Transcendency Degree One Function Fields Over a Finite Field with Many Automorphisms

Abstract

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 transcendency 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 a result of Stichtenoth, $K$ has genus $g(K)=(qm-1)(q-1)$, $p$-rank (Hasse-Witt invariant) $γ(K)=(q-1)^2$ and a $\mathbb{K}$-automorphism group of order at least $2q^2m(q-1)$. In this paper we prove that this subgroup is the full $\mathbb{K}$-automorphism group of $K$; more precisely $Aut_{\mathbb {K}}(K)=Q\rtimes D$ where $Q$ is an elementary abelian $p$-group of order $q^2$ and $D$ has a index $2$ cyclic subgroup of order $m(q-1)$. In particular, $\sqrt{m}|Aut_{\mathbb{K}}(K)|> g(K)^{3/2}$, and if $K$ is ordinary (i.e. $g(K)=γ(K)$) then $|Aut_{\mathbb{K}}(K)|>g^{3/2}$. On the other hand, if $G$ is a solvable subgroup of the $\mathbb{K}$-automorphism group of an ordinary, transcendency degree $1$ function field $L$ of genus $g(L)\geq 2$ defined over $\mathbb{K}$, then by a result due to Korchmáros and Montanucci, $|Aut_{\mathbb{K}}(K)|\le 34 (g(L)+1)^{3/2}<68\sqrt{2}g(L)^{3/2}$. This shows that $K$ hits this bound up to the constant $68\sqrt{2}$. Since $Aut_{\mathbb{K}}(K)$ 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 $Aut_{\mathbb{K}}(K)$ is large enough. This possibility is worked out for subgroups of $Q$.