Transcendence degree one function fields over a finite field with many automorphisms

Let $\mathbb{K}$ be the algebraic closure of a finite field ${\mathbb{F}}_q$ of odd characteristic p. For a positive integer m prime to p, let $F=\mathbb{K}(x,y)$ be the transcendence degree 1 function field defined by $y^q+y=x^m+x^{?m}$. Let $t=x^{m(q?1)}$ and $H=\mathbb{K}(t)$. 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 $\mathfrak{g}(K)=(qm?1)(q?1)$, p-rank (Hasse–Witt invariant) $\gamma (K)={(q?1)}^2$ and a $\mathbb{K}$-automorphism group of order at least $2q^{2}m(q?1)$. In this paper we prove that this subgroup is the full $\mathbb{K}$-automorphism group of K; more precisely ${\text{Aut}}_{\mathbb{K}}(K)=\mathrm{\Delta }?D$ where Δ is an elementary abelian p-group of order $q^2$ and D has an index 2 cyclic subgroup of order $m(q?1)$. In particular, $\sqrt{m}|{\text{Aut}}_{\mathbb{K}}(K)|>\mathfrak{g}{(K)}^{3/2}$, and if K is ordinary (i.e. $\mathfrak{g}(K)=\gamma (K)$) then $|{\text{Aut}}_{\mathbb{K}}(K)|>{\mathfrak{g}}^{3/2}$. On the other hand, if G is a solvable subgroup of the $\mathbb{K}$-automorphism group of an ordinary, transcendence degree 1 function field L of genus $\mathfrak{g}(L)\ge 2$ defined over $\mathbb{K}$, then $|{\text{Aut}}_{\mathbb{K}}(K)|\le 34{(\mathfrak{g}(L)+1)}^{3/2}<68\sqrt{2}\mathfrak{g}{(L)}^{3/2}$; see [15]. This shows that K hits this bound up to the constant $68\sqrt{2}$.Since ${\text{Aut}}_{\mathbb{K}}(K)$ has several subgroups, the fixed subfield $F^N$ of such a subgroup N may happen to have many automorphisms provided that the normalizer of N in ${\text{Aut}}_{\mathbb{K}}(K)$ is large enough. This possibility is worked out for subgroups of Δ.