Computational Linear and Commutative Algebra

Computational Linear and Commutative Algebra Lorenzo Robbiano University of Genoa, Italy Department of Mathematics

Lorenzo Robbiano (University of Genoa, Italy)

Computational Linear and Commutative Algebra

Gandhinagar, February 2016

1 / 92

Standard References

Books are like imprisoned souls till someone takes them down from a shelf and frees them. (Samuel Butler, 1875–1941)

M. Kreuzer – L. Robbiano: Computational Commutative Algebra 1,

Springer (2000)

M. Kreuzer – L. Robbiano: Computational Commutative Algebra 2,

Springer (2005)

M. Kreuzer – L. Robbiano: Computational Linear and Commutative Algebra Springer, to appear!

Software CoCoA 5. Lorenzo Robbiano (University of Genoa, Italy)



2 / 92

Topics I • One Endomorphism • Characteristic and Minimal Polynomial • Big Kernels, Small Images, and Generalized Eigenspaces • The Module Structure • Commendable Endomorphisms • Some Anticipation about Commuting Families

TUTORIAL 1 Introducing CoCoA Computing Characteristic and Minimal Polynomials Computing Kernels Computing Eigenfactors The Buchberger Möller Algorithm




3 / 92

Topics II • Commuting Families of Endomorphisms • Dimension • Kernels and Big Kernels of Ideals • Joint Eigenspaces • Splitting Endomorphisms • Cyclic Vector Spaces • Unigenerated Families • Commendable Families • Dual Families

TUTORIAL 2 Computing Dimension Ker(I) and BigKer(I) Computing 1-dimensional joint Eigenspaces Cyclicity Test Lorenzo Robbiano (University of Genoa, Italy)



4 / 92

Topics III • Zero-dimensional Affine Algebras • Multiplication Endomorphisms • Separators, Primitive Elements, Curvilinear Rings • The Canonical Module and the Gorenstein property • Primary Decomposition and Separators • Computing Primary Decomposition in Characteristic Zero • Computing Primary Decomposition in Finite Characteristic • Computing Maximal Ideals Directly • Using Radicals • The Separable Subalgebra

TUTORIAL 3 Multiplication Matrices Chinese Remaindering Computing Primary Decompositions Computing Radicals Lorenzo Robbiano (University of Genoa, Italy)



5 / 92

Topics IV

• Solving Systems of Polynomial Equations • Meaning of Solving • Computing the Rational Zeros • Solving over a Finite Field • Representations of Fq • Solving over Q

TUTORIAL 4 The Eigenvalue Method Solving over a Finite Field Solving via Approximation




6 / 92

Sources and Motivation 1

Figure: Daniel Lazard

Stickelberger’s Eigenvalue Theorem (1920) in Number Theory, rediscovered in the context of commutative algebra by Lazard (1981). Let P/I be a zero-dimensional ring, let f ∈ P and ϑf the corresponding multiplication map of P/I. • If there exists a point p ∈ ZK (I) then f (p) ∈ K is an eigenvalue of ϑf . • If λ ∈ K is an eigenvalue of ϑf then there is p ∈ ZK (I) with λ = f (p). Lorenzo Robbiano (University of Genoa, Italy)



7 / 92


Figure: Auzinger,

Stetter,

Möller

Auzinger-Stetter-Möller (1988,...) Rediscover this connection and obtain several results, mostly related to the non-exact solutions of a system of polynomial equations.




8 / 92


Figure: David Cox

Cox (2005) Solving equations via algebras, in: A. Dickenstein and I. Emiris (eds.), Solving Polynomial Equations, Algorithms and Comp. in Math. 14, Springer, Berlin 2005, pp. 63–124. Mainly in the case of an algebraically closed field. Lorenzo Robbiano (University of Genoa, Italy)



9 / 92

LECTURE 1 One Endomorphism

I would like to understand things better, but I dont want to understand them perfectly. (Douglas R. Hofstadter 1945–)




10 / 92

Characteristic and Minimal Polynomials Let K be a field, let V be a finite-dimensional K-vector space, and let ϕ ∈ EndK (V) be a K-endomorphism of V. Definition The polynomial χϕ (z) = det(z idV −ϕ) is called the characteristic polynomial of ϕ. The computation of χϕ (z) is done using any matrix which represents ϕ. Since EndK (V) is a finite-dimensional K-vector space, the kernel of the substitution homomorphism ε : K[z] −→ K[ϕ] given by f (z) 7→ f (ϕ) is a non-zero ideal. Definition The monic generator of the ideal Ker(ε), i.e. the monic polynomial of smallest degree in this ideal is called the minimal polynomial of ϕ, and is denoted by µϕ (z).




11 / 92

An Example Example Let M = −10 defined by

1 0

∈ Mat 2 (Q), and consider the two block matrices A, B ∈ Mat 6 (Q)

A =

M 0 0

0 M 0

I2 0 M

! and

B =

M 0 0

I2 M 0

0 I2 M

!

For the Q-linear map ϕ : Q6 −→ Q6 given by A, we have µϕ (z) = (z2 + 1)2

and

χϕ (z) = (z2 + 1)3

Observe that the map ϕ has only one irreducible factor, namely p(z) = z2 + 1. Now we consider the Q-linear map ψ : Q6 −→ Q6 given by B. We have µψ (z) = χψ (z) = (z2 + 1)3 Thus there is again just one irreducible factor p(z) = z2 + 1.




12 / 92

Invariance Proposition (Invariance of µϕ (z) and χϕ (z)) Let L ⊇ K be a field extension, let VL = V ⊗K L, and let the L-vector space homomorphism ϕL = ϕ ⊗K L : VL −→ VL be the extension of ϕ. (a) In L[z] we have χϕL (z) = χϕ (z). (b) In L[z] we have µϕL (z) = µϕ (z).

Definition Let Ker(ϕ) and Im(ϕ) denote the kernel and the image of the K-linear map ϕ. S (a) The K-vector subspace BigKer(ϕ) = i≥1 Ker(ϕi ) of V is called the big kernel of ϕ. T (b) The K-vector subspace SmIm(ϕ) = i≥1 Im(ϕi ) of V is called the small image of ϕ.




13 / 92

Fitting’s Lemma

Proposition (Fitting’s Lemma) Consider the chains of K-vector subspaces Ker(ϕ) ⊆ Ker(ϕ2 ) ⊆ · · · and

Im(ϕ) ⊇ Im(ϕ2 ) ⊇ · · ·

(a) There exists a smallest number m ≥ 1 such that Ker(ϕm ) = Ker(ϕt ) for all t ≥ m. It is equal to the smallest number m ≥ 1 such that Im(ϕm ) = Im(ϕt ) for all t ≥ m. In particular, we have strict inclusions Ker(ϕ) ⊂ Ker(ϕ2 ) ⊂ · · · ⊂ Ker(ϕm ) (b) The number m satisfies BigKer(ϕ) = Ker(ϕm ) and SmIm(ϕ) = Im(ϕm ). (c) We have V = BigKer(ϕ) ⊕ SmIm(ϕ).




14 / 92

An Example of BigKer Example Let V = Q8 and let ϕ : V → V be the endomorphism represented by the matrix 0 0  0  1  0  0 0 0 

0 0 0 0 1 0 0 0

0 0 0 0 0 1 0 0

0 0 0 0 0 0 1 0

0 0 1 0 0 0 0 0

0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 1

 0 0  1  0  0  0 0 0

with respect to the canonical basis of V. A Q-basis of Ker(ϕ) is (v1 , v2 ) where v1 = (0, 0, 0, 1, 0, −1, 0, 0) = e4 − e6 , v2 = (0, 0, 0, 0, 1, 0, 0, −1) = e5 − e8 . A Q-basis of Ker(ϕ2 ) is (v1 , v2 , v3 , v4 ) where v3 = (1, 0, −1, 0, 0, 0, 0, 0) = e1 − e3 , v4 = (0, 1, 0, 0, 0, 0, −1, 0) = e2 − e7 . Moreover, we have Ker(ϕ2 ) = Ker(ϕ3 ) and we conclude that BigKer(ϕ) = Ker(ϕ2 ). Consequently SmIm(ϕ) = Im(ϕ2 ) and it turns out that a Q-basis of Im(ϕ2 ) is the tuple (e3 , e6 , e7 , e8 ). Lorenzo Robbiano (University of Genoa, Italy)



15 / 92

Cayley-Hamilton

Cayley (1821–1895)

Hamilton (1805–1865)

Theorem (Cayley-Hamilton) Let ϕ : V −→ V be a K-endomorphism of V. (a) The minimal polynomial µϕ (z) is a divisor of the characteristic polynomial χϕ (z). (b) The polynomials µϕ (z) and χϕ (z) have the same irreducible factors, and hence the same squarefree part. Lorenzo Robbiano (University of Genoa, Italy)



16 / 92

Generalized Eigenspaces Let ϕ ∈ EndK (V). We decompose its minimal polynomial and get µϕ (z) = p1 (z)m1 · · · ps (z)ms Then the characteristic polynomial of ϕ factors in this way χϕ (z) = p1 (z)a1 · · · ps (z)as with ai ≥ mi Definition (a) The polynomials p1 (z), . . . , ps (z) are called the eigenfactors of ϕ. (b) If an eigenfactor pi (z) of ϕ is of the form pi (z) = z − λ with λ ∈ K then λ is called an eigenvalue of ϕ. (c) For i = 1, . . . , s, the K-vector subspace Eig(ϕ, pi (z)) = Ker(pi (ϕ)) is called the eigenspace of ϕ associated to pi (z). Its non-zero elements are called pi (z)-eigenvectors, or simply eigenvectors of ϕ. (d) For i = 1, . . . , s, the K-vector subspace Gen(ϕ, pi (z)) = BigKer(pi (ϕ)) is called the generalized eigenspace of ϕ associated to pi (z).




