Article Submitted to Journal of Symbolic Computation
Computing the Galois Group of a Polynomial Using Linear Di erential Equations Olivier Cormier1 , Michael F. Singery2 and Felix Ulmer3 IRMAR, Universite de Rennes 1, F-35042 Rennes Cedex
[email protected] 2 Dept. of Math., Box 8205, NC State University, Raleigh, NC 27695-8205
[email protected] 3 IRMAR, Universit e de Rennes 1, F-35042 Rennes Cedex
[email protected] 1
Abstract
In this paper we show how to compute the geometric Galois group GQ(x) of a polynomial f 2 Q(x)[Y ] by considering the associated linear di erential equation Lf (Y ) = 0 (and constructions of it) satis ed by the roots of f . We use that the di erential Galois group of Lf (Y ) is a faithful linear representation of GQ(x) whose character is a summand of the permutation character of GQ(x) acting on the roots of f . Our approach allows to make full use of representation theory (over a eld of characteristic zero) in order to study permutation representations. In the nal section we show how, if f 2 Q(x)[Y ], our approach via di erential Galois theory helps one to also compute the arithmetic Galois group GQ(x) of f over Q(x) if the later has some special properties.
1. From polynomials to linear di erential equations
In the following we will consider the eld Q, but the results remain valid for any eld of characteristic 0. Let
f
=Ym+
m ?1 X i=0
bi(x)Y i 2 Q(x)[Y ]
(1)
Some of these results were presented in a preliminary form at the International Symposium on Symbolic and Algebraic Computation (ISSAC2000) in St Andrews, Scotland [5]. yPartially supported by NSF Grant CCR-9731507.
1
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
2
be a polynomial of degree m. We assume that f is an irreducible element of
Q(x)[Y ] (i.e. that f is absolutely irreducible) and denote K the splitting eld of f over Q(x) and z1; : : :; zm the roots of f (Y ) = 0 in K .
We now describe the well known construction of a linear di erential equation with coecients in Q(x) whose solution space is spanned, as Q-vector space, by the zeros zi of f . On the rational function eld Q(x) we consider the usual derivation given by (Q) = 0 (resp. (Q) = 0) and (x) = 1. There is a unique extension of to the eld K given by
(zj ) = ?
Pm?1
0 i i=0 bi(x) zj P mzjm?1 + mi=1?1 ibi(x)zji?1
obtained by di erentiating f (zj ) = 0. Let zi be a zero of f , then by the above formula all derivatives of zi also belong to Q(x)(zi). Thus, with this derivation, K is a di erential eld extension of Q(x). In the following we denote (t) by t0 and i(t) simply t i . Since Q(x)(zi) is a Q(x)-vector space of dimension m, we get that the m + 1 elements zi; zi0; : : :; zi m must be linearly dependent over Q(x), i.e. zi must satisfy a linear di erential equation of order at most m over Q(x). Let Lf (Y ) be the linear di erential equation corresponding to the linear dependent set zi; zi0; : : :; zi n having the property that n is minimal. Since the computations involve only division by f , all roots of f will be solutions of Lf which thus is the linear di erential equation of lowest order over Q(x) satis ed by all zi. The construction of Lf (Y ) can be done using linear algebra. In particular, if f 2 Q(x)[Y ], then the coecients of Lf are in Q(x). ( )
(
)
( )
We note that one can give an alternate construction of Lf using data from the Newton Polygon of f [24] at various points. We recall that the Newton Polygon of f at x = is the lower convex hull of the points (i; (bi)), where (bi) is the order of bi at x = . Furthermore, the order of a solution of f = 0 at x = is given by the negative of a slope of one of the sides of the Newton Polygon. We shall assume that the order of Lf (Y ) is maximal, i.e., m. One can ensure this (without changing the Galois group) using Lemma 2. We note that the process used in Lemma 2 uses Tschirnhaus transforms to replace f with a new polynomial whose degree in y is the same as that of f and whose degree in x may increase but only by an amount bounded (a priori) by a polynomial in the degree of f in x. Furthermore, if f has coecients in Q the bit size of the coecients of the new polynomial has also increased by, at most, a polynomial amount. In general the linear di erential equation will be of the form Lf (y) = y n + anA?(x()x) y n? + : : : + Aa ((xx))n y where A(x) is a squarefree polynomial. We can write A(x) = A (x)A (x) where ( )
1
(
1)
0
1
2
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
3
the zeros of A (x) are the nite "honest" singular points and the zeros of A (x) are the apparent singular points. We can let A (x) = b(x) r(x) where b(x) is the common denominator of the bi(x) and r(x) = Resultanty (b(x)f; b(x)fy) is the discriminant, since outside the zeros of these polynomials we have n analytic solutions of f = 0. Since at the apparent singularities we have positive integer exponents, one can bound the number of these (and their exponents) using Fuchs' relation (c.f., [15]) and lower bounds on the exponents at singular points. Since any solution of Lf (y) = 0 is a linear combination of the zeros of f = 0, all exponents of Lf are bounded from below by the negative of the largest slope of the Newton Polygon. This, in turn, can be bounded by the total degree of bf . Once one has a bound on the degree of A(x), one can bound the degree of the ai using the fact that the point at in nity must be a regular singular point as well. To determine the an?i =Ai, we let x = be a point where A 6= 0, i.e., where f = 0 has n analytic solutions, and let y ; : : : ; yn be the solutions of f = 0 at x = . We note that Wr(Y; y ; : : :; yn)) Lf (y) = det( det(Wr(y ; : : :; yn )) where Wr is the wronskian matrix. Therefore, each coecient an?i=Ai is the ratio of power series that we can compute. If Ni is a bound on the degrees of the numerator and denominator of an?i =Ai (we can take Ni = deg(Ai) because deg(an?i ) deg(Ai) ? i), then we can determine this rational function from the rst 2Ni +1 coecients of the corresponding ratio of power series. Note that the time complexity of this computation is polynomial in the degree of f and the bit-size of the coecients. 1
2
1
1
1
1
1
Example 1: Take f = y 3 + xy 2 + (x ? 3)y ? 1.
The discriminant of f is equal to r(x) = (x2 ? 3x + 9)2 , so the singularities of f are the roots of x2 ? 3x + 9 and in nity. At in nity we have 3 distinct Puiseux expansions of f whose rst terms are 2 1 2 1 x + x + , ?1+ x + and ?x +1 ? x + , so the exponents of Lf at in nity are f?1; 0; 1g. Let i be the roots of r. At the point x = i we have a Puiseux expansion for f whose rst terms are : 1 + (x ? )1=3 + 1 (2 ? 1)(x ? )2=3 ? 1 (x ? ) + i i i 3 i 3 i 3 and P so the exponents of Lf at i are f0; 1=3; 2=3g. Let Eqi be the sum of the exponents of Lf at a point qi and p1 ; : : :; ps the (possible) apparent singularities of Lf . Therefore Fuchs' relation has the form : 2
2 X X
i=1
E i +
In our example we have
X
Ps
E1 +
i=1
P
s X X i=1
Epi = 3(3 ? 1)(22 + s ? 1)
Epi = 3s +1. Since pi are apparent singularities,
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
4
they have positive integer exponents whose sum is greater than 4 (else they are regular points with exponents f0; 1; 2g). We then have 4s 3s + 1 i.e. s 1 and thus Lf has at most one apparent singularity, let's say at x = . We therefore have A(x) = (x ? )(x2 ? 3x + 9) and Lf is of the form :
Lf (y) = y + a A(x) y00 + aA(x) y0 + aA(x) y where deg(a ?i) 2i. Computing the Puiseux expansions of f at x = 0, which is a regular point for f , we have that the expansion of the coecient a Ax is : 1 + 2 ? 2x ? 4 x ? 2 x + x 3 9 27 81 so we have = 0 and by identi cation we obtain a (x) = 3(x ? x ? 3). By expanding the other coecients of Lf we nd that : Lf (y) = y + x3((xx ??3xx?+3)9) y00 ? (x ? 36x + 9) y0 + x(x ? 63x + 9) y = 0 (3)
1
2
0
2
3
3
2( )
2
3
2
2
(3)
2
2
2
2
2
2
We refer to [13, 20] for an introduction to di erential eld extensions, PicardVessiot extensions and di erential Galois groups. Let k be a di erential eld and f 2 k[Y ]. A Picard-Vessiot extension F of k for Lf is the di erential analogue of a splitting eld and has the following properties: 1. F is the smallest di erential eld extension obtained by adjoining a fundamental solution set fy ; : : :; yng of Lf (y) = 0 to k. 2. The constants of F and k are the same. So F is an extension of k with no new constants. If the eld of constants of the ground eld is algebraically closed, then F always exists. This is the reason for considering f in k[Y ] = Q(x)[Y ] and explains why our approach produces the Galois group of f over Q(x). The di erential Galois group of Lf (over Q(x)) is the subgroup of the automorphism group of F=Q(x) whose elements commute with , i.e. = . 1
Proposition 1: Let K be the splitting eld of f over Q(x), GQ(x) the galois group of K=Q(x), F a Picard-Vessiot extension of Q(x) for Lf (Y ) = 0 and G the di erential Galois group of F=Q(x). Then K = F and GQ(x) = G .
