COMPUTING GALOIS GROUPS WITH GENERIC RESOLVENT POLYNOMIALS

COMPUTING GALOIS GROUPS WITH GENERIC RESOLVENT POLYNOMIALS JOHN KOPPER 1. Introduction Given an arbitrary irreducible polynomial f with rational coefficients it is difficult to determine the Galois group of the splitting field of that polynomial. When the roots of f are easy to calculate, there are a number of tricks that can be employed to calculate this Galois group. If the roots of f are solvable by radicals, for example, it is often easy to calculate by hand the Q-fixing automorphisms of Q that permute the roots of f. In the general case, though, this calculation can be very difficult. Indeed one hopes for an algorithm for computing Galois groups using only the coefficients of the given polynomial. Such algorithms exist; in these notes we will examine one developed by Staduhar [3] and Soicher-McKay[5]. In Section 2 we will describe the algorithm in full generality. Sections 3-5 deal with particular cases and have many examples of the relevant computations. The appendix describes methods of performing the computations in Mathematica. For the purposes of this document we will usually assume our fields have characteristic 0 unless we make an explicit exception. In fact, little will be lost by assuming all fields are extensions of Q. If f k[x] is an irreducible polynomial, we use the phrase the Galois group of f to refer to the Galois group of the extension F/k, where F is the splitting field of f. We will use the shorthand x to mean the ordered n-tuple (x 1,..., x n ). Hopefully the number n will be clear from context. We will get a lot of mileage out of the following standard theorem from algebra: Theorem 1.1. If k is a field and f k[x] an irreducible polynomial of degree n then the Galois group of f is a transitive subgroup of the symmetric group S n. As a final caution, we will engage in the surprisingly common abuse of language in which we will call the Galois group G of a polynomial a subgroup of H for some group H any time G is isomorphic to a subgroup of H. This is worth thinking about in the context of Galois theory. Indeed if f = (x r 1 ) (x r n ) with the r i k then the Galois group of f is some subgroup of S n which acts on the roots r i by permuting the indices. A different choice of labeling the roots of f yields a conjugate subgroup of S n ; we can therefore only make distinctions up to conjugacy. This fact weakens, for example, the conclusion of Theorem 2.5. 2. Generic resolvent polynomials Let k be a field and k(x 1, x 2,..., x n ) = k(x) (i.e. the field of rational functions in n variables). Consider the polynomial P n (t) = (t x 1 )(t x 2 ) (t x n ) = t n x i t n 1 + + ( 1) n x i. 1

2 JOHN KOPPER Definition 2.1. The coefficients of P n (t) are called the elementary symmetric functions and are denoted s 1,..., s n so that P n (x) = t n + s 1 (x 1,..., x n )t n 1 + + ( 1) n s n (x 1,..., x n ). Lemma 2.2. The extension k(s) = k(s 1,..., s n ) k(x) is finite and Galois with Galois group S n Notice that the action of the Galois group S n = Gal(k(x 1,..., x n )/k(s 1,..., s n )) is precisely what you would expect it to be. That is, if σ is an automorphism of F which fixes K, then σ(a 0 + a 1 x 1 + + a n x n ) = a 0 + a 1 x σ(1) + + a n x σ(n). where a 0,..., a n K. Suppose now H S n is a subgroup. Then the elements of F fixed by H form a finite separable extension of K with Galois group H (as a consequence of basic Galois theory). Lemma 2.3. If K F is a finite separable extension, then F = K(α) for some α F. Applying this to our situation, we get some α F = k(x 1,..., x n ) with k(x 1,..., x n ) = k(s 1,..., s n )(α). Definition 2.4. The element α above is called the generic resolvent for H. Suppose now we have an irreducible separable polynomial f k[x]. Then in the algebraic closure of k we may write f(x) = (x r 1 )(x r 2 ) (x r n ) where r 1,..., r n are the (distinct) roots of f(x). Then the coefficients of f(x) are given by symmetric functions on the r i. This is a trivial but crucial remark so it bears repeating: the coefficients of f are elementary symmetric polynomials in its roots. Explicitly, if f(x) = x n + a n 1 x n 1 + + a 0 then a i = s i (r 1,..., r n ). Let H be a subgroup of S n and let α be the generic resolvent for H. Then α has a minimal polynomial m α whose coefficients are in k(s 1,..., s n ). That is, the coefficients of m α are polynomials in k and the elementary symmetric functions. Thus if we evaluate the symmetric functions at r 1,..., r n we obtain a polynomial m k[x] (recall that s i (r 1,..., r n ) = a i k). Let us study the roots of m α. This is the minimal polynomial of α k(x 1,..., x n ) over the field k(s 1,..., s n ). Since α generates the fixed field of H, we have σ S n fixes α if and only if σ H. The roots of m α are precisely the Galois conjugates of α. Here s the point: Theorem 2.5. With notation as above, the polynomial m has a root in k if and only if the Galois group of f is contained in a conjugate of H in S n. Proof. Suppose the Galois group of f is contained in a conjugate σhσ 1 of H. Then the Galois group of f fixes σα(r 1,..., r n ), hence σα(r) k. Thus σα(r) is a k-rational root of m. Conversely if m has a root β k then β = σα(r 1,..., r n ) for some σ. Further, β is fixed by every element of the Galois group. But by design α is fixed only by H, so the Galois group is contained in σhσ 1. This theorem gives a wonderful tool to work with, but is not constructive. The next proposition shows that it is not difficult to construct polynomials with the property described in Theorem 2.5. However, it can be difficult to construct useful