17 / 92

Generalized Eigenspaces Decomposition Main Theorem (Generalized Eigenspace Decomposition) Let ϕ ∈ EndK (V) and let µϕ (z) = p1 (z)m1 · · · ps (z)ms . (a) The vector space V is the direct sum of the generalized eigenspaces of ϕ, i.e. we have V = Gen(ϕ, p1 (z)) ⊕ · · · ⊕ Gen(ϕ, ps (z)) (b) For i = 1, . . . , s, the restriction of the map pi (ϕ) to Gen(ϕ, pi (z)) is nilpotent and its index of nilpotency is mi . (c) For i, j ∈ {1, . . . , s} with i 6= j, the restriction of the map pi (ϕ) to Gen(ϕ, pj (z)) is an isomorphism. (d) For i = 1, . . . , s, we have Gen(ϕ, pi (z)) = BigKer(pi (ϕ)) = Ker(pi (ϕ)mi ) In particular, if mi = 1 then we have Gen(ϕ, pi (z)) = Ker(pi (ϕ)) = Eig(ϕ, pi (z)). Lorenzo Robbiano (University of Genoa, Italy)



18 / 92

The Module Structure Let ϕ ∈ EndK (V) be an endomorphism of the vector space V. The vector space acquires a module structure over the ring K[ϕ] via ψ · v = ψ(v) for all ψ ∈ K[ϕ] and all v ∈ V. Using the canonical epimorphism K[z] −→ K[ϕ] given by z 7→ ϕ, it follows that V acquires also a K[z]-module structure.

Main Theorem (Cyclic Module Decomposition) Let ϕ ∈ EndK (V) be an endomorphism of V. There exists a tuple (v1 , v2 , . . . , vr ) of vectors in V such that Lr (a) V = i=1 hvi iK[z] , (b) hµϕ (z)i = Ann(v1 ) ⊆ Ann(v2 ) ⊆ · · · ⊆ Ann(vr ).




19 / 92

Commendable Endomorphisms Definition The K-linear map ϕ : V −→ V is called commendable (or non-derogatory) if, for every i ∈ {1, . . . , s}, the eigenfactor pi (z) of ϕ satisfies dimK (Eig(ϕ, pi (z))) = deg(pi (z)) Equivalently, we require that dimK[z]/hpi (z)i (Eig(ϕ, pi (z))) = 1 for i = 1, . . . , s. Main Theorem (Characterization of Commendable Endomorphisms) Let ϕ : V −→ V be a K-linear map. Then the following conditions are equivalent. (a) The endomorphism ϕ is commendable. (b) We have µϕ (z) = χϕ (z). (c) The vector space V is a cyclic K[z]-module via ϕ.




20 / 92

Anticipation Definition Given a set of commuting endomorphisms S of V, we let F = K[S] be the commutative K-subalgebra of EndK (V) generated by S and call it the family of commuting endomorphisms, or simply the commuting family, generated by S. Since EndK (V) is a finite-dimensional K-vector space, also F is a finite-dimensional vector space hence a zero-dimensional K-algebra. The dimension of F as a K-vector space will be denoted by dimK (F). Definition Let Φ = (ϕ1 , . . . , ϕn ) be a system of K-algebra generators of F, and let the K-algebra homomorphism π : P −→ F be defined by letting π(1) = idV and π(xi ) = ϕi for i = 1, . . . , n. Then the kernel of π is called the ideal of algebraic relations of Φ and denoted by RelP (Φ).




21 / 92

LECTURE 2 Commuting Families of Endomorphisms

They are strange types of families, with no fathers, no mothers, no children. Their only concern is to be commutative.




22 / 92

Dimension

Schur (1875–1941)

Jacobson (1910–1999)

Definition Let F be a family of commuting endomorphisms of V. The dimension of F as a ring is zero while dimK (F) is the dimension of the family F as a K-vector space. Example If ϕ ∈ EndK (V) and F = K[ϕ], we have dimK (F) = deg(µϕ (z)). Therefore, it is ≤ dim(V), with equality if and only if ϕ is commendable. • The maximal dimension of a commuting family was determined by J. Schur and N. Jacobson a long time ago: it coincides with bd2 /4c + 1 for d = dimK (V). Lorenzo Robbiano (University of Genoa, Italy)



23 / 92

Dimension II Example Let V = K 6 , and let F be the L-algebra generated by {idV } and the set of all endomorphisms of V whose matrix with respect to the canonical basis of V is of the form 00 A0 with a matrix A of size 3 × 3. Then the family F is commuting, since we have 00 A0 00 B0 = 00 B0 00 A0 = 00 00 for all matrices A, B of size 3 × 3. Here we have dimK (V) = 6 and dimK (F ) = 10 = 62 /4 + 1, the maximal possible dimension.

This example will be checked in the SECOND TUTORIAL. In 1961 Murray Gerstenhaber proved that if the family F is generated by two commuting matrices, the sharp upper bound for the dimension of F is dimK (V). A sharp upper bound for the dimension of a family generated by three commuting matrices is apparently not known.




24 / 92

BMForMat

As already anticipated, if we let Φ = (ϕ1 , . . . , ϕn ) be a system of K-algebra generators of F, then we have the ideal of algebraic relations of Φ denoted by RelP (Φ). It can be computed using a nice algorithm called Buchberger-Möller Algorithm for Matrices In the second tutorial you will learn how to use it.




25 / 92

Kernels of Ideals I Definition Let I be an ideal in the commuting family F. T (a) The K-vector subspace Ker(I) = ϕ∈I Ker(ϕ) of V is called the kernel of I. T (b) The K-vector subspace BigKer(I) = ϕ∈I BigKer(ϕ) of V is called the big kernel of I. Proposition Let I be an ideal in the commuting family F. (a) The vector space Ker(I) has a natural structure of an F/I-module. (b) For ϕ1 , . . . ,Tϕn ∈ F and the ideal I = hϕ1 , . . . , ϕn i, weThave n n Ker(I) = j=1 Ker(ϕj ) and BigKer(I) = j=1 BigKer(ϕj ) (c) The vector spaces Ker(I) and BigKer(I) are invariant for F.




26 / 92

Kernels of Ideals II We have some information about the size of the kernel of a maximal ideal in F. Proposition Let m be a maximal ideal in F. (a) There exists a non-zero map ϕ ∈ F such that m = AnnF (ϕ). (b) For a map ϕ ∈ F as in (a), and for every v ∈ Im(ϕ) \ {0}, we have AnnF (v) = m. In particular, we have v ∈ Ker(m) 6= {0}. (c) We have dimK (Ker(m)) = dimF /m (Ker(m)) · dimK (F/m). In particular, if we have dimK (Ker(m)) = 1 then F/m ∼ = K. Example Let K = Q,let V =Q3 , and let ϕ be the endomorphism of V defined by 0 0 1 the matrix 00 00 00 with respect to the standard basis. Since µϕ (z) = z2 and χϕ (z) = z3 , the endomorphism ϕ is not commendable. Next, we consider the commuting family F = K[ϕ] ∼ = K[z]/hz2 i. It has only one maximal ideal m = hϕi and satisfies dimK (F/m) = 1, but dimK (Ker(m)) = dimK (Ker(ϕ)) = 2. Lorenzo Robbiano (University of Genoa, Italy)



27 / 92

An Algorithm

ALGORITHM (The Kernel of an Ideal in a Commuting Family) Let I = hψ1 , . . . , ψs i be an ideal of F, and for i ∈ {1, . . . , s} let Mi ∈ Matd (V) be the matrix which represents ψi with respect to a fixed K-basis of V. Then the following algorithm computes the kernel of I. • Form the block column matrix C = Col(M1 , . . . , Ms ). • Compute generators of Ker(C) and return the corresponding vectors of V.




28 / 92

Comaximal Ideals

Theorem (Kernels and Big Kernels of Comaximal Ideals) Let I1 , . . . , Is be pairwise comaximal ideals in the family F, and let I = I1 ∩ · · · ∩ Is . • We have Ker(I) = Ker(I1 ) ⊕ · · · ⊕ Ker(Is ). • We have BigKer(I) = BigKer(I1 ) ⊕ · · · ⊕ BigKer(Is ).




29 / 92

Eigenfactors Definition For an endomorphism ϕ ∈ F and a maximal ideal m of F, the monic generator of the kernel of the K-algebra homomorphism ε : K[z] −→ F/m defined by z 7→ ϕ + m is called the m-eigenfactor of ϕ and is denoted by pm,ϕ (z). Every eigenfactor is an m-eigenfactor. Proposition Let Φ = (ϕ1 , . . . , ϕn ) be a system of K-algebra generators of F, and let m be a maximal ideal of F. Then the following conditions are equivalent. (a) The maximal ideal m is linear, i.e. dimK (F/m) = 1. (b) We have m = hϕ1 − λ1 idV , . . . , ϕn − λn idV i for some λ1 , . . . , λn ∈ K. (c) For i = 1, . . . , n, we have pm,ϕi (z) = z − λi with λi ∈ K. Corollary Let m = hϕ1 − λ1 idV , . . . , ϕn − λn idV i be a linear maximal ideal of F, and let ψ ∈ F be of the form ψ = f (ϕ1 , . . . , ϕn ) with f ∈ K[x1 , . . . , xn ]. We have pm,ψ (z) = z − f (λ1 , . . . , λn ). Lorenzo Robbiano (University of Genoa, Italy)



30 / 92

Joint Eigenspaces Definition For every endomorphism ϕ ∈ F, let pϕ (z) be an eigenfactor of ϕ, and let C = {pϕ (z) | ϕ ∈ F}. The setT C is called a coherent family of eigenfactors of F if we have a non-trivial intersection ϕ∈F Eig(ϕ, pϕ (z)) 6= {0}. (a) Given a coherent family of eigenfactors C of F, the corresponding intersection T ϕ∈F Eig(ϕ, pϕ (z)) 6= {0} is called a joint eigenspace of F. Its non-zero elements are called joint eigenvectors of F. T (b) If ϕ∈F Gen(ϕ, pϕ (z)) 6= {0}, it is called a joint generalized eigenspace of F. Main Theorem (Joint Generalized Eigenspace Decomposition) Let F be a family of commuting endomorphisms of V, and let m1 , . . . , ms be the maximal ideals of F. Ls (a) We have V = i=1 BigKer(mi ). (b) The joint eigenspaces of F are Ker(m1 ), . . . , Ker(ms ). (c) The joint generalized eigenspaces of F are BigKer(m1 ), . . . , BigKer(ms ). Lorenzo Robbiano (University of Genoa, Italy)



31 / 92

Joint Eigenspaces ALGORITHM (Computing 1-Dimensional Joint Eigenspaces) Let ϕ1 , . . . , ϕn ∈ F be a system of K-algebra generators of F. Consider the following sequence of instructions. 1

2

3 4

5 6

7

Compute all linear eigenfactors p11 (z), . . . , p1` (z) of ϕ1 and find K-bases B1 , . . . , B` of the corresponding eigenspaces Eig(ϕ1 , p1i (z)) = hBi i of ϕ1 . If ` = 0, return S = ∅ and stop. Form the tuples S and T whose elements are the Bi of cardinality one, and bigger than one respectively. For i = 2, . . . , n, perform the following steps. Then return S and stop. Let L = (B01 , . . . , B0m ) be the current tuple T. For j = 1, . . . , m, remove B0j from T and perform the following steps. Compute a matrix representing the restriction ϕij of ϕi to hB0j i. Compute the linear eigenfactors pij1 (z), . . . , pijk (z) of ϕij and K-bases B001 , . . . , B00k of the eigenspaces Eig(ϕij , pijκ (z)) = hB00κ i of the map ϕij . Append the 1-element tuples B00κ to S. Append the tuples in {B001 , . . . , B00k } which have more than one entry to the tuple T.

