EVEN MONODROMY GROUPS OF POLYNOMIALS JIAOWEN YANG AND MICHAEL E. ZIEVE Abstract. For each field K, we determine all positive integers n for which there exists a degree-n polynomial f(X) in K[X] such that, for t transcendental over K, the Galois group of f(X) − t over K(t) is An. We likewise determine all n for which some degree-n polynomial f(X) 2 K[X] satisfies Gal(f(X) − t; K(t)) =∼ Sn and Gal(f(X) − t; K(t)) =∼ An. 1. Introduction Let f(X) 2 K[X] be a degree-n polynomial over a field K, and suppose that the derivative f 0(X) is nonzero. Let t be transcendental over K, and let K denote an algebraic closure of K. The Galois groups of the polynomial f(X)−t over the fields K(t) and K(t) are called the arithmetic and geometric monodromy groups of f(X), and denoted A and G. Monodromy groups play a key role in the study of value sets of polynomials over finite fields [4, 13, 15, 19, 23, 26], non-uniqueness of functional decomposition of polynomials [17, 24, 28], numbers of terms in compositions of polynomials [12], reducibility of polynomials in one or two variables [5,6,7, 10], Diophantine equations with infinitely many solutions [3], refinements of Hilbert's irreducibility theorem [8, 21, 22], and other topics. For many questions about polynomials, it suffices to analyze the case of indecomposable polynomials, namely polynomials which cannot be writ- ten as the composition of lower-degree polynomials. There has been much progress on determining the possibilities for G in this case. In particular, if char(K) = 0 then M¨uller[18] (building on work of Feit) showed that any indecomposable polynomial in K[X] of degree n > 31 has G being Cn, Dn, An or Sn. Moreover, if G is Cn or Dn then one can determine all possibilities for f(X)[27, 28], after which it is straightforward to determine the possibil- ities for A. In case G = An, of course A must be either An or Sn, but it is not clear what are the situations in which each possibility can occur. Fried [9] showed that if n is a non-square then there exist degree-n polynomials The second author thanks the NSF for support under grant DMS-1162181. 1 2 JIAOWEN YANG AND MICHAEL E. ZIEVE over Q with (G; A) = (An;Sn). He also made several attempts to prove the analogous result for square n. Subsequently M¨uller[20] showed that no such polynomials exist if n is a square. More generally, for any degree-n polynomial over a field of characteristic zero, if the geometric monodromy group is contained in An then M¨ullerdetermined exactly when the arith- metic monodromy group is contained in An. Our first result generalizes M¨uller'stheorem to fields of arbitrary characteristic. Theorem 1.1. Let K be a field of characteristic p ≥ 0, and let f(X) 2 K[X] be a polynomial of degree n > 0 such that f 0(X) 6= 0, where if p = 2 then n is odd. Let eXn and dXr be the leading terms of f(X) and f 0(X), respectively. Suppose that Gal(f(X) − t; K(t)) is contained in An. Then r is even. Moreover, Gal(f(X)−t; K(t)) is contained in An if and only if one of the following holds: • p 6= 2 and (−1)(n2−n)=2(d=e)n is a square in K • p = 2 and either n ≡ ±1 (mod 8) or K ⊇ F4. We note that if p - n then the conditions in the conclusion of the above result depend only on n and K, since in that case r = n − 1 and d=e = n. In case p j n and p is odd, it is perhaps surprising that the conditions depend only on n, K, and the ratio between the leading coefficients of f 0(X) and f(X). The analogous assertion is not true when p = 2, since for instance 8 5 8 6 5 the polynomials X + X and X + X + X over F2 have identical deriva- tives, and both have G = A6, but the first polynomial has A = S6 while the second has A = A6. Our proof yields a version of Theorem 1.1 when p = 2 and n is even, but the formulation is much less elegant than that in The- orem 1.1, and in particular the conclusion