivations
b11 b12 b13 b14 b15 b16 b17 b18 b19 b20 b21 b22 b23 b24 b25
= = = = = = = = = = = = = = =
b26
=
b27
=
73
x22 y4 − 2x3 y22 + y3 y4 , x2 x3 y4 + x2 y2 y3 − 3x4 y22 + y2 y5 , x22 y3 − 6x2 x4 y2 + 2x2 y5 + 4x23 y2 − y32 , x1 x2 y3 + x1 y5 + x2 x3 y2 − 3x4 y4 , x2 x4 y2 y4 − 1/3x2 y4 y5 − 2/3x23 y2 y4 − 1/3x3 y22 y3 + 1/3y32 y4 , x2 x4 y22 − 1/3x2 y2 y5 − 2/3x23 y22 − 1/3x3 y3 y4 + 1/3y2 y32 , x2 x3 y22 + x2 y3 y4 − 3x4 y2 y4 + y4 y5 , x22 y5 + 2x2 x23 y2 − 2x2 y32 − 6x3 x4 y4 + 6x4 y2 y3 − y3 y5 , x1 x2 y5 − x1 y32 − 3x2 x4 y4 + 2x23 y4 + x3 y2 y3 , x2 x4 y42 − 1/3x2 y22 y5 − 2/3x23 y42 − 1/3x3 y2 y3 y4 + 1/3y22 y32 , x2 x23 y2 y3 + 3x2 x4 y2 y5 − x2 y33 − x2 y52 − 2x23 y2 y5 − 3x3 x4 y3 y4 + 3x4 y2 y32 , x1 y33 + x1 y52 − 2x33 y22 − 3x23 y3 y4 + 9x24 y2 y4 − 6x4 y4 y5 , x33 y4 + 3/2x23 y2 y3 − 9/2x24 y22 + 3x4 y2 y5 − 1/2y33 − 1/2y52 , x2 x3 y2 y5 + 3x2 x4 y3 y4 + 2x33 y22 + x23 y3 y4 − x3 y2 y32 − 9x24 y2 y4 + 3x4 y4 y5 , x2 x4 y2 y33 + x2 x4 y2 y52 − 1/3x2 y33 y5 − 1/3x2 y53 − 2/3x43 y22 y3 + 5/6x23 y2 y33 −2/3x23 y2 y52 + 3x3 x24 y2 y3 y4 − 2x3 x4 y3 y4 y5 − 9/2x24 y22 y32 + 3x4 y2 y32 y5 −1/6y35 − 1/6y32 y52 , x2 x3 y33 y52 + x2 x3 y54 + 9x2 x24 y34 y4 + 9x2 x24 y3 y4 y52 + 2x53 y22 y3 y5 +6x33 x4 y22 y33 + 6x33 x4 y22 y52 − 5/2x33 y2 y33 y5 + 2x33 y2 y53 −9x23 x24 y2 y3 y4 y5 + 3x23 x4 y34 y4 + 9x23 x4 y3 y4 y52 + 27/2x3 x24 y22 y32 y5 −3x3 x4 y2 y35 − 12x3 x4 y2 y32 y52 + 1/2x3 y35 y5 + 1/2x3 y32 y53 − 27x34 y2 y33 y4 −27x34 y2 y4 y52 + 9x24 y33 y4 y5 + 9x24 y4 y53 , x2 x23 y33 y5 + x2 x23 y53 − 3x2 x4 y35 − 3x2 x4 y32 y52 + 2x63 y22 y3 + 6x43 x4 y22 y5 −5/2x43 y2 y33 + 2x43 y2 y52 + 27x23 x24 y22 y32 − 63/2x23 x4 y2 y32 y5 + 1/2x23 y35 +1/2x23 y32 y52 − 27x3 x34 y2 y4 y5 − 9x3 x24 y33 y4 + 9x3 x24 y4 y52 − 81/2x44 y3 y42 +189/2x34 y22 y3 y5 + 9/2x24 y2 y34 − 81/2x24 y2 y3 y52 + 9/2x4 y34 y5 + 9/2x4 y3 y53 .
Looking at the above elements we see that only two from them, namely b1 and b10 , belong to k[y1 , . . . , y5 ], i. e., GB(I)∩k[y1 , . . . , y5 ] = {b1 , b10 }. For us only b10 (y2 , y4 ) = y23 − y42 is an interesting element. Set v = b10 (z2 , z3 ). Then v = z23 − z32 = −6t2 w, where 6w = 6x31 x3 − 3x21 x22 − 18tx1 x2 x3 + 8tx32 + 9t2 x23 . Observe that w = z2 p2 + 13 tp3 . This means that w ∈ B. So B = B and therefore, by Proposition 6.6.3, R5d5 = B = k[t, z2 , z3 , p2 , p3 ]. In [26] we may find a finite set of polynomials which is proposed as a generating set of Rndn . But, as shows L. Tan in [108], it is not a generating set. Full lists of generators, for n 6 4, are described (in a slightly modified form) in [108]. The generators from Example 6.8.4 can be found in the tables of Andre Cerezo [10]. Cerezo also computed a system of generators of minimal length for the case n = 6 (the minimal length is 23, the degrees of these generators go up to 18).
Part III. Finiteness and properties of Ad
74
The next two examples and proposition are related to the nonbasic Weitzenb¨ock derivations. Example 6.8.5. Let R = k[x1 , x2 , y1 , y2 , y3 ] and d(y1 ) = 0 d(x1 ) = 0 d(y2 ) = y1 d(x2 ) = x1 d(y3 ) = y2 . Then Rd = k[x1 , y1 , x1 y2 − x2 y1 , y22 − 2y1 y3 , 2x21 y3 − 2x1 x2 y2 + x22 y1 ]. Example 6.8.6. Let R = k[x1 , x2 , x3 , y1 , y2 , y3 ] and d(y1 ) = 0 d(x1 ) = 0 d(y2 ) = y1 d(x2 ) = x1 d(y3 ) = y2 . d(x3 ) = x2 Then Rd = k[x1 , y1 , f1 , f2 , f3 , f4 , f5 , f6 ], where f1 = 2x1 x3 − x22 , f2 = x1 y2 − x2 y1 , f3 = x1 y3 − x2 y2 + x3 y1 ,
f4 = 2y1 y3 − y22 , f5 = x1 y22 − 2x2 y1 y2 + 2x3 y12 , f6 = 2x21 y3 − 2x1 x2 y2 + x22 y1 .
Proposition 6.8.7. Let R = k[x1 , . . . , xn , y1 , . . . , yn ] and let d be the k-derivation of R such that d(x1 ) = 0, d(x2 ) = x1 , . . . , d(xn ) = xn−1 and = 0, d(y2 ) = y1 , . . . , d(yn ) = yn−1 . Then the polynomial Pn−1d(y1 ) p+1 xp yn−p belongs to Rd . p=1 (−1)
6.9
Weitzenb¨ ock derivations with 2 × 2 cells
Let Sn denote the polynomial ring k[x1 , y1 , x2 , y2 , . . . , xn , yn ] and let δn be the k-derivation of Sn defined as δ n = y1
∂ ∂ + · · · + yn . ∂x1 ∂xn
(6.9)
It is a nonbasic Weitzenb¨ock derivation and all cells of its Jordan matrix have the same dimension equals 2. We will present finite generating sets of Snδn for n 6 4. The derivation δn is an A-derivation of Sn such that δn (xi ) ∈ A, for i = 1, . . . , n, where A = k[y1 , . . . , yn ]. Consider a more general case. Let A be a k-domain containing k and let d be an A-derivation of A[x1 , . . . , xn ] such that d(xi ) ∈ A. Of course such a derivation is locally nilpotent. If A is a field, then following proposition describes the ring of constants of d.
Chapter 6. Rings of constants for locally nilpotent derivations
75
Proposition 6.9.1. Let A be a field containing k and let d be a nonzero A-derivation of A[x1 , . . . , xn ] such that d(x1 ) = a1 , . . . , d(xn ) = an , where a1 , . . . , an ∈ A and a1 6= 0. Then A[x1 , . . . , xn ]d = A[u2 , . . . , un ], where up = ap x1 − a1 xp , for p = 2, . . . , n. Proof. Put b = a−1 1 x1 . Then d(b) = 1 and our proposition follows from Theorem 6.4.4 and (6.2). Consider the following Question 6.9.2. Is the above proposition true in the case when A is a k-domain? If n = 2 and A is a UFD, then we have a positive answer to this question. Proposition 6.9.3. Let A[x, y] be the polynomial ring in two variables over a unique factorization k-domain A. Assume that d is an A-derivation of A[x, y] such that d(x) = a, d(y) = b, where a, b are coprime elements of A. Then A[x, y]d = A[bx − ay]. Proof. Put z = bx − ay. It is clear that A[z] ⊆ A[x, y]d (since d(z)=0). Assume that F is an element of A[x, y] with d(F ) = 0. We must show that F ∈ A[z]. As d is homogeneous, one may assume that F is homogeneous of degree s > 1. By Proposition 6.9.1 we know that F ∈ Ao [z], where Ao is the field of fractions of A. So, by homogeneity, F = pq z s for some coprime p, q ∈ A. Then qF = pz s and hence q | z = bx − ay. But a and b are coprime, so q is invertible in A and we have F = pq −1 z s ∈ A[z]. The following example shows that Question 6.9.2 has a negative answer in general. Example 6.9.4. Let A = Q[t], R = A[x, y, z]. Consider the A-derivation d of R such that d(x) = 2t, d(y) = 1 + t, d(z) = 1 − t and let F = x − y + z. Then d(F ) = 0 and F 6∈ A[u2 , u3 ], where u2 = (1 + t)x − 2ty and u2 = (1 − t)x − 2tz. Let us return to the derivation δn defined by (6.9). It is clear that S1δ1 = k[y1 ] and, as a consequence of Proposition 6.9.3, we get Proposition 6.9.5. S2δ2 = k[y1 , y2 , x1 y2 − x2 y1 ].
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76
The variables y1 and x1 play the same role as elements h and c from Section 6.5, i. e., δn (x1 ) = y1 6= 0, δn2 (x1 ) = 0. Thus, by (6.3), y˜i = yi for i = 1, . . . , n and moreover, x ˜1 = 0 and x ˜i = y1 xi − yi x1 for i = 2, . . . , n. If i < j then we put uij = yi xj − yj xi . In particular, x ˜2 = u12 , . . . , x ˜n = u1n and hence, by Proposition 6.5.1, we get Proposition 6.9.6. Let B = k[y1 , . . . , yn , u12 , . . . , u1n ]. Then B ⊆ Snδn ⊆ B[1/y1 ]. Let B be the algebra as in Proposition 6.9.6. In this case Question 6.9.2 reduces to the question: ”Is Snδn equal to B?”. Now we will show that it is not true in general. First observe that all elements of the form uij belong to Snδn . Indeed; δn (uij ) = yi yj − yj yi = 0. Example 6.9.7. u23 6∈ k[y1 , y2 , y3 , u12 , u13 ]. Proof. Suppose that u23 ∈ k[y1 , y2 , y3 , u12 , u13 ]. Then there exists a polynomial β in five variables over k such that y2 x3 − y3 x2 = β(y1 , y2 , y3 , y1 x2 − y2 x1 , y1 x3 − y3 x1 ). Putting x1 = 0, y1 = 0 we get the equality y2 x3 − y3 x2 = γ(y2 , y3 ), for some polynomial γ in two variables. It is a contradiction. Using A. van den Essen’s algorithm (and a computer) we get Example 6.9.8. S3δ3 = k[y1 , y2 , y3 , u12 , u13 , u23 ]. Example 6.9.9. S4δ4 = k[y1 , y2 , y3 , y4 , u12 , u13 , u14 , u23 , u24 , u34 ]. We end this section with the following Conjecture 6.9.10. Snδn = k[y1 , . . . , yn , u12 , u13 , . . . , un−1,n ].
6.10
Comments and remarks
6.1 N. Onoda in [88] announces the following theorem concerning to the rings of constants of Weitzenb¨ock derivations. Theorem 6.10.1 ([88]). Let dn be the Weitzenb¨ ock derivation defined by (6.5) over an algebraically closed field k (of characteristic zero) and let C = Rndn . Then (1) C is a Gorenstein ring. (2) (a) If n = 1 then C ∼ = k[x].
Chapter 6. Rings of constants for locally nilpotent derivations
77
(b) If n = 2 then C ∼ = k[x, y]. (c) If n = 3 then C ∼ = k[x, y, z, u]/(x2 u + y 3 + z 2 ). (d) If n = 4 then C ∼ = k[x, y, z, u, v]/(x3 v + y 3 + z 2 + x2 yu). (3) If n 6 4 then C is a complete intersection. (4) If n = 5 then C is not a complete intersection. Note that (3) is a consequence of (2). For a proof of (4) Onoda calculates the Poincar´e series of R5d5 . Observe that Rn is a graded ring with natural gradation and C = Rndn is a graded subring of R homogeneous part Ln . Denote by CpPthe ∞ of degree p of C. Then we have C = Cp and P (t) = p=0 (dimk Cp )tp is, by definition, the Poincar´e series of C. Onoda proves that if n = 5 then P (t) is equal to a(t)/b(t), where a(t) b(t)
1 + t2 + 3t3 + 3t4 + 5t5 + 4t6 + 6t7 + 6t8 +4t9 + 5t10 + 3t11 + 3t12 + t13 + t15 , = (1 − t)(1 − t2 )(1 − t4 )(1 − t6 )(1 − t8 ). =
Next he deduces that P (t) has a root which is not a root of unity. It is known (see [105]) that if C is a complete intersection, then every root of P (t) must be a root of unity. So R5d5 is not a complete intersection. 6.2 Let us return to the algebraically independent polynomials z2 , z3 , . . . defined by (6.6) on page 70. If w is an polynomial in Q[z2 , . . . , zn ] then w ∈ Q[t, x1 , . . . , xn ] and there exist the unique elements A0 (w), A1 (w), · · · ∈ Q[x1 , . . . , xn ] such that w = A0 (w)t0 + A1 (w)t1 + A2 (w)t2 + · · · . It is easy to check the following Proposition 6.10.2. If w = zps , where p ∈ {2, . . . , n} and s > 1, then A0 (w)
= xps 1 ,
A1 (w)
= −spxsp−2 x2 , 1
A2 (w)
=
1 2 s(s
− 1)p2 xsp−4 x22 + sp(p − 2)xsp−3 x3 . 1 1
Using this fact we may prove Proposition 6.10.3. Let s > 1 and let w = z2α2 · · · znαn , where 2α2 + 3α3 + · · · + nαn = s. Then A0 (w) = xs1 , A1 (w)
= −sx1s−2 x2 ,
A2 (w)
x3 , = ax1s−4 x22 + bxs−3 1
where a = 12 (s2 − c), b = c − 2s with c =
Pn
p=2
αp p2 .
Part III. Finiteness and properties of Ad
78
As a consequence of Proposition 6.10.3 we get Proposition 6.10.4. Let w = z2α2 · · · znαn − z2β2 · · · znβn , where 2α2 + 3α3 + · · · + nαn = 2β2 + 3β3 + · · · + nβn = s. Then there exists the unique polynomial v ∈ k[t, x1 , . . . , xn ] such that w = t2 v. Moreover, dn+1 (v) = 0 and 1 v ≡ ax1s−4 (x1 x3 − x22 ) (mod t), 2 where a =
Pn
p=2 (αp
− βp )p2 .
Making use of this proposition we may construct new elements which belong to Rndn .
7
Rings of constants for small n
Let d be a k-derivation of the polynomial ring k[X] = k[x1 , . . . , xn ], where k is a field of characteristic zero. In this chapter we examine the ring k[X]d for small n. In Section 7.1 we prove (using the results of Zariski [113] and Zaks [112]). that if n 6 3, then k[X]d is finitely generated and, if n = 2, then k[X]d is of the form k[f ], for some f ∈ k[X] Section 7.2 is devoted to the case n = 2. We study here closed polynomials in k[x, y] and we present some new descriptions of all the subrings of k[x, y] which are the rings of constants with respect to derivations (see Theorems 7.2.2, 7.2.12 and 7.2.13). An application of the results contained in Section 7.2 can be found in the next Section 7.3, where we present four examples of derivations with the trivial rings of constants. If n = 2 and d 6= 0, then the minimal number of polynomials in k[X]rk generating k[X]d is equal to 0 or 1 (see Corollary 7.1.5). In Section 7.4 we prove that if n > 3, then the minimal number of generators is unbounded. This chapter is based on the results of Nagata, Strelcyn and the author published in [85], [81] and [87].
7.1
Finiteness for n = 1, 2 and 3
We start with the following Proposition 7.1.1. If d is a nonzero k-derivation of R = k[x1 , . . . , xn ], then tr.degk (Rd ) 6 n − 1. Proof. Let s = tr. degk (Rd ), L = k(x1 , . . . , xn ) and K = Ld , where d is the natural extension of d to L. It is clear that s 6 n. Suppose now that s = n. Then K ⊆ L is an algebraic field extension and d is a K-derivation of L so, by Theorem 1.4.1, d = 0. This implies that d = 0 and we have a contradiction. Let us recall a result due to Zariski ([113], see [66] p. 41) concerning the fourteenth problem of Hilbert. Theorem 7.1.2 ([113]). Let K be a subfield of k(x1 , . . . , xn ) containing k. If tr.degk K 6 2 then the ring K ∩ k[x1 , . . . , xn ] is finitely generated over k. 79
Part III. Finiteness and properties of Ad
80
As a consequence of Theorem 7.1.2, Propositions 3.1.2 and 7.1.1 we obtain the following Theorem 7.1.3. If d is a k-derivation of k[x1 , . . . , xn ], where n 6 3, then the ring k[x1 , . . . , xn ]d is finitely generated over k. The next result follows from the facts presented in Sections 3.1 and 5.2. Theorem 7.1.4. Let d be a k-derivation of k[X] = k[x1 , . . . , xn ]. If tr.degk (k[X]d ) 6 1, then there exists a polynomial f ∈ k[X] such that k[X]d = k[f ]. Proof. Let R = k[X]d and s = tr.degk (k[X]d ). If s = 0 then R = k, so R = k[f ], where for example f = 1. If s = 1 then, by Proposition 3.1.3 and Theorem 7.1.3, R is a Dedekind subring of k[X] containing k and hence, by Theorem 5.2.2, R = k[f ], for some f ∈ k[X]. Corollary 7.1.5. If d is a nonzero k-derivation of k[x, y], then there exists a polynomial f ∈ k[x, y] such that k[x, y]d = k[f ].
7.2
Two variables
This section is based on the author’s paper [81]. We present here some additional properties and applications of closed polynomials (see Section 5.2) in k[x, y]. Moreover, we present a description of all the subrings of k[x, y] which are rings of constants for k-derivations of k[x, y]. The following two facts one can deduce from the previous section. Corollary 7.2.1. Let D be a nonzero family of k-derivations in k[x, y]. Put R = k[x, y]D . If f ∈ R r k, then R is the integral closure of the ring k[f ] in k[x, y]. Proof. If f ∈ R r k then R 6= k and, by Corollary 7.1.5 and Theorem 5.2.3, R = k[h], for some closed polynomial h ∈ k[x, y]. Hence k[f ] ⊆ k[h], k[h] is integrally closed in k[x, y] and k[h] is integral over k[f ]. This means that R = k[h] is the integral closure of k[f ] in k[x, y]. Theorem 7.2.2. Let A be a subring of k[x, y] containing k, such that A is integrally closed in k[x, y]. If Krull-dim(A) 6 1 then there exists a kderivation d of k[x, y] such that A = k[x, y]d .
Chapter 7. Rings of constants for small n
81
Proof. Let s = Krull-dim(A). If s = 0 then A = k and we have A = k[x, y]d , where, for example, d is the k-derivation of k[x, y] such that d(x) = x and d(y) = y. Assume that s = 1. Then A is a Dedekind subring of k[x, y] containing k (see [19] Theorem 1) hence, by Theorem 5.2.2, A = k[h] for some closed polynomial h ∈ k[x, y] r k. Consider the k-derivation d of k[x, y] such that d(x) = ∂h/∂y and d(y) = −∂h/∂x. Then h ∈ k[x, y] r k and we see, by Corollary 7.2.1, that A = k[x, y]d . If f is a polynomial in k[x, y] then we write fx and fy instead of ∂f /∂x and ∂f /∂y, respectively. If f, g ∈ k[x, y] then we denote by [f, g] the jacobian of (f, g), that is, [f, g] = fx gy − fy gx . Let f ∈ k[x, y]. We denote by df the k-derivation of k[x, y] defined by df (g) = [f, g], for all g ∈ k[x, y]. Denote also by Ck (f ) the ring of constants for df . Note the following obvious proposition Proposition 7.2.3. (1) Ck (f ) is a subring of k[x, y] containing k[f ]. (2) Ck (f ) = k[x, y] if and only if f ∈ k. We see, by the above proposition, that the case ”f ∈ k” is not interesting. In this case the derivation df is equal to zero. Now we will consider only polynomials in k[x, y] r k. Proposition 7.2.4. If f is a polynomial in k[x, y] r k then f is closed if and only if Ck (f ) = k[f ]. Proof. If Ck (f ) = k[f ] then k[f ] is (by Proposition 3.1.3) integrally closed in k[x, y] and hence (by Proposition 5.2.1), f is closed. Assume now that f is closed. In view of Corollary 7.1.5 we know that Ck (f ) = k[h] for some h ∈ k[x, y]. Then k[f ] ⊆ k[h] and, by the maximality of k[f ], we have k[f ] = k[h] = Ck (f ). Certain examples of closed polynomials may be obtained by the following two propositions. Proposition 7.2.5. Let f, g ∈ k[x, y]. If [f, g] ∈ k r {0} then f and g are closed. Proof. Without loss of any generality we may assume that f and g have no constant terms and that [f, g] = 1. Consider the k-endomorphism F of the ring k[[x, y]] (the power series ring) defined by F (x) = f , F (y) = g.