This is an algorithm which computes a tuple S of K-bases of the 1-dimensional joint eigenspaces of F. Lorenzo Robbiano (University of Genoa, Italy)



32 / 92

Splitting Endomorphisms Can we find a single endomorphism such that its generalized eigenspaces are the joint generalized eigenspaces of the family?

Definition Let m1 , . . . , ms be the maximal ideals of F, and let ϕ ∈ F. (a) If we have the equalities Gen(ϕ, pmi ,ϕ (z)) = BigKer(mi ) for i = 1, . . . , s then ϕ is called a splitting endomorphism for F. (b) If we have the equalities Eig(ϕ, pmi ,ϕ (z)) = Ker(mi ) for i = 1, . . . , s then ϕ is called a strongly splitting endomorphism for F. Proposition Let F be a commuting family, let m1 , . . . , ms be the maximal ideals of F, and let ϕ ∈ F. The map ϕ is a splitting endomorphism for F if and only if pmi ,ϕ (z) 6= pmj ,ϕ (z) for i 6= j. Equivalently, ϕ is a splitting endomorphism if and only if it has s eigenfactors. Proposition If we have card(K) ≥ dimK (F) then there exists a splitting endomorphism for F. Lorenzo Robbiano (University of Genoa, Italy)



33 / 92

Splitting Endomorphisms II Example Let K = Q, V = Q4 , and consider the commuting family F = Q[ϕ1 , ϕ2 ] generated by the endomorphisms represented by the matrices  0 1 A1 =  0 0

2 0 0 0

0 0 0 1

 0 0 2 0

and

 0 0 A2 =  1 0

0 0 0 1

8 0 0 0

 0 8 0 0

µϕ1 (z) = z2 − 2 and µϕ2 (z) = z2 − 8 are irreducible. The endomorphism ψ = ϕ1 + ϕ2 ∈ F satisfies µψ (z) = z4 − 20z2 + 36 = (z2 − 2)(z2 − 18) = p1 (z) · p2 (z) where p1 (z) = z2 − 2 and p2 (z) = z2 − 18. In particular, ϕ1 and ϕ2 are not splitting. The corresponding eigenspaces of ψ are Eig(ψ, p1 (z)) = Ker(ψ 2 − 2 idV ) = hv1 , v2 i, where v1 = (0, 2, −1, 0) and v2 = (4, 0, 0, −1), and Eig(ψ, p2 (z)) = Ker(ψ 2 − 18 idV ) = hw1 , w2 i, where w1 = (0, −2, −1, 0) and w2 = (4, 0, 0, 1). The dimensions of the two eigenspaces of ψ add up to four which is the dimension of V. Therefore these eigenspaces are also the generalized eigenspaces of ψ. Using Algorithm-Computing 1-Dimensional Joint Eigenspaces, we see that there are no 1-dimensional joint eigenspaces of F. Hence the map ψ is a splitting endomorphism for the family F. Lorenzo Robbiano (University of Genoa, Italy)



34 / 92

F- cyclic Vector Spaces Let S ⊆ EndK (V) be a set of commuting matrices, and let F = K[S]. Recall that the vector space V has a natural structure as an F-module given by ϕ · v = ϕ(v) for all ϕ ∈ F and v ∈ V. Proposition Let F be a commuting family of endomorphisms of V. (a) If V is F-cyclic and v ∈ V is an F-module generator of V then AnnF (v) = 0. (b) For a vector v ∈ V, the following conditions are equivalent. (1) The vector space V is a cyclic F-module with generator v. (2) The K-linear map η : F −→ V which sends ϕ to ϕ(v) for every ϕ ∈ F is an isomorphism.

(c) If V is F-cyclic then we have dimK (F) = dimK (V).




35 / 92

Test ALGORITHM (Cyclicity Test) Let S = {ϕ1 , . . . , ϕr } ⊆ EndK (V) be a set of commuting endomorphisms, let F be the family generated by the set S, and let Φ = (ϕ1 , . . . , ϕr ). Then let B = {v1 , . . . , vd } be a basis of V, and, for i = 1, . . . , r, let Ai ∈ Matd (K) be the matrix representing ϕi in the basis B. Consider the following instructions. (1) Using the BM-Algorithm for Matrices compute a tuple of terms O = (t1 , . . . , ts ) whose residue classes form a K-basis of K[x1 , . . . , xr ]/ RelP (Φ). (2) If s 6= d then return "Not cyclic" and stop. (3) Let z1 , . . . , zd be indeterminates. Form the matrix C ∈ Matd (K[z1 , . . . , zd ]) whose columns are ti (A1 , . . . , Ad ) · (z1 , . . . , zd )tr for i = 1, . . . , d. (4) Compute the determinant ∆ = det(C). If ∆ 6= 0, find a tuple (c1 , . . . , cd ) ∈ K d such that ∆(c1 , . . . , cd ) 6= 0, return the vector v = c1 v1 + · · · + cd vd , and stop. (5) Return "Not Cyclic" and stop. This is an algorithm which checks whether V is a cyclic F-module and, in the affirmative case, computes a generator.




36 / 92

Unigenerated Families Definition Let F be a commuting family of endomorphisms of V. (a) Let ϕ, ψ ∈ EndK (V). If there exists a polynomial f (z) ∈ K[z] such that ψ = f (ϕ), we say that ψ is polynomial in ϕ. (b) If the family F contains an endomorphism ϕ such that all endomorphisms in F are polynomial in ϕ, it is called unigenerated by ϕ, and ϕ is called a generator of F. Theorem Let F be a family of commuting endomorphisms, and let ϕ ∈ F. Then the following conditions are equivalent. (a) The endomorphism ϕ is commendable. (b) The K-algebra homomorphisms K[z]/hµϕ (z)i −→ F induced by z 7→ ϕ and K[z]/hµϕ (z)i −→ K[z]/hχϕ (z)i induced by z 7→ z are isomorphisms. (c) The family F is unigenerated by ϕ and the vector space V is F-cyclic. Lorenzo Robbiano (University of Genoa, Italy)



37 / 92

An Example Example Let ϕ1 , ϕ2 be two Q-endomorphisms of V = Q7 which are given by the matrices 

A1 =

           

0 0 0 1 0 0 0

0 0 0 0 1 0 0

0 0 0 0 0 0 0

0 0 0 0 0 1 0

0 0 0 0 0 0 1

0 −12 0 0 7 0 3

0 0 −12 0 0 0 0





           

           

A2 =

0 1 0 0 0 0 0

0 0 1 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 1 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 1

0 0 0 0 0 0 0

            

respectively. We check that A1 A2 = A2 A1 . Hence {ϕ1 , ϕ2 } generates a commuting family F = Q[ϕ1 , ϕ2 ]. We have χϕ1 (z) = µϕ1 (z) = z7 , while µϕ2 (z) = z3 , hence ϕ1 is commendable and ϕ2 is not. The theorem shows that F is unigenerated by ϕ1 , and that ϕ2 is polynomial in ϕ1 . Using Buchberger-Möller for Matrices we find 847 A2 = − 20736 A61 −

85 1728

A51 −

7 144

A41 −

1 12

A31

Please do not try by hand! Lorenzo Robbiano (University of Genoa, Italy)



38 / 92

Commendable Families

Definition Let F be a family of commuting endomorphisms of the K-vector space V. The family F is called commendable if we have the equality dimK (Ker(m)) = dimK (F/m), equivalently dimF /m (Ker(m)) = 1 for every maximal ideal m of F. If the family F is not commendable, we also say that it is derogatory.




39 / 92

Dual Families As usual, let K be a field and V a finite dimensional K-vector space of dimension d. Recall that the dual vector space of V is V ∗ = HomK (V, K), i.e. the set of all K-linear maps ` : V −→ K. For a map ϕ ∈ EndK (V), the dual endomorphism of ϕ is the K-linear map ϕˇ : V ∗ −→ V ∗ given by ϕˇ(`) = ` ◦ ϕ. Definition Let F be a family of commuting endomorphisms in EndK (V). (a) We call F ˇ = K[ϕˇ | ϕ ∈ F] the dual family of F. (b) Given an ideal I of F, we call I ˇ = hϕˇ | ϕ ∈ Ii the dual ideal of I. There is a canonical isomorphism ϑ : V −→ V ∗∗ given by v 7→ (` 7→ `(v)) for v ∈ V and ` ∈ V ∗ . It is called the canonical isomorphism to the bidual. Proposition Let B = (v1 , . . . , vd ) be a basis of V, let ϕ ∈ EndK (V), and let MB (ϕ) be the matrix which represents ϕ with respect to B. Then the dual map ϕˇ is represented by MB∗ (ϕˇ) = (MB (ϕ))tr with respect to the dual basis B∗ . Lorenzo Robbiano (University of Genoa, Italy)



40 / 92

Fundamental Results I

Theorem Let F be a family of commuting endomorphisms of V. (a) If V is a cyclic F-module, then F ˇ is commendable. (b) If the family F is commendable, then V ∗ is a cyclic F ˇ-module.