COMPUTING GALOIS GROUPS WITH GENERIC RESOLVENT POLYNOMIALS 3 polynomials whose coefficients are easy to compute. Ultimately, we care about finding polynomials that satisfy the properties of the theorem and aren t particularly concerned with whether the are the minimal polynomials of a primitive element. Definition 2.6. Let f k[x] be an irreducible, separable polynomial. Let r 1,... r n be the roots of f in its splitting field. Let H be a subgroup of the symmetric group S n. Then a polynomial R H k[x] is called a resolvent polynomial for the group H if R H has a root in k if and only if the Galois group of f is contained in H. As we mentioned, coming up with useful resolvent polynomials can be difficult. It can, though, always be done. Proposition 2.7. Given a subgroup H S n, define θ H = σ H x 1 x 2 2 x n n. Let R H k(s)[x] be the k(s)-minimal polynomial of θ H. Let σ 1,..., σ r be representatives of the (right) cosets of H in S n. Then R H (t) = r (t σ i (θ H )). i=1 Further, if f k[x] is a degree n irreducible separable polynomial with roots r 1,..., r n then the specialization of R H to the r i is a resolvent polynomial for the group H. Proof. See [3]. Generically, R H has degree [S n : H] and involves computing some very long products. In Sections 4 and 5 we will show some more usable resolvent polynomials whose roots contain a small number of sums. 3. Discriminants Perhaps the most important invariant of any polynomial is its discriminant. In this section we will construct the discriminant and show that it is a resolvent of the alternating group. This will give us an easy way to detect when the Galois group of a polynomial is contained in A n. Definition 3.1. Let k be a field of characteristic not 2, and let f k[x] a separable polynomial. If the roots of f in the algebraic closure of k are r 1,..., r n, then let = (r i r j ). 1 i<j n The discriminant of f is defined to be D = 2. The discriminant polynomial of f is the polynomial x 2 D. Proposition 3.2. With notation as in the definition (and char k 2), (i) The polynomial x 2 D is in k[x] (ii) An element σ S n fixes if and only if σ S n.

4 JOHN KOPPER Proof. Any permutation of fixes it up to sign, thus every permutation fixes D = 2. Thus D is in the fixed field of the Galois group of f, i.e. D k. This gives (i). Further, if σ S n then σ = (r σ(i) r σ(j) ) = ( 1) σ (r i r j ). 1 i<j n Since ( 1) σ = 1 if and only if σ A n, (ii) follows. 1 i<j n The proposition shows that the discriminant is always a resolvent for the alternating group. We won t go into detail here, but the discriminant can be computed quite easily (both from a computer science standpoint and an emotional standpoint). In fact it can be easily computed from the determinant of a particular matrix. See [4, p. 621]. The discriminant is of particular strength when considering polynomials of degree 3. Indeed, the only transitive subgroups of S 3 are S 3 and A 3 = Z/3Z. We can therefore use the discriminant to compute the Galois group of a cubic. Proposition 3.3. Let k be a field with char k 2, and let f be an irreducible cubic polynomial in k[x]. Then f has Galois group S 3 if and only if its discriminant polynomial x 2 D is irreducible over k, if and only if D is not a square in k. Otherwise, the Galois group of f is A 3. Example 3.4. Let f(x) = x 3 + 3x + 1. Notice this is irreducible because it has no integer roots, hence no roots, hence cannot be factored. Its discriminant is D = 135 which is not a square, hence the Galois group of f is S n. Another nice use of the discriminant is in determining when a polynomial is not solvable by radicals. It is a standard result from field theory that a polynomial is solvable by radicals if and only if its Galois group is a solvable group. Since A n is simple for n 5 we have a criterion for non-solvability by radicals. Proposition 3.5. Suppose f Q[x] is a polynomial of degree n 5. Suppose further that the discriminant of f is a square in Q. Then f is not solvable by radicals. In Section 5 we will see a (much) better way of detecting solvability of quintics. 4. Quartics Recall that the Galois group of a degree d polynomial must be a transitive subgroup of the symmetric group S d. In this section we study quartic polynomials, and our first task is to enumerate the transitive subgroups of S 4. They are, up to isomorphism: Name Order Number of conjugates Contained in A 4 S 4 24 1 No A 4 12 1 Yes D 8 8 3 No Z/2Z Z/2Z 4 4 Yes Z/4Z 4 3 No In Section 3 we saw that the discriminant R A4 (x) = x 2 D is a resolvent polynomial for A 4 in S 4. The next transitive subgroup we ll consider is the dihedral group