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Part III. Finiteness and properties of Ad
This mapping is a k-automorphism of k[[x, y]] (see [74]). Let d be the kderivation of k[[x, y]] such that d(x) = −fy and d(y) = fx , and let C be the ring of constants for d. Observe that k[[x, y]] = F (k[[x, y]]) = k[[f, g]] = (k[[f ]])[[g]] and hence, it is easy to show that C = k[[f ]]. Now we have Ck (f ) = C ∩ k[x, y] = k[[f ]] ∩ k[x, y] = k[f ] and so, by Proposition 7.2.4, f is closed and, by symmetry, g is closed too. Let (p, q) ∈ Z2 be a direction. A (p, q)-form f ∈ k[x, y] r k is called reduced if there is no (p, q)-form h such that f = ahn with a ∈ k r {0} and n > 2. For example, the (1, 1)-forms x, y, xy, x2 + y 2 , x3 + xy 2 + 2y 3 are reduced. Proposition 7.2.6. Let (p, q) be a direction such that p > 0 and q > 0, and let f be a reduced (p, q)-form. Then f is a closed polynomial. Proof. Let s be the degree of f . We will show that Ck (f ) = k[f ]. Assume that g ∈ Ck (f ) and let g = g0 +g1 +· · ·+gn be the (p, q)-decomposition of g, that is, each gi , for i = 1, . . . , n, is a (p, q)-form of degree i or is equal to zero, and g0 is a constant. Then [f, gi ], for i = 1, . . . , n, is a (p, q)-form of degree sP+ i − p − q (or is equal to zero), and hence the equality 0 = [f, g] = [f, gi ] is the (p, q)-decomposition of zero. Hence [f, g1 ] = · · · = [f, gn ] = 0 and so, g1 , . . . , gn ∈ k[f ] (see [72]). This implies that g ∈ k[f ]. Therefore k[f ] = Ck (f ) and hence, by Proposition 7.2.4, f is closed. Now we will present farther properties of the ring Ck (f ). Proposition 7.2.7. Let f, g ∈ k[x, y] r k. If g ∈ Ck (f ) then Ck (f ) = Ck (g). Proof. Assume that g ∈ Ck (f ). Then [f, g] = 0 and hence gx df = fx dg and gy df = fy dg . Since f and g do not belong to k, fx 6= 0 or fy 6= 0, and also gx 6= 0 or gy 6= 0. Assume that fx 6= 0 and gy 6= 0 (in the next cases we do the same procedure). Let h ∈ Ck (f ). Then fx dg (h) = gx df (h) = gx 0 = 0 and so, h ∈ Ck (g). If h ∈ Ck (g) then gy df (h) = fy dg (h) = 0, that is, h ∈ Ck (f ). In view of Corollary 7.1.5 we know that the ring Ck (f ) is of the form k[h] for some h ∈ k[x, y]. As a consequence of this fact and Proposition 7.2.7 we get
Chapter 7. Rings of constants for small n
83
Theorem 7.2.8. Let f, g ∈ k[x, y] r k. If [f, g] = 0 then there exist a polynomial h ∈ k[x, y] and polynomials u(t), v(t) ∈ k[t] such that f = u(h) and g = v(h). If f is a polynomial in k[x, y] then Sf denotes the support of f , that is, Sf is the set of integer points (i, j) such that the monomial xi y j appears in f with a nonzero coefficient. We denote by Wf the convex hull (in the real space R2 ) of Sf ∪ {(0, 0)}. The set Wf is called (see [2]) the Newton’s polygon of f . The following theorem is a consequence of Theorem 7.2.8. Theorem 7.2.9. Let f, g ∈ k[x, y] r k be such polynomials that [f, g] = 0. (1) If Wf is a line segment then Wg too. (2) Let Wf be a polygon. Then Wg is also a polygon, the polygons Wf and Wg are similar and the ratio of similarity is equal to deg(f )/ deg(g). Proof. See [81]. Let us note that the same result is valid for polynomials f, g ∈ k[x, y] such that [f, g] ∈ k r {0} ([86]). Note also that another application of Theorem 7.2.8 one can find in [70]. If d and δ a k-derivations of k[x, y] then we write d ∼ δ in the case where ad = bδ, for some nonzero a, b ∈ k[x, y]. Let us observe now that if u(t) ∈ k[t] r k, then df ∼ du(f ) . Thus, as a consequence of Theorem 7.2.8 we get Corollary 7.2.10. Let f, g ∈ k[x, y] r k. If [f, g] = 0 then df ∼ dg . Let d be a k-derivation of k[x, y]. Put d(x) = P , d(y) = Q and consider the ring of constants k[x, y]d . It would be of considerable interest to find necessary and sufficient conditions on P and Q for k[x, y]d to be nontrivial (that is, k[x, y]d 6= k). The problem seems to be difficult. Now we will present some theorems concerning this problem. We denote by d∗ the divergence of d, that is, d∗ = d(x)x + d(y)y . A derivation d is called special if d∗ = 0. Every derivation of the form df is special. Indeed: d∗f = df (x)x + df (y)y = −fyx + fxy = 0. The converse of this fact is also true: Proposition 7.2.11. If d is a special k-derivation of k[x, y], then there exists h ∈ k[x, y] such that d = dh .
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Part III. Finiteness and properties of Ad
Proof. Let d(x) = P , d(y) = Q and assume that Px + Qy = 0. Put f = Q and g = −P . Then fy = gx and hence, by Lemma 2.5.3, there exists h ∈ k[x, y] such that hx = f and hy = g, i. e., d = dh . Theorem 7.2.12. If d is a nonzero k-derivation of k[x, y] then the following two conditions are equivalent: (1) k[x, y]d 6= k; (2) d ∼ δ, where δ is a special k-derivation of k[x, y]. Proof. Since k[x, y]d = k[x, y]hd for every nonzero polynomial h in k[x, y], we may assume that the polynomials d(x) and d(y) are relatively prime. (1) ⇒ (2) Assume that k[x, y]d 6= k and let F ∈ k[x, y]d r k. Put d(x) = P , d(y) = Q and h = gcd(Fx , Fy ). Then P Fx + QFy = 0, h 6= 0 and there exist relatively prime polynomials A, B ∈ k[x, y] such that Fx = Ah and Fy = Bh. Hence AP = −BQ and hence, A | Q, Q | A, B | P and P | B. This implies that there exists an element a ∈ k r {0} such that aA = Q and aB = −P . Let δ = hd. Then d ∼ δ and δ is special. Indeed, δ ? = (hP )x + (hQ)y = −(ahB)x + (ahA)y = −aFyx + aFxy = 0. (2) ⇒ (1). Assume that d ∼ δ, where δ is a special k-derivation of k[x, y]. Since the polynomials d(x) and d(y) are relatively prime, δ = hd for some nonzero h ∈ k[x, y]. Let P = δ(x), Q = δ(y). We may assume that P 6= 0 and Q 6= 0 (if P = 0 or Q = 0 then k[x, y]d 6= k). Put f = Q and g = −P . Then fy = qx and hence, by Lemma 2.5.3, there exists H ∈ k[x, y] such that Hx = f and Hy = g. It is clear that H 6∈ k. Moreover, we have δ(H) = Hx δ(x) + Hy δ(y) = f P + gQ = QP − P Q = 0. Thus, H ∈ k[x, y]δ = k[x, y]d and H 6∈ k. Now we may prove the following Theorem 7.2.13. Let d and δ be k-derivations of k[x, y] such that k[x, y]d 6= k and k[x, y]δ 6= k. Then k[x, y]d = k[x, y]δ if and only if d ∼ δ. Proof. The implication ”⇐” is obvious. If k[x, y]d = k[x, y]δ = k[x, y], then d = δ = 0 and hence d ∼ δ. Assume now that k[x, y]d = k[x, y]δ 6= k[x, y]. In view of Proposition 7.2.11 and Theorem 7.2.12 we know that d ∼ df and δ ∼ dg for some f, g ∈ k[x, y] r k. Then Ck (f ) = k[x, y]df = k[x, y]d = k[x, y]δ = k[x, y]dg = Ck (g),
Chapter 7. Rings of constants for small n
85
and hence f ∈ Ck (g), that is, [f, g] = 0. Now, by Corollary 7.2.10, d ∼ df ∼ dg ∼ δ.
7.3
Examples of derivations with trivial ring of constants
In this section we present four examples which show that the facts described in this paper are useful to the proofs that some derivations have no nontrivial polynomial constants. Consider the following two-dimensional system of differential equations ( dx = ey 3 − xy 2 − cx + a dt dy dt
= −ey 3 + xy 2 − y + b,
where a, b, c, e ∈ R. This system is called the Selkov system ([99], see [97]). It is known ([97] Theorem 1) that if c 6= 0 and e 6= 0 then the Selkov has no polynomial first integral. We present a short proof of this fact based on the results from Section 7.2. We do not assume that e 6= 0. In terms of derivations this fact has the following formulation Example 7.3.1. Let d be a k-derivation of k[x, y] of the form d(x) = ey 3 − xy 2 − cx + a,
d(y) = −ey 3 + xy 2 − y + b,
a, b, c, e ∈ k.
If c 6= 0, then k[x, y]d = k. Proof. Suppose that there exists f ∈ k[x, y] r k such that d(f ) = 0. Let F be the nonzero form of f of the highest degree m > 1. Then (ey 3 − xy 2 )Fx −(ey 3 −xy 2 )Fy = 0 and hence, Fx −Fy = 0. Let δ be the k-derivation of k[x, y] such that δ(x) = 1 and δ(y) = −1. Then F ∈ k[x, y]δ . It is clear that x + y ∈ k[x, y]δ and hence, by Proposition 7.2.6 and Corollary 7.1.5, k[x, y]δ = k[x + y]. This means that F is of the form F = p(x + y)m , for some nonzero p ∈ k. Now comparing the coefficients of xm in the equality d(f ) = 0 we obtain the contradiction pc = 0. Consider now the Guerilla Combat system described in [97] (see also [1]). It is given by ( dx = bxy + ax dt dy dt
= hy + cx + e,
where a, b, c, e, h ∈ k. If b 6= 0 and h 6= 0 then the system has not a polynomial first integral ([97] Theorem 3). We present a short proof of this fact.
Part III. Finiteness and properties of Ad
86
Example 7.3.2. Let d be a k-derivation of k[x, y] of the form d(x) = bxy + ax,
d(y) = hy + cx + e,
a, b, c, e, h ∈ k.
If b 6= 0 and h 6= 0, then k[x, y]d = k. Proof. Suppose that there exists f ∈ k[x, y]rk such that d(f ) = 0. Let F be the nonzero form of f of the highest degree m > 1. Then bxyFx = 0 and hence, Fx = 0. Thus, F is of the form F = py m , for some nonzero p ∈ k. Now comparing the coefficients of y m in the equality d(f ) = 0 we obtain the contradiction mp = 0. Note also the following two examples. Example 7.3.3. Let d be the k-derivation of k[x, y] given by d(x) = x + y and d(y) = x. Then k[x, y]d = k. Proof. In view of Proposition 5.1.9 we may assume that k √ = C. Observe that√the matrix of d has two different eigenvalues a = (1 + 5)/2 and b = (1 − 5)/2. Thus, using a linear transformation, we may assume that d(x) = ax and d(y) = by. Suppose now that C[x, y]d 6= C. Then there exists a nonzero homogeneousPpolynomial F ∈ C such that d(F ) = 0. Put m = deg F > 1 and let F = p+q=m rpq xp y q , where each rpq is in C. Then we have 0 = d(F ) =
X
(pa + qb)rpq xp y q .
p+q=m
This implies that pa + qb √ = 0, for some √ nonnegative integers p, q with p+q = m > 0. Hence, p(1+ 5)+q(1− 5) = 0 and we have a contradiction 0 = p + q > 0. Example 7.3.4. Let d be the k-derivation of k[x, y, z] such that d(x) = xz, d(y) = yx and d(z) = zy. Then k[x, y, z]d = k. Proof. In view of Proposition 5.1.9 we may assume that k = C. Suppose that C[x, y, z]d 6= C and let H be a nonzero homogeneous polynomial in C[x, y, z] of degree m > 1 such that d(H) = 0. Then xzHx +yxHy +zyHz = 0 and hence, by Euler’s formula, x(y − z)Hx + y(y − x)Hy = myH.
Chapter 7. Rings of constants for small n
87
Let ϕ : C[x, y, z] −→ C[x, y] be the C-homomorphism of C-algebras such that ϕ(x) = x, ϕ(y) = y, ϕ(z) = 1 and let G = ϕ(H) = H(x, y, 1). Then G is a nonzero polynomial in C[x, y] and, by Proposition 1.2.4, ϕ ◦ ∂/∂x = ∂/∂x ◦ ϕ and ϕ ◦ ∂/∂y = ∂/∂y ◦ ϕ. Thus, we have x(y − 1)Gx + y(y − x)Gy = ϕ(x(y − z)Hx + y(y − x)Hy ) = ϕ(myH) = myG. Consider the C-automorphism α of C[x, y] given by α(x) = x + 1 and α(y) = y + 1. It follows from Proposition 1.2.4 that α is a differential homomorphism (that is, α ◦ ∂/∂x = ∂/∂x ◦ α and α ◦ ∂/∂y = ∂/∂y ◦ α) and hence, by the above equality, we get (x + 1)yUx + (y + 1)(y − x)Uy = m(y + 1)U, where U (x, y) = α(G) = G(x + 1, y + 1). Observe that U 6= 0 and U (0, 0) = 0. Let F be the nonzero homogeneous component of U of the smallest degree; say deg F = s. Then 1 6 s 6 m and yFx + (y − x)Fy = mF . Repeating now similar arguments as in the proof of the previous example, one easily deduces a contradiction. In the next chapters of this paper (see Chapters 10, 11 and 12) we give stronger and more general methods which are useful for problems considered in the above examples (see, for example, Theorem 12.3.1).
7.4
Minimal generators
Let d be a k-derivation of k[X] = k[x1 , . . . , xn ]. If k[X]d 6= k and k[X]d is finitely generated over k then we denote by γ(d) the minimal number of polynomials in k[X] r k which generate k[X]d over k. Moreover, we assume that γ(d) = 0 iff k[X]d = k, and γ(d) = ∞ iff k[X]d is not finitely generated over k. It is known (see Section 4.2) that there exist a natural number n and a k-derivation d of k[X] such that γ(d) = ∞. If n = 1 and d 6= 0 then of course γ(d) = 0. If n = 2 and d 6= 0 then, by Corollary 7.1.5, γ(d) 6 1. If n = 3 then we know (Theorem 7.1.3) that γ(d) < ∞. In this section we show that, for n > 3, the set {γ(d); d ∈ Derk (k[X])} is unbounded. We present here the following result of J. -M. Strelcyn and the author published in [87]: Theorem 7.4.1. If n > 3 and r > 0 then there exists a k-derivation d of k[X] such that γ(d) = r.
88
Part III. Finiteness and properties of Ad Let us start our proof with the following two simple lemmas:
Lemma 7.4.2. If d(x1 ) = x1 , . . . , d(xn ) = xn , then γ(d) = 0. Lemma 7.4.3. Let 1 6 r < n and d(x1 ) = · · · = d(xr ) = 0, d(xr+1 ) = xr+1 , . . . , d(xn ) = xn . Then γ(d) = r. Consequently, in the remaining part of the proof we will assume that r > n. Define m = n − 3 and p = r − n + 2 = r − m − 1; p > 2. Let x = x1 , y = x2 , z = x3 and if n > 3, y1 = x4 , . . . , ym = xn . To prove our theorem let us consider a k-derivation of k[X] defined as follows: d(x) = x, d(y) = y, d(z) = −pz and, if n > 3, d(y1 ) = · · · = d(ym ) = 0. Consider now r polynomials f0 , f1 , . . . fp+m : f0 = xp z, f1 = xp−1 y 1 z, . . . , fi = xp−i y i z, . . . , fp−1 = x1 y p−1 z, fp = y p z which are defined for every n > 3 and fp+1 = y1 , fp+2 = y2 , . . . , fp+m = ym when n > 3. If n = 3 then k[X] = k[x, y, z] and p + 1 = r. If n > 3 then k[X] = k[x, y, z, y1 , . . . , ym ] and p + 1 + m = r. Let us observe that these polynomials belong to k[X]d . Indeed; if 0 6 i 6 p then d(fi ) = d(xp−i y i z) = (p − i + i − p)xp−i y i z = 0 and if i = p + j > p, then d(fi ) = d(yj ) = 0. Lemma 7.4.4. The polynomials f0 , . . . , fr−1 generate k[X]d over k. Proof. Let w ∈ k[X] be such that d(w) First let us consider P = 0. the case when w ∈ k[x, y, z]. Thus, w = aije xi y j z e with all aije ∈ k. Since d is diagonal, all monomials xi y j z e from the above sum are such that d(xi y j z e ) = 0. Let us pick such a monomial w0 = xi y j z e . Then i + j = pe. Let a, b, u, v be the nonnegative integers such that i = ap + u, j = bp + v, u < p, v < p. Then either u + v = 0, or u + v = p. If u + v = 0, then e = a + b and consequently w0 = xi y j z e = xap y bp z a+b = (xp z)a (y p z)b = f0a fpb .
Chapter 7. Rings of constants for small n
89
If u + v = p, then e = a + b + 1 which implies w0 = xi xj z e = xap+u y bp+v z a+b+1 = (xu y v z)(xp z)a (y p z)b = fv f0a fpb . Therefore, if w ∈ k[x, y, z] and d(w) = 0, then w ∈ k[f0 , . . . , fp ]. Assume now that w ∈ k[X] = k[x, y, z, y1 , . . . , ym ]. Then X im w= ai1 ...im y1i1 . . . ym where all coefficients ai1 ...im belong to k[x, y, z]. Since d(w) = 0 we have X im , 0 = d(w) = d(ai1 ...im )y1i1 . . . ym and hence d(ai1 ...im ) = 0. From the first step of our proof, we know that ai1 ...im ∈ k[f0 , . . . , fp ] and therefore w ∈ k[f0 , . . . , fp , y1 , . . . , ym ] = k[f0 , . . . , fr−1 ]. Now we will prove that {f0 , . . . , fr−1 } is a minimal set of generators of k[X]d . For this aim suppose that for some s < r there exist polynomials g1 , . . . , gs such that k[X]d = k[g1 , . . . , gs ]. Then k[f0 , . . . , fr−1 ] = k[g1 , . . . , gs ] so that there exist polynomials α0 , . . . , αr−1 ∈ k[u1 , . . . us ] such that fi = αi (g1 , . . . , gs ) for i = 0, 1, . . . , r − 1. Moreover, there exist polynomials β1 , . . . , βs ∈ k[v0 , . . . vr−1 ] such that gj = βj (f0 , . . . , fr−1 ) for j = 1, . . . , s. Denote F = (f0 , . . . , fr−1 ) and G = (g1 , . . . , gs ) = (β1 (F ), . . . , βs (F )). Then in the ring k[X] the following identities are satisfied: fi = αi (G) = αi (β1 (F ), . . . , βs (F )), for i = 0, 1, . . . , r − 1, that is, F = (α ◦ β)(F ), where α = (α0 , . . . , αr−1 ) and β = (β1 , . . . , βs ). Let us introduce the notations: ω = (0, 0, 1, 0, . . . , 0) ∈ k n , Aie =
∂αi (G), ∂ue
Beq =
∂βe (F ), ∂vq
aie = Aie (ω),
beq = Beq (ω),
for any i, q = 0, 1, . . . , r − 1 and e = 1, . . . , s. Moreover, define Ciq =
s X
Aie Beq ,
ciq = Ciq (ω) =
e=1
s X
aie beq ,
e=1
where i, q = 0, 1, . . . , r − 1. Finally let us introduce the matrices: A = [aie ],
B = [beq ],
C = [ciq ],
where 0 6 i 6 r − 1, 1 6 e 6 s and 0 6 q 6 r − 1. By δiq we denote usual Kronecker delta. Now, using the above notations we will prove the following two lemmas.
Part III. Finiteness and properties of Ad
90
Lemma 7.4.5. If 0 6 i 6 r − 1 and p + 1 6 q 6 r − 1 then ciq = δiq . Proof. Differentiating the identity F = (α ◦ β)(F ) one obtains that r−1
X ∂fi ∂fl = Cil ∂xj ∂xj
(7.1)
l=0
for i = 0, 1, . . . , r − 1 and j = 1, . . . , n. Let q be as in our lemma and denote t = q − p, j0 = t + 3. Then t > 0, j0 > 3 and by virtue of our choice of f0 , . . . , fr−1 we see that ∂f ∂fi fq = yt = xj0 , so ∂xjq = 1 and, if i 6= q, then ∂x = 0, that is, j 0
0
∂fi = δiq . ∂xj0
(7.2)
Therefore, by (7.1) and (7.2): r−1
X ∂fq ∂fl ∂fi Ciq = Ciq · 1 = Ciq = Cil = = δiq , ∂xj0 ∂xj0 ∂xj0 l=0
and consequently ciq = Ciq (ω) = δiq . Lemma 7.4.6. If 0 6 i 6 p and 1 6 q 6 p then ciq = δiq . ∂ ∂ , dy = ∂y . Given a natural number t > 2 we define Proof. Let dx = ∂x Mt as the ideal in k[X] generated by all elements of the form dax dby (fe ), where e = 0, 1, . . . , p and 1 6 a + b 6 t − 1. It is clear that dx (Mt ) ⊆ Mt+1 and dy (Mt ) ⊆ Mt+1 . Moreover every element in Mp is of the form zh, where h ∈ k[X] is such that h(ω) = 0. By successive differentiations of the identity F = (α ◦ β)(F ) one easily sees that, for a > 0, b > 0, a + b > 0 and 0 6 e 6 p, r−1 X dax dby (fe ) = Cel dax dby (fl ) + Ee,a,b , l=0
where Ee,a,b ∈ Ma+b . From the above identity and Lemma 2 one deduces that p X a b dx dy (fe )(ω) = cel dax dby (fl )(ω), (7.3) l=0
because every element of the ideal Ma+b vanishes at ω. Now, observe that if 0 6 e, q 6 p then q dp−q (7.4) x dy (fe ) = (p − q)!q!δeq z.
Chapter 7. Rings of constants for small n
91
Therefore, by (7.3) and (7.4), one obtains that (p − q)!q!δeq =
q dp−q x dy (fe )(ω)
=
p X
cel dxp−q dqy (fl )(ω) = (p − q)!q!ceq
l=0
and consequently ciq = δiq . Now we can conclude the proof of Theorem 7.4.1. By Lemmas 7.4.5 and 7.4.6, the matrix C is invertible. Let D = C −1 A. Then D is an r ×s matrix and I = DB, where I is the r × r identity matrix. Therefore there exist two k-linear mappings B : k r −→ k s , D : k s −→ k r such that D ◦ B = id. Then B is injective, but it is a contradiction because s < r. This proves that {f0 , . . . , fr−1 } is a minimal set of generators of k[X]d over k, that is, γ(d) = r. This completes the proof of Theorem 7.4.1.
7.5
Comments and remarks
7.1 In the proof of Theorem 7.4.1 we never used the assumption that {gi }, {fj }, {αe } are polynomials. Note that the same proof gives the following Proposition 7.5.1. Let k be the field of real or of complex numbers. Let n, p be natural numbers such that n > 3 and 2 6 p 6 n. Denote x = (x1 , . . . , xn ) and let i fi (x) = xp−i 1 x2 x3 , f or 0 6 i 6 p fi (x) = xn−p+i , f or p + 1 6 i 6 p + n − 3.
If g1 , . . . , gs ∈ C ∞ (k n ) and fi = αi (g1 , . . . , gs ), gj = βj (f0 , . . . , fp+n−3 ), for some functions αi ∈ C ∞ (k s ) for 0 6 i 6 p + n − 3 and βj ∈ C ∞ (k p+n−2 ) for 1 6 j 6 s, then s > r. If k is the field of real numbers instead of C ∞ functions it suffices to consider functions of class C p . 7.2 Using the same methods as in the proof of Theorem 7.4.1 we may prove the following Proposition 7.5.2. Let f1 , . . . , fs and g1 , . . . , gt be polynomial in k[x1 , . . . , xn ]. Assume that f1 , . . . , fs are monomials such that fi - fj for all i 6= j. If k[f1 , . . . , fs ] = k[g1 , . . . , gt ] then t > s. 7.3 Let A = k[x1 , . . . , xn ] and B = k[x1 , . . . , xn , y1 , . . . , ym ] be the polynomial rings. Assume that d is a k-derivation of A with γ(d) = s, and let δ be the k-derivation of B such that δ|A = d and δ(y1 ) = · · · = δ(ym ) = 0. Is γ(δ) equal to s + m? We may prove that it is true if s = 1 or 2.
92
Part III. Finiteness and properties of Ad
7.4 We assumed that k is a field of characteristic zero. If char(k) = p > 0 then it is easy to prove that the rings of constants with respect to families of kderivations in k[X] = k[x1 , . . . , xn ] are finitely generated (see [85]). Some results of the author concerning rings of constants in positive characteristics are contained in [80] and [85]. The paper [85] is common with M. Nagata. Applications and extensions of some results of this paper one can find, for example, in [53] and [54] (see also [62]).
