Exercise 2.3

Proof. Let $𝔪\subseteq A$ be the maximal ideal, and $k=A∕𝔪$ the residue field. Then, $M_k=k{\text{⊗}}_A\mathit{M\cong M}∕\mathit{𝔪M}$ by Exercise 2.2. By Nakayama’s lemma (Prop. 2.6), $M_k=0$ implies $M=0$. But $M{\text{⊗}}_{A}N=0$ implies ${(M{\text{⊗}}_{A}N)}_k=0$, so $M_k{\text{⊗}}_k{N}_k=M{\text{⊗}}_{A}k{\text{⊗}}_{k}k{\text{⊗}}_{A}N=0$. Thus, $M_k=0$ or $N_k=0$ since $M_k,N_k$ are both vector spaces over $k$, and so by Nakayama’s lemma again, $M=0$ or $N=0$. □