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) ∈
K[X] satisfies Gal(f(X) − t, K(t)) =∼ Sn and Gal(f(X) − t, K(t)) =∼ An.
1. Introduction Let f(X) ∈ 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) ∈ 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 | 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 | 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 | 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) ∈ 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 | n, and (−1)n/2 is a nonsquare in K (3) p = 2, n ≡ ±3 (mod 8), and K 6⊇ F4 (4) p | 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 |f(Fq)|/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) ∈ 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) ∈ 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 fˆ(X) := f(X) − t is squarefree, since it plainly 0 has no roots in K while K contains every root of fˆ (X). Let x1, . . . , xn ˆ ˆ Qn be the roots of f(X) in K, so that f(X) = i=1(X − xi). Note that 0 ˆ0 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 ∈ K and yk ∈ K. The discriminant of fˆ(X) is defined to be Y 2 ∆ := (xi − xj) . i 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. 3. Fields of characteristic not 2 In this section we prove Theorem 1.1 in case char(K) 6= 2. Our proof is based on the argument used by M¨uller[20] to prove the result when char(K) = 0. Let K be a field of characteristic p 6= 2, and let f(X) ∈ K[X] be a monic polynomial of degree n such that f 0(X) 6= 0. Suppose that the Galois group of f(X) − t over K(t)) is contained in An, so that (since f(X) − t is squarefree) the discriminant ∆ of f(X) − t is a square in K(t)[16, Cor. EVEN MONODROMY GROUPS OF POLYNOMIALS 5 of Thm. 4.13]. Writing dXr for the leading term of f 0(X), it follows from Section2 that ∆ is a polynomial in K[t] of degree r with leading coefficient (−1)(n2−n)/2+r(n+1)dn. Since ∆ is a square in K(t), its degree r must be even. Moreover, since ∆ is an element of K[t] which is a square in K(t), it is also a square in K[t]. Letting c be a square root of its leading coefficient, it follows that ∆ has a unique square root in K(c)[t] with leading term ctr, which can be obtained by successively solving for the coefficients of tr−1, tr−2, . . . , t0 in this square root. Therefore ∆ is a square in K(t) if and only if (−1)(n2−n)/2+r(n+1)dn is a square in K, or equivalently (−1)(n2−n)/2dn is a square in K. This proves the result, since ∆ is a square in K(t) if and only if the Galois group of f(X) − t over K(t) is contained in An. 4. Fields of characteristic 2 In this section we prove Theorem 1.1 in case char(K) = 2. We begin by recalling Berlekamp’s characteristic 2 analogue of the discriminant [2]. Let L be a field of characteristic 2, and let g(X) ∈ L[X] be a monic polynomial of degree n > 0 which has n distinct roots x1, x2, . . . , xn in L. Then the quantity X xixj C(g) := x2 + x2 i n−1 value of c when K 6⊇ F4. Here c is the coefficient of t in N, so also c n−1 n−1 ˆ is the coefficient ofa ˆn in N(g). Recall that N(g) = caˆn + N0 where ˆ N0 is a polynomial in F2[ˆa1,..., aˆn] in which every term is divisible bya ˆi for some i < n. Next consider the homomorphism ψ : F2[ˆx1,... xˆn] → K(t) i which maps xi 7→ ζ s for some fixed n-th root of unity ζ ∈ K and some s n Qn i n satisfying s = t. Since i=1(X + ζ s) = X + t, it follows that ψ(ˆan) = t n−1 and ψ(ˆai) = 0 for i < n, so that ψ(N(g)) = ct . Since the Galois group of Xn + t over K(t) is a cyclic group of order n which is generated by an n-cycle, in particular this group is contained in An. Thus we conclude that n if K 6⊇ F4 then c = 0 if and only if the Galois group of X + t over K(t) is contained in An. Next, if s is a root of Xn + t in