Part IV
Locally finite derivations In Chapter 6 we presented locally nilpotent derivations. Now we study a bigger class of derivations; locally finite derivations. This class contains locally nilpotent and semisimple derivations. Every locally finite derivation d is a sum of two derivations ds and dn , where ds is semisimple and dn is locally nilpotent. This part of the present paper consists of two chapters: Chapter 8, devoted to the farther properties of locally nilpotent derivations, and Chapter 9, on locally finite and semisimple derivations. Consider a derivation ∆ (of an algebra A) of the form ∆ = ad + bδ, where a, b ∈ A and where d, δ are locally nilpotent derivations of A which commute. Section 8.2, based on the paper [28] by Ferrero, Lequain and the author, is devoted to the question of finding necessary and sufficient conditions on a and b for ∆ to be locally nilpotent. Theorem 8.2.1, which is the main result of Section 8.2, gives a partial answer to this question. We present our second proof based on some properties of the function degd (see Section 6.1). The first proof (without degd ) was very long and had fifteen technical lemmas. Using degd we also prove (Theorem 8.1.4) that every Darboux polynomial of a locally nilpotent k-derivation d of k[X] = k[x1 , . . . , xn ] belongs to the ring of constants k[X]d . This implies that the field k(X)d is equal to (k[X])0 , the field of fractions of k[X]d In Chapter 9 we prove some equivalent conditions for a derivation to be locally finite. Moreover, the following three facts will be proved (among other things) in this chapter: (1) (Proposition 9.5.9) If k is an algebraically closed field of characteristic zero, then every semisimple k-derivation of k[x, y] is diagonalizable. (2) (Corollary 9.4.7) If d is a linear homogeneous k-derivation of k[x1 , . . . , xn ], then the algebra Nil(d) is finitely generated over k. (3) (Theorem 9.7.5) Let k be a reduced ring containing Q. If d is a locally nilpotent k-derivation of k[X], then the divergence of d is equal to 0. In Section 9.6 we present a very short proof of the result of Coomes and Zurkowski [13] concerning polynomial flows and locally finite derivations (Theorem 9.6.4). 93
8
Farther properties of locally nilpotent derivations
8.1
On the equality d(a) = ua
Let d be a locally nilpotent k-derivation of A and assume that d(a) = ua for some a, u ∈ A with a 6= 0. In [91] M. R. Rentschler shows that if k is a field (of characteristic zero) and A is such a k-domain that every invertible element of A belongs to k, then u = 0. He presents a sketch of his proof (see 3o in [91]) based on some properties of the automorphism ed (see Section 6.1). We present a new proof of this fact in a more general case. Theorem 8.1.1 ([91]). Let d be a locally nilpotent k-derivation of a reduced k-algebra A. If d(a) = ua for some u, a ∈ A, then d(a) = 0. Proof. We first assume that A is a domain. Suppose that d(a) 6= 0. Then a 6= 0, u 6= 0, degd d(a) = degd a − 1 and, by Proposition 6.1.1, degd a − 1 = degd d(a) = degd (ua) = degd a + degd u > degd a, which is a contradiction. Assume now that A is reduced.TThen there exists a family {Pi ; i ∈ Λ} of minimal prime ideals of A with i∈Λ Pi = 0. We know that in this case d(Pi ) ⊆ Pi , for every i ∈ Λ ([42], p. 13). So, we have the induced locally nilpotent derivation d on A/Pi which is a domain. Moreover, d(a) = u · a. Therefore, d(a) = 0 and consequently, d(a) ∈ Pi , for all i ∈ Λ, that is, d(a) = 0. Now we will show some consequences of Theorem 8.1.1. Proposition 8.1.2. Let d be a k-derivation of a reduced k-algebra A. Assume that a ∈ A and consider the derivation ad. The following two conditions are equivalent: (1) ad is locally nilpotent, (2) a ∈ Ad and d is locally nilpotent. Proof. The implication (2) ⇒ (1) is obvious. (1) ⇒ (2). Put d0 = ad and u = d(a). Then d0 (a) = ua and hence, by Theorem 8.1.1, ad(a) = 0. Suppose that d(a) 6= 0 and put n = degd a. Then n > 1 and we have a contradiction: 0 = d2n−1 (0) = X d2n−1 (ad(a)) = hi, jidi (a)dj+1 (a) = hn, n − 1i(dn (a))2 6= 0. i+j=2n−1
94
Chapter 8. Properties of locally nilpotent derivations
95
Thus, d(a) = 0 and the local nilpotence of d follows from the equality (ad)m = am dm . The implication (2) ⇒ (1) of the above proposition is true for any algebra A. However, for the inverse implication, the hypothesis ”A is reduced” (as shows the example below) is needed. Example 8.1.3. Let A = Q[x]/(x3 ) = Q[x]. Let d be the derivation of A defined by d(x) = x2 . Then d and xd are locally nilpotent derivations, and d(x) 6= 0. As an immediate consequence of Theorem 8.1.1 we get Theorem 8.1.4. Let k be a reduced ring containing Q and let d be locally nilpotent k-derivation of k[X] = k[x1 , . . . , xn ]. If f ∈ k[X] is a Darboux polynomial of d, then f ∈ k[X]d . Let us recall that if R is a domain then R0 denotes its field of fractions. Recall also (see Example 3.1.1) that there exist derivations d with the property (R0 )d 6= (Rd )0 . However, from Theorem 8.1.4 and Proposition 2.2.2 we get Proposition 8.1.5. If k is a field (of characteristic zero) and d is a locally nilpotent k-derivation of k[x1 , . . . , xn ], then k(x1 , . . . , xn )d = (k[x1 , . . . , xn ]d )0 . The next proposition is an unpublished result of Y. Lequain. A special case of this result we may find in [74]. Theorem 8.1.6 (Lequain). Assume that d is a k-derivation of a kdomain A (of characteristic zero). Let a ∈ Nil(d) and let 0 6= f ∈ A. If d(f ) = af , then f is transcendental over Nil(d). Proof. Assume the contrary. Let cr f r + · · · + c0 = 0 be an algebraic equation with c0 , . . . , cr ∈ Nil(d) and r minimal. Applying d to this equation we get (d(cr ) + rcr a)f r + · · · + (d(c1 ) + c1 a)f + d(c0 ) = 0. By Theorem 8.1.1 the coefficient d(cr ) + rcr a is a nonzero element of Nil(d) and, by the minimality of r, d(c0 ) 6= 0. Applying d repeatedly we see that dn (c0 ) 6= 0 for any n ∈ N. But it is a contradiction, because c0 ∈ Nil(d).
96
Part IV. Locally finite derivations
Corollary 8.1.7. Let k be a field of zero. If a(t) is a polynoPcharacteristic 1 p is transcendental over k[t]. mial in k[t] then the series ea(t) = ∞ a(t) p=0 p! ∂ Proof. Let A = k[[t]], d = ∂t and f = ea(t) . Then Nil(d) = k[t] and d(f ) = a(t)f . Therefore, the corollary follows from Theorem 8.1.6.
8.2
The derivation ad + bδ
As we have already mentioned, several problems in commutative algebra and algebraic geometry may be formulated using locally nilpotent derivations. There are many such problems even for polynomial rings in two variables. In this case we know, of course, that every k-derivation ∆ has the form ∆ = ad + bδ, where a, b are uniquely determined polynomials and d, δ are the partial derivatives. The derivations d and δ are locally nilpotent and they commute. In [28], Ferrero, Lequain and the author consider the question of finding necessary and sufficient conditions on a and b for ∆ to be locally nilpotent of A, where A is a ring. We give the following partial answer that includes the case b = 0 and b = 1. Theorem 8.2.1. Let A be a reduced Z-torsion free ring, d and δ two locally nilpotent derivations which commute, and b ∈ A such that d(b) = δ(b) = 0. Let ∆ be the derivation ad + bδ with a ∈ A. Then ∆ is locally nilpotent if and only if d(a) = 0. We start our proof with the following Proposition 8.2.2. Let A be a domain of characteristic zero, d and δ two locally nilpotent derivations of A which commute, b ∈ A such that d(b) = δ(b) = 0. Let ∆ = ad + bδ, where a ∈ A and d(a) 6= 0. Let α be an element of A such that d(α) 6= 0. Then ∆n (α) 6= 0 for every n > 0. Proof. More precisely, we will prove that d(∆n (α)) 6= 0 for every n > 0. First note that it suffices to prove the result for n = 1; indeed, we can apply it inductively to the element ∆n−1 (α). Now we consider two cases: Case I: Suppose that degd (a) > degd (b) + 2. (This includes the case b = 0. ) Look at the d-degree in the equality ∆(α) = ad(α) + bδ(α). We have: degd (ad(α)) = degd (a) + degd (d(α)) = degd (a) + degd (α) − 1 > max{1, degd (b) + degd (α) + 1}.
Chapter 8. Properties of locally nilpotent derivations
97
Since d and δ commute, we have degd (bδ(α)) = degd (b) + degd (δ(α)) = degd (b) + degd (α) 6 max{1, degd (b) + degd (α) + 1}. So, by Proposition 6.1.1, we have degd (∆(α)) = degd (ad(α)) > 1, and in particular, d(∆(α)) 6= 0. Case II: Suppose that degd (a) < degd (b) + 2. Then we have b 6= 0, degd (a) = 1 and d2 (a) = 0. Let i > 1; if we apply di to the equality ∆(α) = ad(α) + bδ(α), we obtain di (∆(α)) = id(a)di (α) + adi+1 (α) + bdi (δ(α)). In particular, if r > 1 denotes the d-degree of α, we have dr (∆(α)) = rd(a)dr (α) + bdr (δ(α)) = rd(a)dr (α) + bδ(dr (α)). Now look at the δ-degree in the last equality. We have degδ (rd(a)dr (α)) = degδ (rd(a)) + degδ (dr (α)) and degδ (bδ(dr (a))) 6 degδ (dr (α)) − 1. So, by Proposition 6.1.1 we get degδ (dr (∆(α))) = degδ (rd(a)dr (α)) > 0, hence dr (∆(α)) 6= 0 and consequently d(∆(α)) 6= 0. We are now in a position to prove the theorem. Proof of Theorem 8.2.1. Assume that d(a) = 0. Then ∆ and d commute. Let α ∈ A and let k be an integer such that ∆s (dk (α)) = 0 for some integer s > 0. (There exists indeed such a k, since d, being locally nilpotent, we can even have dk (α) = 0.) Set β = ∆s (dk−1 (α)). We have d(β) = 0, hence ∆(β) = ad(β) + bδ(β) = bδ(β) and, by induction, ∆i (β) = bi δ i (β) for every i > 0. Since δ is locally nilpotent, there exists n > 0 such that 0 = δ n (β), hence such that 0 = bn δ n (β) = ∆n (β) = ∆n+s (dk−1 (α)). Thus ∆r (dk−1 (α)) = 0 for some r > 0. Repeating the argument k times we obtain that ∆t (α) = 0 for some t > 0. Consequently ∆ is locally nilpotent. For the converse we first assume that A is a domain of characteristic zero. If ∆ is locally nilpotent, d(a) = 0 follows from Proposition 8.2.2. In fact, if d(a) 6= 0, we have ∆n (a) 6= 0 for every n > 1, a contradiction.
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Part IV. Locally finite derivations
Assume now that A is a reduced, Z-torsion free ring and ∆ is locally nilpotent. In this case, by extending the derivations to Q ⊗Z A we may assume that A is a reduced Q-algebra. Thus there exists a family {Pi ; i ∈ I} T of minimal prime ideals with i∈I Pi = 0. We know that in this case D(Pi ) ⊆ Pi , for every derivation D of A and i ∈ I (see [42]). So, we have the induced derivations d, δ and ∆ = a · d + b · δ on A/Pi which is a domain of characteristic zero. Therefore, d(a) = 0, since ∆ is locally nilpotent. Consequently, we have d(a) ∈ Pi for all i ∈ I, and so d(a) = 0. As an immediate consequence of Theorem 8.2.1 we have the following corollaries: Corollary 8.2.3. Let A be a reduced, Z-torsion free ring, d and δ locally nilpotent derivations of A and ∆ = ad+bδ, where a, b ∈ A, d(b) = δ(b) = 0. Then, the following conditions are equivalent: (1) ∆ is locally nilpotent, (2) d(a) = 0, (3) d ◦ ∆ = ∆ ◦ d, (4) (d ◦ ∆)(a) = (∆ ◦ d)(a). Corollary 8.2.4. Let k[x, y] be the polynomial ring in two variables over a reduced, Z-torsion free ring k. Let f ∈ k[x, y], b ∈ k and let ∆ be the k-derivation of k[x, y] defined by ∆(x) = f and ∆(y) = b. Then ∆ is locally nilpotent if and only if f ∈ k[y].
8.3
The theorems of Rentschler
The following two theorems are due to Rentschler [91]. Theorem 8.3.1 ([91]). If d is a locally nilpotent k-derivation of k[x, y] (where k is a field of characteristic zero), then there exist a ∈ k[x, y] and d1 ∈ Derk (k[x, y]) such that (1) d = ad1 , (2) d(a) = d1 (a) = 0, (3) d1 is locally nilpotent, (4) the ideal in k[x, y] generated by d1 (k[x, y]) is equal to k[x, y]. Theorem 8.3.2 ([91]). Let d be a locally nilpotent k-derivation of k[x, y] (where k is a field of characteristic zero). Then there exists a k-automor∂ phism σ of k[x, y] such that σdσ −1 = h ∂x , for some h ∈ k[y].
Chapter 8. Properties of locally nilpotent derivations
8.4
99
Comments and remarks
8.1 If d is a derivation of a ring R then d(R), the image of d, is not an ideal in R (in general). So, the condition (4) of Theorem 8.3.1, does not mean that there exists a principal element of d1 . However, we may show, in this case, that it is true. More precisely, we may deduce Theorem 8.3.1 from Theorem 8.3.2. Indeed, ∂ , for if d ∈ Derk (k[x, y]) is locally nilpotent then, by Theorem 8.3.2, σdσ −1 = h ∂x −1 −1 ∂ some h ∈ k[x, y] and σ ∈ Autk (k[x, y]). Put d1 = σ ∂x σ, a = σ (h). Then we have the conditions (1), (2), (3) and (4) of Theorem 8.3.1. Moreover, putting b = σ −1 (x), we see that d1 (b) = 1. Now it is easy to deduce the following Corollary 8.4.1. Let d be a locally nilpotent k-derivation of A = k[x, y], where k is a field of characteristic zero. If Ad(A) = A then there exists b ∈ A such that d(b) = 1. 8.2 Let ϕ be a k-automorphism of k[X] = k[x1 , . . . , xn ]. Then ϕ is called triangular if ϕ(xi ) − xi ∈ k[xi+1 , . . . , xn ] for all i = 1, . . . , n − 1 and ϕ(xn ) = xn . If furthermore ϕ(xi ) = xi for all i ∈ {1, . . . , n − 1} with at most one exception, ϕ is called elementary. Every triangular k-automorphism of k[X] is a finite product of elementary ones. A k-automorphism of k[X] which is a finite product of linear k-automorphisms and elementary k-automorphisms is called tame. It is well known that every k-automorphism of k[x, y] (where k is a field) is tame. This result is due to Jung [41] and van der Kulk [51]. There exist various proofs of this fact. A short proof is given by Rentschler in [91]. He showed that this fact is a simple consequence of Theorem 8.3.1, because, by his proof, the automorphism σ from Theorem 8.3.1 is tame. 8.3 Theorem 8.3.2 states that any action of the additive group of a field k of characteristic zero on the affine plane is equivalent to an action of the form t (x, y) = (x, y + th(x)), where t ∈ k, (x, y) ∈ k 2 and h ∈ k[x]. If char(k) = p > 0 then each n such an action is of the form t (x, y) = (x, y+tf0 (x)+tp f1 (x)+· · ·+tp fn (x)), where t ∈ k, (x, y) ∈ k 2 , and f0 , . . . , fn ∈ k[x]. This result is due to M. Miyanishi [60]. 8.4 If d(b) = δ(b) = 0, then it was shown in Theorem 8.2.1 that ∆ = ad+bδ is locally nilpotent if d(a) = 0. If we have only δ(b) = 0, then in general, the condition d(a) = 0 is not sufficient to obtain that ∆ is locally nilpotent. For ∂ ∂ example, the derivation ∆ = y ∂x + x ∂y of the polynomial ring Q[x, y] is not 2n+1 locally nilpotent since ∆ (x) = y for every n > 1. Based on numerous specific examples, we conjecture that if δ(b) = 0, then the condition d(a) = 0 is necessary for ∆ to be locally nilpotent. For ∆ = ad + bδ to be locally nilpotent, it is certainly not necessary that δ(b) = d(a) = 0. For example, 0 = ad − ad with d(a) 6= 0 is locally nilpotent. Less trivially, ∆ = ad + bd with d(a) = −d(b) 6= 0 is locally nilpotent provided that d is locally nilpotent, since ∆ = (a + b)d and d(a + b) = 0.
9 9.1
Local finiteness Locally finite endomorphisms
Let k be a ring, M a k-module, and let ϕ : M −→ M be a kendomorphism of M . M is said to be finite if M is finitely generated over k. A k-submodule M 0 of M is said to be ϕ-invariant if ϕ(M 0 ) ⊆ M 0 . The module M becomes a k[t]-module, where k[t] is the ring of polynomials in the variable t if, for a polynomial f = a0 +a1 t+· · ·+an tn and an element x ∈ M , we set f x = a0 x + a1 ϕ(x) + · · · + an ϕn (x). Every k[t]-module can thus be obtained from the k-endomorphism given by the formula ϕ(x) = tx. A submodule M 0 of M is ϕ-invariant iff M 0 is a k[t]-submodule of the k[t]module M . An element x ∈ M is called ϕ-integral if there exists a monic polynomial f ∈ k[t] such that f x = 0, that is, x is ϕ-integral if and only if ϕn (x) = a0 x + a1 ϕ(x) + · · · + an−1 ϕn−1 (x), for some n ∈ N and a0 , . . . , an−1 ∈ k. Let us denote by Mxϕ the smallest ϕ-invariant submodule of M containing x, that is, Mxϕ is the k-submodule of M generated by the set {x, ϕ(x), ϕ2 (x), . . . }. Proposition 9.1.1. Let x ∈ M . The following conditions are equivalent: (1) Mxϕ is finite. (2) There exists a finite ϕ-invariant submodule of M containing x. (3) x is ϕ-integral. (4) Mxϕ = kx + kϕ(x) + · · · + kϕn−1 (x), for some n ∈ N. If k is a field then the above proposition is evident. For a proof in a general case we need the following well known lemma (see, for example, [5] p. 21). Lemma 9.1.2. Let k be a ring, M 0 be a finite k-module and let ψ be a k-endomorphism of M 0 . Then ψ satisfies an equation of the form ψ n + an−1 ψ n−1 + · · · + a0 = 0, where a0 , . . . , an−1 ∈ k. Proof of Proposition 9.1.1. The implications (1) ⇒ (2) and (4) ⇒ (1) are clear. (2) ⇒ (3). If M 0 is a finite ϕ-invariant submodule of M containing x then put ψ = ϕ|M 0 and use Lemma 9.1.2. (3) ⇒ (4). Denote M 0 = kx + kϕ(x) + · · · + kϕn−1 (x). Then M 0 is a k-submodule of M and x ∈ M 0 ⊆ Mxϕ . Since ϕn+1 (x) = a0 ϕ(x) + · · · + 100
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an−1 ϕn (x) and ϕn (x) ∈ M 0 , ϕn+1 (x) ∈ M 0 and inductively, ϕp (x) ∈ M 0 for all p ∈ N. Thus Mxϕ = M 0 . We denote by Fin(ϕ) the set of all ϕ-integral elements of M . This set is (by Proposition 9.1.1) a ϕ-invariant submodule of M . The endomorphism ϕ is called locally finite if Fin(ϕ) = M . Equivalently (by Proposition 9.1.1), S ϕ is locally finite if and only if M = i∈I Vi , where each Vi is a finite ϕ-invariant submodule of M . If ϕ is locally finite and a ∈ k then, the endomorphism aϕ is also locally finite. The following example shows that the sum and the composition of two locally finite endomorphisms are not locally finite in general. Example 9.1.3. Let M be a vector space over a field k such that a set B of the form B = {x1 , x2 , . . . , y1 , y2 , . . . } is a basis of M . Let ϕ and ψ be two k-endomorphisms of M defined as follows: ϕ(xi ) = yi+1 , ϕ(yi ) = 0,
ψ(xi ) = 0, ψ(yi ) = xi+1 .
Then ϕ and ψ are locally finite, but ϕ + ψ and ϕψ are not locally finite. Proposition 9.1.4. Let M be a module over a ring k and let ϕ and ψ be locally finite k-endomorphisms of M . If ϕψ = ψϕ then the endomorphisms ϕ + ψ and ϕψ are locally finite. Proof. Let x ∈ M . Since ϕ is locally finite, there exists a finite ϕinvariant submodule W of M containing x. Let {y1 , . . . , yn } be a finite set of generators of W and let U = Myψ1 + · · · + Myψn . Then U is a finite ψ-invariant submodule of M containing W . Since ϕ and ψ commute, we see (by Proposition 9.1.1) that U is also ϕ-invariant. Therefore, for every x ∈ M , there exists a finite (ϕ+ψ)-invariant and (ϕψ)-invariant submodule U of M containing x. This means that ϕ + ψ and ϕψ are locally finite.
9.2
Equivalent conditions
Let us start with the following Proposition 9.2.1. If d is a k-derivation of a k-algebra R then Fin(d) is a k-subalgebra of R containing Nil(d).
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Part IV. Locally finite derivations
Proof. The inclusion Nil(d) ⊆ Fin(d) is obvious. Let x, y ∈ Fin(d) and let M , M 0 be finite d-invariant k-submodules of R such that x ∈ M and y ∈ M 0 . Let {x1 , . . . , xn } and {y1 , . . . , ym } be finite set of generators over k for M and M 0 , respectively, and let L be the k-submodule of R generated by all the elements of the form xi yj . Then L is a finite d-invariant submodule of R containing xy. This means that xy ∈ Fin(d). If d is a k-derivation of a k-algebra R, then d is locally finite if d, as the k-endomorphism of R, is locally finite. Proposition 9.2.2. Let R = k[r1 , . . . , rn ] be a finitely generated k-algebra and let d be a k-derivation of R. Then the following two conditions are equivalent: (1) d is locally finite. (2) There exists a monic polynomial w ∈ k[t] such that w(d)(ri ) = 0, for all i = 1, . . . , n. Proof. (2) ⇒ (1). It follows from (2) that every element ri is d-integral (Section 9.1) and hence, by Proposition 9.1.1, r1 , . . . , rn ∈ Fin(d). Thus, (by Proposition 9.2.1) R = Fin(d), that is, d is locally finite. (1) ⇒ (2) We know (Section 9.1) that the algebra R is an k[t]-module . Assume that d is locally finite. Then (by Proposition 9.1.1) there exist monic polynomials w1 , . . . , wn ∈ k[t] such that wi · ri = 0, for any i = 1, . . . , n. Put w = w1 · · · wn . Then w is a monic polynomial in k[t] and, for i ∈ {1, . . . , n}, we have w · ri = (w1 · · · wn ) · ri = (w1 · · · w bi · · · wn )wi · ri = (w1 · · · w bi · · · wn ) · (wi · ri ) = (w1 · · · w bi · · · wn ) · 0 = 0. This completes the proof. Proposition 9.2.3. Let d be a k-derivation of the polynomial ring k[X] = k[x1 , . . . , xn ]. The following two conditions are equivalent: (1) d is locally finite; (2) There exists a natural s such that deg dp (xi ) 6 s, for all p ∈ N0 and i = 1, . . . , n. Proof. (1) ⇒ (2). Let M = Mxd1 + · · · + Mxdn . It follows from Proposition 9.1.1 that M is a finitely generated k-submodule of k[X], d(M ) ⊆ M , and x1 , . . . , xn ∈ M . Let {r1 , . . . , rm } be a finite set of generators of M over k and let s be the maximum of the set {max(deg(r1 ), . . . , deg(rm ))}.
