# Why is $\operatorname{Hom}_{\textrm{AG}}(k^\times,k^\times)\cong\mathbb{Z}$?

Suppose $k$ is an algebraically closed field, and $k^\times$ its multiplicative group. I read that $\operatorname{Hom}_{\textrm{AG}}(k^\times,k^\times)\cong\mathbb{Z}$, where the left consists of homomorphisms of affine algebraic groups.

The reason from this is the only algebraic group homomorphisms $k^\times\to k^\times$ are of form $x\mapsto x^m$ for $m\in\mathbb{Z}$.

However, I thought morphisms of algebraic groups also have to be morphisms of affine varieties, so that if $\phi$ is a morphism of affine algebraic groups, $\phi(x)$ has coordinates which are polynomials in the coordinates of $x$. Since there's only one coordinate in this case, doesn't that mean the only possibilities are $x\mapsto x^m$ for $m\geq 0$?

$k^{\times}$ is an affine variety, but it's not an affine subvariety of $k$. Instead it's an affine subvariety of $k^2$, under the embedding
$$k^{\times} \ni x \mapsto (x, x^{-1}) \in k^2.$$
It's the fact that this second coordinate is $x^{-1}$ that allows you to take $m \in \mathbb{Z}$. More abstractly, the coordinate ring of $k^{\times}$ is $k[x, x^{-1}]$.
• Oh right, this is a special case that $GL_n(k)$ is an affine subvariety of $k^{n^2+1}$. Thanks. – Adelaide Dokras Oct 4 '15 at 1:29