Corollary Let F be a family of commuting endomorphisms of V. (a) The vector space V is a cyclic F-module if and only if F ˇ is commendable. (b) The vector space V ∗ is a cyclic F ˇ-module if and only if F is commendable.




41 / 92

Fundamental Results II Theorem Let K ⊆ L be a field extension. Then we have the following equivalences. (a) The F-module V is cyclic if and only if VL is a cyclic FL -module. (b) The family F is commendable if and only if FL is commendable.

ALGORITHM Let F be a commuting family, let {ϕ1 , . . . , ϕr } be a set of K-algebra generators of F, and let B be a K-basis of V. Then the following algorithm checks whether the family F is commendable. (1) For i = 1, . . . , r, represent ϕi by a matrix Ai with respect to B. (2) Apply the Cyclicity Test to the family F ˇ = K[ϕ1ˇ, . . . , ϕrˇ] where ϕiˇ is represented by the matrix Atri with respect to the basis B∗ of V ∗ .




42 / 92

LECTURE 3 Zero-Dimensional Affine Algebras

You haven’t seen nothing until you’ve seen zero. (Jarod Kintz 1982–)




43 / 92

Multiplication Endomorphisms In the following we let K be a field and R a zero-dimensional affine K-algebra i.e a zero-dimensional K algebra of type R = K[x1 , . . . , xn ]/I. Thus R is a finite dimensional K-vector space, and we let d = dimK (R). Definition (a) For every element f ∈ R, the multiplication by f yields a K-linear map ϑf : R −→ R such that ϑf (g) = f · g for all g ∈ R. It is called the multiplication endomorphism by f on R. (b) The family F = K[ϑx1 , . . . , ϑxn ] is called the multiplication family of R. (c) Let B = (t1 , . . . , td ) be a K-basis of R, and let f ∈ R. Then the matrix MB (ϑf ) ∈ Matd (K) which represents ϑf with respect to the basis B is called the multiplication matrix of f with respect to B. Proposition Let F = K[ϑx1 , . . . , ϑxn ] be the multiplication family of R. (a) We have F = {ϑf | f ∈ R}. (b) The map ı : R −→ F given by f 7→ ϑf is an isomorphism of K-algebras. Its inverse is the map η : F −→ R given by ϕ 7→ ϕ(1), i.e. ϑf 7→ f . Lorenzo Robbiano (University of Genoa, Italy)



44 / 92

Primitive Elements Definition Let K ⊆ L be a field extension. An element f ∈ L is said to be a primitive element for K ⊆ L if we have L = K(f ). Proposition Let L be a finite field extension of K, let F be the multiplication family of L, and let f ∈ L. Then the following conditions are equivalent. (a) The element f ∈ L is a primitive element for K ⊆ L. (b) The endomorphism ϑf ∈ F is commendable. Example Let p be a prime, let a, b, x, y be indeterminates, let K = Fp (a, b), and let L = K[x, y]/m where m = hxp − a, yp − bi is a maximal ideal. The multiplication family F of L is commendable. Using Frobenius we conclude that every f ∈ L satisfies f p ∈ K, and hence dimK (K(f )) = dimK (K[f ]) ≤ p. But dimK (L) = p2 , hence there exists no primitive element for K ⊆ L, and so there is no commendable endomorphism in F. Lorenzo Robbiano (University of Genoa, Italy)



45 / 92

Curvilinear Rings

Definition Let R be a zero-dimensional K-algebra, and let q1 , . . . , qs be the primary components of its zero ideal. The ring R is called curvilinear if its maximal ideals are linear and if there exist numbers mi such that R/qi ∼ = K[z]/hzmi i for i = 1, . . . , s.

Corollary Let R be a zero-dimensional K-algebra having s linear maximal ideals m1 , . . . , ms , and assume that card(K) ≥ s. The following conditions are equivalent. (a) The ring R is curvilinear. (b) There exists a commendable endomorphism in the multiplication family of R.




46 / 92

The Canonical Module Definition Let us consider the operation of the ring R on the dual vector space R∗ = HomK (R, K) given by R × R∗

−→

R∗

(f , `)

7−→ (g 7→ `(fg))

It is easy to check that this definition turns R∗ into an R-module. This module is denoted by ωR and called the canonical module (or the dualizing module) of R. Remark Let B = (t1 , . . . , td ) be a K-basis of R. For an element f ∈ R, the endomorphism ϑf : R −→ R is represented with respect to B by the multiplication matrix MB (ϑf ). The matrix which represents the dual map ϑf ˇ : ωR −→ ωR with respect to the basis B∗ is MB∗ (ϑfˇ) = MB (ϑf )tr .




47 / 92

Gorenstein Rings Main Theorem Let R be a zero-dimensional affine K-algebra. The following conditions are equivalent. (a) The canonical module ωR is a cyclic R-module. (b) The K-algebra R is locally Gorenstein. (c) The multiplication family F is commendable. ALGORITHM Let R be a zero-dimensional affine K-algebra. The following algorithm checks whether R is locally Gorenstein or not. (1) Compute a basis O of R as a K-vector space. (2) Compute the multiplication matrices M1 = MO (ϑx1 ),. . . , Mn = MO (ϑxn ). (3) Apply the Cyclicity Test to (M1tr , . . . , Mntr ). (4) If the test is positive return Is locally Gorenstein, otherwise return Is not locally Gorenstein. Lorenzo Robbiano (University of Genoa, Italy)



48 / 92

The Chinese Remainder Theorem

A key ingredient. Theorem (Chinese Remainder Theorem) Let R be a ring, let I1 , . . . , It be comaximal Qt ideals in R, and let π denote the canonical R-linear map π : R/(I1 ∩ · · · ∩ It ) −→ i=1 R/Ii . Q (a) For i = 1, . . . , s, the ideals Ii and Ji = j6=i Ij are comaximal. (b) The map π is an isomorphism of rings. Proof (Hint). Let us prove (b). The map π is clearly injective, so we need to prove its surjectivity. Claim (a) implies that for every i ∈ {1, . . . , t} there are elements ai ∈ Ii and bi ∈ Ji such that ai +Qbi = 1. Now it is easy to check that an element t (c1 + I1 , . . . , ct + It ) of i=1 R/Ii is the image of the residue class of the element c1 b1 + · · · + ct bt = c1 (1 − a1 ) + · · · + ct (1 − at ) under π.




49 / 92

Primary Decomposition Definition Let S be a ring. (a) An ideal I in S is called primary if fg ∈ I for f , g ∈ S implies that we have f ∈ I or g ∈ Rad(I). (b) A representation of an ideal I in S as a finite intersection of primary ideals is called a primary decomposition of I. Example Be careful! Let I = hx2 , xyi. If fg ∈ I, then one of the two is in hxi. Let f be such element. Then f 2 ∈ I hence f ∈ Rad(I). But I = hxi ∩ hx2 , yi hence I is not primary. Lemma Let S be a ring, and let I be an ideal in S. (a) If the ideal I is primary then Rad(I) is a prime ideal. (b) If m is a maximal ideal of S and Rad(I) = m then I is m-primary. If I = q1 ∩ · · · ∩ qs and Rad(I) = m1 ∩ · · · ∩ ms the ideals qi are called the primary components of I and the ideals mi are called the maximal components of I. Lorenzo Robbiano (University of Genoa, Italy)



50 / 92

Separators Definition Let R be a zero-dimensional K-algebra, let m1 , . . . , ms be its maximal ideals, let q1 , . . . , qs be the primary components of its zero ideal, and let i ∈ {1, . . . , s}. (a) An element f ∈ R is called a separator for mi if we have dimK hf i = dimK (R/mi ) and f ∈ qj for every j 6= i. Theorem (Characterization of Separators) Let R be a zero-dimensional K-algebra, let m1 , · · · , ms be its maximal ideals, and let i ∈ {1, . . . , s}. For an element f ∈ R the following conditions are equivalent. (a) The element f is a separator for mi . (b) The image of f is a non-zero element of the socle of the local ring R/qi and, for j 6= i, the image of f is zero in R/qj . (c) The element f is a joint eigenvector of the multiplication family F of R, and it is contained in the joint eigenspace Ker(mi ), where mi is the image of mi under the isomorphism ı : R −→ F. Lorenzo Robbiano (University of Genoa, Italy)



51 / 92

Generically Extended Linear Forms Let us have a look at the first algorithm which computes the primary decomposition. ALGORITHM [Primary Decomposition I] Let I be a zero-dimensional ideal in P = K[x1 , . . . , xn ], and let R = P/I. Consider the following sequence of instructions. (1) Let y1 , . . . , yn be new indeterminates, let L = y1 x1 + · · · + yn xn be the generically extended linear form, and let ` = L + I ∈ R ⊗K K(y1 , . . . , yn ). Compute the minimal polynomial µ` (z) ∈ K(y1 , . . . , yn )[z]. (2) Factor µ` (z) in the form µ` (z) = p1 (z)m1 · · · ps (z)ms , where pi (z) is an irreducible polynomial in K(y1 , . . . , yn )[z]. e i = I + hpi (L)mi i ⊂ P ⊗K K(y1 , . . . , yδ ) and compute a (3) For i = 1, . . . , s, let Q e i . (Notice that Gi ⊂ P.) reduced Gröbner basis Gi of Q (4) For i = 1, . . . , s, let Qi = hGi i ⊂ P. Return the tuple (Q1 , . . . , Qs ) and stop. This is an algorithm which computes the primary components Q1 , . . . , Qs of I. In practice it works only for very small examples! Lorenzo Robbiano (University of Genoa, Italy)



