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Let $G$ and $H$ be affine algebraic groups over a scheme $S$ of characteristic 0 and let $\textbf{Hom}_{S,gp}(G,H)$ be the functor $T \mapsto \text{Hom}\_{T,gp}(G,H)$

Theorem (SGA 3, expose XXIV, 7.3.1(a)): Suppose G is reductive. Then $\textbf{Hom}_{S,gp}(G,H)$ is representable by a scheme.

Can this fail if $G$ is not reductive? I worked out a few example with $G = \mathbb{G}_a$, but they were representable.

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$\operatorname{Hom}(\mathbb{G}_a, \mathbb{G}_m)$ is not representable.

Let $R$ be a $\mathbb{Q}$-algebra. I claim that $\operatorname{Hom}(\mathbb{G}_a, \mathbb{G}_m)(\operatorname{Spec} R)$ is {Nilpotent elements of $R$}. Intuitively, all homs are of the form $x\mapsto e^{nx}$ with $n$ nilpotent.

More precisely, the schemes underlying $\mathbb{G}_a$ and $\mathbb{G}_m$ are $\operatorname{Spec} R[x]$ and $\operatorname{Spec} R[y, y^{-1}]$ respectively. Any hom of schemes is of the form $y \mapsto \sum f_i x^i $for some $f_i$ in $R$. The condition that this be a hom of groups says that $\sum f_k (x_1+x_2)^k = (\sum f_i x_1^i)(\sum f_j x_2^j)$. Expanding this, $f_{i+j}/(i+j)! = f_i/i! f_j/j!$. So every hom is of the form $f_i = n^i/i!$, and n must be nilpotent so that the sum will be finite.

Now, let's see that this isn't representable. For any positive integer $k$, let $R_k = C[t]/t^k$. The map $x \mapsto e^{tx}$ is in $\operatorname{Hom}(\mathbb{G}_a, \mathbb{G}_m)(\operatorname{Spec} R_k)$ for every $k$. However, if $R$ is the inverse limit of the $R_k$, there is no corresponding map in $\operatorname{Hom}(\mathbb{G}_a, \mathbb{G}_m)(\operatorname{Spec} R)$. So the functor is not representable.

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    $\begingroup$ "Characteristic zero" here should be interpreted to mean "$\mathbb{Q}$-algebra," I guess. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 12, 2017 at 22:12
  • $\begingroup$ "n must by nilpotent" → "$x$ must be nilpotent", I think. $\endgroup$
    – LSpice
    Commented Dec 22, 2022 at 23:03
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    $\begingroup$ @LSpice No, I think "$n$ is nilpotent" is correct. You have a unital ring homomorphism $R[y,y^{-1}]\to R[x]$, mapping $y$ to an element $\sum_{i\ge 0}f_ix^i$; this element must be inversible and this is equivalent to: $f_0$ is inversible in $R$ and $f_i$ is nilpotent for all $i\ge 1$. (Then the condition of being a Hopf algebra homomorphism implies the given formula, namely $f_i=f_1^i/i!$, with $f_1$ nilpotent.) $\endgroup$
    – YCor
    Commented Dec 23, 2022 at 3:13
  • $\begingroup$ @YCor, ah, you are right. I thought the $n$ in $f_i = n^i/i!$ was an integer; it didn't occur to me that it was a ring element. $\endgroup$
    – LSpice
    Commented Dec 23, 2022 at 4:16
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    $\begingroup$ @Vincent Sorry for introducing more confusion. I did, indeed, want $n$ nilpotent. $\endgroup$
    – David E Speyer
    Commented Dec 23, 2022 at 10:42
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There is a reasonable salvage, at least if the base scheme is a field: Hom(G,H) is a direct limit of representable subfunctors. See Lemma A.8.13 in the book "pseudo-reductive groups" (where it is used to prove that the scheme-theoretic fixed locus for a linearly reductive group acting on a connected reductive group is always reductive (possibly disconnected) provided the base is a field.

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