= Fixed field of inversion in a rational function field
{title2=$k(t)^{t\mapsto1/t}=k(t+1/t)$}
Let $t$ be transcendental over a field $k$ and let $\tau$ be the <field automorphism> sending $t$ to $1/t$. The rational function $u=t+1/t$ is fixed, and $t$ satisfies $X^2-uX+1=0$. Thus the extension over $k(u)$ has degree at most two. It is not one because the nontrivial automorphism $\tau$ fixes $k(u)$. The <fixed field> of $\tau$ also has index two by the fixed-field theorem, so it equals $k(u)$.
No nonconstant <polynomial> in $k[t]$ is fixed by inversion. For a polynomial of positive degree, substitution of $1/t$ creates negative powers, whereas the original has positive powers. Uniqueness of the coefficients of a <Laurent polynomial> rules out equality. Consequently this fixed field cannot have a generator that is a polynomial in the original variable, even though it has the simple rational-function generator above. The statement is valid in characteristic two as well: $t$ and $1/t$ remain distinct transcendental rational functions.
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