52 / 92

The Heuristic Algorithm After discussing “probability of success”, with some pointers to Monte Carlo and Las Vegas Algorithms, we proceed with an heuristic algorithm. ALGORITHM [Primary Decomposition II] As above, let P = K[x1 , . . . , xn ], let I be a zero-dimensional ideal in P, let R = P/I, let L = c1 x1 + · · · + cn xn be a linear form in P, where c1 , . . . , cn ∈ K, and let ` = L + I be the residue class of L in R = P/I. (1) Compute the minimal polynomial µ` (z) and factor it in the form µ` (z) = p1 (z)m1 · · · ps (z)ms , with pairwise distinct, monic, irreducible polynomials p1 (z), . . . , ps (z) ∈ K[z]. (2) For i = 1, . . . , s, let Qi = I + hpi (L)mi i and check whether Qi is a primary ideal. (3) Return the tuple (Q1 , . . . , Qs ). If all the tests in step (2) yielded the answer TRUE, output the string Primary Decomposition. Otherwise, output the string Partial Decomposition. This is an algorithm which computes a tuple of ideals (Q1 , . . . , Qs ) in P such that we have I = Q1 ∩ · · · ∩ Qs . Moreover, it checks whether this intersection is the primary decomposition of I or merely a partial decomposition and outputs the corresponding string. Lorenzo Robbiano (University of Genoa, Italy)



53 / 92

How do we Check? Idempotents An idempotent of a ring R is an element f ∈ R such that f 2 = f . Clearly, the elements 0 and 1 are idempotents. Proposition Let R = P/I be a zero-dimensional affine K-algebra and let h0i = q1 ∩ · · · ∩ qs be the primary decomposition of the zero ideal of R. For i = 1, . . . , s, we let Ji = ∩j6=i qj . (a) For i = 1, . . . , s, there exist uniquely determined elements ei ∈ Ji and fi ∈ qi such that ei + fi = 1. Qs (b) Under the isomorphism π : R ∼ = i=1 R/qj we have π(ei ) = (0, . . . , 0, 1, 0, . . . , 0) with 1 in i-th position. In particular, the elements e1 , . . . , es are idempotents of R. They are called primitive idempotents. (c) The ring R is local if and only if its only idempotents are 0 and 1.




54 / 92

Primary Test ALGORITHM [Primary Ideal Test] As above, let I be a zero-dimensional ideal in P. The following instructions define an algorithm which checks whether I is a primary ideal and outputs the corresponding boolean value. (1) Compute a tuple B = (t1 , . . . , td ) of terms in Tn such that t1 = 1 and such that the residue classes of the entries of B form a K-basis of P/I. Pd (2) Let y1 , . . . , yd be new indeterminates, let f = i=1 yi ti , and use the expressions Pd of the products ti tj in terms of the basis B to write f 2 = i=1 gi ti + h with gi ∈ K[y1 , . . . , yd ] and h ∈ I · P[y1 , . . . , yd ]. (3) In the polynomial ring K[y1 , . . . , yd ], let J = hg1 − y1 , . . . , gd − yd i. Find the K-rational zeros of J. (4) If the only K-rational zeros of J are (0, 0, . . . , 0) and (1, 0, . . . , 0) then return TRUE. Otherwise, return FALSE. Later we shall examine how to find the rational zeros. Lorenzo Robbiano (University of Genoa, Italy)



55 / 92

The Case of Finite Fields If K is “small”, the heuristic method can fail completely. For instance if K = F2 there only two constants, hence very few linear forms. Other methods are available, based on a fundamental tool. In the following we let K be a finite field. Then the characteristic of K is a prime number p and the number of its elements is of the form q = pe with e > 0. Furthermore, it is known that all fields having q elements are isomorphic. In the following we say that a field K with pe elements represents Fq . Definition Let p be a prime number, and let R be a ring of characteristic p. (a) The map φp : R −→ R defined by a 7→ ap is a ring endomorphism of R. It is called the Frobenius endomorphism of R. (b) Suppose that R is an algebra over the field Fq . Then the map φq : R −→ R defined by a 7→ aq is Fq -linear. It is called the q-Frobenius endomorphism of R.




56 / 92

The Frobenius Space Proposition Let R be a zero-dimensional Fq -algebra. (a) For every a ∈ Fq , we have φq (a) = a. (b) The polynomial z − 1 is an eigenfactor of φq . (c) For every f ∈ R, we have φq (f ) − f = f q − f =

Q

a∈Fq (f

− a).

(d) Given a primary ideal q in R and f ∈ R, we have φq (f ) − f = f q − f ∈ q if and only if f − a ∈ q for some a ∈ Fq . Definition The set Frobq (R) = Eig(φq , z − 1) = {f ∈ R | f q − f = 0} i.e., the fixpoint space of R with respect to φq , is called the q-Frobenius space of R.




57 / 92

A Fundamental Theorem Main Theorem (Properties of the q-Frobenius Space) Let R be a zero-dimensional affine Fq -algebra, and let s be the number of primary components of the zero ideal of R. Ps (a) We have Frobq (R) = { i=1 ci ei | c1 , . . . cs ∈ Fq } where e1 , . . . , es are the primitive idempotents of R. (b) We have dim Fq (Frobq (R)) = s. (c) For f ∈ R, we have f ∈ Frobq (R) if and only if µf (z) splits into distinct linear factors. Ps Proof: ToQ prove (a), we first show that Frobq (R) ⊆ { i=1 ci ei | c1 , . . . cs ∈ Fq }. Let s π: R∼ = i=1 R/qi be the decomposition of R into local rings. Given f ∈ R, we write π(f ) = (f1 , . . . , fs ) and get Frobq (R) = {f ∈ R | fiq − fi = 0 for i = 1, . . . , s}. In the rings R/qi the zero-ideals are primary. From (c) of the preceding proposition, we deduce that there exist c1 , . . . , cs ∈ Fq such that fi = ci for i = 1, . . . , s. Hence, ... Next we prove the other inclusion. Every fundamental idempotent ei satisfies e2i = ei , and hence eqi = ei . Moreover, ... Notice that (b) is a direct consequence of (a). The proof of (c) is left as an exercise. Lorenzo Robbiano (University of Genoa, Italy)



58 / 92

Computing the Number of Primary Components

ALGORITHM Let R be a zero-dimensional affine Fq -algebra, let d = dimK (R), and let B = (b1 , . . . , bd ) be an Fq -basis of R. Consider the following sequence of instructions. (1) Compute the matrix MB (φq ). (2) Compute an Fq -basis C of Ker(MB (φq ) − Id ) where Id is the identity matrix of size d. (3) Return the list of all elements of C and return the cardinality s of C. This is an algorithm which computes the elements in the q-Frobenius space of R and the number s of primary components of the zero ideal of R. Using this algorithm one can construct an algorithm which computes the primary decomposition of a zero-dimensional ideal in characteristic p.




59 / 92

Computing Maximal Ideals Directly The idea is to split the minimal polynomials of the indeterminates in a series of successive extension fields. A key ingredient is the following. ALGORITHM [Minimal Polynomials over Extension Fields)] Let M be a maximal ideal in P, let L = P/M, let {y1 , . . . , ym } be a set of further indeterminates, let Q = K[x1 , . . . , xn , y1 , . . . , ym ], and let I be a zero-dimensional ideal in Q which contains M Q. Moreover, assume that we are given an element f ∈ Q. We denote its residue class in Q/IQ ∼ = (P/M)[y1 , . . . , ym ]/¯I = L[y1 , . . . , ym ]/¯I by ¯f . Consider the following instructions. (1) Choose a term ordering σ 00 on T(x1 , . . . , xn ). Extend it to a term ordering σ 0 on T(x1 , . . . , xn , z) which is an elimination ordering for z. Then extend σ 0 to a term ordering σ on T(x1 , . . . , xn , z, y1 , . . . , ym ) which is an elimination ordering for (y1 , . . . , ym ). Compute the reduced σ-Gröbner basis G of the ideal I + hz − f i. (2) Let g ∈ P[z] be the unique polynomial in G which is monic in z. (3) Return the residue class ¯ g of g in L[z]. This is an algorithm which computes the minimal polynomial µ¯f ,L (z) ∈ L[z] of ¯f . The next tool is an algorithm which shows how to factorize such minimal polynomials in polynomial L-algebras. Lorenzo Robbiano (University of Genoa, Italy)



60 / 92

Computing the Radical

Another possibility to get the primary decomposition is the following. ALGORITHM [The Radical of a Zero-Dimensional Ideal] Let K be a perfect field, let I be a zero-dimensional ideal in P = K[x1 , . . . , xn ]. 1

Let J = I.

2

For i = 1, . . . , n, perform steps 3 – 5.

3

Compute gi (xi ) = sqfree(µxi , P/J (xi )).

4

Replace J with J + hgi (xi )i.

5

If deg(gi (xi )) = dimK (P/J), return the ideal J and stop.

6

Return the ideal J.

This is an algorithm which computes Rad(I).




61 / 92

Primary from Maximal

ALGORITHM [Primary Components via Generators of Maximal Components] Let R = P/I be a zero-dimensional affine K-algebra, and let M be a maximal ideal of P whose residue class ideal in R is the maximal ideal m of R. Let M = hp1 , . . . , pn i and let Mi = hpi1 , . . . , pin i for i ∈ N. Consider the following instructions. (1) Compute successive Mi until Ma + I = Ma+1 + I. (2) Let Q = Ma + I and return the ideal q of R, where q is the residue class ideal of Q in R. This is an algorithm which computes the ideal Q in P such that Q is M-primary and the ideal q in R such that q is m-primary.




62 / 92

Primary from Big Kernels ALGORITHM [Primary Components via Big Kernels] Let F be the family of multiplication endomorphisms of R. Via the isomorphism ¯ the image of m in F. Consider the following instructions. ı : R −→ F let m ¯ (1) Compute BigKer(m). ¯ (2) For i = 1, . . . , n, compute the restriction ϑi of ϑxi to BigKer(m). (3) Let Θ = (ϑ1 , . . . , ϑn ), compute RelP (Θ) and let q = RelP (Θ) + I. (4) Return q. This is an algorithm which computes the ideal q in R such that q is m-primary. Remark We deduce that the decomposition of R viewed as a K-vector space as the direct sum of the joint generalized eigenspaces of F corresponds to the decomposition of R into local rings.