K(t), then all roots of this polynomial are ζis where 0 ≤ i < n and ζ ∈ K is a fixed primitive n-th root of unity. Thus the splitting field of Xn + t over K(t) is L := K(ζ, s), and Gal(L/K(t)) is generated by Gal(L/K(ζ, t)) and Gal(L/K(s)). We have seen that Gal(L/K(ζ, t)) is contained in An, so it remains to determine whether Gal(L/K(s)) is contained in An. This group is isomorphic as a permutation group to the group Gal(K(ζ)/K) in its action on the set of n-th roots of unity. Moreover, the latter group is isomorphic to Gal(F2(ζ)/F2(ζ)∩ K) in its action on n-th roots of unity. Writing F2(ζ) ∩ K = F2m , it follows that Gal(F2(ζ)/F2m ) is generated by the Frobenius automorphism, which acts on the set of n-th roots of unity as 2m-th powering. The sign of this permutation is the same as the sign of the permutation of Z/nZ induced by multiplication by 2m. By (Frobenius’s generalization of) Zolotarev’s lemma 2m [11, 25], this permutation is even if and only if the Jacobi symbol n equals 2m 1. This Jacobi symbol equals n , so if K 6⊇ F4 then m is odd and thus 2 the Jacobi symbol equals n , which equals 1 if and only if n ≡ ±1 (mod 8). n We conclude that the Galois group of X + t over K(t) is contained in An if and only if either K ⊇ F4 or n ≡ ±1 (mod 8). Therefore if K 6⊇ F4 then n ≡ ±1 (mod 8) if and only if c = 0, which we have shown is equivalent to the Galois group of f(X) + t being contained in An. Finally, although we assumed that f(0) = 0, this assumption may be removed by noting that the Galois groups of f(X) + t and f(X) + b + t over K(t) are isomorphic for every b ∈ K, since the automorphism t 7→ t + b of K(t) extends to an isomorphism between the splitting fields of these two polynomials, which induces the desired isomorphism between the Galois groups. This concludes the proof of Theorem 1.1. 8 JIAOWEN YANG AND MICHAEL E. ZIEVE 5. Construction of polynomials with G = An In this section we prove Theorem 1.2. The “only if” portion of the re- sult follows from Theorem 1.1, so it suffices to exhibit polynomials having each prescribed pair of monodromy groups and satisfying each corresponding condition on p, n and K. Throughout this proof, if f(X) is a polynomial in K[X] having nonzero derivative, then we consider the extension K(x)/K(t), where x is tran- scendental over K and t := f(x). This extension is separable because f 0(X) 6= 0. Let Ω be the Galois closure of this extension, and write Ωˆ := Ω.K. Then the geometric monodromy group G of f(X) is Gal(Ωˆ/K(t)) in its action on the set of conjugates of K(x) over K(t). For any a ∈ K, write Qs ei f(X)−a = c i=1(X −bi) where c is the leading coefficient of f(X), the bi are distinct elements of K, and the ei are positive integers. Then the inertia group of any place of Ωˆ which contains t − a has orbits of lengths e1, . . . , er. If none of the ei are divisible by p, then the inertia group is generated by an element having cycle lengths e1, . . . , er. We call the multiset [e1, . . . , er] the cycle type of this element, and we use exponents to denote multiplicities: thus, for instance, [yi, zj] denotes the multiset consisting of i copies of y and j copies of z. We also note that if p - n then the inertia group at any place of Ωˆ which contains 1/t is generated by an n-cycle. Further, if p - n then G is generated by the inertia groups over finite places of K(t), since the group H generated by these inertia groups has the property that Ωˆ H /K(t) is only ramified over ∞, and is tamely ramified, so that by Riemann–Hurwitz this extension has degree 1 and thus H = G. Next, we recall that the inertia group I at any place has a normal Sylow p-subgroup V , and the quotient group I/V is cyclic; when p = 0 we define V to be 1. We write Kˆf for