Proof: Since any root of f is a solution of Lf (Y ) = 0, we have that K F . Since elements of G are automorphisms they must send a root of f into another root. Thus the Q-vector subspace V of the solution space of Lf (Y ) = 0 generated by fz1; : : :; zmg is G -invariant. We denote n the dimension and y1; : : : ; yn a basis of
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
5
V over Q. According to [20] Proposition 3.5 the linear di erential equation L~ (Y ) obtained by expanding y : : : y Y n 0 0 y0 : : : yn Y 1 jWr(y ; : : :; yn)j ... ... ... ... = 0 y n : : : ynn Y n 1
1
1
( ) 1
( )
( )
along the last column is a linear di erential equation over Q(x) having V as solution space. Since V is a subspace of the solution space of Lf (Y ) = 0, we have that L~ (Y ) is a right divisor of Lf (Y ) (cf. [20] Proposition 3.5). Since Lf (Y ) is a di erential equation of minimal order with respect of having all zi as solutions, we have that, up to multiple in Q(x), L~ (Y ) = Lf (Y ). Thus F K showing that F = K . Since for a nite algebraic extension the classical and the di erential Galois group coincide (cf. [23]), we have that GQx and G are the same groups acting on K = F . 2 ( )
Example 2: We consider the polynomial f = Y 3 ? x
2 Q(x)[Y ]. Here the derivative of a root z of f is given by z 0 = 31x z and the linear di erential equation Lf (y) is y 0 ? 31x y. Note that the roots of f are z1 = x1=3, z2 = x1=3 and z3 = 2 x1=3 where is a primitive cube root of unity. Thus the Q-vector space generated by the roots is of dimension one. Our approach will show that the Galois group of f over Q(x) is cyclic. Thus we do not in general get the Galois group over Q(x) which in this case would be S3 . The Galois group of f acts by permutation of the roots zi and so on the vector space P = mi ziQ having the zi as a formal basis. This gives a faithful linear representation of GQx that is called the permutation representation. The character of this representation is called the permutation character and is denoted by P . Since GQx is also the di erential Galois group of Lf (Y ), the action of GQx on the solution space V gives a faithful character L of GQx . The trivial character of GQx is denoted by 1l. Proposition 2: (see [18]) The character L of GQx is a summand of P . If the order of Lf (Y ) is m, then L = P . =1
( )
( )
( )
( )
( )
( )
Proof: Denote zi the roots of f in the Picard-Vessiot extension F and V the solution space of Lf (Y ) = 0. We consider the map ': mi=1ziQ ! V given by '(zi) = zi. Both spaces are GQ(x)-modules of characters respectively P and L. It is clear that ' is a surjective GQ(x)-morphism. Since GQ(x) is a nite group, Maschke's Theorem implies that V is GQ(x)-isomorphic to a direct summand of mi=1ziQ and so L is a direct summand of P . 2
In particular the trivial character is always a summand of P . To simplify the tables in later section we will only list decompositions of the character P ? 1l.
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
2. Absolute irreducibility
6
In order to compute the geometric Galois group GQx of f we want to assume that the polynomial f is absolutely irreducible (i.e. irreducible over Q(x)). We start with a simple, well-known fact concerning the permutation representation of G (c.f., [9], V. Satz 20.2 a). Let P be the permutation representation of G given above. ( )
Lemma 1: Let P1 = fv 2 P j (v ) = v for all 2 Gg. The number of orbits of
G acting on fzig equals the dimension of P . Therefore the number of orbits of G equals ( P ; 1l). 1
Proof: Let O1; : : : ; Ot be the orbits of G in fz1; : : :; zmg. If v = ci zi 2 P1 , then any two zi in the same orbit must have the same coecient ci. Therefore we may write P
v= P
t X i=1
ci(
X
zj 2Oi
zj ) :
Since the elements wi = zj 2Oi zj lie in P and are linearly independent, we have the conclusion of the lemma. 2 1
This lemma immediately gives a criterion that allows one to test absolute irreducibility of f using Lf (Y ). Let k be a eld such that the f 2 k(x)[Y ]. As we have already noted, the coecients of Lf (Y ) lie in k(x). Proposition 3: Suppose that the order of Lf (Y ) is m. The number of irreducible factors of the polynomial f 2 k(x)[Y ] is ( P ; 1l). Furthermore, the number of these irreducible factors is the dimension of the k-vector space of solutions of Lf (Y ) = 0 in k(x). In particular f 2 k(x)[Y ] is irreducible if and only if the space of rational solutions of Lf (Y ) = 0 is of dimension one and generated by the coecient of Y m? . 1
Proof: The number of irreducible factors is the number of di erent orbits in the set of roots of f in the splitting eld under the permutation action of the Galois group G of f over k(x). By Lemma 1, this number is precisely ( P ; 1l), the dimension of the space of solutions of Lf (Y ) = 0 that are left xed by the Galois group. These are precisely the space of solutions that lie in k(x). From [3], Theorem 9.1 or [16], Proposition 4.3 we can conclude that this space has the same dimension as the k-space of solutions of Lf (Y ) = 0 in k(x). 2
We will now describe how one can determine the number of factors of f over k(x). The above Proposition states that this is equivalent to determining the dimension of the space of solutions in k(x) of Lf (Y ) = 0. Although there are standard algorithms to determine all rational solutions of a linear di erential equation (see [3] and [16]), one can use information from the polynomial f to
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
7
simplify things. Let
g(x) = pq((xx)) 2 k(x)
be a solution of Lf (Y ). There exist constants c ; : : : ; cm 2 k such that g(x) = c z + : : : + cm zm. Let b(x) be the least common multiple of the denominators of the bi(x) in equation (1). If x = 2 k is a pole of g(x) then must be a pole of some zi. Such an must be a zero of b(x) or a zero of r(x) = Resultanty (b(x)f; b(x)fy). Furthermore, the order of g(x) at such a point must be greater than or equal to the orders of the zi at this point. The orders of the zi can be determined from the slopes of the Newton Polygon at x = (c.f., [24], Ch. IV, x3). Note that if is a conjugate of over k, then the Newton Polygon at is the same as that at and this can be determined from F and the minimal equation of over k. Therefore, if p ; : : :; pt are the irreducible factors of r(x) b(x), then t Y q(x) = pni i 1
1 1
1
i=1
where the ni correspond to the integer parts of the largest slopes of the Newton Polygons at the roots of the pi . The construction of the Newton Polygon shows that a weak bound for these numbers is given by the degree of b(x). The next step is to determine the degree of p(x). We note that the order of g(x) at in nity is given by deg(q(x)) ? deg(p(x)). This number is again bounded from below in terms of the slopes of the Newton Polygon at in nity. From this we can bound the degree of p(x). Let N be a bound on the degree of p(x). To determine the possible p(x), let x = be a point such that b( )r( ) 6= 0. One can calculate the rst N Taylor coecients of the zi at x = . Equating coecients of powers of x in p(x) = (c z + : : : + cmzm)q(x) gives a system of linear equations in the coecients of p and the ci. The dimension of the solution space of this system is the dimension of the space of rational solutions of Lf (Y ) = 0. 1 1
We note that the above method computes the number of irreducible factors in Q(x)[Y ] of f in a number of steps that is bounded by a polynomial of the total degree of f and the bit size of the coecients of f . For more results about absolute factorisation see for example [10], [11] and [6].
3. Factorisation of linear di erential equations
It is well known that the computation of the Galois group of a polynomial can be reduced to the factorisation of polynomials. A weaker but similar result holds for di erential equations [22]. In the following we want to consider a linear
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
8
di erential equation:
L(y) =
n X i=0
i
aiy = ( )
n X i=0
!
aii
(y) = (p()) (y)
as a polynomial in (i.e. as a di erential operator) over Q(x). The set D of di erential operators over the eld Q(x) forms a non-commutative ring where addition is the usual addition of polynomials and multiplication is the composition of operators de ned by the rule 8a 2 Q(x), a = a + a0. A di erential equation L(y) = 0 is reducible over Q(x) if and only if L(y) = L (L (y)), i.e. if the di erential operator factors in products of degree > 0. We refer to [21] for the properties of D. In particular the non-commutative ring D is a left and right euclidean ring. The ring D is not a unique factorisation domain, but if p = p p ps and p = p~ p~ p~t are two decompositions of an operator p 2 D into irreducible factors, then s = t and there is a permutation 2 St such that D=Dpi = D=Dp~ i . In particular the representations of G on the solution space of pi(y) = 0 and p~ i (y) = 0 are isomorphic G -modules. In classical Galois theory a polynomial is reducible if and only if its Galois group is intransitive. The di erential analogous is: Theorem 1: (see for example [20]) A linear di erential equation L(y ) = 0 over Q(x) factors if and only if its di erential Galois group G GL(n; Q) is a reducible linear group. The factorisation of di erential operators is a dicult task. But if the group is reductive (which is always the case in our situation since the groups are subgroups of Sm and thus nite) a factorisation can be found using the eigenring f[21, 8] de ned as 1
1 2
2
1 2
( )
( )
E (L) = fR 2 D j ord(R) < ord(L) and LR is divisible on the right by Lg: An element R 2 E (L) of order greater or equal to 1 gives a non trivial factor of L. Indeed for z 2 sol(L) we have that L(R(z)) = 0 which shows that z 7! R(z) is a Q-linear map of V to itself. If c 2 Q is an eigenvalue of this linear map, then (R ? c)(y) and L(y) will have a non trivial common factor which can be found by computing a right gcd. The coecients of R 2 E (L) are rational solutions of a
linear di erential operator, they can be found using linear algebra. Furthermore for reductive groups the character of the representation of G on the solution space of L(y) = 0 will be the sum of the characters on the solution space of the irreducible factors of L(y). The factorisation of Lf gives information about G : 1. The group G is abelian if and only if Lf is a product of operators of order one. 2. The group GQ(x) Sm is doubly transitive if and only if Lf has an irreducible factor of order m ? 1.