COMPUTING GALOIS GROUPS WITH GENERIC RESOLVENT POLYNOMIALS 5 D 8 = (1234), (13). The resolvent polynomial should have order 24/8 = 3. It is not hard to guess that α(x 1, x 2, x 3, x 4 ) = x 1 x 3 + x 2 x 4 is a resolvent for D 8. The Galois conjugates of this are the other possible permutations under the action of S 4 : β = x 1 x 2 + x 3 x 4 γ = x 1 x 4 + x 2 x 3. Notice that β and γ correspond to the resolvents for the other copies of D 8 in S 4. It is not difficult to show (cf. [2, p. 5]) that if f(x) = x 4 + ax 3 + bx 2 + cx + d with roots r 1, r 2, r 3, r 4 then: R D8 (x) = (x (r 1 r 2 + r 3 r 4 ))(x (r 1 r 3 + r 2 r 4 ))(x (r 1 r 4 + r2r 3 )) = x 3 b 2 x + (ac 4d)x (a 2 d + c 2 4bd). Let s try our hand with a few examples. Example 4.1. Let f(x) = x 4 + 2x + 2 over the field Q. Certainly f is irreducible by Eisenstein s Criterion. We compute our resolvents: R A4 (x) = x 2 1616 R D8 (x) = x 3 8x 4. Since 1616 is not a square, we conclude that the Galois group of f is not contained in A n. Since R D8 has no rational root, the Galois group is not contained in D 8. This leaves us still with a number of possibilities. However, since R D8 is irreducible (it has no integer roots, hence no rational ones), its splitting field is a Galois subextension of the splitting field of f. Since R D8 has degree 3, we conclude that the Galois group of f is divisible by 3. Thus f must have Galois group S 4. Example 4.2. Let f(x) = x 4 +8x+12 over the field Q. We compute our resolvents: R A4 (x) = x 2 331776 = x 2 576 2 R D8 (x) = x 3 48x 64. Since 331776 is a square, we conclude that the Galois group of f is contained in A n. A similar analysis as above shows that the Galois group of f must be divisible by 3, hence must be A 4. Example 4.3. It is worth noting that the D 8 resolvent doesn t do everything for us. Let f(x) = x 4 6x 2 + 7 over the field Q. This is irreducible because f(x+1) = x 4 +4x 3 8x+2 is irreducible by Eisenstein s Criterion. The resolvents: R A4 (x) = x 2 7168 R D8 (x) = x 3 + 6x 2 28x 168. In this case, R A4 has no rational root but R D8 does. Thus the Galois group of f is contained in D 8. In fact, R D8 (x) = x 3 + 6x 2 28x 168 = (x + 6)(x 2 28). This clearly only has one root in Q. Recalling our construction of R D8, we see that all three generic roots α, β, γ are fixed by Z/2Z Z/2Z. Thus the Galois group of f is a subgroup of D 8 not equal to Z/2Z Z/2Z, hence is either D 8 or Z/4Z. We are now in need of more machinery. We could either define a resolvent for Z/4Z (hard), or use some more elementary machinery, e.g. [4, p. 615].

