TITLE: Fixed subfield of the field of rational functions
QUESTION [5 upvotes]: Let $K(X)$ be the field of rational functions of $X$ over some field $K$. Let $\phi: K(X) \rightarrow K(X)$ be the $K$-morphism such that $\phi (X)=1-X$. We have $L:=\{ f\in K(X) : \phi (f)=f\}$. Find some element $Y \in K(X)$ such that $K(Y)=L$.

I am fairly certain that $Y=(2X-1)^{2}$ and can easily show that $K(Y) \subset L$. I am having some trouble with showing that $L \subset K(Y)$.

REPLY [5 votes]: Here’s another method. You know that your transformation $\phi$ is of order two, and that the “conjugate” of $X$ is $1-X$. The minimal polynomial for $X$ over the fixed field is accordingly $f(T)=T^2-T+(X(1-X))$. Here I’ve used the sum of the conjugates for the linear coefficient (with the necessary change of sign) and the product of the conjugates for the constant term. So it would seem that the fixed field is $K(X(1-X))$. Indeed, if we call $X(1-X)=\xi$ temporarily, we certainly know that $K(\xi)$ is a good field, and that $X$ is a root of $T^2-T+\xi$, with the property that it generates $K(X)$. Notice that this proof works just as well in characteristic two.