Proposition 4:
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
9
3. The group G is trivial if and only if all solutions of Lf are in Q(x) (which can be computed using linear algebra [3, 1]). Proof: The rst result follows from the fact that a nite group is abelian if and only if all its irreducible representations are of degree 1. If the action of GQx on the roots of f is doubly transitive, then P = 1l + , where is an irreducible character of GQx ([9], V Satz 20.2). We have that L is a summand of P . Therefore if L does not have a summand corresponding to an irreducible representation of degree m ? 1, then L = 1l. This implies that G is trivial and so cannot be doubly transitive. Therefore, the solution space of L(y) = 0 has summand that is an irreducible G -module of dimension m ? 1 and so assertion 2: follows. The last assertion follows from the Galois correspondence in di erential Galois theory. 2 To compute the di erential Galois group in our situation, we proceed using the idea of [22] and we use the fact that a nite group G is determined by its action on \constructions": symmetric and alternating products of the solution space, which are also G -modules. 1. One can construct a di erential equation L s s = 0 (resp. L^s = 0) whose solution space corresponds to the symmetric product Syms(V ) (resp. the Vs alternating product (V )). 2. The corresponding characters of G are denoted Ls s and ^Ls and can be computed directly from L. 3. Factorisation of L s s (resp. L^s) gives us the degrees of the irreducible characters of L that appear in the decompositions of Ls s and ^Ls. 4. Comparing the result with the tables below allows us to determine the group. Tables 1, 2, 3 and 4 give the decomposition into irreducible characters of P ?1l, where P is the permutation character, for all the transitive permutation groups of degree 3 to 10 given in [4]. Similar tables can be constructed for every degree. Some notation: A linear character of order i is denoted by 1i and a character of degree n simply by n. In order to distinguish non-equivalent irreducible characters of the same degree n, we use the notation na; nb; : : : and if the decompositions of the i symmetric (resp. alternating) product of na and nb are the same we simply write n s i (resp. n^i). For example, the second exterior power of the irreducible character 6 (of degree 6) appearing in the decomposition of the permutation character of F (7) is the sum of a linear character of order 2, two non-equivalent linear characters of order 6 and two copies of (the character) 6. We see that we can distinguish each of the transitive groups from the decomposition of their permutation character P and from decompositions of symmetric or alternating powers of irreducible characters appearing in P ?1l. This approach will work for any degree: Using a theorem due to Chevalley, for any nite subgroup H of GL(n; C ) there is a faithful representation : GL(n; C ) ! GL(m; C ) ( )
( )
th
42
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial Table 1: Degree
Degree 3 G p ? 1l
C (3) = A(3) 13a ; 13b
Degree 4 G p ? 1l 3^3
C
Degree 5 G p ? 1l 4^4 4^2
G p ? 1l 4^4 4^2 5^5 5^2
E ; ;
C
D(4) 12 ; 2
(4) 12a 12b 12c
;
;
;
D
;
C
(6) 12 13a 13b 16a 16b
;
;
;
;
D
A(4)
6 (6) 12 2 2
; ;
3 12
A(5)
4 12 14b 4
;
S(4)
3 1l
F (5)
(5) 2a 2b
14a
Degree 6 G ps? 1l 2 2 3^3 s 3 2 s 3 3
2
(5) 15a 15b 15c 15d
;
3 to 6
S(3)
(4) 12 14a 14b
;
S(5)
4 1l
;
D ; ; ;
4 12 6
F ; ; ;
(6) 1 2 2a 2b 1l 2
18 (6) 12 2a 2b 13 2
S4 (6d) 2; 3
S4 (6c) 2; 3
12 1l 2 3 1l 3a 3a 3b
1l 1l 2 3
; ; ; ; ;
F18 (6) : 2 12 ; 4 1l
10
F36 (6) 12 ; 4 12 14a ; 14b ; 4
F36 (6) : 2 12 ; 4 12 2; 4
A5 (6)
S5 (6)
A(6)
S(6)
1l 3a 3b 4
12 4 6
1l 10
12 10
5
; ;
5
;
5
; ;
S
A ;
A ;
2 4 (6) 2 3
4 (6) 13a 13b 3
2 4 (6) 13a 13b 3
12 1l 2 3 12 3a 3a 3b
1l 1l 13a 13b 3
12 1l 13a 13b 3
;
; ; ; ; ;
;
;
;
;
5
for some m such that (H ) is uniquely determined by its set of invariant subspaces in C m and any representation can be constructed from a given faithful representation using the tools of linear algebra, i.e., tensor product, duals, direct sums and subspaces (see [22]). Given a polynomial f irreducible over Q(x), we rst compute the di erential equation Lf = 0 associated to it and determine the degrees of the irreducible factors of Lf . If the order of Lf is too small, we may not directly determine the Galois group of f using above Tables, but after P Tschirnhaus transformation Y 7! mj ? uj Y j where ui 2 Q(x) the degree will become maximal. 1 =0
Lemma 2: Let k be a di erential eld of characteristic zero with eld of constants C . Assume that there is an element x 2 k with x0 = 1. If y1 ; : : :; ym 2 k are distinct elements, then there exist polynomials ui 2 C [x] of degree at most m Pm?1 such that the elements y~i = j=0 uj yij are linearly independent over C . In particular, if y1; : : :; ym are solutions of a polynomial f of degree m over Q(x) (resp. Q(x)), then y~1 ; : : :; y~m are also solutions of a polynomial f~ of degree m over Q(x) (resp. Q(x)), the Galois groups of f and f~ are the same and the equation Lf~ will be of maximal order m.
Proof: Let U1; : : : ; Um be di erential indeterminates. The eldPof constants of K = k < U1; : : : ; Um > is once again C . The elements Vi = mj=0?1 Uj+1 yij are
;
;
;
;
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial Table 2: Degree
Degree 7 G p ? 1l 6^6 6^2
17a
C (7) ; ; 17f
G p ? 1l 2^2 s 2 2
C
(8) 12 14a 14b 18a 18b 18c 18d
;
;
E
;
;
;
;
;
F42 (7)
;
;
;
D ; ; ; ; ; ;
E
; ; ; ; ;
S(4)[1=2]2 12 ; 3a ; 3b 1l ; 2; 3
= :cD ; ; ; ; ;
G p ? 1l 4^4 4^2
1 2[24 ]4 12 14a 14b 4 1l 12a 12b 2a 2b
G p ? 1l 4^4 3^3 4^2 s 4 2
GL(2; 3) 3; 4
G p ? 1l 6^2 s 3 2 s 6 2 4^2
E(8) : A4 12 ; 6 3; 6a ; 6b 1l ; 13a ; 13b ; 3
G p ? 1l 7^7 7^2 s 7 2
E(8) : 7 = F56 (8)
; ;
=
1l 12 12 2 3
; ;
7 1l 7 7 7
; ;
A ;
; ; ; ;
[24 ] (4) 3 4 12 1l 6
;
; ;
1l 3 6
E(4)2 : D6 12 ; 6 3a ; 3b ; 3c ; 6
E(8) : F21 7a
;
;
;
= ;