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Then every polynomial w ∈ M is of the form α1 r1 + · · · + αm rm , for some α1 , . . . , αm ∈ k, and we have: deg w = deg(α1 r1 + · · · + αm rm ) 6 max(deg(r1 ), . . . , deg(rm )) = s. Since each dp (xi ) is in M , deg dp (xi ) 6 s. (2) ⇒ (1). Step 1. Assume that k is noetherian. Denote by M the k-submodule of k[X] generated by all monic monomials of the degrees not greater ten s, and let M 0 be the k-submodule of k[X] generated by all elements of the form dp (xi ), where p ∈ N0 and i = 1, . . . , n. Then x1 , . . . , xn ∈ M 0 , d(M 0 ) ⊆ M 0 , and M 0 ⊆ M , Since k is noetherian, M 0 is finite. Thus, by Proposition 9.1.1, d is locally finite. Step 2. Now let k be an arbitrary ring. Denote by k 0 the smallest subring of k containing all the coefficients of the polynomials d(x1 ), . . . , d(xn ). Let k 0 [X] = k 0 [x1 , . . . , xn ] and let d0 = d | k 0 [X]. Then d0 is a k 0 -derivation of k 0 [X] satisfying (2) and k 0 is noetherian. Thus, by Step 1, d0 is locally finite and, by Proposition 9.1.1, there exists a finite k 0 -submodule M 0 = k 0 w1 + · · · + k 0 wm of k 0 [X], such that x1 , . . . , xn ∈ M 0 and d0 (M 0 ) ⊆ M 0 . Put M = kw1 + · · · + kwm . Then M is finitely generated over k, d(M ) ⊆ M and x1 , . . . , xn ∈ M . Therefore, by Proposition 9.1.1, d is locally finite.
9.3
Examples
Every locally nilpotent derivation is locally finite. There exist, of course, locally Pnfinite ∂derivations which are not locally nilpotent. The derivation d = i=1 xi ∂xi is an example of such a derivation. Note the following two examples of locally finite k-derivations of k[X] = k[x1 , . . . , xn ]. Example 9.3.1. Every linear k-derivation d of k[X] is locally finite. Proof. Put M = k + kx1 + · · · + kxn . Then M is a finite k-submodule of k[X], d(M ) ⊆ M , and x1 , . . . , xn ∈ M . Example 9.3.2. Every triangular k-derivation d of k[X], i. e., every kderivation d of the form: d(x1 ) = a1 x1 + b1 d(x 2 ) = a2 x2 + b2 (x1 ) d(x3 ) = a3 x3 + b3 (x1 , x2 ) ··· d(xn ) = an xn + bn (x1 , x2 , . . . , xn−1 ) where ai ∈ k, bi ∈ k[x1 , . . . , xi−1 ], x0 = 1, i ∈ {1, . . . , n}, is locally finite.
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Part IV. Locally finite derivations
Proof. Let Ai denote the polynomial ring k[x1 , . . . , xi ] and let di = d|Ai (for i = 1, . . . , n). Then each di is a k-derivation of Ai and, by Example 9.3.1, d1 is locally finite. Assume that, for some i, di is locally finite. Then x1 , . . . , xi ∈ Mi , where Mi is a finite k-submodule of Ai (so, of k[X]) such that di (Mi ) = d(Mi ) ⊆ Mi . Put Mi+1 = kxi+1 + Mi . Then Mi+1 is a finite k-submodule od Ai+1 , di+1 (Mi+1 ) ⊆ Mi+1 , and x1 , . . . , xi+1 ∈ Mi+1 . Thus di+1 is locally finite and consequently, d = dn is locally finite. Remark 9.3.3. Let d be a triangular k-derivation of k[X] as in Example 9.3.2. If k is reduced, then d is locally nilpotent if and only if a1 = · · · = an = 0. Proof. If i ∈ {1, . . . , n}, p ∈ N then dp (xi ) = api xi + bip (x1 , . . . , xi−1 ), for some bip ∈ k[x1 , . . . , xi−1 ] (where x0 = 1). So, if d is locally nilpotent then api = 0 for some p ∈ N and so, ai = 0 because k is reduced. Assume now that a1 = · · · = an = 0. Then d2 (x1 ) = 0 so, x1 ∈ Nil(d) and, inductively, x1 , . . . , xn ∈ Nil(d). Two k-derivations, d and δ, of a k-algebra R are said to be equivalent if there is a σ ∈ Autk (R) such that d = σδσ −1 . If d and δ are equivalent k-derivations of R then d is locally finite (resp. locally nilpotent) iff δ is locally finite (resp. locally nilpotent). The following result is due to Bass and Meisters [7], Coomes and Zurkowski [13]. Theorem 9.3.4. If k is a field of characteristic zero, then every locally finite k-derivation of k[x, y] is equivalent to one of the following locally finite k-derivations: b ∈ k; b ∈ k; where p ∈ k[x], monic of degree > 1; (4) d(x) = ax, d(y) = by, a, b ∈ k ∗ ; m (5) d(x) = ax, d(y) = may + x , a ∈ k ∗ , m ∈ N; (6) d(x) = y, d(y) = px + qy, where 0 6= p, q ∈ k and t2 − qt − phas not two different roots in k. (1) d(x) = 0, (2) d(x) = 1, (3) d(x) = 0,
d(y) = by, d(y) = by, d(y) = p(x),
V. Zurkowski [118], gave another proof of this fact (for k = C) based on the so called spectra of derivations. Recently an elegant proof was given by A. van den Essen in [20].
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As a consequence of Theorem 9.3.4 we get Theorem 9.3.5. If k is a field of characteristic zero then every locally finite k-derivation of k[x, y] is equivalent either to a linear derivation or a triangular derivation. If, in addition, k is algebraically closed then every locally finite k-derivation of k[x, y] is equivalent to a triangular derivation. For n > 3 the problem of classification of locally finite k-derivations of k[X] (up to k-automorphisms) is open. In [7] (p. 207), Bass and Meisters gave a list of examples of locally finite k-derivations of k[x, y, z] (for k = C). There also exists such a list given by Zurkowski [120]. It is well known, by a result of Bass [6], that there exists a locally finite (even locally nilpotent) k-derivation d of k[x, y, z] which is not equivalent to a linear or triangular derivation. The derivation d is defined by d(x) = −2(xz + y 2 )y d(y) = (xz + y 2 )z (9.1) d(z) = 0. Here d(xz + y 2 ) = 0 hence, d3 (x) = d2 (y) = d1 (z) = 0, that is, d is locally nilpotent.
9.4
Jordan-Chevalley decomposition
Let us note the following well known Theorem 9.4.1. Let V be a finite dimensional linear space over a field k of characteristic zero, and let ϕ : V −→ V be a k-endomorphism. Then: (1) There exist unique k-endomorphisms ϕs and ϕn of V such that (a) ϕ = ϕs + ϕn , (b) ϕs is semisimple, (c) ϕn is nilpotent, (d) ϕs ϕn = ϕn ϕs . Moreover, the endomorphisms ϕs and ϕn have the following property: (2) ϕs = u(ϕ) and ϕn = v(ϕ), for some polynomials u, v ∈ k[t] without constant terms. In particular, (i) ϕs and ϕn commute with any k-endomorphism of V commuting with ϕ. (ii) If V0 ⊆ V1 ⊆ V are k-subspaces and ϕ maps V1 into V0 then ϕs and ϕn also map V1 into V0 .
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Part IV. Locally finite derivations
The decomposition ϕ = ϕs + ϕn is called the Jordan-Chevalley decomposition of ϕ. If the field k is algebraically closed or if the characteristic polynomial of the endomorphism ϕ is a product of linear factors, then a proof of the above theorem one can find in ([34] p. 17) or ([52] XV exercise 14). A proof, for any field k of characteristic zero, one can find, for example, in [49]. The next theorem is a locally finite version of the Jordan-Chevalley decomposition. Such a version is mentioned in [36] p. 96. Theorem 9.4.2. Let V be a linear space over a field k of characteristic zero and let ϕ : V −→ V be a locally finite k-endomorphism. Then: (1) There exist unique k-endomorphisms ϕs and ϕn of V such that (a) ϕ = ϕs + ϕn , (b) ϕs is semisimple, (c) ϕn is locally nilpotent, (d) ϕs ϕn = ϕn ϕs . (2) The endomorphisms ϕs and ϕn commute with any k-endomorphism (not necessarily locally finite) of V commuting with ϕ. (3) If V0 ⊆ V1 ⊆ V are k-subspaces and ϕ maps V1 into V0 then ϕs and ϕn also map V1 into V0 . We assume that the reader is familiar with the basic facts concerning semisimple modules and semisimple endomorphisms (see, for example, [52] or [43]). A k-derivation d of a k-algebra R is called semisimple if d, as the kendomorphism, is semisimple, that is, if R as the k[t]-module (see Section 9.1) is semisimple. It is clear that if k is a field then every semisimple k-derivation is locally finite. The Jordan-Chevalley decomposition for derivations has the following form Theorem 9.4.3. Let d be a locally finite k-derivation of a k-algebra R, where k is a field of characteristic zero. Let d = ds + dn be the JordanChevalley decomposition of the endomorphism d. Then ds and dn are kderivations of R. The derivation ds is semisimple, and the derivation dn is locally finite. This theorem (and its proof) was presented by A. Tyc in 1987 at Algebraic Seminar of the Institute of Mathematics of N. Copernicus University. A proof of this theorem one can find in [119]. From the above theorems we get
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Corollary 9.4.4. Let R and d be as in Theorem 9.4.3. Then d(Rdn ) ⊆ Rdn 0 and d(Rds ) ⊆ Rds . Moreover, Rd = (Rdn )ds where d0s = ds |Rdn . Let us return to Theorem 6.2.2 which states that every homogeneous linear k-derivation of k[x1 , . . . , xn ] has a finitely generated ring of constants. A. Tyc, in 1987 (during the above mentioned seminar), gave an elegant proof of this fact based on the theorem of Weitzenb¨ock (Theorem 6.2.1) and Corollary 9.4.4. An essential role in his proof plays the following Theorem 9.4.5 (Tyc). Let d be a k-derivation of a noetherian k-algebra R. If d is semisimple then Rd is a noetherian ring. Assume now that d is a locally finite k-derivation of a k-algebra R and let d = ds + dn be its Jordan-Chevalley decomposition. It follows from the proof L of Theorem 9.4.2 that if the field k is algebraically closed then R = α∈k Rα , where Rα = {x ∈ R; ∃n∈N0 (d − αIR )n (x) = 0}. In particular, for α = 0 we get Rds
= {x ∈ R; ds (x) = 0} = R0 = {x ∈ R; ∃n∈N0 dn (x) = 0} = Nil(d).
Therefore, from Theorem 9.4.5 we obtain the following Proposition 9.4.6. If d is a locally finite k-derivation of a noetherian kalgebra R, then the k-algebra Nil(d) is noetherian. Corollary 9.4.7. If d is a linear k-derivation of k[x1 , . . . , xn ] then the kalgebra Nil(d) is finitely generated over k.
9.5
Semisimple derivations
Let k be a field of characteristic zero and let d be a k-derivation of a k-algebra R. If a ∈ Rd then we denote by da the k-endomorphism d − aIR , where IR is the identity map of R. It is easy to check the following Lemma 9.5.1. If a, b ∈ Rd , x, y ∈ R and n ∈ N then X dna+b (xy) = hp, qidpa (x)dqb (y). p+q=n
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Part IV. Locally finite derivations
An element x ∈ R is said to be semisimple with respect to d if there exists a finite d-invariant k-subspace of R containing x such that the kendomorphism d|W is semisimple. We denote by Sem(d) the set of all semisimple (with respect to d) elements of R. This set is a d-invariant subspace of R contained in Fin(d). It is obvious that d is semisimple if and only if Sem(d) = R. Moreover, Sem(d) ∩ Nil(d) = Ad . Proposition 9.5.2. Sem(d) is a k-subalgebra of R. Proof. We may of course assume that the field k is algebraically closed. Let x, y ∈ Sem(d) and let M , N be finite d-invariant k-subspaces of R such that x ∈ M , y ∈ N and the endomorphisms d1 = d|M , d2 = d|N are semisimple. Assume that {x1 , . . . , xm } and {y1 , . . . , yn } are bases of M and N , respectively, such that each xi is an eigenvector of d1 and each yj is an eigenvector of d2 . Then d(xi ) = ai xi , d(yj ) = bj yj , for i = 1, . . . , m, j = 1, . . . , n, where a1 , . . . , am , b1 , . . . , bn ∈ k. Let {c1 , . . . , cr } be the set of all pairwise different elements of the set {ai +bj ; i = 1, . . . , m, j = 1, . . . , n}, and denote by h the polynomial (t − c1 ) · · · (t − cr ). Consider now the k-subspace W of R generated by all the elements of the form xi yj . Since M and N are d-invariant, W is d-invariant too. Moreover, W is finite and xy ∈ W . We will show that h(d)(w) = 0, for all w ∈ W . It suffices to show this only for w = xi yj . Set a = ai , b = bj and h1 = t − (a + b). Then h = gh1 , for some g ∈ k[t], and (by Lemma 9.5.1), h1 (d)(xi yj ) = = = = =
d(xi yj ) − (a + b)xi yj da+b (xi yj ) da (xi )yj + xi db (yj ) (d(xi ) − ai xi )yj + xi (d(yj ) − bj yj ) 0yj + xi 0 = 0.
Thus, h(d)(xi yj ) = g(d)(h1 (d)(xi yj )) = g(d)(0) = 0. Therefore, h(d)(w) = 0, for all w ∈ W . This means that the minimal polynomial of d|W is a divisor of h, and hence, the minimal polynomial is a product of pairwise different linear factors. It implies that d|W is semisimple and consequently xy ∈ Sem(d). As a consequence of Proposition 9.5.2 we get Proposition 9.5.3. Assume that R = k[r1 , . . . , rn ] is finitely generated over k. The following conditions are equivalent: (1) d is semisimple.
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(2) There exist finite d-invariant k-subspaces W1 , . . . , Wn of R such that ri ∈ Wi and d|Wi is semisimple, for all i = 1, . . . , n. (3) There exists a finite d-invariant k-subspace W of R such that r1 , . . . , rn ∈ W and d|W is semisimple. Making use of this proposition we may easily check if a given derivation d is semisimple. Look at the following examples in k[X] = k[x1 , . . . , xn ]. Example 9.5.4. Let d(xi ) = ai xi with ai ∈ k for i = 1, . . . , n. Then d is semisimple. Proof. Put Wi = kxi and use Proposition 9.5.3 (2). Example 9.5.5. k[X] = k[x, y], d(x) = y, d(y) = x. Then d is semisimple. Proof. Since d(x + y) = x + y and d(x − y) = −(x − y), the polynomials x + y and x − y belong to Sem(d). Thus, by Proposition 9.5.2, we have: k[x, y] = k[x + y, x − y] ⊆ Sem(d) ⊆ k[x, y]. (x) Example 9.5.6. k[X] = k[x, y], d(x) = x, d(y) = y + f (x) − x ∂f∂x , where f (x) ∈ k[x] (for example: d(x) = x, d(y) = y + x2 ). Then d is semisimple.
Proof. Since d(x) = x and d(y + f (x)) = y + f (x), the polynomials x and y + f (x) belong to Sem(d). Thus, by Proposition 9.5.2, we have: k[x, y] = k[x, y + f (x)] ⊆ Sem(d) ⊆ k[x, y]. If d is a semisimple k-derivation of R and σ is a k-automorphism of R then the k-derivation σdσ −1 is also semisimple. Example 9.5.7. k[X] = k[x, y, z], d(x) = x, d(y) = y − x2 , d(z) = z − y 2 + 2x2 y. Then d is semisimple. ∂ ∂ ∂ Proof. Let δ = x ∂x + y ∂y + z ∂z . Then δ is semisimple and d = σδσ −1 , where σ is the k-automorphism of k[x, y, z] such that σ(x) = x, σ(y) = y + x2 and σ(z) = z + y 2 .
A k-derivation d of k[X] is called diagonal if d(x1 ) = a1 x1 , . . . , d(xn ) = an xn , for some a1 , . . . , an ∈ k. Moreover, d is called diagonalizable if d is equivalent to a diagonal k-derivation, that is, d is diagonalizable iff there exists a k-automorphism σ of k[X] such that the derivation σdσ −1 is diagonal. It follows from Example 9.5.4 that every diagonalizable derivation is semisimple.
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Proposition 9.5.8. Every semisimple k-derivation of k[t] (the polynomial ring in one variable) is diagonalizable. Proof. If d is a semisimple k-derivation of k[t], then d is locally finite so, there exist a, b ∈ k such that d(t) = at + b. If a = 0 then d = 0 (because in this case d is semisimple and locally nilpotent) and hence, d is diagonalizable. Assume now that a 6= 0 and let δ = σdσ −1 , where σ is the k-automorphism of k[t] such that σ(t) = t − a−1 b. Then δ(t) = ax and hence, d is diagonalizable. Proposition 9.5.9. If k is algebraically closed, then every semisimple kderivation of k[x, y] is diagonalizable. Proof. First observe that if d = ds + dn is the Jordan-Chevalley decomposition of a locally finite derivation d, and δ = σdσ −1 , where σ is an automorphism, then δs = σds σ −1 and δn = σdn σ −1 . Now let d be a semisimple k-derivation of k[x, y]. Then d is locally finite and so, d is equivalent to one of the locally finite derivations (1) – (5) of Theorem 9.3.4. By the uniqueness of the Jordan-Chevalley decomposition, we see that d is equivalent to one of the diagonal derivation (1) or (4) (because the derivations (2), (3) and (5) are not semisimple). The following example shows that if k is not algebraically closed then there exists a semisimple k derivation of k[x, y] which is not diagonalizable. Example 9.5.10. Let k be a field of characteristic zero such that the polynomial p(t) = t2 −t+1 is irreducible in k[t] (for example: k = Q or k = R). ∂ ∂ Consider the linear k-derivation d = y ∂x + (2y − x) ∂y of k[x, y]. Then d is semisimple and non-diagonalizable. Proof. Let W = kx + ky. Then W is a finite d-invariant subspace of k[x, y] and p(t) is the minimal polynomial of d|W . Since p(t) is irreducible, d|W is semisimple and so, by Proposition 9.5.3, d is semisimple. We will show that d is not diagonalizable. Let us suppose that the derivation δ = σdσ −1 is diagonal, where σ is a k-automorphism of k[x, y]. Set α = σ(x), β = σ(y), ax = δ(x), by = δ(y) (a, b ∈ k), and let w be the jacobian of (α, β). Then w ∈ k r {0} and we have: ∂α wax = β ∂β ∂y + (α − β) ∂y , ∂α −wby = β ∂β ∂x + (α − β) ∂x .
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111
Thus, the polynomials α and β have no constant terms. Moreover, from the first equality we deduce that if ux + vy (u, v ∈ k) is the linear part of β, then u2 − uv + v 2 = 0. But it is a contradiction with the irreducibility of p(t). Let us end this section with the following Question 9.5.11. Let d be a semisimple k-derivation of k[x1 , . . . , xn ], where k is an algebraically closed field of characteristic zero. Is d diagonalizable?
9.6
Polynomial flows
The problem presented in this section is rooted in the classical theory of ordinary differential equations. Let K denote either the field R of real numbers or the field C of complex numbers. Consider a class C 1 vector field f : Kn −→ Kn and the associated (autonomous) system of ordinary differential equations (5.6). It is well known (see e. g. [4]) that for each initial condition X(0) = a ∈ Kn there exists a unique maximal local solution (flow) ϕ(t, a) of system (5.6) defined for all real t from a maximal open interval I(a) containing t = 0. Such a flow is called global if I(a) = R for each a ∈ Kn . The flow ϕ(t, a) is called polynomial if ϕ(t, a) is polynomial in a for each fixed t, that is, if ϕ(t, a) depends polynomially on the initial condition X(0) = a. Question 9.6.1. For which f , as above, is the flow ϕ(t, a) polynomial? This question was raised by G. Meisters in [57] and discussed in many papers of Meisters, Olech, Bass, Coomes, Zurkowski and others (see [58], [12], see also [59] for information and a bibliography on this subject). If a flow ϕ(t, a) is polynomial, then the vector field f , corresponding to this flow, must be also polynomial (see [7]), that is, f = (f1 , . . . , fn ), where f1 , . . . , fn are polynomials in K[x1 , . . . , xn ]. Moreover, if a flow ϕ(t, a) is polynomial then it is global ([7]) and even (if K = C) entire ([12]). In 1991 B. Coomes and V. Zurkowski in [13] proved that (if K = C) a polynomial vector field f = (f1 , . . . , fn ) has a polynomial flow if and only if the corresponding derivation f1 ∂x∂ 1 + · · · + fn ∂x∂n is locally finite. A short proof of this fact gave A. van den Essen [22]. We show that the above result is a simple consequence ot Corollary 2.4.3. Moreover, this result is valid for any autonomous system over an arbitrary ring containing Q.
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to our notions. Assume that f = (f1 , . . . , fn ) ∈ k[X]n , d = PnReturn ∂ i=1 fi ∂xi and consider the autonomous system ∂X = f (X), X[0] = (x1 , . . . , xn ). ∂t
(9.2)
In view of Corollary 2.4.3 we know that the series Ed (x1 ), . . . , Ed (xn ) form the flow of (9.2). These series are elements of the power series ring k[X][[t]]. Now one easily deduces that Question 9.6.1 reduces to the following Question 9.6.2. When do series Ed (x1 ), . . . , Ed (xn ) belong to the ring k[[t]][X]? There is a difference between the ring k[X][[t]] and the ring k[[t]][X]. They are both subrings of k[[t, X]] and k[[t]][X] $ k[X][[t]] $ k[[t, X]]. Moreover, it is easy to prove the following P p Proposition 9.6.3. If g = ∞ p=0 wp t is an element of k[X][[t]], then the following two conditions are equivalent: (1) g ∈ k[[t]][X]. (2) There exists a natural number s such that deg wp 6 s, for any p ∈ N0 . Now, by Propositions 9.6.3, 9.2.3 and Corollary 2.4.3, we get Theorem 9.6.4 ([12], [22]). Assume that f = (f1 , . . . , fn ) ∈ k[X]n and consider the autonomous system (9.2). Then the following conditions are equivalent: (1) The formal flow of (9.2) P is polynomial. ∂ (2) The derivation d = ni=1 fi ∂x is locally finite. i
9.7
The divergence of locally finite derivations
We start this section with the following two simple propositions Proposition 9.7.1. If d is a k-derivation of k[X], then d is locally nilpotent if and only if Ed (k[t][X]) ⊆ k[t][X]. Proposition 9.7.2. If d is a k-derivation of k[X], then the following conditions are equivalent: (1) d is locally finite; (2) Ed (k[[t]][X]) ⊆ k[[t]][X].
