Abstract

Let K be the algebraic closure of a finite field Fq of odd characteristic p. For a positive integer m prime to p, let F=K(x,y) be the transcendence degree 1 function field defined by yq+y=xm+x−m. Let t=xm(q−1) and H=K(t). The extension F|H is a non-Galois extension. Let K be the Galois closure of F with respect to H. By Stichtenoth [20], K has genus g(K)=(qm−1)(q−1), p-rank (Hasse–Witt invariant) γ(K)=(q−1)2 and a K-automorphism group of order at least 2q2m(q−1). In this paper we prove that this subgroup is the full K-automorphism group of K; more precisely AutK(K)=Δ⋊D where Δ is an elementary abelian p-group of order q2 and D has an index 2 cyclic subgroup of order m(q−1). In particular, m|AutK(K)|>g(K)3/2, and if K is ordinary (i.e. g(K)=γ(K)) then |AutK(K)|>g3/2. On the other hand, if G is a solvable subgroup of the K-automorphism group of an ordinary, transcendence degree 1 function field L of genus g(L)≥2 defined over K, then |AutK(K)|≤34(g(L)+1)3/2<682g(L)3/2; see [15]. This shows that K hits this bound up to the constant 682.Since AutK(K) has several subgroups, the fixed subfield FN of such a subgroup N may happen to have many automorphisms provided that the normalizer of N in AutK(K) is large enough. This possibility is worked out for subgroups of Δ.

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