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.



Can Hom_gp(G,H) fail to be representable for affine algebraic groups?
$\operatorname{Hom}(\mathbb{G}\_a, \mathbb{G}\_m)$ 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.