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113
The following result is due to H. Bass, G. Meisters [7] and B. Coomes, V. Zurkowski [13]. Another its proof is given in [120]. Theorem 9.7.3. Let k be a reduced ring containing Q. If d is a locally finite k-derivation of k[X] = k[x1 , . . . , xn ], then d? , the divergence of d, is an element of k. Proof. Since d is locally finite, the series Ed (x1 ), . . . , Ed (xn ) are elements of k[[t]][X] (Proposition 9.7.2). By the same reason, since the derivation −d is locally finite, the series E−d (x1 ), . . . , E−d (xn ) are also elements of k[[t]][X]. Hence, the mapping Ed is a k[[t]]-automorphism of the polynomial ring k[[t]][X], and J = [Ed (x1 ), . . . , Ed (xn )], the jacobian of Ed , is an invertible element of k[[t]][X]. But k is reduced, so J ∈ k[[t]]. But by Theorem 2.4.7, J = 1 + d? t1 + · · · , so d? ∈ k. If k is non-reduced then the above property does not hold, in general. Example 9.7.4. Let k = Q[y]/(y 2 ) and let d be the k derivation of k[x] (a polynomial ring in a one variable) defined by d(x) = ax2 , where a = y+(y 2 ). Since d2 (x) = 2a2 x3 = 0, d is locally finite. But d? = 2ax 6∈ k. Now we are ready to prove the main result of this section. Theorem 9.7.5. Let k be a reduced ring containing Q. If d is a locally nilpotent k-derivation of k[X], then d? = 0. Proof. Put b = d? and let J be the jacobian of Ed (x1 ), . . . , Ed (xn ). We know, by Theorem 9.7.3 and Corollary 2.4.8, that b ∈ k and J = etb = P ∞ 1 p p p=0 p! b t . Since d is locally nilpotent, the series Ed (x1 ), . . . , Ed (xn ) are polynomials in k[X][t] and hence J ∈ k[X][t]. Thus bp = 0, for some p, and consequently (since k is reduced), b = 0. The derivation d from Example 9.7.4 is locally nilpotent. This means that if k is non-reduced then there exist locally nilpotent k-derivations of k[X] with a nonzero divergence.
9.8
Comments and remarks
9.1 The following example shows that if d and δ are two locally finite (even locally nilpotent) k-derivations of a k-algebra A, then the derivation [d, δ] = dδ−δd is not (in general) locally finite.
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Part IV. Locally finite derivations
∂ ∂ Example 9.8.1. Let A = k[x, y], d = x2 ∂y , δ = y ∂x . Then [d, δ](x) = x2 . Thus, d and δ are locally nilpotent, and [d, δ] is not locally finite.
9.2 Let k be a ring and let d be a k-derivation of a k-algebra R. In view of Section 9.2 we know that d is locally finite if, and only if, every element r ∈ R satisfies an equality of the form dn (r) + an−1 dn−1 (r) + · · · + a1 d(r) + a0 r = 0,
(9.3)
where a0 , . . . , an−1 are elements of k. Consider now a weaker condition. Assume that the elements a0 , . . . , an−1 belong to R (instead of k). There exist some papers in which we can see that such a condition is also useful. See, for example, [27] (the so called condition (F)). A derivation d is locally integral, if every element r ∈ R satisfies an equality of the form (9.3) with a0 , . . . , an−1 ∈ R. M. Ferrero and the author in [29] studied locally integral derivations. It follows from this paper that if R is a finitely generated kalgebra then every k-derivation of R is locally integral. Moreover, if R is arbitrary, then the set of all elements r ∈ R satisfying (9.3) with a0 , . . . , an−1 ∈ R (an analogy to the set Fin(d)) is not a k-subalgebra of R, in general. Note also that a k-derivation d of R is said to be integral if the condition (9.3) is independent from r ∈ R, that is, if d satisfies a monic polynomial over R. Integral derivations play an essential role only in positive characteristics (see [80]). It is not difficult to observe (see [77] or [82] for details) that if Q ⊂ k and R has not proper nilpotent elements then only the zero mapping is an integral derivation. 9.3 Let d be the k-derivation of k[x, y, z] as in Bass’ example (9.1), and denote by σ the k-automorphism ed defined in Section 8.3. Then we have σ(x) = x − 2(xz + y 2 )y + (xz + y 2 )2 z σ(y) = y + (xz + y 2 )z (9.4) σ(z) = z. This automorphism is the well known automorphism of k[x, y, z], constructed by Nagata [68], which is conjectured to be nontame (see page 99). Bass, in [6], proved that σ is not equivalent to a triangular k-automorphism of k[x, y, z]; a k automorphism α of k[x1 , . . . , xn ] is called triangular if α(xi ) − xi ∈ k[xi+1 , . . . , xn ] for all i = 1, . . . , n − 1 and α(xn ) = xn . As a consequence of this result we obtain the fact, mentioned in Section 9.3, that the derivation d defined by (9.1) is not equivalent to a triangular derivation. Indeed, suppose that there exists a triangular k-derivation δ of k[x, y, z] such that δ = αdα−1 , for some α ∈ Autk (k[x, y, z]). −1 Then eδ is triangular and we have a contradiction: eδ = eαdα = αed α−1 = ασα−1 . It was shown by M. Smith [104] that σ is stably tame. A k-automorphism α of k[x1 , . . . , xn ] is said to be stably tame if for some choice of new variables
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115
y1 , . . . , ym the extension of α to k[x1 , . . . , xn , y1 , . . . , ym ], which fixes y1 , . . . , ym , is tame. More precisely, let τ be the k-automorphism of k[x, y, z, u] defined by τ (x) = x − 2(xz + y 2 )y + (xz + y 2 )2 z τ (y) = y + (xz + y 2 )z τ (z) = z τ (u) = u. M. Smith proved, using locally nilpotent derivations, that τ is tame. 9.4 Let k be a field of characteristic zero and let f1 , . . . , fn be polynomials ∂fi ]. It in k[x1 , . . . , xn ] Denote by w the jacobian of (f1 , . . . , fn ), that is, w = det[ ∂x j is well known and easy to be proved that if k[f1 , . . . , fn ] = k[x1 , . . . , xn ], then w is a nonzero element of k. The famous Jacobian Conjecture [45] states that the converse of this fact is also true: if w ∈ k r {0} then k[f1 , . . . , fn ] = k[x1 , . . . , xn ]. The problem is still open even for n = 2. There exists a long list of equivalent formulations of this conjecture (see for example [110], [2]). Let n = 2. One of the equivalent formulations of the Jacobian Conjecture is as follows Conjecture 9.8.2. Let f ∈ k[x, y] and let d be the k-derivation of k[x, y] defined by ∂f d(x) = − ∂f ∂y , d(y) = ∂x . If there exists g ∈ k[x, y] such that d(g) = 1, then d is locally finite. Let d be as in the above conjecture. It is not difficult to see that if d is locally finite then, in this case, d is locally nilpotent. So, the Jacobian Conjecture says that the derivation d is locally nilpotent. Assume that f, g ∈ k[x, y] and w = Jac(f, g) = 1. Consider the k-derivation δ of k[x, y] defined by ∂ ∂ δ(x) = −g 2 ∂x (f /g), δ(y) = g 2 ∂y (f /g)
(observe that δ(x), δ(y) ∈ k[x, y]). If the Jacobian Conjecture is true then the derivation δ is semisimple. It is difficult to prove that δ is semisimple and the author does not know if it is equivalent to the Jacobian Conjecture. We fixed here n = 2. There are similar problems for n > 2. A basis {d1 , . . . , dn } of Derk (R) (where R = k[x1 , . . . , xn ]) is called locally finite (resp. locally nilpotent) if each di is locally finite (resp. locally nilpotent). Note the following result of the author. Theorem 9.8.3 ([79]). Let k be a ring containing Q and let k[X] = k[x1 , . . . , xn ] be the polynomial ring over k. The following conditions are equivalent. (1) The Jacobian Conjecture is true in the n-variable case. (2) Every commutative basis of the R-module Derk (k[X]) is locally finite. (3) Every commutative basis of the R-module Derk (k[X]) is locally nilpotent.
Part V
Polynomial derivations with trivial rings of constants In Section 7.3 we presented four examples of k-derivations of the polynomial ring k[X] = k[x1 , . . . , xn ] (for n 6 3) such that k[X]d = k. In Section 3.2 we collected several examples of k-derivations of k[X] (for arbitrary n) such that k(X)d = k. We gave a method for a construction of the derivations with this property and we showed that such derivations are important for considerations contained in Chapters 3 and 4. If d is a given k-derivation of k[X] then we already know that it is difficult to describe its ring of constants; to decide whether this ring is finitely generated or to find its generating set. But it is also difficult to decide if the ring of constants is trivial, that is, k[X]d = k or k(X)d = k. Let us recall (see Section 5.3) that this is equivalent to the problem of the nonexistence of polynomial, respectively rational, first integral for the associated system of polynomial ordinary differential equations. In the present part of the paper we study k-derivations of k[X] with k[X]d = k and k(X)d = k. We divided this part into four chapters (Chapters 10 – 13). In Chapter 10 we characterize all the linear k-derivations of k[X] without polynomial, and even rational, constants. This chapter is a copy of the author’s paper [84]. In Chapter 11 we study Jouanolou derivations, that is, the k-derivations of k[x, y, z] of the form z s ∂x + xs ∂y + y s ∂z , where s > 2. The well known theorem of Jouanolou [40] states that every Jouanolou derivation d has no Darboux polynomial. This implies (see Proposition 2.2.2) that k(x, y, z)d = k. Inspired by the beautiful proof of this theorem (sketched on pages 193 – 195 in [40]) which we owe to the referee of the book [40], we describe a method of proving the nonexistence of polynomial (and even rational) constants for some homogeneous k-derivations of k[X]. An application of our method is presented in Chapter 12, where we concentrate on the socalled factorisable derivations ([33], [31]). The results of Chapters 11 and 12 are based on the paper [64] by J. Moulin Ollagnier, J. -M. Strelcyn and the author. 116
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117
Chapter 13, the closing chapter of the paper, deals with simple polynomial derivations. A k-derivation d of k[X] is called simple if k[X] has no differential ideals (with respect to d) other than 0 and k[X]. Every simple k-derivation of k[X] has, of course, the trivial field of constants. We present several new examples of simple k-derivations of k[X] and we describe an effective way to construct such derivations.
10
Rational constants of linear derivations
We present here a description of all the linear k-derivations of k[x1 , . . . , xn ] which do not admit any nontrivial rational constant. In view of the results of Section 5.3 this means that we present a description of all the systems of linear differential equations which do not admit any rational first integral. This chapter is based on the author’s paper [84]. The fundamental Lemma 10.2.2 comes from the paper [64] by J. Moulin Ollagnier, J. M. Strelcyn and the author.
10.1
The main results
Let k be a field of characteristic zero and let d be a k-derivation of k[X] = k[x1 , . . . , xn ] such that d(xi ) =
n X
aij xj for i = 1, . . . , n,
(10.1)
j=1
where each aij belongs to k. Let λ1 , . . . , λn be the n eigenvalues (belonging to an algebraic closure k of k) of the matrix [aij ]. We will prove the following two theorems Theorem 10.1.1. If d is a k-derivation of k[X] of the form (10.1), the following conditions are equivalent: (1) k[X]d = k; (2) The eigenvalues λ1 , . . . , λn are N0 -independent. Theorem 10.1.2. If d is a k-derivation of k[X] of the form (10.1). The following conditions are equivalent: (1) k(X)d = k; (2) The Jordan matrix of the matrix [aij ] has one of the following two forms: λ1 0 .. (a) , . 0
λn
where the eigenvalues λ1 , . . . , λn are Z-independent; or 118
Chapter 10. Rational constants of linear derivations
(b)
119
λ1
0 ..
. λi−1
λi 1 0 λi+1
λi+2 ..
0
. λn
for some i ∈ {1, . . . , n − 1} where λi = λi+1 and the eigenvalues λ1 , . . . , λi , λi+2 , . . . , λn are Z-independent.
10.2
Linear derivations
In the present section we prove three useful lemmas on linear derivations. Let us make precise some notions. Assume that d is a k-derivation of k[X] of the form (10.1), and let λ1 , . . . , λn be the n eigenvalues (belonging to an algebraic closure k of k) of the matrix [aij ]. Moreover, let the Jordan matrix (in k) of the matrix [aij ] be of the form: Jm1 (ρ1 ) 0 .. (10.2) . 0
Jms (ρs )
where s > 1, m1 + · · · + ms = n, {ρ1 , . . . , ρs } = {λ1 , . . . , λn } and each Jmi (ρi ) is the mi × mi block: ρi 1 .. Jmi (ρi ) = (10.3) . 1 . ρi The homogeneity of polynomials d(x1 ), . . . , d(xn ) together with the fact that they are of the same degree 1 has the following consequences. Lemma 10.2.1. (1) If f ∈ k[X]d then all homogeneous components of f belong also to k[X]d . (2) Let f, p ∈ k[X] and assume that f 6= 0 and d(f ) = pf . Then p ∈ k and d(g) = pg for all homogeneous components g of f .
120
Part V. Polynomial derivations with trivial constants Proof. It follows from Proposition 2.2.3.
When the matrix [aij ] is diagonalizable, the following lemma is easy to be proved. We present below a proof of it in the general case. Lemma 10.2.2 ([64]). Let d be the k-derivation of k[X] defined by (10.1). Assume that f is a nonzero homogeneous polynomial in k[X] satisfying the equality d(f ) = pf for some p ∈ k. Then, there exist nonnegative integers i1 , . . . , in such that i1 λ1 + · · · + in λn = p . (10.4) i1 + · · · + in = deg f Proof. It is not difficult to see that, without loss of generality, we may assume that the field k is algebraically closed and that the matrix A = [aij ] has the Jordan’s form. Then A = D + N , where D is the diagonal matrix, N nilpotent one and DN = N D. This decomposition leads to the decomposition d = dD + dN , where dD is a diagonal k-derivation of k[X] and dN is a locally nilpotent k-derivation of k[X]. Let m be the smallest natural number such that g = (dN )m−1 (f ) 6= 0 and dN (g) = 0. Then g is a nontrivial homogeneous polynomial of the same degree as f . As dN dD = dD dN , then dD (g) = pg. Hence, the problem is reduced to the easy diagonal case. Another proof of Lemma 10.2.2 we may find in [64]. As a consequence of the above lemmas we get Lemma 10.2.3. Let d be a k-derivation of k[X] defined by (10.1) with the Jordan matrix (10.2). Assume that the elements ρ1 , . . . , ρs are Zindependent. Then the following three properties hold: (1) If f and g are nonzero homogeneous polynomials in k[X] such that d(f ) = pf and d(g) = pg for some p ∈ k, then deg f = deg g. (2) If f is a polynomial in k[X] such that d(f ) = pf for some p ∈ k, then f is homogeneous. (3) If d(f /g) = 0, where f and g are nonzero relatively prime polynomials in k[X], then f and g are homogeneous of the same degree. Proof. (1). We know that {ρ1 , . . . , ρs } = {λ1 , . . . , λn }. Hence, by Lemma 10.2.2, there exist nonnegative integers i1 , . . . , is and j1 , . . . , js such that i1 ρ1 + · · · + is ρs = p, j1 ρ1 + · · · + js ρs = p,
i1 + · · · + is = deg f, j1 + · · · + js = deg g.
Chapter 10. Rational constants of linear derivations
121
Then (i1 −j1 )ρ1 +· · ·+(is −js )ρs = 0 and hence, i1 = j1 , . . . , is = js because the elements ρ1 , . . . , ρs are Z-independent. Therefore, deg f = i1 +· · ·+is = j1 + · · · + js = deg g. (2). Suppose that f contains two nonzero homogeneous components f1 and f2 with deg f1 6= deg f2 . Then d(f1 ) = pf1 , d(f2 ) = pf2 (by Lemma 10.2.1) and so, using (1), we obtain a contradiction: deg f1 = deg f2 . (3). It is a consequence of (1), (2) and Proposition 2.2.2.
10.3
Proof of Theorem 10.1.1
(2) ⇒ (1). Suppose that there exists f ∈ k[X] such that d(f ) = 0 and f 6∈ k. We may assume (by Lemma 10.2.1) that f is homogeneous. Then d(f ) = 0f and hence, by Lemma 10.2.2, there exist i1 , . . . , in ∈ N0 satisfying (10.4). Since i1 λ1 + · · · + in λn = 0 and λ1 , . . . , λn are N0 -independent, i1 = · · · = in = 0. Therefore, deg f = i1 + · · · + in = 0, which contradicts the fact that f 6∈ k. (1) ⇒ (2). Assume now that k[X]d = k. Let k 0 be an algebraically closed field containing k and let d0 be the k 0 -derivation of k 0 [X] such that 0 d0 (xi ) = d(xi ) for all i = 1, . . . , n. Then (by Theorem 5.1.4) k 0 [X]d = k 0 so, we may assume that the field k is algebraically closed. Moreover, we may assume that [aij ] is the Jordan matrix (10.2). Therefore, there exist a subset {y1 , . . . , ys } contained in {x1 , . . . , xn } such that d(yi ) = ρi yi for all i = 1, . . . , s. Suppose now that the eigenvalues λ1 , . . . , λn are not N0 -indepen- dent. Then there exists a nonzero sequence (j1 , . . . , js ) of non-negative integers such that j1 ρ1 + · · · + js ρs = 0. Put f = y1j1 · · · ysjs . Then f ∈ k[X] r k and d(f ) = (j1 ρ1 + · · · js ρs )f = 0f = 0. But it is a contradiction. This completes the proof of Theorem 10.1.1.
10.4
Proof of Theorem 10.1.2
In view of Proposition 5.1.5 we may assume that k is an algebraically closed field of characteristic zero. Let d be a k-derivation of k[X] defined by (10.1) with the Jordan matrix (10.2). We must show that the following two conditions are equivalent: (1) k(X)d = k; (2) The elements ρ1 , . . . , ρs are Z-independent and s > n − 1.
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Part V. Polynomial derivations with trivial constants
We may of course assume that the matrix [aij ] of the derivation d coincides with the Jordan matrix (10.2). Therefore, there exist a subset {y1 , . . . , ys } contained in {x1 , . . . , xn } such that d(yi ) = ρi yi for all i = 1, . . . , s. (1) ⇒ (2). Assume that k(X)d = k. First we will show that ρ1 , . . . , ρs are Z-independent. Suppose that there exists a nonzero sequence (j1 , . . . , js ) of integers such that j1 ρ1 +· · ·+js ρs = 0. Put ϕ = y1j1 · · · ysjs . Then we have a contradiction: ϕ ∈ k(X) r k and d(ϕ) = (j1 ρ1 + · · · js ρs )ϕ = 0ϕ = 0. Now we will prove that s > n − 1. Suppose that s < n − 1 and consider the numbers m1 , . . . , ms defined in (10.2). We know that m1 +· · ·+ms = n. Therefore, we must consider only two the following cases: Case I: There exists j ∈ {1, . . . , s} such that mj > 3. In this case there exist three variables x, y, z ∈ {x1 , . . . , xn } such that d(x) = ax, d(y) = ay + x and d(z) = az + y, where a = ρj . Put ϕ = f /g, where f = x2 and g = y 2 − 2xz. Then d(f ) = 2af , d(g) = 2ag and hence, d(ϕ) = 0 and ϕ 6∈ k. It is a contradiction with the fact that k(X)d = k. Case II: There exist i, j ∈ {1, . . . , s} such that i 6= j, mi > and mj > 2. In this case there exist four variables x, y, z, t ∈ {x1 , . . . , xn } such that d(x) = ax, d(y) = ay + x, d(z) = bz and d(t) = bt + z, where a = ρi and b = ρj . Then (xt − yz)/xz ∈ k(X)d r k. This completes the proof of implication (1) ⇒ (2). For the proof of implication (2) ⇒ (1) we need the following lemma. Lemma 10.4.1. Let δ be a k-derivation of k[x1 , x2 ] such that
δ(x1 ) = ax1 δ(x2 ) = ax2 + x1
where a ∈ k. If a 6= 0 then k(x1 , x2 )δ = k. Proof. Let δa and δ0 be the k-derivations of k[x1 , x2 ] such that δa (x1 ) = ax1 , δa (x2 ) = ax2 , δ0 (x1 ) = 0 and δ0 (x2 ) = x1 . Then δ = δa + δ0 , δ0 is locally nilpotent, and k[x1 , x2 ]δ0 = k[x1 ]. Suppose that f and g are nonzero relatively prime polynomials in k[x1 , x2 ] such that δ(f /g) = 0. It follows from Lemma 10.2.3 that f and g are homogeneous of the same degree; set m = deg f = deg g. Moreover (by
Chapter 10. Rational constants of linear derivations
123
Proposition 2.2.2), δ(f ) = pfP and δ(g) = pg for some p ∈ k. Observe that δa (f ) = maf . Indeed, if f = i+j=m uij xi1 xj2 with uij ∈ k, then δa (f ) =
uij δa (xi1 xj2 ) =
X i+j=m
X
uij maxi1 xj2 = maf.
i+j=r
Hence, δ0 (f ) = δ(f ) − δa (f ) = pf − maf = (p − ma)f and hence, Theorem 8.1.1 implies that δ0 (f ) = 0. This means that f ∈ k[x1 ], that is, f = uxm 1 for some u ∈ k r {0}. Using the same argument we see that g = vxm with v ∈ k r {0}. Therefore, f /g = u/v ∈ k. 1 Now we may continue our proof of Theorem 10.1.2. (2) ⇒ (1). Assume that ρ1 , . . . , ρs are Z-independent and s > n − 1. We must consider two cases: ”s = n” and ”s = n − 1”. Case (a): s = n. In this case the derivation d is diagonal; d(xi ) = ρi xi for all i = 1, . . . , n. Let f and g be nonzero relatively prime polynomials in k[X] such that d(f /g) = 0. We know from Lemma 10.2.3 that f and g are homogeneous of the same degree. Moreover, d(f ) = pf and d(g) = pg for some p ∈ k (Proposition 2.2.2). Since ρ1 , . . . , ρn are Z-independent, it is easy to see that f and g are monomials. Put f = axi11 · · · xinn and g = bxj11 · · · xjnn , with a, b ∈ k r {0}. Then we have: i1 ρ1 + · · · + in ρn = p = j1 ρ1 + · · · + jn ρn and hence, (i1 − j1 )ρ1 + · · · + (in − jn )ρn = 0. Thus, i1 = j1 , . . . , in = jn , that is, f /g ∈ k. Case (b): s = n − 1. Using an induction on n we will show that k(X)d = k. It is obvious that n > 2. If n = 2 then our assertion follows from Lemma 10.4.1. Let n > 2. Without loss of generality we may assume that the matrix [aij ] of d is of the form
a 1 0 a
b1
0
0 .. . br
(10.5)
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Part V. Polynomial derivations with trivial constants
where a, b1 , . . . , br are Z-independent Thus, the derivation d is of the form: d(x1 ) = d(x 2) = d(x3 ) = d(x4 ) = .. . d(xn ) =
elements of k and where r = n − 2. ax1 ax2 + x1 b1 x3 b2 x4 br xn .
