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...

  • $\begingroup$ Is $\sigma$ supposed to be in the automorphisms of $K(x)$ over $K$, $\text{Aut}(K(x)/K)$? If not, would you explain the notation? If so, then certainly at least any element of $K$ is fixed by $\sigma.$ $\endgroup$ Sep 30, 2012 at 16:42
  • $\begingroup$ I edited the statement,is the fixed field of the cyclic group generated by $\sigma$. Thanks $\endgroup$
    – Daniel
    Sep 30, 2012 at 19:47
  • $\begingroup$ math.stackexchange.com/questions/109344/… $\endgroup$ Sep 30, 2012 at 19:55
  • $\begingroup$ this is more general than that case, in this case $\mathbb{K}$ need not be finite. $\endgroup$
    – Daniel
    Oct 1, 2012 at 2:36

1 Answer 1


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 $L=K(x)^G$.

  • $\begingroup$ Can you please add some more details ? I don't understand a few things: 1) why the degree is $p$ ? 2)how did you deduce that the fixed field by sigma have index p ? (do you mean degree ? because index is thw word for groups, not degree that is for field extensions) 3)How do you get the last line ? (but maybe this would be more clear after I understand 1&2) $\endgroup$
    – Belgi
    Oct 4, 2012 at 0:38
  • $\begingroup$ @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). $\endgroup$
    – user18119
    Oct 4, 2012 at 0:58

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