Ext-vanishing in abelian categories Given an abelian category $A$ with enough projectives and enough injectives such that projectives do not coincide with injectives.

Can we have $Ext^i(I,P)=0$ for any $i>0$ and injective $I$ and projective $P$?

(that is an open problem for $A$ being the module category of a finite dimensional algebra)
 A: Let $k$ be a finite field, $\mathcal{F}$ be the category of functors from finite-dimensional $k$-vector spaces to (all) $k$-vector spaces. This is a Grothendieck category with enough projective (and, of course, injective) objects. Moreover, it is locally noetherian by recent work of Putman-Sam-Snowden. As the source category of this functor category has a zero objet, $\mathcal{F}$ splits as the direct product of the category of $k$-vector spaces (corresponding to constant functors) and a category $\overline{\mathcal{F}}$ (corresponding to reduced functors - that is, functors which are zero on the zero vector space).
Lionel Schwartz proved that $\mathrm{Ext}^*_\mathcal{F}(F,P)=0$ (even if $*=0$!) if $F$ is a polynomial functor in $\mathcal{F}$, $P$ is finitely-generated projective and $F$ or $P$ is reduced (see the appendix of this paper --- indeed it is stated in a dual and slightly less general way, but it is a straightforward corollary of it). As finitely-cogenerated injectives objects of $\mathcal{F}$ are (filtered) colimits of polynomial functors, and all projective objects of $\mathcal{F}$ are direct sums of finitely-generated projectives, one deduces easily that $\mathrm{Ext}^*_{\overline{\mathcal{F}}}(I,P)=0$ (always with $*=0$ included) when $P$ is a projective object and $I$ is a finitely co-generated injective object (or more generally any analytic functor). It does not answer completely your question, because there are "big" (even indecomposable) injectives in $\mathcal{F}$ (for example, the injective hull of any non-constant indecomposable projective object), whose structure remains quite mysterious, and it is not clear that this result extends to these ones, but it is a first step.
