Fixed field of $G=\operatorname{Gal}(k(x)/k)$ in $k$ 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 $k(x)^G$?
 A: 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! ) 
