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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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1 Answer 1

up vote 7 down vote accepted

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.

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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
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An automorphism of a commutative ring induces an isomorphism of its quotient field. Isn't this clear? –  Angelo Jul 31 '12 at 15:07
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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
    
Aha. Thanks for the explanations. (I would like to note that the functoriality of taking fraction fields is not entirely obvious. It is false that the fraction field is left adjoint to the inclusion of fields into the category of integral domains and ring homomorphisms; it is necessary to consider the category of integral domains and injective ring homomorphisms.) –  Qiaochu Yuan Aug 1 '12 at 14:12

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