Birational morphism of the affine line

Question

Let $k$ be an algebraically closed field. Suppose $K$ has characteristic nonzero, can we characterize the birational morphism of $A^1$ the way we characterize isomorphisms(all linear maps)? I know it will not be as specific, but do we have any general ideas of what they are? For instance, is the Frobenius morphism birational? Thanks in advance.

Answer 1

One way you can solve this is by recognizing that a morphism is a birational equivalence iff it is dominant and the induced extension of function fields is an isomorphism. Here, our morphism is given by $t\mapsto f(t)$ for some polynomial $f\in k[t]$, and this is dominant iff $f$ is nonconstant. The map on function fields is then $k(t) \to k(t)$ by $t\mapsto f(t)$.

To determine the degree of this field extension and whether it is an isomorphism, here is an elementary abstract algebra exercise: if $\varphi:k(t)\to k(t)$ is a homomorphism of fields, then it is of the form $t\mapsto \frac{f(t)}{g(t)}$ for polynomials $f,g$, and the degree of the field extension is $\max(\deg f,\max \deg g)$. In our case, $g=1$, so the degree of the field extension is $\deg f$, so our map is a birational map iff $f$ is a degree one polynomial in $t$.