the algebraic closure of K in Ω. Since Ω/K(t) is Galois, it follows that Kˆf /K is separable. Recall that a subgroup of Sn is primitive if it does not preserve any partitions of {1, 2, . . . , n} whose number of parts lies strictly between 1 and n. It is well-known that if f(X) ∈ K[X] has nonzero derivative, then the geometric monodromy group G of f(X) is primitive if and only if f(X) is indecomposable over K, in the sense that f(X) cannot be written as the composition of polynomials of degree less than deg(f). We will use ramification arguments to prove that the polynomials we construct in this section are indecomposable over K. This implies that G is primitive, which EVEN MONODROMY GROUPS OF POLYNOMIALS 9 often forces G to contain An in light of the following classical group-theoretic theorems. • (Marggraf) A primitive subgroup of Sn which contains a k-cycle for some k with 1 < k < n must be (n − k + 1)-transitive. • (Jordan) A primitive subgroup of Sn which contains a k-cycle for some prime k with k < n − 2 must contain An. • (Jordan) A primitive subgroup of Sn which contains an element with 2 n−2 cycle type [2 , 1 ] must contain An if n > 8. When convenient, we combine these theorems with consequences of the classification of finite simple group, specifically with the classification of 3-transitive subgroups of Sn. Case 1: p 6= 2 and p - n. In this case, if f(X) ∈ K[X] has degree ∼ n and G = An, then Theorem 1.1 implies that n is odd and that Kˆf = K(p(−1)(n−1)/2n). Therefore it suffices to exhibit a degree-n polynomial ∼ f(X) ∈ K[X] having G = An, for each p 6= 2 and each odd n > 1 with p - n. For n = 3 we can take f(X) = X3. For any odd n > 3 for which p - n(n − 2)(n − 4), consider f(X) = Xn−4(X2/n + 2X/(n − 2) + 1/(n − 4))2, for which f 0(X) = Xn−5(X2/n + 2X/(n − 2) + 1/(n − 4))(X + 1)2. Since f(−1) 6= 0 and X2/n + 2X/(n − 2) + 1/(n − 4) has two distinct roots, it follows that the extension K(x)/K(t) is only ramified over ∞, 0 and f(−1), and that the inertia group at any places of Ωˆ lying over 0 or f(−1) is generated by an element having cycle type [n − 4, 22] or [3, 1n−3], respectively. Note that p 6= 3, since otherwise p - n(n − 2)(n − 4). Therefore K(x)/K(t) is tamely ramified. Since G is generated by the inertia groups over 0 and f(−1), it follows that G ≤ An. It remains to show that G ≥ An. First we show that f is indecomposable over K: for, if f = g ◦h with g, h ∈ K[X] and deg(h) ≥ 2, then since f(−1) has a unique ramified f-preimage, and this preimage has prime ramification index, it follows that g is not ramified over f(−1). Hence g can only ramify over ∞ and 0, and since g is tamely ramified it follows by Riemann–Hurwitz that each of these points has a unique g-preimage. But then every f-preimage of 0 has ramification index in f being divisible by deg(g), so deg(g) divides the greatest common divisor of the f-ramification indices of the points in f −1(0), whence deg(g) = 1. Thus f is indecomposable over K, so G is a primitive subgroup of Sn which contains a 3-cycle, whence G ≥ An. It remains to exhibit examples when both p and n are odd and p | (n − 2)(n−4). We first give a construction which covers all cases except n = p+2. 