Let $k$ be an algebraically closed field with characteristic $p$. All group schemes are over $k$. Suppose $G=G_a\times G_a$ is the affine group scheme with $G_a$ the usual additive affine group scheme. Suppose $H_1$ be the affine subgroup scheme of $G$ where for each $R$ we project onto the first component. Let $H_2$ be the affine subgroup scheme of $G$ where for each $k$-algebra $R$ we obtain $H_2(R)=\{(x,y):x^p=y\}\subset G(R)$. Then it can be shown that $H_1\cap H_2$ is an affine subgroup scheme.

I will call a (Zariski) closed subset of $k^2$ with a group structure where addition and inversion are given by polynomial maps an affine algebraic group. Given an affine algebraic group $S$, we can construct a group functor by taking $A$ be the coordinate ring of $S$ and setting $S(R)=\text{Hom}_k(A,R).$ In this way, every affine algebraic group gives rise to an affine group scheme represented by $A$.

My aim: I am asked to show that the converse is false. Namely, given some affine group scheme, here $H_1\cap H_2$, I wish to see that $H_1\cap H_2$ does not arise from an affine algebraic group.

My issue: I find this claim dubious. Taking $S$ to be the origin inside $k^2$ with addition and inversion given trivially, we have an affine algebraic group. Then the coordinate ring of $S$ is just the zero ring, and $S$ determines the functor taking $R$ to $\text{Hom}_k(A,R)$ (which is always just the set of the zero map). But this is also exactly the functor $H_1\cap H_2$, since it takes a $R$ to the subset of $R\times R$ such that $y=0$ and $x^p=y$, which also forces $x=0$. So then is it not the case that $H_1\cap H_2$ arises from an affine algebraic group?

Some remarks: I clearly am missing some part of the theory. The particular choices of $H_1,H_2$ should play a role, but it seems my argument works for any such $H_1$ and $H_2$. Also, the fact that $k$ is an algebraically closed field doesn't become relevant in my argument, whereas surely it plays a role in the disproof of the claim.

  • $\begingroup$ Your claim about $x^p = 0$ implying $x = 0$ fails if $R$ has nontrivial nilpotents. $\endgroup$ Oct 27, 2020 at 5:44

1 Answer 1


$H_1 \cap H_2$ is the affine group scheme $\alpha_p = \text{ker} \left( \mathbb{G}_a \xrightarrow{x \mapsto x^p} \mathbb{G}_a \right)$, with functor of points

$$\alpha_p(R) = \{ x \in R : x^p = 0 \}.$$

If $R$ has no nontrivial nilpotents, and in particular if $R = k$, then $\alpha_p(R) = 0$. However, $\alpha_p$ is not the zero group scheme, because for example it has nontrivial points over $k[x]/x^p$ (which is in fact the underlying affine scheme of $\alpha_p$). An affine algebraic group is determined by its $k$-points so this shows that $\alpha_p$ is not an affine algebraic group.

The assumption that $k$ is algebraically closed is only used to define what an affine algebraic group is.


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