depends on a number obtained from a complicated computation involving all coefficients of f(X). We do not know whether there is a version of Theorem 1.1 in this case in which the conclusion only depends on K and some quantity which can be determined at a glance from f(X). Theorem 1.1 was proved by M¨uller[20] when p = 0, and our proof for odd p is based on his arguments. The case p = 2 is much more difficult, and requires different methods. As a consequence of the above result, we determine all fields K and all positive integers n for which there exists a degree-n polynomial in K[X] which has nonzero derivative and has (G; A) = (An;An), and likewise for (G; A) = (An;Sn). EVEN MONODROMY GROUPS OF POLYNOMIALS 3 Theorem 1.2. Let K be a field of characteristic p ≥ 0, and let n > 2 be an integer. There exists a degree-n polynomial in K[X] having monodromy groups (G; A) = (An;An) if and only if one of the following holds: (n−1)=2 (1) p 6= 2, n is odd, p - n, and (−1) n is a square in K (2) p 6= 2, n is even, p j n, and (−1)n=2 is a square in K (3) p = 2, n is odd, and either n ≡ ±1 (mod 8) or K ⊇ F4 (4) p j n and n is odd (5) p = 2 and n is even with n > 4 (6) p = 2, n = 4 and K ⊇ F4. There exists a degree-n polynomial f(X) 2 K[X] for which the monodromy groups (G; A) equal (An;Sn) if and only if K has a separable degree-2 ex- tension and one of the following holds: (n−1)=2 (1) p 6= 2, n is odd, p - n, and (−1) n is a nonsquare in K (2) p 6= 2, n is even, p j n, and (−1)n=2 is a nonsquare in K (3) p = 2, n ≡ ±3 (mod 8), and K 6⊇ F4 (4) p j n and n is odd (5) p = 2 and n is even with n > 4 (6) p = 2, n = 4 and K 6⊇ F4. Moreover, for any such n and K, if Ω denotes the splitting field of f(X) − t over K(t), then the algebraic closure of K in Ω can be an arbitrary separable degree-2 extension of K in cases (4) and (5), whereas in cases (3) and (6) it p (n−1)=2 can only be K:F4, and in cases (1) and (2) it can only be K( (−1) n) and K(p(−1)n=2), respectively. Our interest in these questions arose from the desire to determine the limit points of the set of ratios jf(Fq)j=q, where q varies over all prime powers and f(X) varies over all polynomials in Fq[X] having a prescribed degree n. By a result of Birch and Swinnerton-Dyer [4], these limit points could be determined if one knew all possibilities for the pair of groups (G; A). Guralnick and Saxl [14] have given a manageable list of possibilities for the group G in case f(X) 2 Fq[X] is indecomposable. Theorem 1.2 may be viewed as a first step towards determining the possibilities for A for each group G on the Guralnick{Saxl list. We will return to the question of value sets of indecomposable polynomials in a subsequent paper. 2. Discriminant computation In this section we recall a basic property of discriminants in our setting. 4 JIAOWEN YANG AND MICHAEL E. ZIEVE Let K be a field of characteristic p ≥ 0, and let f(X) 2 K[X] be a monic polynomial of degree n such that f 0(X) 6= 0. Let t be transcendental over K, and note that f^(X) := f(X) − t is squarefree, since it plainly 0 has no roots in K while K contains every root of f^ (X). Let x1; : : : ; xn ^ ^ Qn be the roots of f(X) in K, so that f(X) = i=1(X − xi). Note that 0 ^0 Pn Q 0 Q f (X) = f (X) = i=1 j6=i(X −xj), so that f (xi) = j6=i(xi −xj). Write 0 Qr ∗ f (X) = d k=1(X − yk), where d 2 K and yk 2 K. The discriminant of f^(X) is defined to be Y 2 ∆ := (xi − xj) : i<j We compute n Y 2 (n2−n)=2 Y Y ∆ = (xi − xj) = (−1) (xi − xj) i<j i=1 1≤j≤n j6=i n (n2−n)=2 Y 0 = (−1) f (xi) i=1 n r (n2−n)=2 Y Y = (−1) d (xi − yk) i=1 k=1 r n (n2−n)=2+nr n Y Y = (−1) d (yk − xi) k=1 i=1 r (n2−n)=2+nr n Y = (−1) d (f(yk) − t) k=1 r (n2−n)=2+r(n+1) n Y = (−1) d (t − f(yk)): k=1 Since the yk are in K, it follows that ∆ is a polynomial in K[t] of degree r with leading coefficient (−1)(n2−n)=2+r(n+1)dn.