Observe that d(k[x1 , . . . , xn−1 ]) ⊆ k[x1 , . . . , xn−1 ]. Put R = k[x1 , . . . , xn−1 ] and let δ be the restriction of d to R. Then δ is a linear k-derivation of R. If n = 3 then δ is the derivation as in Lemma 10.4.1. If n > 3 then the matrix of δ is of the form (10.5) with r = n − 1. Thus, by an induction, Rδ = k. Assume now that f and g are nonzero relatively prime polynomials in k[X] such that d(f /g) = 0. Then (by Proposition 2.2.2) d(f ) = pf , d(g) = pg for some p ∈ k, and (by Lemma 10.2.3) f , g are homogeneous of the same degree; set m = deg f = deg g. Denote by y the variable xn and put f = F y u + F1 y u−1 + · · · g = Gy v + G1 y v−1 + · · · , where u > 0, v > 0, and where F, G are nonzero homogeneous polynomials in R with deg F = m − u and deg G = m − v. Comparing the coefficients of y u in the equality d(f ) = pf we see that d(F ) = (p − ubr )F , that is, δ(F ) = (p − ubr )F . Comparing the coefficients of y v in the equality d(g) = pg we have: δ(G) = (p − vbr )G. Use now Lemma 10.2.2 (for the derivation δ and the polynomials F and G). By this lemma there exist nonnegative integers j1 , j2 , i1 , . . . , ir−1 and j10 , j20 , i01 , . . . , i0r−1 such that j1 a + j2 a + i1 b1 + · · · + ir−1 br−1 = p − ubr , j10 a + j20 a + i01 b1 + · · · + i0r−1 br−1 = p − vbr . Then ((j1 + j2 ) − (j10 + j20 ))a + (i1 − i01 )b1 + · · · + (ir−1 − i0r−1 )br−1 + (u − v)br = 0 and hence, u = v (because the elements a, b1 , . . . , br are Z-independent). Thus, δ(F ) = qF and δ(G) = qG where q = p − ubr = p − vbr . This implies that δ(F/G) = 0. Since Rδ = k, F = cG for some nonzero c ∈ k.
Chapter 10. Rational constants of linear derivations
125
Consider now the homogeneous polynomial h ∈ k[X] equal to f − cg. It is obvious that d(h) = ph. We will show that h = 0. Suppose that h 6= 0. Since f and g are nonzero and relatively prime, the polynomials f and h are also nonzero and relatively prime. Repeating the above procedure for the polynomials f and h we see that they have the same degree with respect to y. But degy h < degy f = u because the coefficient of y u in h is equal to F −cG = 0. It is a contradiction. Therefore h = 0, that is, f /g = c ∈ k. This completes the proof of Theorem 10.1.2.
11
A theorem of Jouanolou In Chapter 4 of his book [40], J. -P. Jouanolou gives the following result.
Theorem 11.0.2. Let s > 2 be a natural number and let d be the Cderivation of C[x, y, z] defined by d(x) = z s ,
d(y) = xs ,
d(z) = y s .
(11.1)
Then, for every polynomial P in C[x, y, z], the following equation d(F ) = P F
(11.2)
does not admit a nontrivial solution F in C[x, y, z]. In particular, the field of constants C(x, y, z)d reduces to C, or equivalently, the system of differential equations dx dy dz = zs, = xs , = ys (11.3) dt dt dt does not admit any nontrivial rational first integral. The theorem would fail for s = 1; the subfield of constants does not reduce to C as x3 + y 3 + z 3 − 3xyz for instance is a constant of d. Moreover, in this case, equation (11.2) has very simple solutions with P 6= 0; for example, P = F = x + y + z. Note that the passage from non-solvability of equation (11.2) to the triviality of the subfield of constants C(x, y, z)d follows from Proposition 2.2.2. Assume now that P = 0 and consider the equation d(F ) = 0,
(11.4)
i. e., try to find some nonconstant polynomial, that will be a first integral of system (11.3). At the present time, we do not know any direct proof of the fact that no such first integral does exist, even for the most simple case s = 2. At a first glance, it seems feasible to look for a homogeneous polynomial solution F of a given degree p of equation (11.4) by the method of ”indeterminate coefficients”. A homogeneous polynomial F of degree p in C[x, y, z] can indeed be written F (x, y, z) =
X i+j+k=p
126
aijk xi y j z k ,
(11.5)
Chapter 11. A theorem of Jouanolou
127
so that the right-hand side of (11.5) can be substituted to F in equation (11.4). All that leads to a linear system L(p) for the unknowns aijk . In principle, for a given p, it is possible to write down the system L(p) and to solve it; but, finding a general rule to get L(p) for an arbitrary p is much more difficult. In particular, we have to make use of computer algebra to write down L(10) and no general rule for L(p) appears. In what concerns nonsolvability of equation (11.2), the direct proof for second degree polynomials F is already astonishingly long and complicated. In Jouanolou’s book, two different proofs of his theorem are given. The first one, described on pages 160–192, is due to Jouanolou and the second one, sketched on pages 193–195, is due to the referee of the book. Both of them essentially use some elementary facts from algebraic geometry in their conclusion. Trying to understand the second proof, we have gradually realized that the starting point of it relies on some very clever and general ideas, which can be applied to many other derivations. The second proof of Jouanolou’s theorem is unfortunately written in an extremely concise way and there is a gap at the end of it: in fact, the conclusion only holds for a natural integer s > 1 that satisfy s 6≡ 1(mod 3). Nevertheless, the proof is complete in the crucial case s = 2. We will give a complete proof of Jouanolou’s theorem together with a detailed discussion of all its steps. The proof under consideration divides in two parts, the “local analysis”, which is fairly general and the “global analysis” which relies on elementary algebraic geometry and is very specific to Jouanolou’s example. This is a remarkable fact that in many nontrivial examples, local analysis is sufficient to yield the non-existence of nontrivial constants of derivations. In next Chapter 12 of this paper we will consider three multidimensional examples for which local analysis is a sufficient tool to derive the nonexistence of a nontrivial constant of derivation in k[x1 , . . . , xn ] or even in k(x1 , . . . , xn ). Although all results of this chapter are formulated and proved for Cderivations, they remain valid if instead of C one considers an arbitrary field k of characteristic zero or even an arbitrary commutative ring without zero divisors which contains Z as a subring. This is a consequence of Section 5.1. Let us note that there exist another proofs of Jouanolou’s theorem (see [55], [11], [117]).
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Part V. Polynomial derivations with trivial constants
This chapter is based on some results of J. Moulin Ollagnier, J. M. Strelcyn and the author [64].
11.1
Degree and multiplicities of plane algebraic curves
In the two-dimensional projective case, irreducible homogeneous polynomials (in three variables) define plane algebraic curves. The multiplicities of a curve at various points of P2 (C) and the degree of the curve are related by an inequality which plays a crucial role in the second part of the proof of Jouanolou’s theorem. Let us now make the definition precise and state this inequality. Let F be a homogeneous irreducible polynomial of degree m in C[x, y, z]. Take some point M of the projective plane P2 (C) and let (a, b, c) be a representation of M in homogeneous coordinates. In order to define the multiplicity of F at M , we have to choose local affine coordinates; without lost of generality, we can assume that c 6= 0 and that it can be set to 1. Denote then by f the (non-homogeneous) two-variable irreducible polynomial defined by f (x, y) = F (x, y, 1). Polynomial f is not 0 and its degree is at most m, the degree of F . Consider now the Taylor’s development of f around point (a, b): f=
m X
hi (x − a, y − b),
i=0
where each hi is an homogeneous two-variable polynomial of degree i. Let µ be the lowest degree i for which hi is not 0; this natural number does not depend on the choice of local affine coordinates, but only on polynomial F and point M . Thus, it can be written µM (F ) and defined as the multiplicity of F at M . The multiplicity is strictly positive (µM (F ) > 0) iff F (M ) = 0 i. e., if curve {F = 0} passes through point M . Points at which the multiplicity of a given F is 1 are the ordinary points of the curve and those where µM (F ) > 1 are multiple points of it. It is a well known fact that an irreducible curve has only finitely many multiple points in the projective plane P2 (C) (see for instance [30], p. 69). Moreover, if F and G are relatively prime homogeneous polynomials in C[x, y, z], the set of their common zeroes in P2 (C) is finite. More precisely, according to a theorem due to B´ezout (see [30], p. 112), X µM (F )µM (G) 6 deg(F )deg(G). M ∈ P2 (C)
Chapter 11. A theorem of Jouanolou
129
This result applies to polynomial F together with one of its nonzero partial derivatives to yield X µM (F )(µM (F ) − 1) 6 deg(F )(deg(F ) − 1). (11.6) M ∈ P2 (C) In fact, a stronger inequality holds for an irreducible homogeneous polynomial F (see [30] p. 117): X µM (F )(µM (F ) − 1) 6 (deg(F ) − 1)(deg(F ) − 2). (11.7) M ∈ P2 (C) Strangely enough, inequality (11.6) is not sufficient to conclude the proof of Jouanolou’s theorem; we need the full strength of inequality (11.7).
11.2
Darboux points
Let V1 ,. . . , Vn be n homogeneous polynomials of the same degree s in C[x1 , . . . , xn ] and consider the derivation dV defined by dV (xi ) = Vi ,
1 6 i 6 n.
(11.8)
We will be interested in the following general equation dV (F ) =
n X i=1
Vi
∂F = PF ∂xi
(11.9)
in which F is an unknown polynomial of some degree m > 1, while the “eigenvalue” P is some unknown element of C[x1 , . . . , xn ]. We may assume (by Section 2.2) that F is a homogeneous irreducible nontrivial polynomial of some degree m and P is a homogeneous polynomial of degree s − 1. Using Euler’s formula n X
xi
i=1
∂F = mF, ∂xi
(11.10)
we get from (11.9) an equation in which the partial derivative of F with respect to the last variable xn no longer appears: n−1 X i=1
(xn Vi − xi Vn )
∂F = (xn P − mVn )F. ∂xi
(11.11)
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Part V. Polynomial derivations with trivial constants
A point Z ∈ Pn−1 (C) will be called a Darboux point of derivation dV if vector V (z) = (V1 (z), . . . , Vn (z)) is proportional to vector z = (z1 , . . . , zn ) for every system z of homogeneous coordinates of Z. Let then Z be a Darboux point of derivation dV ; without lost of generality, we can suppose that the last coordinate zn of z = (z1 , . . . , zn ) is equal to 1. By the very definition of a Darboux point, all the differences Vi (z1 , . . . , 1) − zi Vn (z1 , . . . , 1) vanish so that [P (z1 , . . . , 1) − mVn (z1 , . . . , 1)]F (z1 , . . . , 1) = 0. Let us stress the fact that we cannot a priori exclude the possibility that F (z1 , . . . , 1) 6= 0. Choose now the local affine coordinates y1 , . . . , yn−1 defined by x1 = z1 + y1 , . . . , xn−1 = zn−1 + yn−1 . This change of coordinates sends the studied Darboux point Z to the origin of our new coordinate system. In what follows, we will adopt the following convention: if some homogeneous polynomial in n variables z1 , . . . , zn is denoted by a capital letter, we denote by the corresponding small letter the nonhomogeneous polynomial in n − 1 variables y1 , . . . , yn−1 , that we get from the homogeneous polynomial in n variables. For instance, we define f by f (y1 , . . . , yn−1 ) = F (z1 + y1 , . . . , zn−1 + yn−1 , 1).
(11.12)
In this local system of coordinates, equation (11.11) becomes n−1 X i=1
(vi − (zi + yi )vn )
∂f = (p − mvn )f. ∂yi
(11.13)
The study of this equation will be called the local analysis of our derivation dV . Looking simultaneously at many or all such equations in various Darboux points and at their relationships will be called a global analysis of the derivation. Note now the following lemma which is easy to be proved. Lemma 11.2.1. Let f1 , . . . , fr , p, g be polynomials in C[x1 , . . . , xr ] such that (a) f1 (0) = · · · = fr (0) = 0, (b) g 6= 0, ∂g ∂g (c) f1 ∂x + · · · + fr ∂x = pg. r 1 ˜ ˜ Let f1 , . . . , fr be the linear homogeneous components of f1 , . . . , fr , respectively, and let h be the nonzero homogeneous component of the lowest degree of g. Then ∂h ∂h f˜1 ∂x + · · · + f˜r ∂x = p(0)h. r 1
Chapter 11. A theorem of Jouanolou
131
We are interested in equation (11.13), that we need study around the point (0, . . . , 0) of Cn−1 . The involved polynomials are in general nonhomogeneous polynomials in n−1 variables and can be decomposed into their P homogeneous components: φ = φ(i) , where each polynomial φ(i) is homogeneous of degree i; in particular, φ(0) is the constant term of polynomial φ. Let µZ (F ) be the lowest integer such that f(i) 6= 0, i. e., the multiplicity of F at point Z. Using now Lemma 11.2.1 we get n−1 X
(vi − (zi + yi )vn )(1)
i=1
∂h = (p − mvn )(0) h, ∂yi
(11.14)
where h is the nontrivial homogeneous component f(µZ (F )) of lowest degree of f . In equation (11.14), the partial derivatives of h are multiplied by linear homogeneous polynomials and h by a constant. Therefore, the homogeneous h is a nontrivial eigenvector of the linear C-derivation Pn−1 polynomial ∂ l of C[y , 1 . . . , yn−1 ], where each li is equal to (vi − (zi + yi )vn )(1) . i=1 i ∂yi
11.3
Proof of Jouanolou’s theorem: initial part
We will prove that, for s > 2, the equation zs
∂F ∂F ∂F + xs + ys = PF ∂x ∂y ∂z
(11.15)
does not admit a non-constant homogeneous polynomial solution F in C[x, y, z] for any homogeneous polynomial eigenvalue P of degree s − 1 in C[x, y, z]. As we already know from Section 2.2 the limitation to the homogeneous F is not restrictive. Thanks to Lemma 2.2.1, the unknown F can be supposed to be irreducible without any restriction of generality. The first step consists in finding all Darboux points of Jouanolou’s derivation defined by (11.1), i. e., points of the projective complex plane where vectors (x, y, z) and (z s , xs , y s ) are proportional. That leads to the following three equations xs+1 = yz s ;
y s+1 = zxs ;
z s+1 = xy s
whose corresponding nontrivial solutions represent the coordinates of S = s2 + s + 1 different points of P2 (C). An easy computation shows that the z-coordinate can be chosen equal to 1 for all these points and that they are represented by all the triples (ξ, ξ s+1 , 1), where ξ runs over the set US of all S-roots of unity.
132
Part V. Polynomial derivations with trivial constants In this particular situation, equation (11.11) writes (z s+1 − xy s )
∂F ∂F + (zxs − y s+1 ) = (P z − my s )F ∂x ∂y
(11.16)
where m > 1 is the degree of the sought homogeneous polynomial F . For such an F , equations (11.15) and (11.16) are equivalent.
11.4
Local analysis
Suppose that equation (11.16) admits a non-constant solution F in C[x, y, z] and fix such a solution. We have now to perform a local analysis of equation (11.16) around every Darboux point (ξ, ξ s+1 , 1) of our derivation. Choose the following local affine system (u, v) of coordinates x = ξ(1 + u);
y = ξ s+1 (1 + v).
Although these coordinates are slightly different from those used in Section 11.2, all considerations from this section remain valid here with unessential modifications. In this new system of coordinates, equation (11.16) becomes (1 − ξ(1 + u)[ξ s+1 (1 + v)]s )ξ −1 ∂f ∂u s s+1 s+1 (11.17) +([ξ(1 + u)] − [ξ (1 + v)] )ξ −(s+1) ∂f ∂v = (p − m[ξ s+1 (1 + v)]s )f where (
f (u, v) = F (x, y, 1) = F (ξ(1 + u), ξ s+1 (1 + v), 1) p(u, v)
= P (x, y, 1) = P (ξ(1 + u), ξ s+1 (1 + v), 1).
According to equation (11.14), we have to compute homogeneous components of degree 1 of the factors by which the partial derivatives of f are multiplied and the constant term of the right-hand side factor in equation (11.17). (1 − ξ(1 + u)[ξ s+1 (1 + v)]s )ξ −1 (1) = −ξ −1 (u + sv) ([ξ(1 + u)]s − [ξ s+1 (1 + v)]s+1 )ξ −(s+1) (1) = ξ −1 (su − (s + 1)v) (p − m[ξ s+1 (1 + v)]s ) = ξ −1 (ξP (ξ, ξ s+1 , 1) − m). (0) Call now µ(ξ) the multiplicity of F at the Darboux point (ξ, ξ s+1 , 1) and h the nonzero homogeneous component of f of degree µ(ξ) 6 m.
Chapter 11. A theorem of Jouanolou
133
Polynomial h would satisfy equation (11.14). After a multiplication by the factor −ξ −1 , this becomes [u + sv]
∂h ∂h + [−su + (s + 1)v] = [−ξP (ξ, ξ s+1 , 1) + m]h ∂u ∂v
(11.18)
The left-hand side of this equation does not dependent on ξ. Let ρ1 and ρ2 be the eigenvalues of the matrix corresponding to the linear differential operator of the left-hand side of equation (11.18). They are equal to the √ √ two complex conjugate numbers ρ1 = (s+2−is 3)/2 and ρ2 = (s+2+is 3)/2. According to Lemma 10.2.2, there then exists two non-negative integers i1 (ξ) and i2 (ξ) such that i1 (ξ) + i2 (ξ) = µ(ξ),
ρ1 i1 (ξ) + ρ2 i2 (ξ) = −ξP (ξ, ξ s+1 , 1) + m. (11.19)
The arithmetic relations (11.19) give the starting point for the global analysis of our derivation.
11.5
Global analysis
Denote now be I1 and I2 the sums of numbers i1 (ξ) and i2 (ξ) at various Darboux points of the studied derivation and call M the sum of the multiplicities of the Darboux polynomial F at all these points: X X X I1 = i1 (ξ), I2 = i2 (ξ), M = µ(ξ). ξ∈ US
ξ∈ US
ξ∈ US
Summing now all identities (11.19) for all ξ in US yields I1 + I2 = M,
ρ1 I1 + ρ2 I2 = Sm = (s2 + s + 1)m.
(11.20)
Indeed, P is an homogeneous polynomial in three variables and its degree is s − 1. The corresponding one-variable polynomial ξP (ξ, ξ s+1 , 1) has a degree at most s2 and its constant term is 0 so that it writes 2
ξP (ξ, ξ
s+1
, 1) =
s X
βj ξ j
j=1
and consequently 2
X ξ∈ US
ξP (ξ, ξ
s+1
, 1) =
s X j=1
βj (
X
ξ∈ US
ξ j ) = 0.
134
Part V. Polynomial derivations with trivial constants
P Indeed, for every positive integer j smaller than S, ξ∈ US ξ j = 0. As I1 and I2 are integers, as eigenvalues ρ1 and ρ2 are complex conjugate numbers, and as m > 1, the second identity of (11.20) implies that I1 = def
I2 = I and equations (11.20) write 2I = M,
(s + 2)I = (s2 + s + 1)m.
(11.21)
Applying inequality (11.7) to F yields X X µ(ξ)(µ(ξ) − 1) 6 µQ (F )(µQ (F ) − 1) 6 (m − 1)(m − 2) Q ∈ P2 (C)
ξ∈ US
(11.22) so that X
(µ(ξ))2 6 (m − 1)(m − 2) + M.
(11.23)
ξ∈ US
Now, as the total number of all ξ is s2 + s + 1, 2
M2 =
X ξ∈ US
X
µ(ξ) 6 (s2 + s + 1)
(µ(ξ))2
ξ∈ US
which, together with (11.21) and (11.23) gives the following inequality involving I, m and s: 4I 2 6 (s2 + s + 1)[(m − 1)(m − 2) + 2I].
11.6
(11.24)
Conclusion of the proof: first case
According to (11.21), numbers I, m and s are also related by equality (s + 2)I = (s2 + s + 1)m
(11.25)
Numbers s + 2 and s2 + s + 1 may be relatively prime or not; if they are so, it will be rather easy to conclude that inequality (11.24) cannot hold, which will achieve the proof of Jouanolou’s theorem for such s. If they are not, this proof needs supplementary arguments. In the first case, where s + 2 and s2 + s + 1 are relatively prime, there exists a positive integer r such that I = r(s2 + s + 1) and m = r(s + 2). After substitutions, the inequality (11.24) becomes Ts (r) = 3s2 r2 − (2s2 − s − 4)r − 2 6 0.
Chapter 11. A theorem of Jouanolou
135
It is easy to see that the quadratic polynomial Ts (r), where s is a strictly positive integer, takes strictly positive values for all strictly positive integers r. This contradiction completes the proof of Jouanolou’s theorem in the case where numbers s + 2 and s2 + s + 1 are relatively prime, and then in particular in the crucial case s = 2.
11.7
The second case
The greatest common divisor of s + 2 and s2 + s + 1 is easily seen to be either 1 or 3. It remains to be supposed that it is 3. In this case, there exists a natural number q > 1 such that s = 3q + 1
s + 2 = 3(q + 1) S = s2 + s + 1 = 3(3q 2 + 3q + 1) (11.26)
and numbers q + 1 and 3q 2 + 3q + 1 are relatively prime. Thus, by (11.25), there would exist some positive integer r such that I = r(3q 2 + 3q + 1) and m = r(q + 1) and inequality (11.24) would be T˜q (r) = (9q 2 + 6q + 1)r2 − 3(6q 2 + 3q − 1)r − 6 6 0. For every natural number q, T˜q (1) < 0 and T˜q (r) > 0 for every natural number r > 2. Thus we get a contradiction for r > 2; the unique case which is not excluded up to now is the one of a Darboux polynomial of degree q + 1. We need supplementary arguments to prove that such a polynomial cannot exist. When r = 1, I = 3q 2 + 3q + 1 and the total multiplicity M of F at all Darboux points of the derivation, which, according to (11.21), is equal to M = 2I = 2(3q 2 + 3q + 1),
(11.27)
can be decomposed as follows X X M= 1+ (µ(ξ) − 1). {ξ ∈ US , µ(ξ) > 1} {ξ ∈ US , µ(ξ) > 2} The first term is simply the total number R of points of curve {F = 0} among all Darboux points. Thanks to inequality (11.22), the second term is bounded above by (m − 1)(m − 2)/2 = q(q − 1)/2, which, together with (11.27), yields a lower bound for R: R > 2(3q 2 + 3q + 1) −
q(q − 1) . 2
(11.28)
136
Part V. Polynomial derivations with trivial constants
On the other hand, consider the decomposition of the nontrivial homogeneous polynomial F as a sum of monomials: X F = fi,j,k xi y j z k i+j+k=m=q+1
The one-variable polynomial φ defined by (see (11.26)) φ(ξ) = F (ξ, ξ s+1 , 1) = F (ξ, ξ 3q+2 , 1) =
X
fi,j,k ξ (3q+2)i+j
i+j+k=m=q+1
vanishes for R different values of ξ in the finite set US . The map (i, j, k) −→ (i + (3q + 2)j) is easily shown to be injective from the set of all triples of non-negative integers whose sum is q + 1 to the interval [0, (3q +2)(q +1)] of integers. As at least one of the coefficients fi,j,k is not 0, polynomial φ is not zero. Its degree is then at most (3q + 2)(q + 1). As φ vanishes at least in R different points, we get an upper bound for R: R 6 (3q + 2)(q + 1). (11.29) The lower (11.28) and upper (11.29) bounds on R are contradictory: for a natural number q, the double inequality 2(3q 2 + 3q + 1) −
q(q − 1) 6 R 6 (3q + 2)(q + 1) 2
leads to the inequality (3q + 2)(q + 1) − 2(3q 2 + 3q + 1) +
q q(q − 1) = − (5q + 3) > 0. 2 2
But this last inequality holds for q = 0 only. This contradiction concludes the proof of Jouanolou’s theorem. Note that the case q = 0 is not excluded by the proof. It corresponds to s = 1 for which we know that some Darboux curves do exist, as explained in the introduction to this Chapter.
11.8
Comments and remarks
11.1 In algebraic terms, Jouanolou’s theorem is equivalent to the fact that the C-derivation d of C[x, y, z] defined by (11.1) has no nontrivial principal differential ideal. Nevertheless, derivation d has some nontrivial differential ideals, as for instance, the two-generator ideal A = (y − x, z − x).
