# Order of ring automorphisms of localizations of polynomial rings over finite fields

Suppose that $F$ is a finite field and $S\subset F[t]$ is a (finite) set of primes. Is is true that any ring automorphism of $R:=F[t][S^{-1}]$ has finite order?

A ring automorphism of $R$ is uniquely determined by an automorphism of $F$ ($F$ is the set of ring elements which are algebraic over the prime ring, so $F$ has to be mapped to itself) and the image of $t$.

Say I want to pick $id_F$ and some image $x\in R$. Under which conditions does this choice really give an automorphism (and not only a homomorphism)?

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Every such automorphism is contained in the automorphism group of the field or rational functions $F(t)$ over $F$, which equals $\mathrm{PGL}_2(F)$, and so is a finite group.
[Edit:] upon further reflection, my answer is incomplete, because $\mathrm{PGL}_2(F)$ is the group of isomorphisms of $F(t)$ as an $F$-algebra, not as a ring. Call $G$ the group of automorphisms of $F(t)$ as a ring. Since $F$ is the algebraic closure of the prime field $\mathbb F_p$ in $F(t)$, every element of $G$ induces an automorphism of $F$; this induces a homomorphism of $G$ onto the automorphism group of $F$, which is finite, with kernel $\mathrm{PGL}_2(F)$. Both groups are finite, so $G$ is finite.
This is not clear to me. An automorphism of $R$ extends to an endomorphism of $F(t)$ but why is this extension necessarily an automorphism? – Qiaochu Yuan Jul 31 '12 at 14:51
The extension is functorial, since $F(t)$ is the field of fractions of $R$. So the extension of the inverse of the automorphism of $R$ will be the inverse of the extension. – Konstantin Ardakov Jul 31 '12 at 15:08