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)?



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


