# Vanishing of $\text{Ext}^2$ sheaf from abelian variety to multiplicative group

Does anyone know a proof or reference for the following statement? Or if it's false (which seems unlikely to me), a counterexample?

Let $k$ be a field (maybe we need it to be perfect) and $A$ an abelian variety over $k$. Then the sheaf of abelian groups $\underline{\text{Ext}}^2(A,\mathbb{G}_m)$ on the fppf site of $k$-schemes is trivial.

For abelian schemes $\mathscr{A}/S$, $S$ being regular, $\mathscr{Ext}^i_{(\text{Sch}/S)_{\text{fppf}}}(\mathscr{A}, \mathbf{G}_m)$ is torsion for all $i\ge 2$. I recall this is in Breen's Inventiones paper "Extensions of abelian sheaves...", $\S$7. On the other hand if $p$ is a prime such that $p!$ is invertible locally on $S$, then for $1<i<2p-1$ such higher ext's actually vanish. This is discussed in Breen's paper "Un théorème d'annulation [...]", 1975, and relies on the former.
• For $n > 0$, if we apply $\mathbf{R}\mathscr{H}om(\cdot, \mathbf{G}_m)$ to the $n$-torsion Kummer sequence of $\mathscr{A}$ then we get an exact sequence of sheaves $\cdots \rightarrow \mathscr{A}^{\vee} \stackrel{n}{\rightarrow} \mathscr{A}^{\vee} \rightarrow \mathscr{E}xt^1(\mathscr{A}[n], \mathbf{G}_m) \stackrel{\delta}{\rightarrow} \mathscr{E}xt^2(\mathscr{A}, \mathbf{G}_m)[n] \rightarrow 0$, so $\delta$ is an isomorphism. But the source of $\delta$ vanishes (SGA7, VIII, 3.3.1). So this $\mathscr{E}xt^2$-sheaf is torsion-free, and hence vanishes whenever it is torsion. – nfdc23 Aug 10 '16 at 1:30
• @JustinCampbell: Sheafifying Ext$^i$ gives $\mathscr{E}xt^i$, by an erasable $\delta$-functor argument (since it holds for $i=0$). But the points of this sheaf valued in the base are generally not at all the original Ext$^i$, so is your use of the phrase "field-valued points" what you meant? – nfdc23 Aug 10 '16 at 1:33