# Intersection of two affine algebraic groups is not an affine algebraic group

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.

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

$$H_1 \cap H_2$$ 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.