Finding the fixed field of an automorphism.

Question

Suppose $K$ is a field. Let $K(x)$ be the field of rational functions in the variable $x$ over $K$. Let $G$ be the subgroup of the automorphism group of $K(x)$ over $K$ generated by $\sigma$, where $\sigma(x)=1-\dfrac{1}{x}$. Find a specific $u \in K(x)$ such that the fixed field of $G$ is $K(u)$, and find the minimal polynomial of $x$ over $K(u)$.

I’m not sure where to start on the problem. I’m looking for hints and not a full solution. Thanks!

I think your question is a duplicate of this. Because I have the dupehammer for the tag [tag:abstract-algebra] and I answered that one, I refrain from casting a vote. — Jyrki Lahtonen, Jun 01 '20 at 16:39
Before you look at that, do follow Angina Seng's excellent suggestion. Can you calculate the order of $\sigma$? — Jyrki Lahtonen, Jun 01 '20 at 16:41

score 0 · Answer 1 · answered Jun 01 '20 at 16:28

0

$F=K(x)^G$. What is the $F$-minimal polynomial of $x$ ? Prove that $F=E$, where $E$ is the field containing $K$ and the coefficients of this polynomial, by comparing $[K(x):F]$ with $[K(x):E]$.

answered Jun 01 '20 at 16:28

reuns

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Finding the fixed field of an automorphism.

1 Answers1