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Let $K$ be a field oh characteristic $p$. Let's take $\sigma \in \operatorname{Aut}(K(x),K)$ where $x$ is trascendental over $K$, where $\sigma(x)=x+1$. Find a primitive element of the fixed field of $ \left\langle {\sigma} \right\rangle $.

I have no idea how to attack this problem. Maybe one step it's to note that $ \left\langle {\sigma} \right\rangle $ is finite, in fact has order p (the characteristic of the field). I was trying with particular cases to note something general, but I could not find even one element fixed by the automorphism...

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    this is more general than that case, in this case $\mathbb{K}$ need not be finite.2012-10-01

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Clearly $x^p-x$ is a fixed element.

  1. $[K(x):K(x^p-x)]\le p$. Indeed, write $L=K(x^p-x)$. Then $x$ is root of the polynomial $T^p-T-(x^p-x)\in L[T]$. So $x$ is algebraic over $L$ and $K(x)=L[x]$ has degree at most $p$ over $L$.

  2. By the fundamental theorem of Galois, $[K(x):K(x)^G]=|G|=p$ where $G$ is the group generated by $\sigma$.

  3. As $L\subseteq K(x)^G \subseteq K(x)$, we get $L=K(x)^G$.

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    @Belgi: I rewrite and simplified the proof. The index of $L$ in $K(x)$ is the degree of $K(x)$ over $L$ (even if the meaning is different if they were considered as groups, I think there should no be ambiguity here. Neverthless, I removed it).2012-10-04