Generic Polynomials of Degree Three

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1 Generic Polynomials of Degree Three Benjamin C. Wallace April Introduction In the nineteenth century, the mathematician Évariste Galois discovered an elegant solution to the fundamental problem of determining the solvability by radicals of a given polynomial. To do this, Galois formulated the notion of a permutation group and established the connection between such groups and polynomial equations in what is now known as the fundamental theorem of Galois theory or the Galois correspondence. In particular, Galois showed that any given polynomial is associated to what is now known as a Galois group in such a way that the polynomial is solvable by radicals if and only if this group has a special structural property, aptly named solvability. In its modern incarnation, this correspondence is between the field extension generated by the roots of the polynomial and the Galois group. The importance of this correspondence stems from the fact that groups are relatively easy to understand and work with in comparison to fields. This is evidenced in the monumental classification of finite simple groups. Moreover, while a typical field is infinite, a typical Galois group is finite. In sum, Galois theory reduces the otherwise difficult or even insurmountable task of determining solvability by radicals to a straightforward computational procedure. Such a reduction is often the hallmark signifying the completion of a mathematical problem. However, beyond being a useful tool, Galois theory has stimulated the development of and has found applications in many modern mathematical disciplines. Such extensions arise naturally out of questions such as are there Galois groups for infinite field extensions?, can the ideas of Galois theory be applied to other types of equations?, and is the Galois correspondence an instantiation of a larger pattern? These have led, respectively, to the generalizations of Galois theory to the infinite and differential Galois theories and the discovery of Galois-like correspondences in algebraic topology. One particularly natural question to ask is which permutation groups are the Galois groups of a polynomial? The present paper is concerned with some of the developments related to this question. 1

2 2 The Inverse Galois Problem The inverse Galois problem asks, given a finite group G and a field K, whether there exists a Galois extension M/K such that Gal(M/K G. If such an extension does exist, then we call it a G-extension over K and say that G is realizable over K. In its most general form, this problem is still open. However, several important cases have been solved and we list some of these below. Every finite group is realizable over C(t (this follows from the Riemann existence theorem. The symmetric and alternating groups S n and A n are realizable over Q for each n. All finite solvable groups are realizable over Q (this was shown by Igor Shafarevich. Note that these statements merely assert the existence of field extensions that realize certain Galois groups. It would be more interesting, however, to explicitly find these fields and the polynomials that generate them. This is the constructive aspect of the inverse Galois problem. In its most general form, this strengthened problem asks for all the polynomials with a given group as Galois group. Since there are likely infinitely many such polynomials, they must be given in some kind of general algebraic form. Hence, the problem arises of determining the circumstances under which such a general form exists and when it does finding it. In order to solve this problem, we must be precise about what we mean by a general algebraic form for the polynomials that generate a given Galois group. Our treatment of this subject follows [1]. Generic Polynomials Definition. Let P (t, X K(t[X] be a monic polynomial, where t = (t 1,..., t n and X are indeterminates, and let M be its splitting field over K(t. If, for some finite group G, 1. M/K(t is a G-extension and 2. every G-extension over K is the splitting field of a specialization P (a, X K[X] of P (t, X for some a = (a 1,..., a n K n, then we call P (t, X a parametric polynomial for G-extensions (or say that P (t, X parametrizes G-extensions over K. If P (t, X K(t[X] is parametric for G-extensions over any field containing K, then we call P (t, X a generic polynomial for G-extensions (or say that P (t, X is generic for G-extensions over K. While the notion of a parametric polynomial essentially captures the idea of the general algebraic form we mentioned earlier, it must be refined to that of a generic polynomial as we do not want Gal(P (t, X/K(t to depend on any particular properties of the ground field K(t 2

3 that do not carry over to its extensions. Though it is certainly not obvious, it can be shown that the definition of a generic polynomial is not in vain, as there exist parametric polynomials that are not generic, e.g. that realize Z/8Z as a Galois group over Q. Notice also that while we are really interested in Galois groups of extensions over fields K, this definition of a parametric polynomial requires that it realize G as a Galois group over K(t. This is justified in the following important theorem. Theorem.1 (Hilbert s irreducibility theorem. If P (t, X K(t[X] is irreducible, then there exist infinitely many a K n such that P (a, X K[X] is well-defined and irreducible over K and such that Gal(P (t, X/K(t Gal(P (a, X/K. 4 Generic Polynomials of Degree Three In all that follows, let K be a field of characteristic not equal to 2. Recall that for such fields, if f(x K[X] is a degree polynomial that is separable and irreducible over K, then Gal(f/K S, { A, if Disc(f is a square in K otherwise. In the following two theorems, we exhibit generic polynomials for both S - and A -extensions. In doing so, we make use of the fact above, as our attention is restricted to fields of characteristic not equal to 2. Note, however, that this simplification is not essential; that is, the following theorems can also be proved for fields of characteristic 2. Theorem 4.1. The polynomial f(t, X = X + tx + t K(t[X] is generic for S. Proof. First we compute 1 0 t t t t Disc(f = 0 t 0 0 = 4t 27t 2, 0 0 t t which is not a square in K(t. Moreover, f(t, X is easily seen to be irreducible over K(t (by the Eisenstein criterion with irreducible element t K[t] and separable since f (t, X = X 2 + t 0 K(t. Therefore, Gal(f/K(t S as required. Now let L/K be an S -extension, so that L is the splitting field of an irreducible, separable, third-degree polynomial which we can take to have the form g(x = X + ax + b K[X] (otherwise, perform the shift X X a. Note that by the assumption of irreducibility, b 0.

