On $\mathrm{Ext}_R^n(V,M)$ and $\mathrm{Tor}^R_n (V,M)$, where $M$ is an $R$-module with non-zero annihilator and $V$ is a $Q(R)$-vector space

Question

Let $R$ be an integral domain with fraction field $Q(R)$. Let $M$ be an $R$-module such that $\mathrm{Ann}_R (M)\ne \{0\}$. If $V$ is a $Q(R)$-vector space (hence also an $R$-module), then how to show that $\mathrm{Ext}_R^n(V,M)=\mathrm{Tor}^R_n (V,M)=\{0\}$ ?

I can easily see that $V \otimes_R M=\{0\}=\mathrm{Hom}_R(V,M)$, but I can't see what happens with the higher derived functors.

Please help.

Answer 1

This is immediate from the fact that Ext and Tor are $R$-linear in each variable. Let us look at Ext; Tor works that same way. Let $r\in R$ be a nonzero element that annihilates $M$. The multiplication by $r$ map $\operatorname{Ext}^n_R(V,M)\to \operatorname{Ext}^n_R(V,M)$ can be described in two different ways: it is induced by applying the functor $\operatorname{Ext}^n_R(V,-)$ to the map $r:M\to M$, and it is also induced by applying the functor $\operatorname{Ext}^n_R(-,M)$ to $r:V\to V$. Since $r:M\to M$ is the zero map, this means $r:\operatorname{Ext}^n_R(V,M)\to \operatorname{Ext}^n_R(V,M)$ is $0$. But $r:V\to V$ is an isomorphism, so $r:\operatorname{Ext}^n_R(V,M)\to \operatorname{Ext}^n_R(V,M)$ is an isomorphism. That is, the zero map $\operatorname{Ext}^n_R(V,M)\to \operatorname{Ext}^n_R(V,M)$ is an isomorphism, so $\operatorname{Ext}^n_R(V,M)=0$.