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Let $k$ be a field and $k(x)$ be rational field of k and $G=\operatorname{Gal}(k(x)/k)$

$$k(x)^G=\{a \in k(x): \sigma(a)=a\text{ for all }\sigma \in G \}$$

How can we calculate $k(x)^G$?

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What have you tried? What elements do you know $G$ has, and what are their fixed fields? –  Qiaochu Yuan Feb 14 '12 at 18:57
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Well, I know that elements of $G$ take $x$ to $(ax+b)/(cx+d)$ where $ad-bc \neq 0$ –  Mohan Feb 14 '12 at 19:02
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Okay. Can you pick a relatively simple element of that form and compute its fixed field? –  Qiaochu Yuan Feb 14 '12 at 19:05
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1 Answer

up vote 3 down vote accepted

1) Although the fixed field is relatively easy to compute in the case of an infinite field $k$ (and I'll leave that to you), it is much more difficult to calculate it in the case that $k=\mathbb F_q$, a field with $q$ elements .
The completely unintuitive result, developed in Lang's Algebra exercise 36 of Chapter VI, is that $$k(x)^G=k(\phi(x)) \quad \text {where} \quad\phi(x)=\frac {(x^{q^2}-x)^{q+1}}{(x^q-x)^{q^2+1}}$$
2) We have a Galois extension $k(x)/k(x)^G$, with Galois group $G$.
It is interesting to check directly that $G$ has order $(q+1)q(q-1)$ by counting the automorphisms $x \mapsto \frac{ax+b}{cx+d}$.
And even more interesting to confront that calculation to the following result:
given a rational function $\phi(x)=\frac{P(x)}{Q(x)}$ , with $P$ and $Q$ relatively prime polynomials in $k[x]$, the degree of the extension $k(x)/k(\phi(x))$ is $deg \;\phi(x)=max(deg P, deg Q))$.
In our case the degree of $\frac {(x^{q^2}-x)^{q+1}}{(x^q-x)^{q^2+1}}$ is also $(q+1)q(q-1)$, as it should.
(Beware of the insidious trap that the numerator and denominator as written are not relatively prime! )

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