; ;
;
;
12a
1 2[23 ]4 12 14a 14b 2a 2b 14 14a 14b 14c
;
A
E(8) ; ; 12g
6 12 15
D
8 (8) 12a 12b 12c 2 2 12
;
;
; ;
Q
8 (8) 12a 12b 12c 2 2 1l
;
;
; ;
[22 ]4 12 14a 14b 2a 2b 12 1l 12a 12b
;
;
; ;
;
;
E E ; ; ; ; ; ; ;
(8) : 4 12 2a 2b 2c 1l 12a 12b 1l 12a 12b
; ; ; ; ; ;
SL ; ; ;
2 4 (8) = (2 3) 2a 2b 3
7 1l 7b 7c
; ;
=
eD ; ; ; ; ;
=
dD ; ; ; 1l ; 12 ; 14a ; 14b ; 2; 4
1 2[24 ] (4) 12 2 4 1l 12a 12b 2a 2b
= ;
E ; ; ; ; ;
S ;
S
1 2[24 ] (4) 3 4 1l 12 3 a 3b
;
;
; ; ;
1l 3a 3b 3c
E(8) : S4 12 ; 6 3; 6a ; 6b 1l ; 2; 3
A
[ (4)2 ]2 12 6 6 9
; ;
; ; ; ; ;
1l 12 2a 2b 6 9
PSL(2; 7)
PGL(2; 7)
; ; ;
; ;
3a
D ; ; ;;
(8) : 4 12a 12b 12c 4 1l 12a 12f
=
[23 ] (4) 3 4 12 12 6
7 1l 3b 7 8
E(8) : D8 12 ; 2; 4 12 2 a ; 2b ; 2c
1 2[24 ] (4) 12 2 4 12 2 4
E ;
1 2[24 ] (4) 12a 12b 12c 4 12 14a 14b 2 2
A ;
[23 ] (4) 3 4 1l 1l 3 a 3b
;
S(7)
6 1l 15
12 12
[23 ]4 12 14a 14b 4 12 14a 14b 2a 2b
; ;
x
A(7)
6 1l 7 8
; ;
[42 ]2 12 2a 2b 2c 14 2 1l 12a 12b
E(8) : 4 12 ; 2; 4 12 14a ; 14b ; 2a ; 2b
;
; ; ; ;
;
E(8) : D6 12 ; 3a ; 3b 1l ; 2; 3
1l 12
[1 4 (4)2 ]2 12 2 4 1l 12a 12b 12c 12d 2
;
;
2 8 (8) 12 2a 2b 2c 12 2 1l 12a 12b
G p ? 1l 4^4 4^2 s 4 2
;
; ;
D ; ; ; ; ; ;
(8) 1 2 2a 2b 2c 1l 2 1l 12a 12b
(8) : 3 1 2 3a 3b 1l 13a 13b 3
;
Q ;
;
L(7) = L(3; 2)
6 12 12 16a 16b 6 6
8 :2 12a 12b 12c 2a 2b 12 12a 12b 12c
; ;
7 and 8
4[ ]2 12a 12b 12c 14a 14b 14c 14d
;
(8) : 2 12a 12b 12c 2a 2b 12 1l 12a 12b
G p ? 1ls s
2 ;s 2b 2 2a 2c 2 G ps? 1l 3 2 3a ^3 3b ^3
F21 (7) 3a ; 3b
;
Degree 8 G p ? 1l 2^2
D ; ;
(7) 2a 2b 2c
11
7 1l 6 7 8
[24 ]4 12 14a 14b 4 12 2 4
;
;
;
;
=
cD ; ; ; ;
1 2[24 ] (4) 12 2 4 12 14a 14b 4
E ; ; ; ;
[24 ] (4) 12a 12b 12c 4 12 2 a 2 b 2c
;
S ;
[24 ] (4) 3 4 12 12 6
; ;
1l 3 6 [1 2 (4)2 ]2 12 6 6 9
= :S ; ;
1 2[ (4)2 ]2 12 6 6 9
[ (4)2 ]2 12 6 6 9
; ; ; ; ; ; ;
1l 12 4 6 9 14a 14b 4
; ; ; ; ; ;
1l 12 4 6 9 2 4
1l 12 4 6 9 12a 12b 2a 2b
E(8) : L7 = AL(8) 7 1l 21 1l 6 7 14
; ; ;
= S ; ;
A(8)
7 1l 21 1l 7 20
; ;
S
; ;
; ; ; ; ;
S(8) 7 12
D ; ; ; 1l ; 12 ; 2a ; 2b ; 4 [24 ] (4) 12 2 4 12 2 4
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial Table 3: Degree
Degree 9 G p ? 1l s3
2a s3
2b
C ; ; : : : ; 19f
(9) 13a 13b 19a
;
G p ? 1l 3^3 s 3 2 4^2
1 3[33 ]3 13a 13b 3a 3b 13 3 3
G p ? 1l 3^3 6^6 6^2
[32 ] (3) 2 3 a 3b 12
G p ? 1l 6^2 6^3
S ; 12 ; 16a ; ; ; ; ; ;
G p ? 1l 6^2 6^6 3^3
[33 : 2] (3) 2 6 12 2 6 6
G p ? 1l 8^8 8^2 G p ? 1l 8^8 8^2
;
=
13a
E ; ; ;
;
[1 2 (3)3 ]3 13a 13b 6
= :S ; ;
[ (3)3 ]3 13a 13b 6
12 12 16a 16b 2a 2b 2c 6
12 12 16a 16b 6a 6b
1l
12 3 12
;
= :S(3)3 ]S(3) 2; 6 3; 12
;
; ;
;
;
E(9) : 8
8 12 18c 18d 83
;
; ;
;
S
;
;
; ;
E(9) : D12 2; 6 12 ; 2a ; 2b ; 4; 6
;
S
;
2 4
S
;
;
;
=: S 2; 6 12 ; 2; 3a ; 3b ; 6
1 2 [33 : 2] (3)
S
[ (3)3 ] (3) 2 6 3 12 12 12
; ;
E(9) : 2D8 2a
E(9) : D8 4a ; 4b
12a 12b 4
;
; ; ; ;
= S
1l
E(9) : 4 4 a ; 4b
;
1 2[ (3)3 ] (3) 2 6 3 12 12 1l
[1 2
8 1l 28
;
E(9) : 6 2; 6 12 ; 16a ; 16b ; 2a ; 2b ; 2c ; 6 1l ; 12 ; 63
; ; ; ; 18a ; 18b ; ; L(9) : 3 A(9) S(9) 8 1l 7 21
S(3)[1=2]S(3) 2a ; 2b ; 2c ; 2d 1l ; 12 ; 2 1l ; 12 ; 2
[33 : 2]3 13a 13b 6
2 6 16b 2a 2b 2c 6 2 63
8 1l 2 2 8 8 8
D ; ; ; 1l ; 12 ; 2 2a ; 2b
(9) 2a 2b 2c 2d
D ;
E
[32 ] (3)6
M (9)
;
(9) : 6 13a 13b 6
;
S ; ; ; ;
x ; ; ;
(3)[ ]3 13a 13b 2a 2b 2c
S(3)[x]S(3) 2a ; 2b ; 4
S ; ;
[33 ] (3) 2 3 a 3b 16
S
9
[33 ]3 13a 13b 3a 3b 13 3 a 3b
(9) : 3 13a 13b 3a 3b 1l
; ;
S ; ;
E(9) ; : : : ; 13h
12
8 12 2b 8a 8 2 b
; ; ;
E(9) : 2A4
E(9) : 2S4
; ; ; ;
; ;
2a 2b
8 1l 8a 8b 8c
8 12 4 8 16
L(9) 8 1l
; ; ;
7 a 7b 7c 7d
8 12 28
linearly independent over C . To see this, note that (V ; : : :; Vm ) = (U ; : : :; Um )V where V isPthe van der Monde matrix of y ; : : : ; yn. Let c ; : : :; cn be constants such that ciVi = 0. We then have that (U ; : : : ; Un)V (c ; : : :; cn )T = 0 and so V (c ; : : : ; cn)T = 0. Since V is nonsingular, we must have that (c ; : : : ; cn)T = 0. We can therefore conclude that the wronskian determinant W of the Vi is a nonzero di erential polynomial in the Yi. This polynomial has order n ? 1 in the Ui and so by a result of [17] p.35, there exist polynomials ui of degree at mostP n such that the substitution Ui 7! ui keeps W nonzero. Thus the elements y~i = mj ? uj yij are linearly independent over C . If y ; : : : ; ym are zeros of a polynomial f over Q(x), then the permutation action of the Galois group G of f on the yi coincides with the permutation action of G on the y~i. Thus y~ ; : : : ; y~m are also solutions of a polynomial f~ of degree m over Q(x), and the Galois group of f and f~ coincide. Being solutions of a polynomial of degree m over Q(x), y~ ; : : :; y~m are solutions of a linear di erential equation Lf of degree at most m over Q(x). Since y~ ; : : :; y~m are linearly independent over the eld of constants Q, the equation Lf is of degree m. The same result holds for Q(x). 2 1
1
1
1
1
1
1
1
1 =0
1
1
1
~
1
~
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial Table 4: Degree
Degree 10 G p ? 1l s 2 2 G p ? 1l 5^5 5^3 5 ^5 5^ a ; 5b G p ? 1l 4^4 4^2 s 4 2 6^2 G p ? 1l 4^4 5^5 4^2 5^2 s 5 5 G p ? 1l 4^4 5^5 4^2 5^2 s 5 2 G p ? 1l 8^8 8^2 G p ? 1l 8^8 8^2 s 2 2 8^4 G p ? 1l 9^9 9^2
C
D ; ;
(10) 12 15a 15b 15c 15d 110a 110b 110c 110d
;
;
;
;
= F 1 2 ; 42
1 2[ (5)]2
F
x ; ;
;
[24 ]5
;
;
;
S5 (10d) 4; 5
; ; ;
;
3a 3b 3a 3b 4
;
A ;
=
;
;
;
;
= :F
[1 2 (5)2 ]2 12 8 1l 2a 2b 8a 8b 8c
;
; ; ; ;
L(10)
9 1l 102 8a 8b
; ;
= F
1 2[ (5)2 ]2 12 8 12 2a 2b 8 16 14a 14b 14c
; ; ; ; ; ;
L(10) : 2 8a
9 12 8b 10a 10b
; ;
;
= S ; ;
D ; ;
S
[ (5)2 ]2 12 4a 4b 12 2 4
1 2[ (5)]2 12 4a 4b 12 6
2 a 2b 5 12 5a 5b 12
x ; ;
(5)[ ]2 12 4a 4b 12 6
;
; ;
4c 5 6
F ;
;
;
;
;
12 515 105
S ;
; ; ; ;
D(5) ; ; ;
[25 ]
[25 ] (5) 4 5 12 12 14a 14b 4
;
=
;
S
S ;
1 2[25 ] (5) 4 5 12 1l 6 10
;
1l 54 104 153 20
;
D(5) ; ; ;
2 a 2b 5 12 5a 5b 1l
; ;
[24 ] (5) 4 5 12 12 6
[52 : 4]2 12 8 1l 14c 14d 4a 4b 8a 8b
=
4a 5 6
F
;
; ; ; ; ;
1 2[25 ]
2 a 2b 5 1l
;
;
; ;
D(5) ; ;
[24 ]
;
x ; ;
1 2[25 ] (5) 4 5 12 1l 14a 14b 4
;
14a 14b
A
1l 515 105
