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.
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Additionally, the automorphisms of a split torus $\mathbb{G}_m^n$ of rank $n$ is the group $\mathrm{GL}(n,\mathbb{Z})$ (note by rigidity the automorphism group cannot itself be algebraic). This is seen by using the equivalence of categories between diagonalizable groups and finitely-generated abelian groups (without $p$ torsion in characteristic $p$), under which tori of rank $n$ correspond to $\mathbb{Z}^{\oplus n}$. The autorphism group of $\mathbb{Z}^{\oplus n}$ in the category of finitely-generated abelian groups is easily seen to be $\mathrm{GL}(n,\mathbb{Z})$.
Stefan Dawydiak
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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).
Matt E
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