63 / 92

The Separable Subalgebra Definition Let K be a perfect field and let R be a zero-dimensional affine K-algebra. Then an element a ∈ R is said to be separable if its minimal polynomial is square-free, equivalently if it is coprime with its derivative. Proposition Let K be a perfect field and let R be a zero-dimensional affine K-algebra. (a) An element a ∈ R is separable if and only if there exists a reduced K-subalgebra S of R such that a ∈ S. (b) The set S of the separable elements of R is a K-subalgebra of R, called Rsep . Proposition Let K be a perfect field, let R be a zero-dimensional affine K-algebra, let a ∈ R, let µa (z) be the minimal polynomial of a, and let f (z) = sqfree(µa (z)). (a) The element f 0 (a) ∈ R is invertible. (b) There exist elements b, r ∈ R such that we have a = b + r with r ∈ Rad(0) and µb (z) = f (z), hence b ∈ R sep . Lorenzo Robbiano (University of Genoa, Italy)



64 / 92

Computing the Decomposition R = R sep ⊕ Rad(0) Corollary We have a decomposition R = Rad(0) ⊕ Rsep as a direct sum of K-subvector spaces.

ALGORITHM Let K be a perfect field, let R be a zero-dimensional affine K-algebra, let a ∈ R, let µa (z) be the minimal polynomial of a, and finally let f (z) = sqfree(µa (z)). Consider the following sequence of instructions. (1) Let i = 0, bi = a, ri = 0 (2) Let i = i + 1, compute

f (bi ) f 0 (bi ) ,

and let bi+1 = bi −

f (bi ) f 0 (bi ) , ri+1

= ri +

f (bi ) f 0 (bi ) .

(3) Repeat steps (2) until f (bi ) = 0. (4) Return b = bi and r = ri . This is an algorithm which computes an element b with µb (z) = f (z) , and hence b ∈ R sep , and an element r ∈ Rad(0) such that a = b + r. Lorenzo Robbiano (University of Genoa, Italy)



65 / 92

An Example

Example Let P = Q[x], let I = h(x2 − 2)4 i, and let R = P/I. We apply he above Algorithm to the element ¯x and get ¯x = ¯ b + ¯r where 5 7 35 3 35 5 x + 21 b = − 128 64 x − 32 x + 16 x 5 7 21 5 35 3 19 r = 128 x − 64 x + 32 x − 16 x

We check that x = b + r, that µ¯b (z) = z2 − 2 = sqfree(µ¯x ), and µ¯r (z) = z4 which shows that ¯r ∈ Rad(0).




66 / 92

LECTURE 4 Solving Zero-Dimensional Systems of Polynomial Equations

A tool knows exactly how it is meant to be handled, while the user of the tool can only have an approximate idea. (Milan Kundera 1929–)




67 / 92

Meaning of Solving: Examples I As usual, we let K be a field. 1

2

3

4

5

A linear system In this case the existence of solutions depends only on rank of matrices. Extending the base field does not change the nature of the problem. x2 + 1 = 0 In this case there are no solutions in R, there are in C. x2 − y = 0 In this case there are many solutions. An infinite number, if K is infinite, as many as #(K), if K is finite. Moreover, they can be parametrized as (t, t2 ), for t ∈ K. x2 − 4x + 1 = 0 √ No solutions in Q. The solution in R are 2 ± 3. x5 + x − 1 = 0 Using CoCoA we get x5 + x − 1 = (x2 − x + 1)(x3 + x2 −√1) Moreover x2 − x + 1 = 0 has two complex solutions 21 ± 23 i. There exist also a formula for solving x3 + x2 − 1 = 0 using radicals. It is due to Cardano (1545).




68 / 92

Meaning of Solving: Examples II

6

x5 − x − 1 = 0 Using CoCoA we check that x5 − x − 1 is irreducible. Following Galois we know that there is no formula which represents the solutions via radicals. But there are solutions in C.

So how and where do we represent them? Using CoCoA we get that there is only one real solution which is approximately 1.16730499267578 But approximation leads to different kinds of problems. Lorenzo Robbiano (University of Genoa, Italy)



69 / 92

Meaning of Solving: Examples III

Back to the equation x5 − x − 1 = 0. Let f (x) = x5 − x − 1. 1

K = F2 :

f (x) = (x2 + x + 1)(x3 + x2 + 1).

2

K = F3 :

f (x) irreducible.

3

K = F17 :

f (x) = (x − 8)(x − 6)(x3 − 3 x2 − 5 x + 6).

4

K = F19 :

f (x) = (x3 + 7 x2 − 6 x − 9)(x + 6)2 .

5

K = F101 :

f (x) = (x + 40)(x4 − 40 x3 − 16 x2 + 34 x − 48).

For instance, where do we find the other three solution in the case of F17 ?




70 / 92

Polynomial Systems Let K be a field, let P = K[x1 , . . . , xn ] be a polynomial ring over K, and let f1 , . . . , fr ∈ P. We want to solve the polynomial system   f1 (x1 , . . . , xn ) = 0 .. .   fr (x1 , . . . , xn ) = 0 What does it mean “solve this system”? If K ⊆ L is a field extension then the set of all tuples (a1 , . . . , an ) ∈ Ln such that fi (a1 , . . . , an ) = 0 for i = 1, . . . , r is called the zero set of the ideal I = hf1 , . . . , fr i in Ln and is denoted by ZL (I). The elements of ZL (I) are also called the solutions of the polynomial system in Ln . Solving makes most sense if the zero set of I is finite. The Finiteness Criterion says that ZK (I) is finite if and only if dimK (P/I) is finite, i.e. if and only if P/I is a zero-dimensional affine algebra over K. This is our usual setting, and we are ready to apply the techniques developed so far. Lorenzo Robbiano (University of Genoa, Italy)



71 / 92

General Considerations In general, it may happen that there are no solutions of the polynomial system in K n . On the other hand, by Hilbert’s Nullstellensatz there exists a finite algebraic extension field L of K such that ZL (I) is not empty, if I is a proper ideal. Therefore, it is true that if we allow a base field extension we have a better chance to find a solution. It is also true that if K = Q there are approximate solutions. We know that a rational solution (λ1 , . . . , λn ) ∈ K n corresponds to a linear maximal ideal hx1 − λ1 , . . . , xn − λn i of P which contains I. On the other hand, a solution with coordinates in an extension field of K corresponds to a non-linear maximal ideal M of P which contains I. Consequently, a first way to represent this solution is simply to write generators of M. If we are more sophisticated, we can even compute and write generators of the corresponding primary component Q of I. Is there a better method to represent this situation, more similar to the description via coordinates as in rational case? Lorenzo Robbiano (University of Genoa, Italy)



72 / 92

Rational Zeros Proposition (Eigenvalues and Rational zeros) Let f ∈ P, and let ϑf : R −→ R be the corresponding multiplication map. (a) If there exists a point p ∈ ZK (I) then f (p) ∈ K is an eigenvalue of ϑf . (b) If λ ∈ K is an eigenvalue of ϑf then there exist an algebraic field extension L ⊇ K and a point p ∈ ZL (I) such that λ = f (p). Proof. Let us first prove (a). We let p = (a1 , . . . , an ) be a point of ZK (I). Then we let (x1 − a1 , . . . , xn − an ) be the corresponding maximal ideal in P which contains I, let m be its isomorphic image in R, and let ı(m) be its image in F. Then we deduce that f (p) is an eigenvalue of ϑf . Now we prove (b). Let λ ∈ K be an eigenvalue of ϑf . Then z − λ is an eigenfactor of ϑf . There is a maximal ideal m of R such that f − λ ∈ m. Let M be the preimage n of m in P. By Hilbert’s Nullstellensatz, there exists a tuple p = (a1 , . . . , an ) ∈ K such that M K[x1 , . . . , xn ] ⊆ hx1 − a1 , . . . , xn − an i. Then the field L = K(a1 , . . . , an ) is an algebraic extension of K for which we have the inclusion m ⊗K L ⊆ h¯x1 − a1 , . . . , ¯xn − an i, and therefore p = (a1 , . . . , an ) ∈ ZL (I). Finally, we note that f − λ ∈ m implies λ = f (p). Lorenzo Robbiano (University of Genoa, Italy)



73 / 92

Some Remarks Example Let P = Q[x, y], let I = hx2 , y2 + 1i, and let R = P/I. For the polynomial x, the multiplication map ϑx : R −→ R is nilpotent and has the eigenvalue λ = 0. However, there is no point p = (a1 , a2 ) ∈ ZQ (I) such that a1 = x(p) = 0. In fact, the zeros of I 2 in Q are (0, i) and (0, −i). Corollary For i ∈ {1, . . . , n}, the i-th coordinates of the points of ZK (I) are the eigenvalues of the multiplication map ϑxi . Remark (The “Bad” Algorithm) For i = 1, . . . , n, compute the eigenvalues λi1 , . . . , λiνi ∈ K of the multiplication map ϑxi : R −→ R. Then check for every tuple (λ1j1 , . . . , λnjn ) with 1 ≤ ji ≤ νi whether it solves the system. The ones who do form the set ZK (I) of K-rational solutions of I. Qn This method has a major drawback, namely that we have to check for i=1 νi tuples whether they solve the system. Lorenzo Robbiano (University of Genoa, Italy)



74 / 92

The “Good” Algorithm ALGORITHM (The Eigenvalue Method) Let K be a field, let P = K[x1 , . . . , xn ], and let I be an ideal in P such that P/I is a zero-dimensional K-algebra. Consider the following sequence of instructions. 1

T = {I}.

2

For i = 1, . . . , n, perform steps 2 – 7.

3

S = ∅.

4

For each J ∈ T perform steps 5 – 6.

5

Compute the set L of linear eigenfactors of ¯xi ∈ P/J. For each ` ∈ L append the ideal J + hì to S.

6 7

Replace T with S.

8

Return T.

This is an algorithm which computes a tuple S consisting of tuples of generators of all linear maximal ideals in P which contain I.