[25 ] (5) 4 5 1l 12
;
;
(5)[ ]2 12 4a 4b 1l 3 a 3b
; ; ; ; ;
F ;
;
;
= D ; ;
[24 ] (5) 4 5 12 12 14a 14b 4
12 1l 6 4 6
[25 ]5
;
[52 ]2 12 2a 2b 2c 2d 15 2
10 (10) 12 2a 2b 2c 2d 1l 2
15a 15b 15c 15d 5 12
1 2[ (5)2 ]2 12 4a 4b 12 14a 14b 4 1l 12 4a 4b
; ;
D ; ; ; ; ;
(10) 2 12 22 a 2b
;
;
= :D ; ;
A5 (10) 4; 5
[24 ] (5) 4 5 1l 1l 3a 3b 10
;
[1 2 (5)2 ]2 12 4a 4b 1l 12a 12b 2a 2b
; ; ; ;
A ;
;
15a 15b 15c 15d 5 1l
(5)[ ]2 1 2 4a 4b 12 14a 14b 4 1l 12 42
1l 1l
;
10 and 11
;
[25 ] (5) 4 5 12 12 6
; ;
[52 : 4]22 12 8 12 18c 18d 8a 8b 8c
;
;
;
F
[ (5)2 ]2 12 8 12 2a 2b 8 16 14 2
; ; ; ; ;
M (10) 10a
;
[ (5)2 ]2 12 8 1l 6a 6b 16
;
; ;
;
;
; ; ; ; ;
= :S ; ;
; ; ;
= S
[1 2 (5)2 ]2 12 8 1l 12 16
1 2[ (5)2 ]2 12 8 12 12 16
;
;
;
;
2 12 4 16 36
S6 (10)
9 1l 10b 16
[52 : 42 ]22 12 8 1l 2 2 8a 8b 8c
;
; ; ; A
;
[52 : 42 ]2 12 8 1l 22 4a 4b 4c 4d 8
;
18a 18b
;
12 54 104 153 20
10a
;
9 12 10b 16
;
L(10):22 9 12 16 20
;
A(10) 9 1l 36
S
[ (5)2 ]2 12 8 12 12 16
;
;
; ;
2 32 36
S(10) 9 12 36
Degree 11 G p ? 1l 10^10 10^2 10^3
C (11) ; ; 111j
111a
D(11) ; ; 2e
2a
F55 (11) 5a ; 5b
F110 (11)
10 12 4 12 14 10 10
;
;
L(11)
10 1l 10 11 122
; ;
M (11)
10 1l 45 102 45 55
; ;
A(11) 10 1l 45 120
S(11) 10 12 45
13
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
14
Example 3: Consider the polynomial f = Y 4 (Y 4 ? 8 Y 2 + 18) + 81 x2 (from [14], p. 405, f8;32) which is irreducible over Q(x). Although the corresponding
di erential equation could have order 8, a calculation shows that it is : x4 ? 15x2 ? 1) y(3) + 2997x6 ? 4600x4 ? 995x2 + 6 y00 Lf (y) = y(4) + 2(12 x(x2 ? 1)(3x2 + 1) 24x2(x2 ? 1)(3x2 + 1)2 6 27(5x2 ? 21) y = 0 x4 ? 97x2 ? 6 y0 ? + 81024xx3?(x1863 2 ? 1)(3x2 + 1)2 256(x2 ? 1)(3x2 + 1)2 Using the Eigenring, we can show that it is irreducible. Therefore the permutation representation has an irreducible summand of dimension 4. According to the Table 2, the Galois group GQ(x) of f belongs to a family of 19 groups (from x ?15x ?1) [1=4:cD(4)2 ]2 to [24]S (4)). The 4th exterior power of Lf is y0 + 2(12 x(x ?1)(3x +1) y = 0 x ?1) . So there remain 7 possible groups. The and has the rational solution x ((1+3 x) 2nd exterior power of Lf is of order 6 and factors as a product of two irreducible di erential equations of order 3, where one is : 6 4 2 4 2 7425 x ? 1951 x ? 613 x 2(60 x ? 15 x ? 1) 00 (3) L3(y) = y + x(5x2 ? 1)(1 + 3x2) y + 12x2(1 + 3x2)2(5x2 ? 1)+ 3 y0 x6 ? 1269x4 ? 17x2 ? 3 y = 0 + 2025 12x3 (5x2 ? 1)(1 + 3x2)2 Therefore, GQ(x) is either [23]A(4) or 1=2[24 ]S (4). The 3rd exterior power of x ?1) as a rational solution, so G L3 has x (5(1+3 Q(x) is [23]A(4). x) If one obtains a decomposition of Lf like the decomposition of the permutation character of D6 (6) or D(6) (let us forget the linear characters), this means that in the latter case Lf is the product of two irreducible isomorphic equations of order 2 and in the former case Lf is the product of two irreducible non-isomorphic equations of order 2. We can distinguish the two cases using the Eigenring: consider Lc the Least Common Left Multiple (LCLM) of the two irreducible equations L1 and L2 of order 2. A simple consideration (see for instance Section 2.2 in [21]) yields that in the rst case the Eigenring of Lc , ED (Lc ), is of dimension 1 or 4 (depending on whether L1 and L2 are equal or not) whereas in the second case ED (Lc) is of dimension 2. 4
2
2
2
2
2
2 4
2
2
2 4
Example 4: Consider the polynomial f = Y 2 (Y 2 + 3)2 + 4x (from [14], p. 404, f6;3) which is irreducible over Q(x). By the transformation Y 7! (Y +1)=(Y ? 1),
f is changed in f~ = (x+4)Y ?6xY +15xY +(8?20x)Y +15xY ?6xY +x+4. The di erential equation Lf = 0 is of order 6 and factors into two equations of order 1 and two irreducible equations L and L of order 2. Hence, GQx is equal to D (6); D(6) or F (6). We have : 5x + 2 5x ? 4 + 4x y0 + L (y) = y00 + 2? x(x + 4)(x ? 1) 36x(x + 4)(x ? 1) y = 0 6
5
4
3
2
~
1
6
18
2
1
2
( )
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
15
213x + 703x + 3684x ? 1760 y0 L (y) = y00 + 362xx(?x + 4)(x ? 1)(3x ? 21x + 88) x + 16147x + 21381x ? 28240x + 3168 y = 0 + 336x ? 3342 18(x + 4)x (x ? 1) (3x ? 21x + 88) The LCLM of L and L is of degree 4 and it's Eigenring is of dimension 2, so GQx is equal to D(6) or F (6). The 2nd symmetric power of L has a rational solution xx so GQx = D(6). 4
3
2
2
2
5
4
3
2
1
( )
2
2
2
2
18
5 +4 ( +4)2
1
( )
4. Galois group over Q(x)
As mentioned before, our approach computes the geometric Galois group GQx of f over Q(x). This is due to the fact that the Picard-Vessiot extension is a no new constants extension. To illustrate this we look at the following example:
( )
Example 5: Consider f = Y 3 ? x
2 Q(x)[Y ]. The roots of f (Y ) = 0 are z = x = , z = x = and z = x = where is a primitive cube root of unity. If we consider the splitting eld Q(x)(z ; z ; z ), we see that it contains the new constant = z =z . This explains why the di erential Galois group that we compute using di erential equations turns out to be A(3), which is the Galois group over Q(x). However there is a connection between the di erential Galois group GQx of f over Q(x) and GQx . We consider the following diagram : 1
1 3
1 3
2
2
3
1 3
1
2
2
3
1
( )
( )
Q(x)(y ;: : :; yn )
??
1
Q(x)(y ;: : :; yn ) 1
@ @
@@
? ?
Q(x)
Q(x)(y ; : :: ; yn ) \ Q(x) 1
Q(x)
Thus GQx , which is isomorphic to the Galois group of Q(x)(y ; : : : ; yn) over Q(x)(y ; : : :; yn ) \ Q(x), is a normal subgroup of GQx . The next corollary shows that the groups are distinct if and only if new constants appear. 1
( )
1
( )
Lemma 3: Let k0 be a eld of characteristic 0 with algebraic closure k0 . Let F
be a eld with k0 (x) F k0 (x). If k0(x) 6= F , then there exists a c 2 F \ k0 with c 2= k0 . Furthermore F = k(x) for some k0 k k0 . Proof: F is algebraic over k0 (x). We begin by assuming that [F : k0 (x)] < 1. The primitive element theorem implies that F = k0(x)(z) for some z 2 k0(x).
