Suppose $Q$ is the field of fractions for a domain $R$ and $A$ is an $R$-module such that $rA = 0$ for some $0 \ne r \in R$. Why is it the case that $\text{Ext}_R^n(Q,A) = 0$ for all $n \ge 0$?
I have a feeling this isn't too hard, and maybe relies on a fact about injective modules I'm forgetting. The case $n = 0$ is pretty simple, because it's easy to see why $\text{Hom}(Q,A) = 0$.
If you can give me a useful hint without a full answer to the problem, I'd be grateful.
So I did find an answer. The correct approach is to view multiplication by $r$ in two different ways, from the $\text{Ext}(Q,\_)$ perspective and alternately from the $\text{Ext}(\_,A)$ perspective. In particular, multiplication by $r$ in $Q$ and multiplication by $r$ in $A$ induce the same map on $\text{Ext}(Q,A)$. The latter is easily shown to be a $Q$-vector space, so this map must be an isomorphism; but on the other hand, multiplication by $r$ is zero on $A$, so must induce zero. The only resolution (pardon the pan) is that the Ext is zero.