Automorphisms of $\mathbb{G}_m$.

Question

Let $\mathbb{G}_m$ be the multiplication group whose underlying set is $k^*$, where $k$ is a field. How to show that as an algebraic group there are only two automorphisms of $\mathbb{G}_m$? How many automorphisms are there of $\mathbb{G}_m$? I think that if $\phi: \mathbb{G}_m \to \mathbb{G}_m$ is an automorphism, then the induced map $\phi^*: k[t, t^{-1}] \to k[t, t^{-1}]$ is also an automorphism. Thank you very much.

Answer 1

The character lattice of $\mathbb G_m$ is the ring of endomorphisms of $\mathbb G_m$, and it isomorphic to $\mathbb Z$; the element $n \in \mathbb Z$ corresponds to the character $t \mapsto t^n$.

The only units in $\mathbb Z$ are $\pm 1$, corresponding to the identity, and to the involution $t \mapsto t^{-1}$, so these are the only automorphisms of $\mathbb G_m$.

Note that in terms of rings, the induced map $\phi^*: k[t,t^{-1}] \to k[t,t^{-1}]$ has not only to be an automorphism of $k$-algebras, but it has to respect the Hopf algebra structure (so that it corresponds to an automorphism of $\mathbb G_m$ as an algebraic group, and not just as a variety).