Proposition

Let $A$ be a local ring, $M$ and $N$ finitely generated $A$-modules. Prove that if $M \otimes_A N = 0$, then $M = 0$ or $N = 0$.

Solution

Suppose $M \ne 0$ and $N \ne 0$. We will show that $M \otimes_A N \ne 0$.

Let $m$ be $A$’s maximal ideal. Let $k = M / m$. Then $M_k = k \otimes_A M = A / m \otimes_A M = M / mM$ as shown in this post. Since $M \ne 0$, $mM \ne M$ by Nakayama’s Lemma. Therefore, $M_k = M / aM \ne 0$. Similarly, $N_k \ne 0$.

Then $M_k, N_k$ are vector spaces over a field $k$. Since $M$ and $N$ are finitely generated $A$-modules, $M_k$ and $N_k$ must be finitely generated as well. In other words, $M_k$ and $N_k$ are finite-dimensional vector spaces. Let $\{ m_1, \cdots, m_l \}, \{ n_1, \cdots, n_{l’} \}$ be bases of $M_k$ and $N_k$. Since $M_k \ne 0$ and $N_k \ne 0$, $l \geq 1$ and $l’ \geq 1$.

Define $\phi: M_k \times N_k \rightarrow k$ by $\phi(\sum a_im_i, \sum b_in_i) = a_1b_1$. Then $\phi$ is $k$-bilinear. By Proposition 2.12[Atiyah], there exists a unique $k$-linear mapping $\phi’: M_k \otimes_k N_k \rightarrow k$ such that $\phi(x, y) = \phi’(x \otimes y)$ for all $x, y \in M_k, N_k$.

Then $\phi’(m_1 \otimes n_1) = \phi(m_1, n_1) = 1 \cdot 1 = 1 \ne 0 = \phi’(0 \otimes 0)$. Therefore, $M_k \otimes_k N_k$ contain at least two distinct elements, so $M_k \otimes_k N_k \ne 0$.

If $M \otimes_A N = 0$, then $k \otimes_A (M \otimes_A N) = 0$. Therefore, $M \otimes_A N \ne 0$.