Tensor product of two finitely-generated modules over a local ring is zero

Solution 1:

From $M\otimes_RN=0$ we get $R/\mathfrak m\otimes_R (M\otimes_R N)=0$. Then it follows that $(R/\mathfrak m\otimes_R M)\otimes_{R/\mathfrak m}(R/\mathfrak m\otimes_R N)=0$. But $R/\mathfrak m\otimes_R M\simeq M/\mathfrak mM$ and similarly $R/\mathfrak m\otimes_R N\simeq N/\mathfrak mN$. Since $M/\mathfrak mM$ and $N/\mathfrak mN$ are $R/\mathfrak m$-vector spaces, it follows that $M/\mathfrak mM=0$ or $N/\mathfrak mN=0$, that is, $M=\mathfrak mM$ or $N=\mathfrak mN$, so by Nakayama we have $M=0$ or $N=0$.

Edit. For proving $$R/\mathfrak m\otimes_R (M\otimes_R N)\simeq(R/\mathfrak m\otimes_R M)\otimes_{R/\mathfrak m}(R/\mathfrak m\otimes_R N)$$ use the associativity and the following property of tensor product: $L\otimes_SS\simeq L$, where in this case $S=R/\mathfrak m$ and $L=R/\mathfrak m\otimes_RM$.