10 JIAOWEN YANG AND MICHAEL E. ZIEVE In fact this construction works more generally when p and n are odd and n n−p n−p p n > p + 2 and p - n. Consider f(X) = X + X = X (X + 1) . The inertia group I at any place of Ωˆ lying over 0 has orbits of lengths n − p and p. Hence the stabilizer in I of any point in the orbit of length n − p must be a subgroup of I having index n − p, and hence must contain the unique Sylow p-subgroup V of I. Therefore V embeds into the group of permutations of the points in the orbit of length p, so |V | ≤ p. Since I has an orbit of length p, we know that p divides |I|, so finally we have |V | = p, and moreover V is generated by a p-cycle. Now we show that f is indecomposable over K: for, if f = g ◦ h with g, h ∈ K[X] and deg(h) ≥ 2, then since f −1(0) consists of two points with coprime ramification indices in f, it cannot be that both these points are totally ramified in h, so they must have the same h-image which therefore is unramified in g, whence 0 has a unique g-preimage and this preimage is unramified, so deg(g) = 1. Thus f is indecomposable over K, so G is a primitive subgroup of Sn which contains a p-cycle. Since p < n − 2, a theorem of Jordan’s implies that G ≥ An. But our formula from Section2 shows that the discriminant of f(X) − t is (n−1)/2 n n−1 (−1) n t , which is a square in K(t), so that G ≤ An. Therefore G = An for this polynomial. Finally, consider the case that p is odd and n = p + 2. If p < 7 n p then f(X) = X + X has G = An. The case p ≥ 7 is covered by the following construction, which works more generally when p is odd, n ≡ p + 2 (mod 2p), and n > 7. Consider f(X) = Xn−4(X2 + 2X − 1)2, for which f 0(X) = Xn−5(X2 + 2X − 1)(2X2 + 2). Note that if a2 = −1 then (a2 + 2a − 1)2 = (−2 + 2a)2 = −8a, so f(a) = −8an−3, whence f(a) = f(−a) 6= 0. Therefore the branch points of f are ∞, 0 and f(a), with corresponding inertia groups generated by elements of cycle type [n], 2 2 n−4 [n − 4, 2 ], and [2 , 1 ]. Since all of these are contained in An, therefore G ≤ An. Next we show that f is indecomposable over K. For, suppose f = g ◦ h where g, h ∈ K[X] and deg(h) ≥ 2. Since n is odd, both g and h have odd degree. Therefore g cannot ramify over f(a), due to the ram- ification type of f over f(a). Since g is tamely ramified, and has at most two branch points, in fact g must be totally ramified over both ∞ and 0. Therefore deg(g) divides the ramification index in f of each point in f −1(0), whence deg(g) | gcd(n−4, 2) = 1, so f is indecomposable over K. Therefore G is a primitive subgroup of Sn which contains an element of cycle type EVEN MONODROMY GROUPS OF POLYNOMIALS 11 2 n−4 [2 , 1 ], so since n > 7 a result of Jordan’s implies that G ≥ An, whence G = An. Case 2: p 6= 2 and p | n. First assume n is even. If f(X) ∈ K[X] has ∼ p n/2 degree n and G = An, then Theorem 1.1 implies that Kˆf = K( (−1) ). Therefore it suffices to exhibit a degree-n polynomial f(X) ∈ K[X] having ∼ G = An, for each odd prime p and each even n > 1 with p | n. Let ` be the smallest prime with ` > n/2. By Bertrand’s postulate, we have ` < n, and if n > 6 then ` < n − 2. It follows that gcd(`, n) = 1. Consider f(X) = Xn−`(X − 1)`, for which f 0(X) = −`Xn−`−1(X − 1)`−1. The only finite branch point of f is 0, for which a corresponding inertia group is generated by an element having cycle type [n−`, `]. Since gcd(n−`, `) = gcd(n, `) = 1, the (n − `)-th power of this element is an `-cycle. Next we show that f is indecomposable over K. Suppose that f = g ◦ h where g, h ∈ K[X] with deg(h) > 1. If h(0) 6= h(1) then both 0 and 1 are totally ramified under h, which is impossible since a polynomial of degree at least 2 has at most one finite totally ramified point. Therefore h(0) = h(1), so h(0) must be totally ramified in g, whence the degree of g divides the ramification indices in f of both 0 and 1, so that deg(g) | gcd(n − `, `) = 1. Thus f is indecomposable over K, so G is a primitive subgroup of Sn which contains an `-cycle, which implies by Jordan’s theorem that G ≥ An if n > ` + 2, which occurs unless n = 6. Finally, if n = 6 then p = 3, and one can check directly that the Galois group of f(X) − t over F3(t) is S6, so the Galois group of f(X) − t over F3(t) is a transitive normal subgroup of S6 and hence