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Fixed field of automorphisms of $k(x)$, with $k$ a field, induced by $I(x)=x$, $\varphi_1(x) = \frac{1}{1-x}$, $\varphi_2 (x)=\frac{x-1}{x}$?

Since $I(x)=x$, $\varphi_1(x)=\frac{1}{1-x}$, $\varphi_2 (x)=\frac{x-1}{x}$ form a group of order 3 the group is cyclic, so it is generated by $\varphi_1$ then I have to find the fixed field of $\varphi_1$.

If $a(x)=\frac{f(x)}{g(x)} \in k(x)$ with $(f,g)=1$ and $\varphi_1$ fix to $a(x)$ then $a(x)=a(\frac{1}{1-x}) \Rightarrow \frac{f(x)}{g(x)} = \frac{f(\frac{1}{1-x})}{g(\frac{1}{1-x})} \Rightarrow f(x) \mid f(\frac{1}{1-x)})$ and by th same reason $ f(\frac{1}{1-x}) \mid f(x)$ so $f(\frac{1}{1-x})=f(x)$ so $\varphi_1$ fix to $a(x)$, $f(x)$ then $\varphi_1$ fix to $g(x)$. So $\varphi_1$ fix to $\frac{f(x)}{g(x)}$.

Someone can tell me if it is correct.

user26857
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Oscarblablabla
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1 Answers1

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Hint:

  • The sum $$t=x+\frac1{1-x}+\frac{x-1}x=\frac{x^3-3x+1}{x^2-x}$$ is stable under the action of your group of transformations.
  • Can you show that $[k(x):k(t)]=3$?
Jyrki Lahtonen
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  • yes I did it. So you are telling me that the fixed field is the polynomials of degree three??? – Oscarblablabla May 25 '14 at 06:34
  • No. The fixed field is $$k(t)={\frac{a(t)}{b(t)}\mid a,b\in k[t]}.$$ A finite group of $n$ field automorphisms has such a fixed field that the extension degree is also $n$. Here the fixed field is isomorphic to $k(x)$. Actually a Theorem of Lüroth says that all the intermediate fields properly between $k$ and $k(x)$ are isomorphic (but not equal) to $k(x)$. – Jyrki Lahtonen May 25 '14 at 06:43
  • @JyrkiLahtonen Thank you for your answer. I was working on a problem and stumbled across this useful answer. Where can I find a proof of 'A finite group of $n$ field automorphisms has such a fixed field that the extension degree is also $n$'? I guess we can generalize this by saying than $n$ is the order and apply it to infinite case as well. – Spock Nov 24 '14 at 22:24
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    @Spock the proof of this is found in the Fundamental Theorem of Galois Theory, which is not trivial enough to explain in a comment. It says that the degree of a (certain type of) field extension equals the order of the automorphism group that fixes the base field. But you probably know this by now, if you've continued doing math the past six years :p – Jos van Nieuwman May 30 '20 at 22:19
  • @JyrkiLahtonen thanks for providing the stable element. I assume the next step is to find an $f ∈ k\left(x + \frac{1}{1-x} + \frac{x-1}{x}\right)[y] = k(t)[y]$ that has $x$ as a root, in order to conclude that $[k(x):k(t)] ≤ 3$ with equality if we can show it to be irreducible. I have not succeeded yet, can you help? – Jos van Nieuwman May 31 '20 at 15:32
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    @JosvanNieuwman $x$ is a root of $$f(y)=(y^3-3y+1)-(y^2-y)t\in k(t)[y].$$ – Jyrki Lahtonen May 31 '20 at 15:59
  • Ah yes, I should have looked at the other representation you gave of it... Makes it trivial. Irreducibility proves less so. Is that a necessity though? Or is there a quick argument that shows the degree cannot be 2? – Jos van Nieuwman May 31 '20 at 17:54
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    @JosvanNieuwman I might argue as follows. By Galois theory (the result you described a few coomments up) we know that the field of invariants, call it $L$, satisfies $[k(x):L]=3$. We also know by construction that $t\in L$, hence that $k(t)\subseteq L$. So $[k(x):k(t)]\ge3$ is immediate. The cubic $f(y)$ implies that $[k(x):k(t)]\le3$. Therefore we must have equality. And we also get $L=k(t)$ as a by-product :-) – Jyrki Lahtonen May 31 '20 at 18:01