Hom(G<sub>a</sub>, G<sub>m</sub>) is not representable.

Let R be a ring of characteristic zero. I claim that Hom(G<sub>a</sub>, G<sub>m</sub>)(Spec R) is {Nilpotent elements of R}. Intuitively, all homs are of the form $x\rightarrow e^{nx}$ with n nilpotent.

More precisely, the schemes underlying G<sub>a</sub> and G<sub>m</sub> are 
Spec R[x] and Spec $R[y, y^{-1}]$ respectively. Any hom of schemes is of the form $y \rightarrow \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 by 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 $x \rightarrow e^{tx}$ is in Hom(G<sub>a</sub>, G<sub>m</sub>)(Spec $R_k$) for every k. However, if R is the inverse limit of the $R_k$, there is no corresponding map in Hom(G<sub>a</sub>, G<sub>m</sub>)(Spec R). So the functor is not representable.