Does Repf($G$) have enough injectives?

Question

Let $G$ be a group scheme over a field $k$. My question is

Does Repf(G), the category of finite-dimensional linear representations of $G$, have enough injectives?

It is well-known that Rep(G), the category of linear representations of $G$, does have enough injectives (it is explained in Section 4.2, Representations of Algebraic Groups - J.C.Jantzen).

Probably the main difficulty for this question is that $k[G]$ is not necessarily finite-dimensional. For example, following Jantzen's proof, from a finite-dimensional representation $M$, we embed it as a $k-$vector subspace of a finite-dimensional $k$-vector space $I$ (in this case, we can take $I=M$) and then show that $M$ is embedded as a $k-$subrepresentation of $I\otimes k[G]$, the latter might not be finite-dimensional.

Further, I think Rep($G$) is a Grothendieck category, which has enough injectives. However, it seems that it is not true for Repf($G$) because the regular representation $k[G]$ is not a generator in this category.

Thanks!

Answer 1

Nope, and you don't even need to do anything particularly scheme-y or characteristic $p$-ish.

Indecomposable finite dimensional representations of the additive group $\mathbb{G}_a$ over $\mathbb{C}$ correspond to unipotent Jordan blocks, with each one including into the next (and being a quotient of it). So we see there are no finite dimensional injective (or projective) objects.