contains A6. Since this latter Galois group is isomorphic to G, it follows that G ≥ A6. Returning to an arbitrary even n which is divisible by an arbitrary odd prime p, the calculation in Section2 implies that the discriminant of f(X) − t is n/2 n n−2 (−1) (−`) t , which is a square in K(t), so that G ≤ An and thus G = An. Next assume n is odd. If n = 3 then f(X) = X3 − cX with c ∈ K∗ 3 √ has G = A3, and f(X) − t has discriminant c , so that Kˆf = K( c) can be an arbitrary extension of K having degree at most 2. If n = 5 then f(X) = X5 + 2cX3 with c ∈ K∗ has G = A , and f(X) − t has discriminant √ 5 5 2 b t , so that Kˆf = K( b) can be an arbitrary extension of K having degree at most 2. Now assume n ≥ 7, and put f(X) = Xn − cXn−4 with c ∈ K∗, so that f 0(X) = 4cXn−5. In this case an inertia group at a place over 0 is generated by an element with cycle type [n − 4, 14]. Suppose that f = g ◦ h where g, h ∈ K[X] with deg(h) > 1. Since n is odd, both g and h have 12 JIAOWEN YANG AND MICHAEL E. ZIEVE odd degrees. It follows that at least one unramified root of f(X) must have h-image being h(0), since otherwise there is no way to arrange the h-images of the four unramified roots of h(X). Therefore g is unramified at h(0), and moreover g−1(0) has at most one other point, and if this point exists then it is unramified, so since deg(g) is odd we conclude that deg(g) = 1. Therefore f is indecomposable over K, so G is a primitive subgroup of Sn which contains an (n − 4)-cycle. Marggraff’s theorem implies that G is 5- transitive, and since n is odd this implies that G ≥ An. The discriminant of (n−1)/2 n n−5 f(X) − t is (−1) (4c) t , which is a square in K(t), so that G ≤ An p (n−1)/2 and thus G = An. Moreover, Kˆf = K( (−1) c) can be an arbitrary extension of K having degree at most 2. Case 3: p = 2 and n is odd. In this case, if f(X) ∈ K[X] has degree n ∼ and G = An, then Theorem 1.1 implies that Kˆf equals K if n ≡ ±1 (mod 8), and equals K.F4 if n ≡ ±3 (mod 8). Therefore it suffices to exhibit a degree- ∼ n polynomial f(X) ∈ K[X] having G = An, for each odd n > 1. If n = 3 3 ∼ 7 5 5 then f(X) = X has G = A3. If n = 7 then f(X) = X + X + X has ∼ n n−4 n−4 4 G = A7. If n > 3 then put f(X) = X + X = X (X + 1) . By [1, 2.28] we know that G ≤ An. The inertia group I at a place over 0 has orbits of lengths 4 and n − 4. Thus the stabilizer in I of any point in the orbit of length n − 4 is a subgroup of I having index n − 4, and hence contains the unique Sylow 2-subgroup V of I. Therefore V acts faithfully on the points in the four-element orbit of I, so V ≤ A4, whence |V | ≤ 4. Since I has an orbit of length 4, we know that |I| ≡ 0 (mod 4), whence |V | = 4. Thus G contains an element having cycle type [22, 1n−4]. Suppose that f = g ◦ h where g, h ∈ K[X] with deg(h) > 1. If f(0) 6= f(1) then both 0 and 1 are totally ramified in h, which is impossible. Thus f(0) = f(1), and since the ramification indices of 0 and 1 in f are coprime, we see that f(0) is unramified in g, and also that f(0) is the unique point in g−1(0). Therefore deg(g) = 1, so that f is indecomposable over K, and thus G is a primitive 2 n−4 subgroup of An which contains an element have cycle type [2 , 1 ], so by Jordan’s theorem G = An if n > 7. Finally, if n = 5 then G is a transitive subgroup of A5 which contains an order-4 subgroup, so G = A5. Case 4: p = 2 and n ≡ 0 (mod 4). First consider n = 4. Then 4 3 ∼ f(X) = X + X has G = A4. I claim that if n = 4 and G = A4 then ˆ Kf = K.F4. For, firstly if a degree-4 polynomial with nonzero derivative has no degree-3 term, then G is a non-cyclic group of order 4. Thus if ∼ a degree-4 polynomial f(X) with nonzero derivative has G = A4, then EVEN MONODROMY GROUPS OF POLYNOMIALS 13 f(X) has