Chapter 11. A theorem of Jouanolou
137
11.2 Note that from Jouanolou’s theorem (by considering the terms of highest degree on the both sides of (11.9)) one deduces immediately the non-existence of Darboux polynomials for any C-derivation of the form d(x) = z s + f (x, y, z) d(y) = xs + g(x, y, z) d(z) = y s + h(x, y, z), where s > 1, f, g, h ∈ C[x, y, z], deg(f ) < s, deg(g) < s, and deg(h) < s. 11.3 Let d be a C-derivation of C[x1 , . . . , xn ] and let σ be a C-automorphism of C[x1 , . . . , xn ]. Define the derivation δ = σdσ −1 of C[x1 , . . . , xn ]. Then the nonexistence of nontrivial constants or of Darboux polynomials for d is equivalent to their non-existence for δ. As an example, let us apply this remark to Jouanolou’s derivation d (for s = 2) and to the linear C-automorphism σ(x, y, z) = (y + z, x + z, x + y). Then we obtain the non-existence of Darboux polynomials for δ1 = σdσ −1 and δ2 = σ −1 dσ. The computations gives δ2 (x) = z 2 + (x − y)2 δ1 (x) = z 2 + xz − xy + yz δ1 (y) = x2 + yx − yz + zx δ2 (y) = x2 + (y − z)2 . δ1 (z) = y 2 + zy − zx + xy δ2 (z) = y 2 + (z − x)2 11.4 In the proof of Jouanolou theorem, the nontrivial homogeneous component of lowest degree is used. In some cases, the consideration of components of highest degree may also lead to the non-existence proof. As an example, consider C-derivation d˜ of C[x, y] defined by ˜ = 1 − xy s , d(x)
˜ = xs − y s+1 , d(y)
where s > 1 is a natural number. This derivation arises from the left-hand side of equation (11.16) when z = 1. Let us now prove that, for s > 1, derivation d˜ does not admit any nontrivial ˜ constant, i. e., that C[x, y]d = C. Indeed, suppose that some f ∈ C[x, y]\C satisfies ˜ ) = 0 and denote by f + the homogeneous component of highest degree of f . d(f Then f + 6= 0 and ∂f + ∂f + xy s + y s+1 = 0. ∂x ∂y The Euler’s theorem on homogeneous functions yields 0 = y s (x
∂f + ∂f + +y ) = y s (degf + ), ∂x ∂y
which implies that f + = 0. This contradiction concludes the proof.
138
Part V. Polynomial derivations with trivial constants
Nevertheless, for s = 1, derivation d˜ has a nontrivial Darboux polynomial. ˜ ) = pf , where f = x + y + 1 and p = 1 − y. Indeed d(f 11.5 The analogue of Jouanolou’s theorem fails in positive characteristics. Let indeed R be any commutative ring of prime characteristic p > 0 and take s = p. In this case, d(x + y + z) = xp + y p + z p = (x + y + z)p = (x + y + z)p−1 (x + y + z), where the derivation d is defined by (11.1). 11.6 A natural question arises: what happens when more than three variables are considered. More precisely, consider the polynomial ring C[x1 , . . . , xn ] and its derivation d defined by d(x1 ) = xsn
and d(xi ) = xsi−1
for 2 6 i 6 n.
What about the solvability of equation d(F ) = P F for F and P in C[x1 , . . . , xn ] ?
12
Some applications of the local analysis
In this chapter we continue our presentation of some results of J. Moulin Ollagnier, J. -M. Strelcyn and the author published in [64]. We consider three multidimensional examples for which the local analysis described in the previous chapter is a sufficient tool to derive the non-existence of a nontrivial constant of derivation in k[x1 , . . . , xn ] or even in k(x1 , . . . , xn ). In the vocabulary of differential equations, this denote the nonexistence of polynomial, or even rational, first integrals for the corresponding systems of ordinary differential equations. All the results of this chapter are formulated and proved for Cderivations. However, it follows from Section 5.1 that they remain valid if instead of C one considers an arbitrary field k of characteristic zero.
12.1
Factorisable derivations
Let n > 2 and let W1 , . . . , Wn ∈ C[x1 , . . . , xn ] be homogeneous Zindependent polynomials of the same degree s > 1. The C-derivation d(xi ) = xi Wi , 1 6 i 6 n,
(12.1)
as well as the corresponding system of ordinary differential equations is called factorisable. The factorisable systems of ordinary differential equations was intensively studied from a long time; see for example [33] and [31], where many references on this subject can be found. One of the main features of a factorisable derivation is the fact that the polynomials x1 , . . . , xn are always Darboux polynomials of it. Consequently any polynomial of the form n Y C xαi i , (12.2) i=1
where C 6= 0 and α1 , . . . , αn ∈ N0 , is also a Darboux polynomial of it. First we prove the following statement which will be useful in the next section. Proposition 12.1.1. Let d be a factorisable derivation defined by (12.1). Suppose that all its homogeneous Darboux polynomials are of the form (12.2). Then: (1) All its Darboux polynomials are also of this form; (2) C(x1 , . . . , xn )d = C. 139
140
Part V. Polynomial derivations with trivial constants
Proof. (1). Let F ∈ C[x1 , . . . , xn ] be a Darboux polynomial, P i. e., d(F ) = P F for some homogeneous P ∈ C[x1 , . . . , xn ]. Let F = Fi be the homogeneous decomposition of F . If Fi 6= 0 for only one i, our conclusion is evident. If this is not the case, one can find two different indices i and j, such that Fi 6= 0 and Fj 6= 0. From Proposition 2.2.3 we know that d(Fi ) = P Fi and d(Fj ) = P Fj . (12.3) In virtue of our assumptions we know that Fi = axα1 1 . . . xαnn , a 6= 0 and that Fj = bxβ1 1 . . . xβnn , b 6= 0, where (α1 , . . . , αn ) 6= (β1 , . . . , βn ).
(12.4)
As Fi and Fj are Darboux polynomials of d, then one immediately obtains that X X d(Fi ) = a( αk Wk )xα1 1 . . . xαnn , d(Fj ) = b( βk Wk )xβ1 1 . . . xβnn . P P P Now, (12.3) implies that P = αk Wk = βk Wk and thus, (αk − βk )Wk = 0. From the assumption on Z-independence of W1 , . . . , Wn one deduces that αk = βk for 1 6 k 6 n. This contradicts (12.4). Then Fi 6= 0 for only one i and (1) is proved. (2). Suppose that F/G ∈ C(x1 , . . . , xn )d , where F, G ∈ C[x1 , . . . , xn ] and F, G are relatively prime. Then (by Proposition 2.2.2) d(F ) = P F and d(G) = P G for some P ∈ C[x1 , . . . , xn ]. We know from (1) that Q Q F = C1 xαi i and G = C2 xβi i for some C1 6= 0, C2 6= 0 and some nonnegative integers α1 , . . . , αn , β1 , . . . , βn . From the proof of (1) we know that αi = βi , 1 6 i 6 n, and thus the polynomials F and G are not relatively prime. This contradiction finishes our proof. Let us now introduce a new notation. If W is a homogeneous polynomial of degree s, then W (k) denotes the coefficient of the monomial xsk which appears in W . Theorem 12.1.2. Let d be a factorisable derivation defined by (12.1). (k) (k) (1) If for some k, 1 6 k 6 n, the numbers W1 , . . . , Wn are N0 independent, then C[x1 , . . . , xn ]d = C. (k) (k) (2) If for some k, 1 6 k 6 n, the numbers W1 , . . . , Wn are Zindependent, then C(x1 , . . . , xn )d = C. Proof. (1). Suppose that for some F 6= 0, F ∈ C[x1 , . . . , xn ], deg(F ) = m > 1 one has d(F ) = 0. As we have previously noticed, we can assume that
Chapter 12. Some applications of the local analysis
141
F is homogeneous of degree m. Moreover, with no restriction of generality (n) (n) we can suppose that k = n, i. e., that the numbers W1 , . . . , Wn are N0 -independent. Consider the Darboux point z = (0, . . . , 0, 1) ∈ Pn−1 (C) of the derivation d and introduce local affine coordinates (y1 , . . . , yn−1 ) around z, defined by: xi = yi for 1 6 i 6 n − 1 and xn = 1. Equation (11.13) now writes n−1 X
yi (Wi (˜ y ) − Wn (˜ y ))
i=1
∂f = −mWn (˜ y )f, ∂yi
where y˜ = (y1 , . . . , yn−1 , 1), and equation (11.14) writes n−1 X
(n)
(Wi
− Wn(n) )yi
i=1
∂h = −mWn(n) h. ∂yi
In virtue of Proposition 10.2.2, there exist non-negative integers i1 , . . . , in−1 such that n−1 X (n) ij (Wj − Wn(n) ) = −mWn(n) , (12.5) j=1
0
0 so that (12.5) yields
(n) j=1 ij Wj
Pn
= 0. This equal(n)
(n)
ity together with (12.6) contradicts the N0 -independence of W1 , . . . , Wn , which achieves the proof of (1). (2). From Section 2.2 it follows that, in order to prove (2), it suffices to prove that, for a given P ∈ C[x1 , . . . , xn ], if two nonzero polynomials F1 , F2 ∈ C[x1 , . . . , xn ] are such that d(Fi ) = P Fi , i = 1, 2, then they are proportional. First we will consider the particular case when F1 and F2 are nontrivial homogeneous polynomials. In agreement with the notations from Chapter 11 we will denote by hi the nontrivial homogeneous component of the lowest degree of fi , where (see (11.12)) fi (y1 , . . . , yn−1 ) = Fi (y1 , . . . , yn−1 , 1), i = 1, 2. Exactly in the same way as in the proof of (1), passing through the equality (11.13) we obtain the existence of non-negative integers i1 , . . . , in and
142
Part V. Polynomial derivations with trivial constants
k1 , . . . , kn such that n X
(n)
ij Wj
j=1
=
n X
(n)
kj Wj
= P (0, 0, . . . , 0, 1),
(12.7)
j=1 n X
ij = deg F1
j=1
,
n X
kj = deg F2 .
j=1
Indeed, this follows from Lemma 10.2.2 applied to h1 and h2 respectively. P (n) From (12.7) it follows that nj=1 (ij − kj )Wj = 0. Our assumption of (n)
(n)
Z-independence of W1 , . . . , Wn implies that ij = kj , for 1 6 j 6 n, and that these numbers are the unique non-negative integers satisfying (12.7). Consequently h1 and h2 are two proportional monomials, that is, h1 = rh2 for some r ∈ C \ {0}. Consider now the polynomial F3 = F1 − rF2 . We have two possibilities; either F3 6= 0 or F3 = 0. In the first case F3 is a homogeneous Darboux polynomial of d with the same P as for F1 and F2 . It is easy to see that the degree of the lowest homogeneous component h3 of corresponding polynomial f3 is greater then deg(h1 ). Repeating now the same arguments as above, but with respect to the polynomials F1 and F3 we conclude that deg(h1 ) = deg(h3 ). This contradiction proves that F3 = 0, i. e., that F1 and F2 are proportional. Let us pass now to the general case when F1 and F2 are not supposed to be homogeneous. In virtue of Proposition 2.2.1 all nonzero homogeneous components of F1 and F2 are Darboux polynomials for d with the same P as above. Thus from the first part of our proof we deduce that all these homogeneous components are of the same degree and mutually proportional. This concludes the proof. Before finishing this section note that the assumptions of Theorem 12.1.2, even in strengthened form, cannot exclude the existence of Darboux polynomials which are not of the form (12.2). Indeed, let A = −At be a skew-symmetric n × n matrix with complex entries; A = [aij ]. Let p1 , . . . , pn ∈ C and let bij = aij + pj , 1 6 i, j 6 n. Consider linear polynomials n X Wi = bij xj , 1 6 i 6 n j=1
together with the corresponding factorisable derivation defined by (12.1). Then an easy reasoning proves that one can always find a nontrivial Pn homogeneous linear polynomial F such that d(F ) = P F with P = i=1 pi xi .
Chapter 12. Some applications of the local analysis
12.2
143
A useful determinant
The following elementary fact, whose proof is omitted, is well known (see for example Section 60 of [56]). Let n > 2 and let α0 , . . . , αn−1 belong to C. Consider the n × n matrix A: 0 1 0 ··· 0 0 0 1 ··· 0 .. . . . . . . A= . . . . 0 0 0 ··· 1 −α0 −α1 −α2 · · · −αn−1 Its characteristic polynomial P (λ) = det(A − λI) equals P (λ) = (−1)n (α0 + · · · + αn−1 λn−1 + λn ) This immediately implies the following special case, which will be useful in the sequel: Let n > 2. Consider the n × n matrix Mn : −1 1 0 · · · 0 0 −1 0 1 · · · 0 0 .. .. Mn = ... ... ... (12.8) . . −1 0 0 · · · 0 1 −1 0 0 · · · 0 0 Its characteristic polynomial P (λ) equals n
n
P (λ) = (−1) (λ + λ
n−1
1
n
+ · · · + λ + 1) = (−1)
n X
λi
i=0
so that the eigenvalues of matrix Mn are all (n + 1)-th roots of 1, except 1 itself.
12.3
An example of factorisable system
In this section, we describe a factorisable derivation for which Theorem 12.1.2 of the preceding section cannot be applied. Nevertheless the above method, together with specific arguments, leads to the proof that this derivation has no polynomial constants and even no rational constants. Consider the C-derivation d of C(x1 , . . . , xn ) defined for n > 2 by d(xi ) = xi xi+1
1 6 i 6 n,
(12.9)
144
Part V. Polynomial derivations with trivial constants
where the index n + 1 is identified with the index 1, i. e., xn+1 = x1 . For n = 2, polynomial x1 − x2 is a nontrivial constant of the derivation d, and thus the subring C[x1 , x2 ]d of constants of d is larger than C. For n > 3, we will now prove the following Theorem 12.3.1. Let d be the derivation defined in (12.9) where n > 3; then d does not admit any other Darboux polynomial than the products of powers of coordinate functions, i. e., the equation d(F ) = P F
(12.10)
P Q has no other solution than P = λi xi+1 , F = C xλi i for some element C ∈ C and some n-tuple (λ1 , . . . , λn ) of non-negative integers. As a consequence, the subfield C(x1 , . . . , xn )d of constants of d is equal to C. Proof. The proof divides in two parts. We will first prove that the subring C[x1 , . . . , xn ]d of constants of d reduces to C, i. e., that equation (12.10) has no nontrivial solution F for the eigenvalue P = 0. In the second part we will prove that any Darboux polynomial of d is in fact a constant. The first part of the proof relies on the local analysis around the Darboux point (1, . . . , 1) and yields the non-existence of a nontrivial homogeneous polynomial first integral. Introduce local affine coordinates (y1 , . . . , yn−1 ) around the Darboux point (1, . . . , 1) of Pn−1 (C). These coordinates are defined by xi = 1 + yi
for 1 6 i 6 n − 1, and xn = 1.
(12.11)
According to (11.12), a homogeneous polynomial first integral F would become a (non-homogeneous) polynomial f (y1 , . . . , yn−1 ) of degree at most m that satisfies equation (11.13), which writes n−1 X
[(1 + yi )(1 + yi+1 ) − (1 + yi )(1 + y1 )]
i=1
∂f = −m(1 + y1 )f, ∂yi
where yn = 0. After cancellations, this becomes n−2 X i=1
(1 + yi )(yi+1 − y1 )
∂f ∂f − y1 (1 + yn−1 ) = −m(1 + y1 )f. (12.12) ∂yi ∂yn−1
Chapter 12. Some applications of the local analysis
145
Let h be the nontrivial homogeneous component of lowest degree of f ; call then µ the degree of h. Polynomial h has to satisfy the following equation n−2 X
(yi+1 − y1 )
i=1
∂h ∂h − y1 = −mh. ∂yi ∂yn−1
(12.13)
By Lemma 10.2.2, there would exist non-negative integers α1 , . . . , αn−1 such that n−1 X αi ρi = −m, (12.14) i=1 n−1 X
αi = µ
(12.15)
i=1
where ρ1 , . . . , ρn−1 are the eigenvalues of the corresponding square matrix of size n − 1. This matrix coincides with the matrix Mn−1 defined by (12.8), so that its eigenvalues are all n-th roots of unity except 1. As µ = deg(h) 6 deg(f ) 6 m, from (12.14) and (12.15) one obtains that m=|
n−1 X i=1
αi ρi | 6
n−1 X
αi = µ,
(12.16)
i=1
and consequently that µ = m, i. e., that deg(h) = deg(f ), and finally that h = f . We will consider separately two cases; when n is even and when n is odd. If n is odd, then ρi 6∈ R for 1 6 i 6 n − 1, and thus from (12.16) one deduces than m < µ, which contradicts µ 6 m. If n is even, as among the eigenvalues ρ1 , . . . , ρn−1 there is exactly one real which is equal −1, say ρi0 = −1, then from (12.16) one deduces that αi = 0 for all i 6= i0 . Passing to the basis in which the matrix Mn−1 is diagonal, one easily deduces that h = f = (l1 y1 + · · · + ln−1 yn−1 )m ,
(12.17)
where l1 , . . . , ln−1 ∈ C. Substituting (12.17) to (12.13) yields l1 = −l2 = l3 = · · · = (−1)n ln−1 .
(12.18)
On the other hand, substituting (12.17) to (12.12) one obtains that n−2 X i=1
(1 + yi )(yi+1 − y1 )li − y1 (1 + yn−1 )ln−1 = −(1 + y1 )h.
146
Part V. Polynomial derivations with trivial constants
Putting here y1 = y2 = · · · = yn−1 = 1, one deduces that ln−1 = 0. Taking in account (12.18), we obtain that h = 0, which contradicts h 6= 0. This finishes the first part of the proof. Let us pass now to the second part of the proof; i. e., to the proof that a Darboux polynomial of d, which is not divisible by anyone of the coordinate polynomials xi , has to be a constant of the derivation d. We will prove that all nontrivial Darboux polynomials of the derivation d are of the form (12.2). In virtue of Proposition 12.1.1.(1), it is sufficient to consider only the homogeneous Darboux polynomials F . As the factors of Darboux polynomials are also Darboux polynomials then, without any restriction of generality, one can suppose that F is divisible by no one of the polynomials xi . PnThe polynomial P , such that d(F ) = P F , is a linear form p = j=1 λj xj with coefficients λi in C. Denote by Gi the polynomial obtained from F by setting xi = 0. For every value of index i, the defining identity n n X X ∂F d(F ) = xj xj+1 = PF = ( λj xj )F ∂xj j=1
j=1
can be ordered with respect to variable xi and the corresponding constant term yields the following equation for polynomial Gi (in all variables except xi ) X X ∂Gi =( λj xj )Gi xj xj+1 ∂xj j 6= i j 6= i − 1, j 6= i Both members of this equation can now be ordered with respect to variable xi+1 ; looking to the homogeneous part of highest degree in xi+1 of the last equation, one deduces λi+1 = 0. And, as that is true for all indices, the polynomial P has to be 0, which concludes the second part of the proof.
12.4
Another example
Let s > 1 be a natural number. In this section we will consider the C-derivation d of C[x1 , . . . , xn ] defined for n > 2 by d(xi ) = (xi + xi+1 )s , 1 6 i 6 n,
(12.19)
where the index n + 1 is identified with the index 1, i. e., xn+1 = x1 . For n = 2, x1 − x2 ∈ C[x1 , x2 ]d .
Chapter 12. Some applications of the local analysis
147
Theorem 12.4.1. Let d be the derivation defined by (12.19). C[x1 , . . . , xn ]d = C, for all s > 1 and n > 3.
Then
Proof. The proof is along the same line as the proof of Theorem 12.3.1. Thus we will only sketch it. It is based on the local analysis around the Darboux point (1, . . . , 1) of Pn−1 (C). Let homogeneous PnF be a nontrivial ∂F s polynomial of degree m > 1 such that i=1 (xi + xi+1 ) ∂xi = 0. Consequently (using Euler’s formula) n−1 X
(xn (xi + xi+1 )s − xi (xn + x1 )s )
i=1
∂F = −m(xn + x1 )s F. ∂xi
(12.20)
In the local coordinates (y1 , . . . , yn−1 ) defined by (12.11), the equation (11.13) applied to (12.20) writes n−2 X ∂f ((yi + yi+1 + 2)s − (1 + yi )(2 + y1 )s ) ∂yi i=1
+((yn−1 + 2)s − (1 + yn−1 )(2 + y1 )s ) = −m(2 + y1 )s f.
∂f ∂yn−1
(12.21)
Let h be the nontrivial homogeneous component of lowest degree of f , deg(h) = µ > 0. Polynomial h satisfies the following equation (see (11.14)) n−2 X ∂h ∂h + (−sy1 + (s − 2)yi + syi+1 ) (−2y1 + sy2 ) ∂y1 ∂yi i=2 (12.22) ∂h = −2mh. +(−sy1 + (s − 2)yn−1 ) ∂yn−1 Applying one time more Euler’s theorem on homogeneous functions, one deduces from (12.22) that n−2 X i=1
(yi+1 − yi )
∂h 2m + (s − 2)µ ∂h + y1 =− h. ∂yi ∂yn−1 s
(12.23)
Now we will apply Proposition 10.2.2 to the equation (12.23). The corresponding matrix coincides with the matrix Mn−1 defined by (12.8), whose eigenvalues are ρ1 , . . . , ρn−1 . For some non-negative integers α1 , . . . , αn−1 one has n−1 X 2m + (s − 2)µ αi ρi = − . (12.24) s i=1
148
Part V. Polynomial derivations with trivial constants
as well as the equality (12.15). As 0 6 µ 6 m, taking in account (12.24) and (12.15) one obtains that n−1 n−1 X X 2m + (s − 2)µ µ6 =| αi ρi | 6 αi = µ. s i=1
i=1
Consequently µ = m and thus h = f . To conclude we proceed now exactly in the same way as in the previous section. We distinguish two cases of n even and of n odd. When n is even we repeat word for word the argument from that section. When n i odd, we obtain the formula (12.17). Substituting (12.17) to (12.21) and putting y1 = −2, one obtains that for all y2 , . . . , yn−1 ∈ C one has l1 y2s
+
n−2 X
li (yi + yi+1 + 2)s + ln−1 (yn−1 + 2)s = 0,
i=2
which easily implies that l1 = l2 = · · · = ln−1 = 0. Thus h = 0 which is a contradiction. Note that for s = 1, the derivation (12.19) admits F = x1 + · · · + xn as a Darboux polynomial. Indeed d(F ) = 2F .
