I am working through Humphreys' Linear algebraic groups, and I am stuck on the following exercise ( ex 4 on pg 57)

I need to show that the only automorphisms of $G_m$ (as an algebraic group) is $x \mapsto x$ and $x \mapsto x^{-1}$. The thing is I don't see why the second one is a morphism. Since $G_m$ is just $k^* \subset \mathbb{A}^1$ under multiplication, and therefore shouldn't a morphism be a polynomial function?

Thank you.


1 Answer 1


This is why I dislike the definition of morphisms of varieties as polynomial maps: it's confusing for the notationally-minded. It is a polynomial map once you embed $\mathbb{G}_m$ into affine space, i.e. choosing the coordinate ring to be a quotient of a polynomial algebra. The coordinate ring of $\mathbb{G}_m$ is the $k$-algebra $k[x,x^{-1}]$, but you can recast it as a quotient of a polynomial ring as $k[x,y]/(xy - 1)$, showing a specific embedding into affine space. On $k$-points, the morphism becomes $(a,b)\mapsto (b,a)$, and of course $b = a^{-1}$. So you see that it is a polynomial map.


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