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
16
P i Since z 2 k (x), we may write z = Pji abjixxj with ai; bi 2 k . Therefore F k (x)(faig; fbj g) and, again by the primitive element theorem, this latter eld may be written as k (x)( ) with 2 k . We now extend the zero derivation on k via x0 = 1 to a derivation of k (x). In particular the set of constants is k and P so 0 = 0. An element w 2 F can be written uniquely as w = si !i i with !i 2 k (x) and s < [k ( ) : k ]. Consider now some w 2 F which is not in k (x) and having the property thatPs is minimal. For this w we may assume !s = 1. Di erentiating we get w0 = si ? !i0 i and, by the minimality of s, w0 2 k (x). We therefore have that !i0 = 0 for i 2 f1; : : : ; s ? 1g and !0 = g 2 k (x). Therefore (w ? ! )0 = 0 and w ? ! 62 k (x). This proves the rst claim under the assumption that [F : k (x)] < 1. One removes this assumption by noting that any F as described in the lemma is a direct limit of elds nitely generated over k (x). To prove the second claim, let k = F \ k . If F 6= k(x), then the rst part, applied to k(x) F k(x) = k (x), implies that there exists an element c 2 F \ k such that c 62 k, a contradiction. 2 Corollary 1: If GQx 6= GQx , then the eld extension Q(x)(y ; : : :; yn ) \ Q(x) of Q(x) is generated by constants (i.e. is obtained by adjoining new constants to Q(x)). 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
( )
( )
1
For the rest of the chapter, we assume that f 2 Q[x;y] and is absolutely irreducible. Note that in this case GQx is a transitive normal subgroup of the transitive group GQx . From Table 5 we see that, in most cases, there are only very few possibilities for GQx once GQx is known. Table 5 illustrates this fact: Given a transitive group G on n letters, we list, for n = 3; 4; 5; 6 and 7, all the proper normal transitive subgroups N of G. Thus once the group GQx is known, there are only few possibilities left for GQx . We now x a possible group G for GQx having GQx as proper normal subgroup and want to decide, using di erential Galois theory, if G = GQx . By Corollary 1 we know that if G = GQx 6= GQx then the extension Q(x)(y ; : : : ; yn) of Q(x) contains new constants. We consider a tentative Q(x)-basis B = fb = 1; b ; : : :; bjGjg (which wePsuppose given) of jGj elements of Q(x)(y ; : : :; yn) over Q(x) and write z = jiGj uibi with ui 2 Q(x). The condition z0 = 0 is therefore equivalent to a certain differential system S of order one, in the variables ui, having solutions in Q(x), which can be decided using [2, 3]z. We assume that we know how to compute the system S . A new constant corresponds to a \non trivial solution", i.e. for which 9i > 1; ui 6= 0. We now want to decide if a non trivial solution ( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
1
( )
1
2
1
z= z
=1
jGj
X
i=1
u i bi
It is important to note that those algorithms work over Q(x)
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial Table 5: Transitive groups G and their normal transitive subgroups
G N G=N
A(3)
G N G=N
C (4) E (4)
G N G=N
C (5) D(5) F (5) A(5) S (5) C (5) C (5) D(5) A(5) C (2) C (4) C (2) C (2)
G N G=N
C (6) D6 (6)
G N G=N
F36 (6)
G N G=N
C (7) D(7) F21 (7) F42 (7) L(3; 2) A(7) C (7) C (7) C (7) D(7) F21 (7) C (2) C (3) C (6) C (3) C (2)
17
N
S (3) A(3) C (2) D(4) A(4) S (4) C (4) E (4) E (4) E (4) A(4) C (2) C (2) C (3) S (3) C (2)
D(6) A4 (6) F18 (6) C (6) D6 (6) D6 (6) C (2) C (2) C (3)
2S4 (6) A4 (6) S4 (6d) S4 (6c) E (4) C (2) C (2)
A C
2 4 (6) (2)
A A C
2 4 (6) 4 (6) (2)
PSL(2; 5)
S4 (6d) S4 (6c) F18 (6) : 2 A4 (6) A4 (6) D6 (6) F18 (6) C (2) C (2) S (3) C (2) F36 (6) : 2 PGL(2; 5) A(6) S (6) F18 (6) : 2 F36 (6) PSL(2; 5) A(6) C (2) C (2) C (2) C (2) S (7) A(7) C (2)
where the bi are monomials in yj is indeed a new constant. For this we will use Taylor series at a point 2 Q. This approach raises the question of nding a tentative Q(x)-basis in an ecient way. This can be done for a class of possible Galois groups GQx over Q(x), the symmetric Sm and the alternating groups Am, due to the fact that those groups are respectively m ? 1 and m ? 2 transitive groups of degree m. The multiple transitivity allows us in both cases to set up a basis by choosing arbitrary roots yi of f (see next proof). This can be done for a larger class of groups. Recall that a Frobenius group is a transitive subgroup of some symmetric group such that the only group element leaving two elements xed is the identity. ( )
Lemma 4: If the Galois group GQ(x) 2 Sm over Q(x) is an s-transitive group and
if the identity is the only element of GQ(x) that leaves s + 1 elements xed, then by choosing arbitrary roots y1; : : :; ys+1 of f , we get a Q(x)-basis of the splitting eld of f by multiplying all elements in the basis
f1; y ; : : :; ym? ; y ; y y ; : : : ; y ym? ; : : :; ysm?s ym? ym? g of Q(x)(y ; : : :; ys)=Q(x) and the basis f1; ys ; : : : ; ys? g of Q(x)(y ; : : :; ys)(ys ) over Q(x)(y ; : : : ; ys) where 2 N is such that m(m ? 1) (m ? s + 1) is the 1
1
1
1
2
2 1
2 1
1
+1
2
1 +1
2
1
1
1
+1
order of GQ(x). In particular for s = 1 this includes the case where GQ(x) is a Frobenius group, for s = 2 this includes the Zassenhaus groups [7] and for s = m ? 1 and s = m ? 2 this includes the symmetric group Sm and the alternating group Am. 1
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
18
Proof: We proceed by induction. Let s = 1. Since GQ(x) is transitive a basis of Q(x)(y1)=Q(x) is f1; y1 ; : : :; y1m?1 g. Let y2 6= y1 be a second root of f and denote K the splitting eld of f over Q(x). The Galois group of K=Q(x)(y1; y2) is a subgroup of GQ(x) whose elements leave y1 and y2 xed. This group must be trivial by assumption and thus K = Q(x)(y1; y2). The degree of Q(x)(y1; y2)=Q(x)(y1) is = jGQ(x)j=m showing that a Q(x)(y1)-basis of K is f1; y2; : : : ; y2?1g. Putting both bases together as above yields a Q(x)-basis of K and gives the result. Assume that the result holds until s = n ? 1 and consider s = n > 1. Since GQ(x) is transitive the basis of Q(x)(y1)=Q(x) is f1; y1; : : : ; y1m?1g. The Galois group of K=Q(x)(y1) is the stabiliser Gy of y1. Since s = n > 1 GQ(x) is 2-transitive and the polynomial fy = f=(Y ? y1) is absolutely irreducible over Q(x)(y1). Its Galois group Gy operates on the roots of fy as a permutation group of the m ? 1 remaining roots. Since Gy is (s ? 1)-transitive and the identity is the only element of Gy that leaves s elements xed, we get a Q(x)(y1)-basis of K by induction. Putting both bases together to get a Q(x)-basis gives the result. The Frobenius groups satisfy the above conditions for s = 1 and the symmetric Sm and the alternating groups Am, due to the fact that those groups are respectively m ? 1 and m ? 2 transitive groups of degree m, also satisfy the above conditions for s = m ? 1 and s = m ? 2. 2 1
1
1
1
1
1
If the group G is of the above type (s-transitive and the identity is the only element that leaves s + 1 elements xed), then if the group is of order m (m ? 1) (m?(s?1)) it is also easy not only to set up a basis but also a multiplication table for the splitting eld (for smaller groups this would involve some choices of elements). Example 6: Suppose that the galois group GQ(x) of g = Y 5 + 4i=1 ai Y i 2 Q(x)[Y ] is the Frobenius group F (5) of order 20 = 5 4. Since the assumptions of the above Lemma are veri ed for s = 2, a Q(x)-basis of the splitP
ting eld is B = f1; y2; y22; y23; y1 ; y1y2 ; y1y22 ; y1y23; y12; y12y2; y12y22; y12y23 ; y13; y13y2 ; y13y22 ; y13y23; y14; y14y2; y14y22; y14y23g where y1; y2 are distinct roots of g. In order to multiply elements in the splitting eld we use g , the minimal polynomial of y1 over Q(x), and
g (Y ) = Y + (y + a )Y + (y + y a + a )Y + (y + y a + y a + a )Y +(y + y a + y a + y a + a ) 1
4
1
4 1
4
3 1 4
3
2 1 3
2 1
1
1
2
4
3
2
3 1
2 1 4
1 3
2
1
the minimal polynomial of y2 over Q(x)(y1). Lemma 5: Let y1; : : :; yn be solutions of Lf (Y ) = 0, s 2 N and u1 ; : : :; un 2 Q(x). There exist a point 2 Q and aPbound M 2 N, depending on , s, u1; : : :; un and y1; : : :; yn , such that: z = jiG=1j uibi 2 Q (where the bi are mono-
mials in yj ) if and only if the coecients of x ? ; (x ? )2 ; : : :; (x ? )M in the Taylor series expansion of z at are 0.