13
Simple polynomial derivations
In this chapter we present a next class of k-derivations d of k[x1 , . . . , xn ] with the property k(x1 , . . . , xn )d = k. Let d be a k-derivation of a k-algebra R. Recall that an ideal A of R is called a d-ideal (or a differential ideal) if d(A) ⊆ A. The k-derivation d is called simple if R has no d-ideals other than 0 and R. Simple derivations are useful to constructions of simple noncommutative rings which are not fields. Let R[t, d] be the Ore extension of R [37], that is, R[t, d] is a noncommutative ring (often called a skew polynomial ring of derivation type [46]) of polynomials over R in an indeterminate t with multiplication subject to the relation tr = rt + d(r) for all r ∈ R. It is well known ([14]) that, in characteristic zero, the derivation d is simple if and only if R[t, d] is a simple ring (that is, R[t, d] has no two-sided ideals other than 0 and R[t, d]). We may use simple derivations to a construction of simple Lie rings. Recall that a Lie ring L is said to be simple if L has no Lie ideals other than 0 and L (see [44]). Denote by R0 the Lie ring whose elements are those of R and whose product is as follows [a, b] = ad(b) − d(a)b, for all a, b ∈ R (see [39], [92]). In this situation C. R. Jordan and D. A. Jordan [39] proved that, if R is noetherian, then R0 is simple if and only if d is simple. The author proved in [78] that it also holds for non-noetherian rings. Assume now that R = k[x1 , . . . , xn ] is the polynomial ring over a field k of characteristic zero. It is obvious that if n = 1 then the derivative ∂/∂x1 is a simple k-derivation of R. If n = 2, then only some sporadic examples of simple derivations of R are known (see for example [38]). Let d(x1 ) = f and d(x2 ) = g. It would be of considerable interest to find necessary and sufficient conditions on f and g for d to be simple in k[x1 , x2 ]. The problem seems to be difficult even if we assume that d(x1 ) = 1. There exists, of course, the same problem for n > 2. In this case only one example of a simple derivation is known (Example 13.4.1). We present some new examples of simple k-derivations of k[x1 , . . . , xn ] and we give an algorithm to decide if a k-derivation of k[x, y] of the form ∂ ∂ ∂x + (ay + b) ∂y , where a, b ∈ k[x], is simple. One of the main tools in our investigation is a result of Shamsuddin [101] mentioned (without proof) in [38]. We also present our proof of this result. 149
150
13.1
Part V. Polynomial derivations with trivial constants
Properties of simple derivations
Proposition 13.1.1. If d is a simple k-derivation of a k-algebra R then Rd is a field. Proof. Let 0 6= r ∈ Rd . Then (r) = R, because (r) is a nonzero d-ideal of R. Hence sr = 1, for some s ∈ R. But d(s) = 1d(s) = srd(s) = sd(rs) = sd(1) = s0 = 0, so r is invertible in Rd . Proposition 13.1.2. Assume that Q ⊆ k. If d is a simple k-derivation of a k-algebra R then R is a k-domain. Proof. Let P be a minimal prime ideal of R. Since Q ⊂ k, P is a d-ideal of R ([42]). Therefore 0 = P is a a prime ideal of R. It follows from the above propositions that if Q ⊆ k and d is a simple k-derivation of a k-algebra R, then we may assume that k is a field of characteristic zero and R is a k-domain with Rd = k. The next proposition is useful to a reduction to the case when the field k is algebraically closed. Proposition 13.1.3. Let k ⊂ k 0 be a field extension and let d be a kderivation of a k-algebra R with Rd = k. Consider the k 0 -derivation d ⊗ 1 of the k 0 -algebra R ⊗k k 0 (see Section 5.1). Then d is simple if and only if d ⊗ 1 is simple. Proof. Assume that d is simple and let A0 be a nonzero (d ⊗ 1)-ideal of R ⊗k k 0 . Put A = {a ∈ R; a ⊗ 1 ∈ A0 }. It is clear that A is a d-ideal of R. Hence A = R or A = 0. If A = R then 1 ⊗ 1 ∈ A0 so A0 = R ⊗k k 0 . Suppose now that A = 0 and let {wi ; i ∈ I} be a basis of k 0 over k. Every nonzero element of R ⊗k k 0 has then a unique presentation of the form r1 ⊗ w1 + · · · + rn ⊗ wn , where n is a natural number, r1 , . . . , rn are nonzero elements of R and w1 , . . . , wn are pairwise different elements from the set {wi ; i ∈ I}. Since A0 6= 0, there exists a nonzero u ∈ A0 . Let u = r1 ⊗ w1 + · · · + rn ⊗ wn , with n and r1 , . . . , rn as above, and assume that n is minimal. If n = 1 then r1 ⊗ 1 = (1 ⊗ w1−1 )(r1 ⊗ w1 ) ∈ A0 and we have a contradiction: 0 6= r1 ∈ A = 0. Thus, n > 2. Let [r1 ] be the smallest d-ideal of R containing r1 . Since d is simple, [r1 ] = R and hence, s0 r1 + s1 d(r1 ) + s2 d2 (r1 ) + · · · + sp dp (r1 ) = 1
Chapter 13. Simple polynomial derivations for some s0 , . . . , sp ∈ R. Let v = δ(u), where δ is the map δ = 1)j . Then v is a nonzero element of A0 of the form
151 Pp
j=0 sj (d ⊗
v = 1 ⊗ w1 + t2 ⊗ w2 + · · · + tn ⊗ wn , for some t2 , . . . , tn ∈ R. But (d ⊗ 1)(v) = d(t2 ) ⊗ w2 + · · · + d(tn ) ⊗ wn is also an element of A0 so, by the minimality of n, the elements t2 , . . . , tn belong to Rd = k. Thus, we have v = 1 ⊗ w, where w = w1 + t2 w2 + · · · + tn wn . It is obvious that w 6= 0. Hence, 1 ⊗ 1 = (1 ⊗ w−1 )v ∈ A0 , and hence A0 = R ⊗k k 0 . This completes the first part of our proof. Assume now that d ⊗ 1 is simple, and let A be a nonzero d-ideal of R. Then we have the exact sequence i⊗1
p⊗1
0 −→ A ⊗k k 0 −→ R ⊗k k 0 −→ R/A ⊗k k 0 −→ 0 of k 0 -modules, where i(a) = a for a ∈ A and p : R −→ R/A is the natural map. Put A0 = (i⊗1)(A⊗k k 0 ). Since A0 is a nonzero (d⊗1)-ideal of R⊗k k 0 and d ⊗ 1 is simple, A0 = R ⊗k k 0 . Hence, R/A ⊗k k 0 ≈ (R ⊗k k 0 )/A0 = 0, and hence A = R. This means that d is simple.
13.2
Shamsuddin’s result
In [38] the following result of Shamsuddin [101] is mentioned without proof. Since this result plays an essential role in the next sections, we present our proof here. Theorem 13.2.1 ([101]). Let R be a ring containing Q and let d be a simple derivation of R. Extend the derivation d to a derivation d˜ of the ˜ = at + b where a, b ∈ R. Then the polynomial ring R[t] by setting d(t) following two conditions are equivalent. (1) d˜ is simple. (2) There exist no elements r of R such that d(r) = ar + b. Proof. (1) ⇒ (2). If r is an element of R such that d(r) = ar + b then ˜ (t − r) is a proper d-ideal. ˜ (2) ⇒ (1). Suppose that there exists a nonzero d-ideal A of R[t] such that A 6= R[t]. Then A ∩ R is a d-ideal of R and hence, since d is simple, A ∩ R = 0. Let n = min{deg f ; 0 6= f ∈ A} and let σ(A) = {0} ∪ {r ∈ R; ∃f ∈A deg f = n, cf = r},
152
Part V. Polynomial derivations with trivial constants
where cf denotes the leading coefficient of a polynomial f . It is clear that n > 1 and σ(A) is an ideal of R. Now we will show that σ(A) is a d-ideal. Let r ∈ σ(A) and let f be a polynomial in R[t] such that f ∈ A, deg f = n and cf = r. If d(r) = 0 then ˜ ) − naf . obviously d(r) ∈ σ(A). Assume that d(r) 6= 0 and put g = d(f Then g ∈ A, deg g = n and cg = d(r). Hence d(r) ∈ σ(A), that is, σ(A) is a d-ideal of R. The simplicity of d and the fact that σ(A) 6= 0 imply that σ(A) = R. Thus, there exists a monic polynomial f ∈ A such that deg f = n. Let f = tn + rn−1 tn−1 + · · · + r1 t + r0 , where r0 , . . . , rn−1 ∈ R, and consider the ˜ ) − naf . Then g ∈ A and g = stn−1 + sn−1 tn−2 + · · · + polynomial g = d(f s1 t + s0 , where s0 , . . . , sn−2 are some elements of R and s = nb + d(rn−1 ) − arn−1 . Now by the minimality of n we deduce that s = 0 and we get the equality d(r) = ar + b, where r = −n−1 rn−1 . It is a contradiction with (2).
13.3
Derivation D(a, b)
Let k be a field of characteristic zero and let k[x, y] be the polynomial ring over k. If a, b ∈ k[x] then we denote by D(a, b) the k-derivation of k[x, y] defined by D(a, b)(x) = 1 D(a, b)(y) = ay + b. In this section we present an algorithm to decide if the derivation D(a, b) is simple. If f ∈ k[x] then f 0 denotes the derivative of f . We write a∆b if the derivation D(a, b) is not simple. ∂ Since the derivative ∂x is a simple k-derivation of k[x], from Theorem 13.2.1 we get Proposition 13.3.1. Let a, b ∈ k[x]. Then a∆b if and only if there exists f ∈ k[x] such that f 0 = af + b. Note some properties of the relation ∆. Proposition 13.3.2. Let a, b ∈ k[x]. Then: (1) a∆0. (2) 0∆b.
Chapter 13. Simple polynomial derivations
153
(3) If a∆b and deg b < deg a then b = 0. (4) If a∆b and deg b = deg a then b = αa where α ∈ k r {0}. Proof. (1). The ideal (y) is differential. (2). Let h be a polynomial in k[x] such that h0 = b. Then the ideal (y − h) is differential. (3). Assume that deg b < deg a. If b = 0 then, by (1), a∆b. Now let a∆b and suppose that b 6= 0. Let f be a polynomial in k[x] such that f 0 = af + b (Proposition 13.3.1). Then f 6= 0 and we have a contradiction: deg f − 1 = deg f 0 = deg a + deg f > deg f . (4). Let deg a = deg b. If a = αb with 0 6= α ∈ k then the ideal (y + α) is differential and hence a∆b. Assume now that a∆b and let f 0 = af + b, where f ∈ k[x]. If f = 0 then b = a0 + b = 00 = 0 hence a = 0 (because deg a = deg b), and hence b = 1a. If f = −α ∈ k r {0} then 0 = (−α)0 = −αa + b, that is, b = αa. Suppose now that deg f > 1. Then a 6= 0, b 6= 0 and we have a contradiction: deg f −1 = deg f 0 = deg(af +b) = deg(af ) = deg a+deg f > deg f . Proposition 13.3.3. Let a, b, b1 , b2 ∈ k[x] and α ∈ k. Then: (1) If a∆b1 and a∆b2 then a∆(b1 + b2 ). (2) If a∆b then a∆αb. Proof. Let f1 , f2 , f be polynomials in k[x] such that f10 = af1 + b1 , f20 = af2 + b2 and f 0 = af + b. Then (f1 + f2 )0 = a(f1 + f2 ) + (b1 + b2 ) and (αf )0 = a(αf ) + (αb). Now the proposition follows from Proposition 13.3.1. Immediately from Proposition 13.3.2 we obtain the following examples of simple k-derivations in k[x, y]. Example 13.3.4. Let d(x) = 1, d(y) = xy + 1. Then d is a simple kderivation of k[x, y]. Example 13.3.5. Let d(x) = 1, d(y) = (x2 + x)y + x2 . Then d is a simple k-derivation of k[x, y]. If deg a > deg b then, using Proposition 13.3.2, we may easily decide if the derivation D(a, b) is simple. The next proposition is useful in the case when deg a < deg b.
154
Part V. Polynomial derivations with trivial constants
Proposition 13.3.6. Let a, b ∈ k[x] with a 6= 0. Assume that b = ca + r, where c, r ∈ k[x] and deg r < deg a. Then a∆b if and only if a∆(c0 + r). Proof. First observe that if u ∈ k[x] then u0 = au + (−ua + u0 ) and hence a∆(−au + u0 ). Thus, if a∆b then a∆(ca + r) and a∆(−ca + c0 ) and so, by Proposition 13.3.3, a∆(c0 + r). If a∆(c0 + r) then Proposition 13.3.3 implies a∆b, because a∆(ca − c0 ). Now we are ready to decide, in a finite number of steps, if a derivation of the form D(a, b) is simple. Example 13.3.7. If a = x3 +1, b = x8 +3x5 +1 then the derivation D(a, b) is simple. Proof. Suppose that D(a, b) is not simple. Then a∆b and, using Proposition 13.3.6, we get a∆b1 and next a∆b2 , where b1 = 5x4 −2x2 +4x+1 and b2 = −2x2 − x + 6. But b2 6= 0 and deg b2 < deg a so we have a contradiction with Proposition 13.3.2. Note also the following two propositions. Proposition 13.3.8. Let n ∈ N. The derivation D(x, xn ) is simple if and only if n is even. Proof. Assume that n = 2p. We will use Theorem 13.2.1. Suppose that there exists f ∈ k[x] such that f 0 = xf + x2p . Then we see that f has a form f = −x2p−1 + a2p−2 x2p−2 + a2p−3 x2p−3 + · · · + a1 x + a0 , where a0 , . . . , a2p−2 ∈ k. So, we have −(2p − 1)x2p−2 + (2p − 2)a2p−2 x2p−3 + · · · + 2a2 x + a1 = a2p−2 x2p−1 + · · · + a1 x2 + a0 x. Therefore, a1 = a3 = · · · = a2p−3 = 0, and −(2p − 1) = a2p−3 = 0 what is a contradiction. Thus, D(x, x2p ) is simple by Theorem 13.2.1. Assume now that n = 2p + 1. Put u(x) =
p X i=0
2i
p! x2(p−i) . (p − i)!
Then (y − u(x)) is a d-ideal of k[x, y].
Chapter 13. Simple polynomial derivations
155
Proposition 13.3.9. Let 0 6= a ∈ k[x]. Then for every b ∈ k[x] there exists a unique c ∈ k[x] such that a∆(b − c) and deg c < deg a. Proof. If b ∈ k[x] then we have the following sequence of equalities b = u 1 a + r1 u 0 = u 2 a + r2 1 .. . 0 un = 0a + rn+1 , where u1 , . . . , un , r, . . . , rn+1 ∈ k[x] and deg ri < deg a for i = 1, . . . , n + 1. Put c = r1 + · · · + rn+1 . Then, deg c < deg a and, by Proposition 13.3.6, a∆(b − c). Suppose now that c1 , c2 ∈ k[x], a∆(b−c1 ), a∆(b−c2 ), deg c1 < deg a and deg c2 < deg a. Then, by Proposition 13.3.3, a∆(c1 − c2 ) and deg(c1 − c2 ) < deg a, and hence, by Proposition 13.3.2, c1 = c2 .
13.4
Examples of simple derivations
The author knows only one published example of a simple k-derivation of k[x1 , . . . , xn ] for n > 2. Such the example is given by D. A. Jordan in [38]. Example 13.4.1 ([38]). Let n > 2 and let d be the k-derivation of k[x1 , . . . , xn ] defined as follows d(x1 ) d(x2 ) d(x3 ) . .. d(xn )
= 1 − x1 x2 = x31 = x2 = xn−1 .
The derivation d is simple. P ∂ is also mentioned. In [38] another example: δ = ∂x∂ 1 + ni=2 xi xi−1 ∂x i However, there is a mistake in it. Here (x2 ) is a δ-ideal so, δ is not a simple derivation. A small correction: ”(xi xi−1 + 1) instead of xi xi−1 ”, gives a new example of simple k-derivation of k[x1 , . . . , xn ]. It is a consequence of the following
156
Part V. Polynomial derivations with trivial constants
Theorem 13.4.2. Let d be a k-derivation of k[x1 , . . . , xn ] of the form n
X ∂ ∂ d= + , (ai xi + bi ) ∂x1 ∂xi i=2
where a2 , b2 , . . . , an , bn are polynomials satisfying, for any i = 2, . . . , n, the following four conditions (1) ai , bi ∈ k[x1 , . . . , xi−1 ], (2) bi 6= 0, (3) degxi−1 ai > 1, (4) degxi−1 bi < degxi−1 ai . Then d is simple. Proof. If n = 1 then d = ∂x∂ 1 is a simple k-derivation of k[x1 ]. Let n > 1. Put R = k[x1 , . . . , xn−1 ], d0 = d|R, a = an , b = bn , and assume that the k-derivation d0 is simple. Suppose that there exists an element f ∈ R such that d0 (f ) = af + b. Then we have ∂f ∂f ∂f + (a2 x2 + b2 ) + · · · + (an−1 xn−1 + bn−1 ) = an f + bn (13.1) ∂x1 ∂x2 ∂xn−1 and f 6= 0, because bn 6= 0. Set s = degxn−1 f and look at the degree with respect to xn−1 of the polynomials in (13.1). The degree of the left side of (13.1) is not greater than s, but the degree of the right side is greater then s. This contradiction means that there is no polynomial f ∈ R with d0 (f ) = af +b and hence, by Theorem 13.2.1 and by induction, d is a simple k-derivation of R[xn ] = k[x1 , . . . , xn ]. Immediately from Theorem 13.4.2 we get Example 13.4.3. Let d1 and d2 be k-derivations of k[x1 , . . . , xn ] defined as follows d1 (x1 ) = 1 d2 (x1 ) = 1 2 d (x ) = x x + 1 1 2 1 2 d2 (x2 ) = x1 x2 + x1 d1 (x3 ) = x2 x3 + 1 d2 (x3 ) = x22 x3 + x2 . .. .. . d1 (xn ) = xn−1 xn + 1, d2 (xn ) = x2n−1 xn + xn−1 . The derivations d1 and d2 are simple. Using the same argument as in the proof of Theorem 13.4.2 we get
Chapter 13. Simple polynomial derivations
157
Example 13.4.4. Let d be the k-derivation of k[x, y, z] defined as follows d1 (x) = 1 d1 (y) = xy + 1 d1 (z) = y The derivation d is simple. Note also: Example 13.4.5. Let d be the k-derivation of k[x, y] such that d(x) = (x2 − y 2 ) + 2, d(y) = x2 − y 2 . Then d is simple. ∂ ∂ Proof. Let δ = ∂x + (xy + 1) ∂y . Then d = 2σδσ −1 , where σ is the k-automorphism of k[x, y] such that σ(x) = x − y, σ(y) = x + y. Therefore, d is simple because δ is simple (see Section 13.3).
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Index df 111 diagonal 88, 109 diagonalizable 109-110 equivalent 110 factorisable 139-143, 146 homogeneous 21, 23, 86, 119, 126, 140, 142-143, 146, 148 integral 114 Jouanolou 24 linear 22, 62, 103, 110, 118-121, 125 locally finite 102-107, 111-115 locally integral 114 locally nilpotent 60-67, 98, 103-107, 113-115 ad 95 ad + bδ 96-99 of a field 63 monomial 21 of algebraic field extension 11 of field of rational functions 10 of polynomial ring 9 of power series rings 12 semisimple 107-110, 115 simple 149-157 basis of derivations 34 special 18, 24, 31, 34, 83-84, 113 commutative 31-34, 115 triangular 104, 115 locally finite 115 Weitzenb¨ock 70-76 locally nilpotent 115 basic 70-71 noncommutative 34 zero 21 Bass’ example 68, 105, 114 Derksen’s example 49 Bezout theorem 129 Deveney and Finston’s example 62 differential Cauchy theorem 14 algebra 1 closed polynomial 54-55, 79-82 formula Coomes-Zurkowski’s theorem 112 in power series rings 13 d-degree 61, 96 ideal 8, 23, 136, 149 d-ideal 8, 149 principal 23 Darboux point 130 primary decomposition 23 system Darboux polynomial 22-24, 54, 95, 119, 126, 137-139, 146 autonomous 27, 56 irreducible 23 stationary 27 derivation 7 direction 18, 23, 71, 82
action of a group 62, 99 algebra Bn 66-67 B 66-67 differential 8 of quotients 8 generated by monomials 91 of fractions 8 algebraic curve 128 group 48 connected 48 automorphism ed 62, 94, 114 Ed 26-31, 57-58, 63 of k[x, y] 98-99 of k[x1 , . . . , xn ] 99 elementary 99 linear 99 stably tame 115 tame 99, 115 triangular 99, 115
167
168 ordinary 22-23 divergence 18, 24-25, 29-31, 34, 83, 112113 eigenvalue polynomial 23 element d-integral 102 semisimple 108 endomorphism locally finite 101 composition 101 sum 101 nilpotent 106 semisimple 106, 108 equivalent derivations 109 van den Essen’s algorithm 67-69, 76 Euler formula 20, 24 Euler theorem 20 exact sequence 51, 151 extension of scalars 51, 150 family M 66 ϕ-integral element 100 field extension algebraic 11, 37 algebraically closed 41, 44 finite 35 finitely generated 11 purely transcendental 10, 38, 42-44 Fin(d) 102 first integral 56-57, 85, 126 formal 57-59 partial 23 polynomial 2 rational 2 five variables 72-74 flow 27, 56 entire 111 formal 27, 112 global 111 local 111 polynomial 111 formal solution 16, 27 four variables 71, 76, 115
Index fourteenth problem of Hilbert 3, 49, 79 function of class Cp 91 Galois theory of differential equations 1 γ-decomposition 20 γ-degree 20 γ-form 19 reduced 82 γ-gradation 20 γ(d) 87 global analysis 127, 130, 133 Gr¨obner basis 67-69 reduced 69, 72 group of γ-forms 20 Guerilla Combat’s system 85 Hausdorff topology 12 homogeneous component 20, 22 polynomial 19, 71 rational function 20 homomorphism differential 8-10 εd 64 πa 17, 28, 57 σb 63-64 image of derivation 34, 37-38, 99 initial condition 111 integral closure 46, 80 invariant submodule 100 jacobian 11-12, 25, 28, 81, 83, 113, 115 jacobian conjecture 115 Jordan-Chevalley decomposition 105107, 110 Jordan matrix 70, 119 Jouanolou’s theorem 126, 131 Jung theorem 99 k-derivation 7 k-domain 7 Kaplansky 1 Kolchin 1 Krull dimension 81 Kulk theorem 99
Index Leibniz formula 7 Lequain’s theorem 95 Lie algebra 7 simple 149 Lie product 7, 25, 114 local analysis 127, 130, 132, 139, 146148 logarithm 62 M -adic topology 12 membership algorithm 68 method of indeterminate coefficients 126 minimal number of generators 87, 91 multiple point 128 multiplicatively closed subset 8 multiplicity 128 Nagata’s example 49, 114 Newton number 7, 30 Newton polygon 83 Nil(d) 60, 95, 102, 107 one variable 110 ordering admissible 68 lexicographic 68-69, 72 ordinary point 128 Ore extension 149 partial derivative 9, 21, 32, 149 Picard theorem 14 Picard-Vessiot theory 1 Poincare series 77 positive characteristic 92, 99 primary decomposition 23 principal element 63-65, 99 Rentschler’s theorem 94, 98-99 ring 7 complete intersection 77 complete 12 Dedekind 80-81 Gorenstein 76 graded 20, 77 integrally closed 41, 46, 55
169 noetherian 107 of constants 8 of invariants 47 reduced 7 simple 149 UFD 23-24, 61, 75 Z-torsion free 7, 96 Ritt 1 Roberts’ example 49, 62 root of unity 77 Selkov system 85 Sem(d) 108 Shamsuddin’s theorem 151-152 Shannon and Sweedler’s algorithm 68 six variables 74, 76 skew polynomial ring 149 subgroup of finite index 48 submodule Fin(ϕ) 101 Mx 100 S(ϕ) 108 support of a polynomial 83 Suzuki’s lemma 37, 42 system of differential equations 16, 56 autonomous 57, 111 factorisable 139 linear 118 Taylor formula 64, 128 tensor product 51 theory of differential equations 23, 111 three variables 57, 68-69, 71, 75, 86, 105, 114, 126, 137, 157 tr.deg 79-80 two variables 24, 41, 56, 75, 80-86, 9899, 104-105, 110, 115, 137, 152154 Tyc 106 T(f ) 8 vector field 111 polynomial 111 Weitzenb¨ock theorem 62
170 Zaks’ theorem 55 Zariski theorem 79
Index