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Let $F$ be a field of characteristic $p>0$. For $a\in F$, I have to compute the fixed field of the $F$-automorphism of the field of rational functions $F(x)$ such that $x\to x+a$.

Bob Gardner
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    "Add $a$" is not a field isomorphism, unless $a = 0$. Am I missing something? – Arthur Nov 23 '19 at 08:17
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    What Arthur said. My educated guess is that you really want to find the fixed field of an $F$-automorphism of the field of rational functions $F(x)$. Then, $x$ being transcendental, there exists an $F$-homomorphism $\sigma:F(x)\to F(x)$ such that $\sigma(x)=x+a$. As $F(x+a)=F(x)$, $\sigma$ is an automorphism. Because we are in characteristic $p$, $\sigma^p=id_{F(x)}$, and the group of automorphisms it generates is finite. Please check, whether that is what the question really was, and edit it accordingly. – Jyrki Lahtonen Nov 24 '19 at 05:58
  • Assuming I correctly guessed what the question is you may benefit from a study of this old answer of mine, where in the late parts the case $a=1$ is handled. That argument is about the fixed field of a dihedral group of automorphisms of order $2p$ as opposed to a cyclic group of order $p$ here. Anyway, it is kinda well known that the fixed field of $x\mapsto x+1$ is $F(x^p-x)$. Sorry about failing to come up with perfect link for that. – Jyrki Lahtonen Nov 24 '19 at 06:09
  • Also, welcome to the site! This is a wonderful resource. To avoid negative reactions to overly terse questions, please take a look at our guide for new askers. – Jyrki Lahtonen Nov 24 '19 at 06:11

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