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
19
Proof: Since the ui and the yi all satisfy some linear di erential equations, one can construct a di erential equation L (y) satis ed by z ([19]). Now 1 (and thus any element of Q) and z are both solutions of the Least Common Left Multiple L (y) of L (y) and L (y) = y0. We take for a regular point of L (y) and for M the order of L (y) at . If at the Taylor series of two solutions agree on terms of order less than M , then the solutions coincide. The result now follows. 2 1
3
1
2
3
3
We use the notation of the above Lemma and note that 2 Q \ Q(y ; : : : ; yn) is uniquely determined by its Taylor series + O((x ? )M ). In order to test if GQx is one of the possible transitive groups G (for which we can setup a Q(x)-basis and compute in it) strictly containing GQx as a normal subgroup, we proceed as follows: Consider a tentative Q(x)-basis B = fPb = 1; b ; : : :; bjGjg of jGj elements of Q(x)(y ; : : :; yn ) over Q(x) and write z = jiGj ui bi with ui 2 Q(x). The condition z0 = 0 is therefore equivalent to a certain di erential system S of order one, in the variables ui, having solutions in Q(x). if S has no non-trivial rational solution then we have no new constants and we must have GQx 6= G. If G is the only candidate di erent from GQx for GQx , then GQx = GQx . if S has non-trivial rational solutions, then we test if the elements in our basis are linearly independent by testing if among our solutions ui there PjGj is an non-trivial expression i uibi = 0 with ui 2 Q(x) (any P non trivial linear combination of value 0 leads to a new constant, e.g., 1 + jiGj uibi). To test this we evaluate the expression in the series for bi and put all terms up to O((x ? )M ) to zero. If no such combination can be found, then GQx is of order at least jGj. If G is the largest candidate for GQx , then GQx = G. If our set B is linearly independent, then there must be new constants. Those new constants will correspond to a non-trivial rational solution of S whose Taylor series evaluates to + O((x ? )M ) with 62 Q. 1
+1
( )
( )
1
1
2
=1
( )
( )
( )
( )
( )
=1
=1
+1
( )
( )
( )
+1
Example 7: We consider the polynomial f = y 5 ? 5xy 4 +50y 3 ? 50xy 2 +125y ?
25x (G. Malle, private communication). The di erential equation associated to f is: x + 1) y + 12(?1 + 5x ) y + 12(5x ? 8x ? 5) y00 Lf (y) = y + 5(3 x(x ? 5) (x ? 5) x(x ? 5) + 25(x384? 5) y0 ? 25x(x384? 5) y = 0 2
(5)
2
(4)
2
2
2
4
2
2
(3)
4
2
2
3
4
This equation admits a one dimensional subspace of rational solutions (spanned by x), so f is absolutely irreducible. Since Lf factorizes completely into linear factors, we get from Table 1 that the Galois group of f over Q(x) is the cyclic group
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
20
C (5). According to the Table 5 its Galois group over Q(x) is either C (5); D(5) or F (5). Suppose it is F (5). Since F (5) is Frobenius of order 20 = 5 4, a tentative Q(x)basis should be B = f1; y2; y22; y23; y1; y1 y2; y1y22; y1y23; y12; y12y2; y12y22; y12y23 ; y13; y13y2; y13y22; y13y23; y14; y14y2; y14y22; y14y23g where y1 is any root of f and y2 is another root of f whose minimal polynomial over Q(x)(y1) is (c.f. Example 6) : f (y) = y + (y ? 5x)y + (y ? 5xy + 50)y + (y ? 5xy + 50y ? 50x)y +y ? 5xy + 50y ? 50xy + 125 4
1
3
1
4 1
3 1
2 1
2 1
2
1
3 1
2 1
1
1
We can thus express any power y2n for n 4 using y1 and lower powers of y2 . P Suppose that z = bi2B uibi is a constant. Constructing the linear di erential system U 0 = 51 MU associated to z 0 = 0 (see the matrix M in Table 6), and computing the rational solutions, we obtain :
u = x ? x ?x ? ? x ; u = + x x ?x ; u = ? x ? x ; u = x xx ? x ? ; x? ; u = ? x ? x ?x ? x ; u = ? ? x ? x x? u = x xx? ; u = ? x ? x x? ? x ; u = ? x x?? ? x ; u = x xx ? x ? ; u = ? ? x ? x ; u = ? x xx ? x ? ; u = x x x? ; u = ? x ? x ?x ? x ; u = x ? x ?? ?x ; u = ? xx ?? ; u = ? x ? ; u = x x ? ; u = x ? u = ? xx? ; 1 +30 2 +25 3 ) 2 5 2 4) 1 +7 2 +(3 3 2 2 5 2 2 (95 300) 1 +(45 130) 2 +13 3 2 5 (25 2 60) 1 +(20 2 25) 2 +27 3 5( 2 5) 39 1 +23 2 +(3 2 4) 3 8 2 5 15 1 +9 2 +(5 2 4) 3 10 5( 2 5) (25 2 20) 1 +(15 2 10) 2 +3 3 5( 2 5) (5 2 4) 1 2 3 14 5( 2 5)
0
4
6
12
0 12
16
18
(60
(5 1 +3 2 ) 5( 2 5) 5( 2
1
5)
(40 2
1
5
9
300) 1+(15 2 130) 2 13 2 5 20 1 +10 2 +3 3 3 5( 2 5) 43 1 30 2 +(8 2 73) 3 2 5 5 1 +4 2 +7 3 7 5( 2 5) 135 1 +(5 2 175) 2 27 3 5( 2 5) 3 1 +2 2 + 3 11 5( 2 5) 5 1 +14 2 +7 3 13 5( 2 5) 5 1 3 15 25( 2 5) 1 +3 2 17 5( 2 5) 19
3
1
25( 2 5)
where the i are constants. Computing the Puiseux expansions of f at x = 0, we take for y1 the series which has the value 0 at x = 0 and for y2 another series (its value at x = 0 is a solution of Z 4 + 50Z 2 + 125 = 0), we nd that : z = 0 ? 53 ? (601 + 262 )Z + 54 3Z 2 ? 52 (21 + 2 )Z 3 + O(x20) where Z satis es Z 4 + 50Z 2 + 125 = 0. Since z will be zero if and only if all i and thus all ui are zero, the elements of our tentative basis are linearly independent and our group is of order at least 20 (and thus cannot be D5 or C5 ). We conclude that GQ(x) is F (5). In order to exhibit the new constants we note that z is a rational number only if 1 ; 2 and 3 are all equal to 0, so we have a new non-rational constant when one of the i ; i = 1::3; is non-zero.
