# Bijective homomorphism of algebraic groups is isomorphism (characteristic 0)

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 inseparable. 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.

• 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). Commented Oct 6, 2015 at 19:48

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. ]

• Dear Georges: I'm confused, this has kernel $\mu_2$? Commented Oct 6, 2015 at 19:43
• Dear Alex: yes, so what ? The OP never said anything about the kernel! Commented Oct 6, 2015 at 19:45
• Didn't he say bijective? Commented Oct 6, 2015 at 19:45
• No, he didn't ! Commented Oct 6, 2015 at 19:46
• In the title? :S Commented Oct 6, 2015 at 19:46

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.