What are the one-parameter subgroups of GL?

Solution 1:

Assume that $k$ is algebraically closed and let $\lambda : k^\times \rightarrow \mathrm{GL}_n(k)$ be a morphism of algebraic groups. Let $R = \{x \in k : x^m = 1 \text{ for some $m \in \mathbb{N}$ not divisible by char $k$}\}$.

If $x \in R$ satisfies $x^m = 1$ then $\lambda(x)^m = I_n$ and so $\lambda(x)$ is diagonalizable. An infinite family of commuting linear maps on a finite-dimensional vector space may be simultaneously diagonalized, so there is a basis for $k^n$ in which all the maps $\lambda(x)$ for $x \in R$ are diagonal. Since $R$ is Zariski dense in $k^\times$, it follows that all $\lambda(x)$ are diagonal in this basis.

Finally if $\lambda(x) = \mathrm{diag}(\lambda_1(x),\ldots,\lambda_n(x))$ then each $\lambda_i$ is a morphism of algebraic groups $k^\times \rightarrow k^\times$ and so is of the form $x \mapsto x^r$ for some $r \in \mathbb{Z}$. This shows that every one-parameter subgroup is of the form you expected.