0
x2 ?5
5
0
0
0
0
?1 x2 ?5
x2 ?5
0 0
0 0
10 0
?2 x2 ?5 0
375
x2 ?5 ?2150x x ?5 165 x2 ?5 ?15x x2 ?5
0
0
0
0
0
0
x2 ?5
5
0
0
0
0
0
75x x2 ?5
?100 x x2 ?5
0
0
0
0
0
0
5 x2 ?5
0
0
0
0
0
0
0
0
75x x2 ?5
0
5 x2 ?5
0
0
0
0
0
0
0
10 x2 ?5
0
0
75x x2 ?5
0
x2 ?5
10
0
0
0 0
0
0
150
0
0
x2 ?5
10
0
x2 ?5
0
0
?2
?1
x2 ?5
x2 ?5 ?15x x2 ?5
0
0
0
x2 ?5 ?15x x2 ?5
0
0
x2 ?5
0
0
0
x2 ?5 ?15x x2 ?5
0
0
?2150x x ?5
0
0
0
0
0
0
0
0
0
0
75x x2 ?5
5 x2 ?5
0
0
0
0
0
0
0
0
0
0
5
?150 x x2 ?5
0
0
x2 ?5
3
150
3 x2 ?5
x2 ?5 0 0 0
?1
?1 x2 ?5 0 0
0
0
?1 x2 ?5 0
165
150 x2 ?5 ?15x x2 ?5 2 x2 ?5
0
0
0
0
0
0
0
0
0
?15x x2 ?5
0
0
0
x2 ?5
3
0
0
0
?15x x2 ?5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
x2 ?5
3
0
0
0
0
0
0
0
0
0
?100 x x2 ?5
0 0
0
0
?100 x x2 ?5
0
0
0
0
500 x2 ?5
0
0
0
x2 ?5
500
0
0
0
0
0
0
?375 x2 ?5
0
0
0
75x x2 ?5
x2 ?5
10
0
0
0
0
x2 ?5
?375
0
0
x2 ?5
0
0
x2 ?5
10
0
0
0
0
0
0
0
x2 ?5
5
0
0
x2 ?5
15
0
0
75x x2 ?5
?200 x x2 ?5
0
0
0
0
0
15 x2 ?5
0
0
?1 x2 ?5
10 x2 ?5
0
15 x2 ?5
150x x2 ?5
0
0
?200 x x2 ?5
0
15 x2 ?5
0
0
0
0
0
0
0
?2 x2 ?5
0
?2 x2 ?5
0
165 x2 ?5 ?15x x2 ?5
0
?375 x2 ?5
0
0
0
220 x2 ?5
x2 ?5
10
0
0
x2 ?5
220
0
0
15
0
0
x2 ?5
0
0
0
0
0
0
0
?2
5 x2 ?5
0
0
0
0
x2 ?5
x2 ?5
3
0
0
x2 ?5
?2
?15x x2 ?5
0
x2 ?5
0
0
0
0
0
x2 ?5
1
0
0
?2
0
0
0
0
0
0
?3
x2 ?5
x2 ?5 ?15x x2 ?5
x2 ?5
0
0
0
0
3 x2 ?5
0
0
0
0 0
0
0
0
0
0
0
0
0
0
0
0
3 x2 ?5
0
0
0
0
0
0
0
0
0
0
0
0
Table 6: Matrix
M
0
0
0
0
0
?200 x x2 ?5
500
0
?1
?3 x2 ?5
?20x x2 ?5
5
x2 ?5 ?20x x2 ?5
0
0
0
0
?3 x2 ?5
0
0
x2 ?5
0
0
0
0
0
?1
220
0
10 x2 ?5
?20x x2 ?5 ?2
x2 ?5
3 77 77 0 77 77 0 77 ?225x 777 x ?5 7 77 77 0 77 77 0 77 7 75x 7 x2 ?5 7 77 77 75 x2 ?5 7 77 77 0 77 75x 7 7 2 x ?5 7 7 ?2375 777 x ?5 7 7 ?50x 77 x2 ?5 7 77 75x 7 7 x2 ?5 7 7 ?375 77 x2 ?5 7 77 150x 7 7 x2 ?5 7 77 7 70 77 x2 ?5 7 77 0 77 77 77 0 77 7 15 2 x ?5 7 75 0
?20x x2 ?5
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4
21
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
22
5. Testing if the geometric Galois group is abelian
The aim of this section is to show that factorisation of Lf is not always necessary in order to compute properties of GQx since this information can sometimes be found in the eigenring E (L). We begin with a simple group theoretic fact (c.f., [9], I. Satz 5.13). ( )
Lemma 6: Let G Sm be a transitive permutation group and let P be the m-
dimensional vector space on which G acts via the permutation representation. The group G is abelian if and only if P = P1 : : : Pm where the Pi are non-isomorphic one-dimensional G-modules. Proof: If P has the decomposition of the Lemma, then G GL(P ) is diagonalizable and so is abelian. Conversely, if G is abelian, then we can write P = P1 : : : Pm where the Pi are one-dimensional G-modules. We must show that they are pairwise non-isomorphic. Let fe1; : : : emg be the set on which G acts transitively and consider these as basis elements of P . Since G acts transitively on this set, the span of the orbit of e1 has dimension m. Now assume that two of the Pi are isomorphic as P G-modules. We can then write e1 = ti=1 wi where t < m and for any i there is a character i of G such that g(wi) = i(g)wi for all g 2 G. This implies that the orbit of e1 lies in the span of fw1; : : : ; wtg. Since this span has dimension less than m, we have a contradiction. 2 Proposition 5: Suppose that f
2 Q(x)[Y ] is irreducible and let the order of
Lf be n m. The group GQx is abelian if and only if 1. dim(E (Lf )) = n, and 2. There exists a basis R ; : : :; Rn of E (Lf ), each of order n ? 1 such that Lf = SiRi for i 2 f1; : : : ; ng for some Si 2 D. Proof: Let V = Sol(Lf ). Note that V is a direct summand of the vector space P on which G acts via the permutation representation. We then have that GQx is abelian if and only if V = V V Vn as GQx -module, where dim(Vi) = 1 for all i 2 f1; : : :; ng. We shall use the fact that E (Lf ) is naturally isomorphic to EndG (Vf ) = HomG (Vf ; Vf ) where we consider Vf as a G-module, [21]. Assume that GQx is abelian. From the previous lemma, we have that the permutation representation is a direct sum of pairwise non-isomorphic one-dimensional G-modules. Therefore we may write V = V V Vn , with the Vi pairwise non-isomorphic. Thus EndG (Vf ) = mi EndG (Vi). Since each E (Vi ) is one dimensional, we have that the maps i : Vf ! Vi form a basis of EndG (Vf ). Let Ri 2 E (Lf ) correspond to i. Since Ri has an m ? 1 dimensional kernel, the order of Ri is m ? 1. Since ker Ri (V ) ker(Lf ), the operator Ri must divide Lf on the right, i.e. Lf = SiRi. We now prove the converse. Let W be the sum of the one dimensional subspaces im(Ri) V . If W 6= V , then V = W W~ . Consider the projection ( )
1
( )
1
2
( )
( )
1
=1
2
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
23
: V ! W~ and R 2 E (LfP ) be corresponding operator. Since the Ri form a basis of E (Lf ) we have R = mi ciRi, showing that im(R) = W~ W . Thus W is a direct sum of the one dimensional subspaces im(Ri ), showing that GQx is diagonalisable and thus abelian. 2 =1
( )
Although the above result can be made e ective, it is not apparent that this approach leads to a polynomial time algorithm to decide if the Galois group is abelian. Such an algorithm is known using other techniques ([12], Corollary 3.3).
6. Conclusion
We do not claim that our approach to the computation of the Galois group is more ecient than the known algorithms (using resultants or permutation invariants). This is certainly false for polynomials of small degree. However the approach could help to check eciently special properties of the Galois group (cf. Proposition 4) using di erential equations. It also illustrates the role of constants in di erential Galois theory.
7. Acknowledgments
Most computations were realized with the computer algebra systems Gap, Magma, or Maple and using the packages ISOLDE and BERNINAx on the machines of the French CNRS-UMS Medicis.
References
[1] S. A. Abramov. Rational solutions of linear di erential and di erence equations with polynomial coecients. USSR Comput. Math. and Math. Phys., 29(11):1611{1620, 1989. [2] M. Barkatou. On rational solutions of systems of linear di erential equations. J. Symb. Comp., 28(4-5):547{567, 1999. [3] M. Bronstein. Solutions of linear di erential equations in their coecient eld. J. Symb. Comp., 13(4):413{439, 1992. [4] G. Butler and J. McKay. The transitive groups of degree up to 11. Comm. Algebra, 11(8):863{911, 1983. [5] O. Cormier, M.F. Singer and F. Ulmer. Computing the Galois group of a polynomial using linear di erential equations. in Proceedings of the 2000 International Symposium on Symbolic and Algebraic Computation, p. 7885, August 6-10, 2000, St Andrews, ACM Press. x http://www-sop.inria.fr/cafe/Manuel.Bronstein
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
24
[6] S. Gao. Factoring multivariate polynomials via partial di erential equations. Clemson University, 2000. [7] D. Gorenstein. Finite Groups. Harper& Row, New York, 1968. [8] M. van Hoeij. Factorization of di erential operators with rational functions coecients. J. Symb. Comp., 24(5):537{561, 1997. [9] B. Huppert. Endliche Gruppen I. Springer-Verlag, Berlin-New York, 1967. [10] E. Kaltofen. Fast parallel absolute irreducibility testing. J. Symbolic Comput., 1(1):57-67, 1985. Misprint corrections: J. Symbolic Comput., 9:320 (1989). [11] E. Kaltofen. E ective Noether irreducibility forms and applications. J. Comput. System Sci., 50(2):274-295, 1995. [12] H.W. Lenstra, Jr. Algorithms in algebraic number theory. Bull. Amer. Math. Soc., 26(2):211{244, 1992. [13] A. Magid. Lectures on Di erential Galois Theory. University Lecture Series, 7 - American Mathematical Society, 1994. [14] G. Malle and H. Matzat. Inverse Galois Theory. Springer Monographs in Mathematics, Berlin, 1999. [15] E. G. C. Poole. Introduction to the Theory of Linear Di erential Equations. Dover Publications, Inc., New York, 1960. [16] M. van der Put and M. F. Singer. Di erential Galois Theory. preprint, 2000. [17] J. F. Ritt. Di erential algebra. Dover Publications, Inc., New York, 1966. [18] L. Schmiedt-Thieme. Lineare di erentialoperatoren mit endlicher galoisgruppe. IZWR preprint, 99-25, Universitat Heidelberg, 1999. [19] M. F. Singer. Algebraic solutions of n-th order linear di erential equations. In Proceedings of the 1979 Queens Conference on Number Theory, pages 379{420. Queen's Papers in Pure and Appl. Math., 54, 1979. [20] M. F. Singer. An outline of di erential galois theory. In Computer Algebra and Di erential Equations, pages 3{57. Edited by E. Tournier, Academic Press, London-New York, 1990. [21] M. F. Singer. Testing reducibility of linear di erential operators: a group theoretic perspective. Appl. Alg. in Eng. Comm. and Comp., 7(2):77{104, 1996.
Cormier, Singer, Ulmer: Computing the Galois Group of a Polynomial
25
[22] M. F. Singer and F. Ulmer. Galois groups for second and third order linear di erential equations. J. Symb. Comp., 16(1):1{36, 1993. [23] F. Ulmer. On liouvillian solutions of linear di erential equations. Appl. Alg. in Eng. Comm. and Comp., 2(3):171{193, 1992. [24] R. Walker. Algebraic Curves. Dover Publications, Inc, New York, 1962.