a degree-3 term. The monodromy groups of f do not change when we multiply f by a constant, so we may assume that f is monic. We compute the Berlekamp discriminant of f(X) + t to be 1 + a(t)/b(t)2, where ∼ a, b ∈ K[t] with deg(b) = 1 and deg(a) ≤ 1. Since G = A4, we know that a(t)/b(t)2 = c(t)2 + c(t) for some c(t) ∈ K(t), and plainly c(t) = d/b(t) for some d ∈ K. Then d2 +db(t) = a(t), so d is the ratio between the coefficients of t in a(t) and b(t), and hence d ∈ K. Therefore the elements u(t) ∈ K(t) for which u(t)2 + u(t) = 1 + a(t)/b(t)2 are w + d/b(t) where w is a primitive cube root of unity, so the extension of K(t) generated by such elements is ˆ K.F4(t), whence K = K.F4. Next consider n ≡ 0 (mod 4) with n > 4. Put f(X) = Xn+Xn−2+cXn−3 with c ∈ K∗, so that f 0(X) = cXn−4. Then f −1(0) consists of four points, of which three are unramified and one has ramification index n − 3. Let g, h ∈ K[X] satisfy f = g◦h and deg(h) > 1. If 0 is totally ramified in h, then deg(h) divides the ramification index of 0 in f, namely n−3; but also deg(h) divides deg(f) = n, so deg(h) = 3. In this case h(X) = `(X3) for some degree-one `(X) ∈ K[X], so that f(X) ∈ K[X3], contradiction. Therefore some nonzero root α of f(X) satisfies h(0) = h(α). Since α is unramified in f, it follows that h(0) is unramified in g, so that the ramification index of 0 in f equals the ramification index of 0 in h. Thus n − 3 ≤ deg(h) − 1 ≤ n/ deg(g)−1, so that deg(g) ≤ n/(n−2) < 2, whence deg(g) = 1. Therefore f(X) is indecomposable over K, so that G is primitive. Moreover, the inertia group at a place over 0 is generated by an (n−3)-cycle, so Marggraf’s theorem implies that G is 4-transitive. Since n is even, it follows that G ≥ An unless n ∈ {12, 24} and G = Mn; but one can check that Mn does not contain an (n−3)-cycle for these values of n, so G ≥ An. I claim that the Berlekamp discriminant of f(X) + t is 1/c2 if n ≡ 0 (mod 8), and 1 + 1/c2 if n ≡ 4 (mod 8). I have a proof of this, but maybe a simpler proof can be found via Cohen’s or Abhyankar’s methods. In any case, it follows that G ≤ An, so that G = An. Moreover, if L/K is any separable extension of degree at most 2, then L = K(d) where e := d2 + d ∈ K, so that d = d2 + e ∈ K(d2) 2 2 2 2 2 and thus L = K(d ) where (d ) + d = e is a square in K. If K 6= F2 then 2 we can choose e to be nonzero, by replacing e by e+i +i for any i ∈ K \F2. ˆ Therefore if K 6= F2 then for suitable c we can make Kf be any separable extension of K having degree at most 2. Moreover, if K = F2 then c = 1, ˆ ˆ and Kf is F4 if n ≡ 0 (mod 8), while Kf is F2 if n ≡ 4 (mod 8). Finally, suppose K = F2 and n > 4 with n ≡ 0 (mod 4). Then let ` be the smallest 14 JIAOWEN YANG AND MICHAEL E. ZIEVE prime such that ` > n/2. By Bertrand’s postulate, it follows that ` < n − 2. Now consider f(X) = Xn + X`. Since the ramification index of 0 under f is prime and greater than n/2, it cannot be written as the product of two positive integers both of which are at most n/2, so f(X) is indecomposable over K, whence G is primitive. Since G contains an `-cycle with ` < n − 2 and ` prime, Jordan’s theorem implies that G ≥ An. Now, [1, 2.27] implies that if n ≡ 0 (mod 8) then A ≤ An and thus A = An; and if n ≡ 4 (mod 8) then G ≤ An but A 6≤ An, so that G = An and A = Sn. 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M¨uller, On Ritt’s polynomial decomposition theorems, arXiv:0807.3578 [math.AG].1 Jiaowen Yang, Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China Michael Zieve, Department of Mathematics, University of Michigan, Ann Arbor, MI 48109–1043, USA Mathematical Sciences Center, Tsinghua University, Beijing 100084, China E-mail address: [email protected] URL: www.math.lsa.umich.edu/∼zieve/