# R integral domain, Q its field of fraction, M R-module with nontrivial annihilator. Is Ext(Q,M) always 0? [duplicate]

Let $R$ be an integral domain, let $Q$ be its field of fractions, and let $M$ be an $R$-module with nontrivial annihilator. Determine if $\mathrm{Ext}_{R}^{n}(Q,M) = 0$ for all $n \geq 0$.

Intuitively, I believe it is true, because the most natural example $R = \mathbb{Z}$, $Q = \mathbb{Q}$, and $M = \mathbb{Z}_m$ looks promising. However, this example is "bad", because $\mathbb{Z}$ is a PID, and therefore divisible is equivalent to injective.

My first thought is constructing a short exact sequence, for example, $0 \to Ann_R(M) \to Q \to M \to 0$ and look at its induced long exact sequence on the cohomology. However, I never succeed. My professor suggests that I should somehow use the fact that $Q$ is the field of fractions, maybe though the fact that $Q$ will kill all the torsions (considering the induced cohomology on $Tor$).

Can anyone give me some more suggestions? Thanks!

Let $a \in R$ non-zero with $aM=0$. Since Ext is linear in both variables, we have that multiplication by $a$ on $\operatorname{Ext}^i(Q,M)$ is both an isomorphism and the zero map (since it is an isomorphism on $Q$ and zero on $M$). Of course the Ext group must be zero itself then.