75 / 92

Hint at the Eigenvector Method Theorem (Characterization of Evaluation Forms) Let K be a field, let P = K[x1 , . . . , xn ], let I be a zero-dimensional ideal in P, let R = P/I, let F be the multiplication family of R, and let p ∈ ZK (I). (a) The evaluation form evp ∈ ωR is a joint eigenvector of the dual family F ˇ and generates a 1-dimensional joint eigenspace. (b) Every 1-dimensional joint eigenspace of F ˇ is generated by an evaluation form. (c) For every f ∈ R, the eigenvalue of ϑf ˇ corresponding to evp is f (p). (d) Given a K-basis B = (t1 , . . . , td ) of R, we have evp = t1 (p) t1∗ + · · · + td (p) td∗ in ωR . ALGORITHM (The Second Eigenvector Method) In the setting of the theorem, let B = (t1 , . . . , td ) be a K-basis of R such that t1 = 1 and ti+1 = xi for i = 1, . . . , n. Consider the following instructions. (1) Compute a set S of bases of all 1-dimensional joint eigenspaces of F ˇ. Let S = {(`1 ), . . . , (`s )} with ì ∈ ωR . (2) For i = 1, . . . , s, write ì = ai1 t1∗ + · · · + aid td∗ with aij ∈ K. Return the set of tuples {(ai2 /ai1 , . . . , ai n+1 /ai1 ) | i = 1, . . . , s}. This is an algorithm which computes the set ZK (I) of all K-rational zeros of I. Lorenzo Robbiano (University of Genoa, Italy)



76 / 92

Solving over Fq

Definition Let K be Qar field, let f (x) ∈ K[x], and let K ⊆ L be a field extension. If f (x) = i=1 (x − ci ) with ci ∈ L, then we say that f (x) splits in L. Moreover, if L = K(c1 , . . . , cr ) then L is called a splitting field of f (x). Let p be a prime number and let q = pe . It is known that there exists a unique, up to isomorphism, finite field with q elements which is denoted by Fq . The uniqueness depends on the fact that a field with q elements is a splitting field of zq − z, and the fact that two splitting fields of the same polynomial are isomorphic. We say that a field K with pe elements represents Fq .




77 / 92

Splitting a Polynomial Proposition (Splitting a Polynomial over a Finite Field) Let p be a prime number, let K = Fp , let f (x) ∈ K[x] be a monic irreducible polynomial of degree e. Then let t be a new indeterminate and let L = K[t]/hf (t)i. (a) The field L has q = pe elements, hence it represents Fq . (b) In the ring L[x] we have f (x) = (x − ¯t)(x − ¯t p ) · · · (x − ¯t p splitting field of f (x).

e−1

), hence L is a

i

(c) Let I = hf (t), f (x)i ⊂ K[t, x]. Then Mi = hf (t), x − tp i is a maximal ideal for every i = 0, . . . , e − 1, and I = ∩e−1 i=0 Mi is the primary decomposition of I. i

i

(d) In claims (b) and (c) we can substitute tp with the normal form NF(tp , hf (t)i) for every i = 0, . . . , e − 1. (e) Let V be a finite-dimensional K-vector space, let ϕ ∈ EndK (V) be such that µϕ (z) is irreducible in K[z], and let ϕL be the extension of ϕ to VL = V ⊗K L. Then ϕL is diagonalizable. There is a nice way to compute the “big powers” of steps (c) and (d). Lorenzo Robbiano (University of Genoa, Italy)



78 / 92

Solving over a Finite Field Main Theorem (Solutions of Polynomial Systems over Finite Fields) Let p be a prime number, let K = Fp , let Mt be a maximal ideal of K[t], and let Mx be a maximal ideal of K[x]. Assume that dimK (K[x]/Mx ) = dimK (K[t]/Mt ), call e this number, and let L = K[t]/Mt . (a) The field L has q = pe elements, hence it represents Fq . (b) The ideal Mt + Mx of K[t, x] is radical. (c) For every i = 1, . . . , e there exists a tuple gi1 (t), . . . , gin (t) ∈ (K[t])n such that, if we let Mi = Mt + hx1 − gi1 (t), . . . , xn − gin (t)i then M1 ∩ · · · ∩ Me is the primary decomposition of Mt + Mx . (d) For i = 1, . . . , e and j = 1, . . . , n, let gij (t) be elements of K[t] as described in (c), and let ¯ gij (t) be their residue classes in L. Then we have the primary decomposition gi1 (t), . . . , xn − ¯ gin (t)i Mx L[x] = ∩ei=1 hx1 − ¯ (e) The polynomial system represented by the maximal ideal Mx of K[x] has e solutions in L. They are ¯ g1n (t) , . . . , ¯ gen (t) . g11 (t), . . . , ¯ ge1 (t), . . . , ¯ (f) Let V be a finite-dimensional K-vector space, let Φ = (ϕ1 , . . . , ϕn ) be endomorphisms of V, and let F = K[ϕ1 , . . . , ϕn ] be such that RelP (Φ) is a maximal ideal in P = K[x1 , . . . , xn ]. Moreover, let ΦL = ((ϕ1 )L , . . . , (ϕn )L ) where (ϕi )L is the extension of ϕi to VL = V ⊗K L for i = 1, . . . , n, and let FL be the family generated by ΦL . Then the family FL is simultaneously diagonalizable. Lorenzo Robbiano (University of Genoa, Italy)



79 / 92

The Algorithm ALGORITHM (Computing Solutions over Finite Fields) In the setting of the theorem, consider the following instructions. (1) Compute the primary decomposition M1 ∩ · · · ∩ Me of Mt + Mx in K[t, x]. (2) Choose a term ordering σ on T(t1 , . . . , tm , x1 , . . . , xn ) which is an elimination ordering for (x1 , . . . , xn ) and compute the reduced σ-Gröbner bases G1 , . . . , Ge of the maximal ideals M1 , . . . , Me respectively. (3) For i = 1, . . . , e, let x1 − gi1 (t), . . . , xn − gin (t) be the polynomials in Gi which are linear in the indeterminates x1 , . . . , xn , and returns the e tuples ¯g11 (t), . . . , ¯ g1n (t) , . . . , ¯ gen (t) . ge1 (t), . . . , ¯ This is an algorithm which computes all the tuples in Ln which are solutions of the polynomial system associated to the maximal ideal Mx . Proof. From claim (c) of the theorem we deduce that the primary components of Mt + Mx are the e maximal ideals ideals Mi = Mt + hx1 − gi1 (t), . . . , xn − gin (t)i. Let σ be a term ordering on T(t1 , . . . , tm , x1 , . . . , xn ) which is an elimination ordering for (x1 , . . . , xn ), let σ 0 be its restriction to T(t1 , . . . , tm ), let Gi be the reduced σ-Gröbner basis of Mi , and let G0i be the reduced σ 0 -Gröbner basis of Mt , so that G0i ⊂ Gi . We observe that NFσ (xj , Mi ) = xj , since otherwise Gi would not contain any polynomial in xj , contradicting the fact that Mi is zero-dimensional. If we let NFσ (gij (t), Mi ) = hij (t) and let G0i = (g1 (t), . . . , gs (t)), it follows that Gi = (g1 (t), . . . , gs (t), x1 − hi1 (t), . . . , xn − hin (t)). The correctedness of the algorithm is then guaranteed by claim (e) of the theorem, and the fact that ¯ hij (t) = ¯ gij (t) in K[t]/Mt . Lorenzo Robbiano (University of Genoa, Italy)



80 / 92

Special Isomorphisms

Theorem (Isomorphisms of K-Algebras) Let K be a field, let T = K[t1 , . . . , tm ], let P = K[x1 , . . . , xn ] and let Q = K[t1 , . . . , tm , x1 , . . . , xn ]. Let f1 , . . . , fm be elements of P, let Φ : T → P be defined by Φ(ti ) = fi , and let A = ht1 − f1 , . . . , tm − fm i. Then let g1 , . . . , gn ∈ T, let Ψ : P → T be defined by Ψ(xi ) = gi , and let B = hx1 − g1 , . . . , xn − gn i. Finally, let J be and ideal of T and I an ideal in P. Then the following conditions are equivalent. (a) We have the equality JQ + B = IQ + A. (b) The two K-algebra homomorphisms Φ and Ψ induce K-algebra isomorphisms ϕ : T/J → P/I and ψ : P/I → T/J which are inverse to each other.




81 / 92

Representations of Fq

Theorem (Isomorphisms of Representatives of Fq ) Let p be a prime number, let K = Fp , let Mt be a maximal ideal of K[t], let Mx be a maximal ideal of K[x], let dim K (K[x]/Mx ) = dimK (K[t]/Mt ), and call e this number. Then let gi1 (t), . . . , gin (t) ∈ (K[t])n be tuples as described in the above theorem. (a) There exist e tuples fi1 (x), . . . , fim (x) ∈ (K[x])m for i = 1, . . . , e such that Mt + hx1 − gi1 (t), . . . , xn − gin (t)i = Mx + ht1 − fi1 (x), . . . , tm − fim (x)i (b) Let Φi : K[t] → K[x] be defined by Φi (tj ) = gij (x), an let Ψi : K[x] → K[t] be defined by Ψi (xj ) = fij (t). Then Φi and Ψi induce ϕi : K[t]/Mt → K[x]/Mx and ψi : K[x]/Mx → K[t]/Mt which are K-algebra isomorphisms inverse to each other. (c) The isomorphisms in (b) are all the K-algebra homomorphisms between the fields K[t]/Mt and K[x]/Mx .




82 / 92

An Algorithm ALGORITHM (Computing Isomorphisms of Representatives of Fq ) Consider the following instructions. (1) Compute the e tuples g11 (t), . . . , g1n (t) , . . . , ge1 (t), . . . , gen (t) using the solutions over a finite field. (2) Let σ be an ordering on T(t1 , . . . , tm , x1 , . . . , xn ) of elimination of (t1 , . . . , tm ), and compute the reduced σ-Gröbner bases G1 , . . . , Ge of the ideals Mx + hx1 − gi1 (t), . . . , xn − gin (t)i for i = 1, . . . , e. (3) For i = 1, . . . , e, let t1 − fi1 (x), . . . , tm − fim (x) be the polynomials in Gi which are linear in the indeterminates t1 , . . . , tm , and return the e tuples f11 (t), . . . , g1n (t) , . . . , ge1 (t), . . . , gen (t) and the e tuples f11 (x), . . . , f1m (x) , . . . , fe1 (x), . . . , fem (x) This is an algorithm which computes the tuples which define the isomorphisms ϕi and ψi such that ϕi and ψi are inverse to each other for i = 1, . . . , e.