4 If, moreover, a 0, then L is the splitting field of a ( ( b b g a X = a b b a X + bx + b = X + a b 2 X + a ( b 2 a = f b 2, X. Otherwise, g(x = X + b, so L = K(α, ω, where α = b and ω = 1. But then L is the splitting field of (X (α + α 2 (X (αω + α 2 ω 2 (X (αω 2 + α 2 ω. Expanding this gives X bx + (b 2 b. By the assumption of irreducibility of g(x, b 1. Moreover, b 0 so the case a = 0 reduces to the case of a 0 discussed above. So L must be the splitting field of a specialization of f(t, X, i.e. f(t, X is parametric for S. That f(t, X is generic is immediate, as K was taken to be any arbitrary field of characteristic not equal to 2. For the next theorem, we need the following simple proposition. Proposition 4.2. If M/K is a Z/nZ-extension and if for some vector space V over K there is a homomorphism ρ : Z/nZ GL K (V, then V can be embedded in M in a way that respects the group action, i.e. there is an embedding π : V M such that σπ(v = π(ρ(σv for σ Z/nZ, v V. See the following diagram. σ Z/nZ ρ GL K (V ρ(σ π(v M π V v K Also, we note that the rational root test applies to any UFD. Proposition 4.. Let n 1 and let f(x = a n X n + + a 0 R[X], where R is a UFD with field of fractions K. If r/s K is a root of f(x with gcd(r, s = 1, then r a 0 and s a n. 4

5 The proof follows exactly as in the case R = Z; this generalization is possible because Gauss s lemma holds for general UFDs. Theorem 4.4. The polynomial f(t, X = X tx 2 + (t X + 1 K(t[X] is generic for A. Proof. By the (generalized rational root test, ±1 are the only possible roots of f(t, X in K(t. However, it is easily verified that neither of these are roots. But f(t, X is a degree polynomial, so it is irreducible in K(t. Moreover, f (t, X = X 2 2tX + (t 0 K(t, so f(t, X is separable. Since 1 t t t t 1 Disc(f = 2t t 0 0 = t 4 6t + 27t 2 54t + 81 = (t 2 t t t t t is a square in K(t, we can conclude that Gal(f/K(t A. Now as before, let L/K be an A -extension. [ Since] A Z/Z, we can write A = σ, where 0 1 σ id. Now it can be seen that the map σ induces a homomorphism A GL 2 (V 1 1 since [ ] [ ] B = = B So by Proposition 4.2, there are x, y L such that [ ] [ ] [ ] [ ] σ(x 0 1 x y = =. σ(y 1 1 y x y Letting z = x y as required. L, we see that L is the splitting field of ( (X z X 1 ( ( X 1 1 z 1 z ( z = X + z 1 z z 2 ( z = X + z 1 z z 2 ( z + z 1 = f z z 2, X X 2 + X 2 +, ( z + z 2 1 z z 2 ( z + z 1 z z 2 X + 1 X + 1 The two theorems above mark the full solution of the inverse Galois problem with regards to S and A. The generic polynomials they provide describe exactly how these groups arise as Galois groups over any given field. 5

6 But why stop there? Notice that we have used only one parameter t in each generic polynomial presented; only one parameter is necessary for these relatively simple groups. In studying generic polynomials for other groups, it may appear hard even impossible to make do with a single parameter. One may therefore seek generic polynomials that are in this sense minimal. Definition. Let K be a field and G a group. The generic dimension for G over K is the minimal number of parameters in a generic polynomial for G over K, or if no such polynomial exists. The problem of finding the generic dimension of a group is highly non-trivial and is a subject of current research in inverse Galois theory. 5 Conclusion The generic polynomials for S and A demonstrated above mark the final word on the inverse Galois problem for these two groups. It can be seen already that there is a bit of a jump in the complexity of both the polynomial and the proof that it is generic in moving from S to A. Similarly, much more sophisticated methods are required in dealing with larger and more complicated groups and lead to generic polynomials that are themselves more complex. In fact, as we have mentioned, this complexity can be quantified by the notion of generic dimension, which itself opens new doors and problems to be solved. Such issues as generic polynomials for other groups and results pertaining to generic dimension may be found in [1]. References [1] Christian U. Jensen, Arne Ledet, and Noriko Yui. Generic Polynomials: Constructive Aspects of the Inverse Galois Problem. Mathematical Sciences Research Institute Publications. Cambridge University Press, [2] Jocelyn Tufts. Generic polynomials of small degree. March

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