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Let $X$ and $Y$ be affine algebraic groups over an algebraically closed field $k$ of characteristic $0$, and $\phi: X \to Y$ be a group homomorphism, which is also a morphism of varieties. I would like to prove that $\phi$ is an isomorphism.

I know that in characteristic $p$ there is Frobenius map $Fr: \mathbb G_a \to \mathbb G_a, x \mapsto x^p$, which is bijective but not an isomorphism. It corresponds to field extension $k(x^p) \subset k(x)$, which is purely transcendental. I suppose that to start I should prove that, as all field extensions in characteristic 0 are separable, the field extension corresponding to $\phi$ is trivial.

After this, one could use the fact that a bijective birational morphism $\phi$ between irreducible affine varieties $X, Y$, such that $Y$ is normal, is an isomorphism.

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    $\begingroup$ You may want to mention bijectivity in the body of the post, since, as of now, it's only mentioned in the title. This can cause some confusion (cf. below). $\endgroup$ – Alex Youcis Oct 6 '15 at 19:48
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This is not true: $$\phi: \mathbb C^*\to \mathbb C^*:z\mapsto z^2$$

Edit
Apparently the OP meant $\phi$ to be bijective.
In that case the morphism $\phi$ is an isomorphism because algebraic groups are normal (even smooth) over a field of characteristic zero and we can apply the Proposition here:

A bijective morphism $X\to Y$ between irreducible algebraic varieties over an algebraically closed field of characteristic zero is an isomorphism as soon as $Y$ is normal.
[The varieties are not assumed to be groups nor to be affine. ]

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  • $\begingroup$ Dear Georges: I'm confused, this has kernel $\mu_2$? $\endgroup$ – Alex Youcis Oct 6 '15 at 19:43
  • $\begingroup$ Dear Alex: yes, so what ? The OP never said anything about the kernel! $\endgroup$ – Georges Elencwajg Oct 6 '15 at 19:45
  • $\begingroup$ Didn't he say bijective? $\endgroup$ – Alex Youcis Oct 6 '15 at 19:45
  • $\begingroup$ No, he didn't ! $\endgroup$ – Georges Elencwajg Oct 6 '15 at 19:46
  • $\begingroup$ In the title? :S $\endgroup$ – Alex Youcis Oct 6 '15 at 19:46
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Here is one way to proceed. Note that since you are in characteristic $0$ your morphism is generically smooth and so, since your map is a group map, actually smooth. But, since the morphism must be relative dimension $0$ it must be étale. But, note that since $K$ is algebraically closed and $X(K)\to Y(K)$ is injective, you know that your morphism is radiciel. But, a morphism which is radiciel and étale is an open embedding. Since your morphism is also surjective it follows that it's an isomorphism.

EDIT: To make the radiciel part a little more rigorous, what one should really say that is that $\ker\phi$ is the trivial group scheme since it's reduced and has trivial $\bar{K}$-points. Then, for any algebraically closed field $L$ over $K$ one has that $\ker(\phi_L)=(\ker \phi)_L$ is trivial, so $\phi:X(L)\to Y(L)$ is injective, which shows that $\phi$ is radiciel.

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