Fixed subfield of the field of rational functions 
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 $L \subset K(Y)$.
 A: 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.
A: If $\operatorname{char}(K)\neq 2$, then $X$ satisfies a polynomial of degree $2$ over $K(Y)$ (namely, $p(t)=(2t-1)^2-Y$), so $[K(X):K(Y)]\leq 2$.  Since $K(Y)\subseteq L\subset K(X)$, it follows that $K(Y)=L$.
If $\operatorname{char}(K)=2$, on the other hand, this doesn't work (the polynomial $p(t)$ used above is identically $0$).  And in fact it is clear that $K(Y)\neq L$, since $Y=1$ so $K(Y)$ is just $K$ in this case.  So you need to find a different $Y$.  As a hint, you can notice that since $\phi\circ\phi$ is the identity, $f\phi(f)\in L$ for any $f\in K(X)$.