83 / 92

Solving over Q ALGORITHM (Computing the Splitting Field) Let K = Q, let P = K[x], and let f (x) be a non linear irreducible polynomial in P. Consider the following sequence of instructions. (1) Let i = 1, let ti be a new indeterminate, and let Li = K[ti ]/hf (ti )i. (2) Factorize f (x) in Li [x], let Lini (f (x)) be the set of linear factors of f (x) in Li [x]. If Lini (f (x)) yields a complete factorization of f (x), then return the field L = Li , the set Lin(f (x)) = Lini (f (x)), and stop. (3) Repeat the following steps until Lini (f (x)) yields a complete factorization of f (x). (4) Increase i by 1, let fi (¯t1 , . . . , ¯ti−1 , x) be a non linear irreducible factor of f (x) in the polynomial ring Li−1 [x], and then let Li = K[t1 , . . . , ti ]/hf (t1 ), f2 (t1 , t2 ), . . . , fi (t1 , . . . , ti )i. Then perform Step 2 with f (x) = fi (¯t1 , . . . , ¯ti−1 , x). (5) Return the field L, the set Lin(f (x)), and stop. This is an algorithm which computes the splitting field L of f (x) and the set Lin(f (x)) of the linear factors of f (x) in L[x]. Lorenzo Robbiano (University of Genoa, Italy)



84 / 92

An Example

Example Let P = Q[x]. The discriminant of f (x) = x3 − 3x + 1 is 81 = 92 which implies that the Galois group of f (x) is F3 . Then we let Mt = hf (t)i and Mx = hf (x)i in Q[t, x]. The primary decomposition of the ideal Mt + Mx is given by Mt + Mx = M1 ∩ M2 ∩ M3 where M1 M2 M3

= Mt + hx − ti = Mt + hx − t2 + 2i = Mt + hx + t2 + t − 2i

Therefore L = Q[t]/Mt is the splitting field of f (x) and dimQ (L) = 3.




85 / 92

An Example (continued) Example The discriminant of the polynomial g(x) = x3 − 4x + 1 is 229. This is not a square which implies that the Galois group of g(x) is the symmetric group S3 . The difference of the Galois groups implies a different size of the splitting fields of f (x) and g(x). To check this claim, we let Mt1 = hg(t1 )i and Mx = hg(x)i in Q[t1 , x]. The primary decomposition of Mt1 + Mx is given by Mt1 + Mx = M1 ∩ M2 where = =

M1 M2

Mt1 + hx − t1 i Mt1 + hx2 + t1 x + t12 − 4i

Then we let M(t1 , t2 ) = Mt1 + ht22 + t1 t2 + t12 − 4i. The primary decomposition of M(t1 , t2 ) + hx2 + t1 x + t12 − 4i is given by M(t1 , t2 ) + hx2 + t1 x + t12 − 4i = M21 ∩ M22 where M21 M22

= =

M(t1 , t2 ) + hx − t2 i M(t1 , t2 ) + hx + t1 + t2 i

We conclude that M(t1 , t2 ) + Mx = M1 ∩ M21 ∩ M22 where M1 M21 M22

= = =

M(t1 , t2 ) + hx − t1 i M(t1 , t2 ) + hx − t2 i M(t1 , t2 ) + hx + t1 + t2 i

and that L = Q[t1 , t2 ]/M(t1 , t2 ) is the splitting field of g(x). We have dimQ (L) = 6. Lorenzo Robbiano (University of Genoa, Italy)



86 / 92

Another Example Example Let P = Q[x, y] and let I = hx2 + x − 1, y3 − x − 1i. Using the ordering Lex with y > x we get the reduced Gröbner basis (y6 − y3 − 1, x − y3 + 1). From the fact that the polynomial y6 − y3 − 1 is irreducible we deduce that I is a maximal ideal. Then we use an algorithm described before to find the appropriate extension of Q where I splits 2 into linear maximal ideals. Let M1 = ht12 + t1 − 1i and Ix = hx + x − 1i. In Q[t1 , x] we get the primary decomposition M1 + Ix = M1 + hx − t1 i ∩ M1 + hx + t1 + 1i . We let L1 = Q[t1 ]/M1 . In L1 [x, y] we get the decomposition M = N1 ∩ N2 where N1 = hx − ¯t1 , y3 − ¯t1 − 1i

N2 = hx + ¯t1 + 1, y3 − ¯t1 − 1i

We have L1 [x, y]/N1 ∼ = L1 [y]/hy3 − ¯t1 − 1i. The next step is to split the polynomial y3 − ¯t1 − 1 ∈ L1 [y]. Let M2 = M1 + ht23 − t1 − 1i and Iy = hy3 − t1 − 1i. The ideal M2 is maximal in Q[t1 , t2 ]. Moreover, in the polynomial ring Q[t1 , t2 , y] we get the primary decomposition M2 + Iy = M2 + hy − t2 i ∩ M2 + hy2 + t2 y + t22 i . Let M3 = M2 + ht32 + t2 t3 + t22 − 1i = ht12 + t1 − 1, t23 − t2 − t1 − 1, t32 + t2 t3 + t22 i




87 / 92

Example continued Example The ideal M3 is maximal in Q[t1 , t2 , t3 ], and in Q[t1 , t2 , t3 , y] we get the primary decomposition M3 + Iy = M3 + hy − t1 i ∩ M3 + hy − t2 i ∩ M3 + hy + t2 + t3 i At this point we let M = M3 , we let L = Q[t1 , t2 , t3 ]/M, and we have three solutions in L of the polynomial system associated to M. They are the residue classes of (t1 , t2 ), (t1 , t3 ), (t1 , −t2 − t3 ) Next, we have L1 [x, y]/N2 ∼ = L1 [y]/hy3 + ¯t1 i. Let us see how much the polynomial y3 + ¯t1 splits in L. We compute the primary decomposition of M + hy3 + t1 i and get M + hy3 + t1 i = M + hy − t1 t2 t3 i ∩ M + hy − t1 t2 t3 − t1 t22 i ∩ M + hy + t1 t22 i We conclude that all the solutions can be found in L. They are (t1 , t2 ) (−t1 − 1, t1 t2 t3 )

(t1 , t3 ) (−t1 − 1, t1 t2 t3 + t1 t22 )

(t1 , −t2 − t3 ) (−t1 − 1, −t1 t22 )

Finally we observe that degQ (L) = 12.




88 / 92

Approximation: an Example Example Over the field K = Q we consider the polynomial system f1 f2 f3

= 3x2 − y2 + 2yz − z2 − 8x − 8y + 5z − 5 = 0 = x3 − 6x2 − 6xy − 4y2 + z2 + 3x + 7y − 7z + 15 = 0 = z3 + 4x2 + 2xy − 3z2 − 13x − 5y + 6z + 5 = 0

In the ring P = Q[x, y, z] we form the ideal I = hf1 , f2 , f3 i. To find the zeros of I we use the Eigenvalue Method. The minimal polynomial of x evaluated at x factors as f (x)(x − 2)3 where f (x) = x15 − 102 x14 + 4254 x13 − 91420 x12 + 1108030 x11 − 7617460 x10 +26368737 x9 − 16709678 x8 − 163401402 x7 + 408143116 x6 −102334132 x5 − 561462068 x4 + 119765981 x3 + 1225188788 x2 −1326547350 x + 524273780 We add hx − 2i3 to I and get Q1 = hx − y − 3, z2 − 2z + 1, yz − y + z − 1, y2 + 2y + 21 z + 12 i. The radical of Q1 is M1 = hx − 2, y + 1, z − 1i. We conclude that I has the unique rational zero (2, −1, 1). Lorenzo Robbiano (University of Genoa, Italy)



89 / 92

Example continued Example Next we add hf (x)i to I and get the ideal M2 = hf1 , f2 , f3 , f4 i where f1 f2 f3 f4

= = = =

x2 − 31 y2 + 23 yz − 13 z2 − 83 x − 83 y + 53 z − 53 2 35 z3 + 2 xy + 43 y2 − 83 yz − 53 z2 − 73 x + 17 3 y − 3z + 3 20 1 2 38 17 13 85 2 xy2 − 2 xyz + xz2 − 10 xy − 46 3 y − 5 xz + 3 yz − 3 z − 3 x − 3 y − 3 z + 3 82 2917 13069 y4 − 10 y2 z2 − 3 y3 − 282 xyz + 50 y2 z + 103 xz2 + 4 yz2 − 3 xy − 9 y2 5033 845 2 13034 10751 3754 20215 − 1438 3 xz + 9 yz + 9 z − 9 x − 9 y − 9 z + 9

To check that M2 is a maximal we compute a Lex-Gröbner basis of M2 . We get (x − g1 (z), y − g2 (z), g(z)) where g(z) = z15 − 15 z14 + 453 z13 − 1723 z12 + 1653 z11 + 39878 z10 − 143949 z9 +44358 z8 + 1493331 z7 − 3545819 z6 + 1822878 z5 + 12465182 z4 −23275915 z3 + 35305982 z2 − 40170300 z + 15978680 and g1 , g2 are polynomials of degree 14 whose coefficients are fractions where numerators and denominators are in the order of 1040 . The verification that g(z) is irreducible confirms that M2 is a maximal ideal. Lorenzo Robbiano (University of Genoa, Italy)



90 / 92

Example continued Example If we do not resort to approximation, apparently to best represent the roots associated to M2 one has simply to write M2 = hf1 , f2 , f3 , f4 i. If we leave the world of exactness and enter the realm of approximation, we can deduce that of the 15 solutions, three of them have real coordinates. They are (approximately) ( −2.375085, 0.866703, ( 9.492307, −17.531086, ( 2.029906, −0.970152,

2.514900 0.723763 0.998203

) ) )

2 is not equal to 3, not even for very large values of 2. (Gabriel’s Law)




91 / 92

Next?

The End or approximately The End?




92 / 92