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Let $k$ be a field, and let $K = k(x)$ be the rational function field in one variable over $k$. Let $\sigma$ and $\tau$ be the automorphisms of $K$ defined by $\sigma(f(x)/g(x)) = f(1/x)/g(1/x)$ and $\tau(f(x)/g(x)) = f(1-x)/g(1-x)$, respectively. Determine the fixed field $F$ of $\{\sigma,\tau\}$, and determine $\mathrm{Gal}(K/F)$. Find an $h \in F$ so that $F = k(h)$.

I tried to equate $\sigma(f), \tau(f)$ and $f$ to come up with some conditions, but it didn't really help. So, what is the goto methodology / thought process when solving these kind of problems? Where do I start?

Edit : From the conversation I had in the comments, it seems that $F$ might be the collection of $f/g$ where $f$ and $g$ are polynomials of the same degree with symmetric coefficients.

Sntn
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    Note that $\sigma (x^n)=x^{-n}$. What does this tell us about the fixed field? – CyclotomicField Feb 03 '20 at 07:09
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    @CyclotomicField That it doesn't have non-constant polynomials? – Sntn Feb 03 '20 at 07:13
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    That's correct but I would just call it the constants rather than not having the non-constant polynomials. Can you find the inverse of this automorphism? – CyclotomicField Feb 03 '20 at 07:17
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    @CyclotomicField It's the inverse of itself. Both are. – Sntn Feb 03 '20 at 07:18
  • @CyclotomicField But there might be non-constant rational functions, no? – Sntn Feb 03 '20 at 07:19
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    Yes but the fixed field doesn't contain any of those as you mentioned before. – CyclotomicField Feb 03 '20 at 07:21
  • @CyclotomicField It doesn't contain non-constant polynomials. The same cannot be said about rational functions in general. For example, $x^2+1 \over x^2+x+1$ is fixed by $\sigma$. – Sntn Feb 03 '20 at 07:25
  • We can say that $f$ and $g$ must have same degree though. – Sntn Feb 03 '20 at 07:26
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    Might it help to observe that $k(x+\frac{1}{x})$ is fixed by $\sigma$? Also, when I look at $\sigma, \tau$ I am reminded of cross-ratios and the six values one gets by permuting the four points. – ancient mathematician Feb 03 '20 at 07:41
  • @ancientmathematician In general, it works for $f$ and $g$ of same degree where both have symmetric coefficients. – Sntn Feb 03 '20 at 08:17
  • Actually, that might just be it. (Check edit) – Sntn Feb 03 '20 at 08:30
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    I think the way I wrote it is better, it gives the generator of the fixed field of $\sigma$ [Surely?]. Likewise for $\tau$ isn't the fixed field $k( (x-\frac{1}{2})^2)$? I think that $\langle\sigma,\tau\rangle$ is $S_3$ and that it is easy by trial and error to work out the six images of $x$ - as I said it's just like doing the cross-ratios calculation. – ancient mathematician Feb 03 '20 at 12:40
  • @JyrkiLahtonen I think yes. Thanks. – Sntn Feb 04 '20 at 05:04
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    And even before that the questions was studied here. – Jyrki Lahtonen Feb 04 '20 at 05:05

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