If $M$ and $N$ are finitely generated $R$-modules such that their tensor product is $0$, then their annihilators over $R$ are comaximal

Question

I am trying to prove this using the local-global principle. Let $I = Ann_R(M) + Ann_R(N)$.

It suffices to show that $I_\mathfrak{m} = R_\mathfrak{m}$ for all maximal ideals $\mathfrak{m}$ containing $I$.

Let $\frac{r}{s} \in R_\mathfrak{m}$. So, it suffices to find $t\not \in \mathfrak{m}$ such that $tr \in I$. How do I proceed?

How do I use the fact that $M \otimes N = 0$?

Answer 1

Suppose that $I$ is a proper ideal. Choose a maximal ideal $\mathfrak{m}$ such that $I \subset \mathfrak{m}$. Then $\operatorname{Ann}_R(M) \subset \mathfrak{m}$ and $\operatorname{Ann}_R(N) \subset \mathfrak{m}$

Note that $0=(M\otimes_R N)_{\mathfrak{m}}=M_\mathfrak{m}\otimes_{R_\mathfrak{m}} N_\mathfrak{m}$, tensoring this with $\kappa(\mathfrak{m})$, we obtain $$0=(\kappa(\mathfrak{m}) \otimes_{R_\mathfrak{m}} M_\mathfrak{m}) \otimes_{\kappa(\mathfrak{m})} (\kappa(\mathfrak{m})\otimes_{R_\mathfrak{m}} N_\mathfrak{m})$$

Note that this is a vector space tensor product. But we know that a tensor product of nonzero vector spaces is nonzero. Thus wlog $0=\kappa(\mathfrak{m}) \otimes_{R_\mathfrak{m}} M_\mathfrak{m}$. By Nakayama, this implies $M_\mathfrak{m}=0$, which contradicts $\operatorname{Ann}_R(M) \subset \mathfrak{m}$.