6 JOHN KOPPER We won t work out the entire calculation here, but it s interesting to see how quickly things get out of hand. Fortunately, we don t need a polynomial quite as bad as the one constructed in Proposition 2.7. For example, we may use θ Z/4Z = x 1 x 2 2 + x 2 x 2 3 + x 3 x 2 4 + x 4 x 2 1 ([3]) as a resolvent for the subgroup Z/4Z. Computing the corresponding polynomial in terms of symmetric functions is an appropriate use of Mathematica. Indeed if our quartic is given by f(x) = x 4 + ax 3 + bx 2 + cx + d then the constant term of the resolvent polynomial is: a 10 d 2 11a 8 bd 2 2a 8 c 2 d + a 7 b 2 cd + 15a 7 cd 2 + 45a 6 b 2 d 2 + 17a 6 bc 2 d + a 6 c 4 5a 6 d 3 12a 5 b 3 cd 133a 5 bcd 2 22a 5 c 3 d + a 4 b 5 d 76a 4 b 3 d 2 6a 4 b 2 c 2 d 12a 4 bc 4 + 32a 4 bd 3 + 128a 4 c 2 d 2 + 29a 3 b 4 cd + 2a 3 b 3 c 3 + 344a 3 b 2 cd 2 + 48a 3 bc 3 d + 16a 3 c 5 48a 3 cd 3 4a 2 b 6 d + 32a 2 b 4 d 2 134a 2 b 3 c 2 d + 36a 2 b 2 c 4 64a 2 b 2 d 3 648a 2 bc 2 d 2 48a 2 c 4 d + 16ab 5 cd 12ab 4 c 3 128ab 3 cd 2 + 296ab 2 c 3 d 96abc 5 + 256abcd 3 + 416ac 3 d 2 + b 6 c 2 28b 4 c 2 d + 16b 3 c 4 + 176b 2 c 2 d 2 224bc 4 d + 64c 6 320c 2 d 3. Needless to say, it is not very instructive to try to compute these things by hand. See the Appendix for details on this computation. Staduhar [3] provides a nice table of resolvents for a great many transitive subgroups of S n for n = 4, 5, 6, 7. Incidentally, the Galois group of x 4 6x 2 + 7 is D 8, though this can be seen by more direct methods. 5. Quintics Our Pavlovian response drives us to first list all transitive subgroups of S 5 : Name Order Contained in A 5 S 5 120 No A 5 60 Yes F 20 20 No D 10 10 Yes Z/5Z 5 Yes The group F 20 is sometimes referred to as the Frobenius group. This particular group has an important feature: Theorem 5.1. An irreducible polynomial f of degree 5 is solvable by radicals if and only if its Galois group is contained in the Frobenius group F 20. A consequence of Theorem 5.1 is a method of detecting solvable quintics: we need only produce a resolvent polynomial for the group F 20 and determine whether it has a root in k. Of course, our general construction doesn t offer a very useful resolvent polynomial (if you read Example 4.3 you don t need to be convinced of this). Dummit [1] has a remarkably simple resolvent for F 20. Let θ =x 2 1x 2 x 5 + x 2 1x 3 x 4 + x 2 2x 1 x 3 + x 2 2x 4 x 5 + x 2 3x 1 x 5 + x 2 3x 2 x 4 + x 2 4x 1 x 2 + x 2 4x 3 x 5 + x 2 5x 1 x 4 + x 2 5x 2 x 3,

COMPUTING GALOIS GROUPS WITH GENERIC RESOLVENT POLYNOMIALS 7 then the stabilizer of θ is evidently F 20. This resolvent has a much more usable minimal polynomial. In fact, if f(x) = x 5 + ax + b then the resolvent polynomial simplifies to R F20 (x) =x 6 + 8ax 5 + 40a 2 x 4 + 160a 3 x 3 + 400a 4 x 2 + (512a 5 3125b 4 )x + (256a 6 937ab 4 ). Appendix A. Doing These Things in Mathematica The techniques we ve described obviously generalize to allow one to detect solvable polynomials of arbitrary degree. Of course, the computational complexity increases quite quickly (what are the solvable subgroups of S 118?) The curious reader should read [6] to see solvable sextics. The adventurous reader should try the case n = 7. W conclude by providing the reader with a brief overview of one of many ways to carry out these computations in practice. Most of our computations, including those of Example 4.3, were performed in Mathematica. Here s a couple of useful commands: Permute[expr, perm] Applies a permutation to the given expression. SymmetricReduction[f, {x1, x2,...,xn}, {s1, s2,...,sn}] Writes the polynomial f in terms of symmetric functions in the variables x1, x2,..., xn (and labels the symmetric functions s1, s2,..., sn). TeXForm[expr] Puts the expression into a format your TeX compiler can parse. In order to apply permutations to a given polynomial, I used the following homebrewed command (thanks to StackExchange user Daniel Lichtblau [7]): groupelementaction[expr_, vars_, perm_] /; Length[perm] == Length[vars] && PermutationListQ[perm] := expr /. Thread[vars -> Permute[vars, perm]] References [1] D. S. Dummit. Solving Solvable Quintics. Mathematics of Computation. Vol. 57, No. 195 (Jul., 1991). [2] K. Conrad. Galois Groups of Cubics and Quartics (Not in Characteristic 2). [3] R. P. Staduhar. The Determination of Galois Groups. Mathematics of Computation. Vol. 27, No. 124 (Oct., 1973). [4] Dummit and Foote [5] L. Soicher and J. McKay. Computing Galois Groups Over the Rationals. Journal of Number Theory. Vol. 20, Issue 3 (Jun., 1985). [6] C. Boswell and M. L. Glasser. Solvable Sextic Equations. arxiv:math-ph/0504001, 2005. [7] http://mathematica.stackexchange.com/questions/79268/symmetric-group-action-onpolynomials