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Assume char $K = 0$ and let $G$ be the subgroup of $Aut_KK(x)$ that is generated by the automorphism induced by $x\rightarrow x + 1_K$. Then G is an infinite cyclic group. Determine the fixed field $E$ of $G$. What is $[K(x) : E]?$

My attempt is the following:
We have $g\in G$ if and only if $g\in Aut_KK(x)$ and for all $a(x)\in K(X)-K$
$g(a(x))=f(a(x+1))$ where $f\in Aut_KK(X)$.
Hence $a(x)\in E$ if and only if $f(a(x+1))=a(x)$ for all $f\in Aut_KK(x)$.

I really have no idea how to proceed with this question any help will be greatly appreciated.

TheGeometer
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  • If $a(x)$ is fixed under $\sigma: x\mapsto x+1$ this means that $a(x)=a(x+1_K)$. Can you prove by induction that we must then have $a(x+n\cdot1_K)=a(x)$ for all natural numbers $n$? Can you then show that $a(x)$ must be a constant (i.e. an element of $K$)? – Jyrki Lahtonen Feb 28 '18 at 06:18
  • In other words, introducing those $f$s and $g$s muddied your waters. Is it clear why $\sigma(a(x))=a(x+1)$? It is probably best that you think that thru before you follow the plan outlined in my first comment. – Jyrki Lahtonen Feb 28 '18 at 06:21
  • Oh, and if you get stuck in the end game, take a peek at the 3rd paragraph of an old answer of mine. There may be cleaner ways (because that answer was to a slightly more complicated question), but you absolutely need the assumption char $K = 0$ for the claim to hold. A few subtleties along the way. Sorting it all out may take a bit of effort, but is ever so enjoyable. Do ask, if you want more help at some steps. This is somewhat different from automorphisms of those subfields of $\Bbb{C}$, but you'll get the hang of it soon enough! – Jyrki Lahtonen Feb 28 '18 at 06:26

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