This exercise states that if $A$ is a local ring with maximal ideal $\mathfrak{m}$ and residue field
$k=A/\mathfrak{m}$ and $M,N$ are finitely generated $A$-modules then $M\otimes_{A}N=0\implies M=0$ or $N=0$.
This is what I have worked so far:
For an $A$-module $L$ we write $L_{k}=k\otimes_{A}L$ for the $k$-module obtained from $L$ by extension of scalars. Assume that $M\otimes_{A} N=0$. Then it holds that $(M\otimes_{A}N)_{k}=0$. If I can prove that
\begin{equation}
M_{k}\otimes_{k}N_{k}=0
\end{equation}
Then I would have proved that $M=0$ or $N=0$. Indeed, for $k$-vector spaces we have that
\begin{equation}
\dim(M_{k}\otimes_{k} N_{k})=\dim(M_{k})\cdot\dim(N_{k})
\end{equation}
and we also have that $M_{k}\cong M/\mathfrak{m}M$. Therefore by Nakayama's Lemma we have that
$M_{k}=0\iff M=\mathfrak{m} M\iff M=0$ and $N_{k}=0\iff N=\mathfrak{m} N\iff N=0$,
as $\mathfrak{m}$ is the Jacobson radical of $A$.
Problem: I cannot find the relationship between $M\otimes_{A}N$ and $M_{k}\otimes_{k}N_{k}$ (although it seems to me obvious). I have tried the following things:
\begin{equation}
M_{k}\otimes_{k}N_{k}=(k\otimes_{A}M)\otimes_{k}(k\otimes_{A}N)=k\otimes_{A}(M\otimes_{k}k)\otimes_{A}N=
k\otimes_{A}(M\otimes_{A}N)=(M\otimes_{A}N)_{k}
\end{equation}
But I cannot see that $M$ is a $k$-module and so the isomorphism $M\otimes_{k}k\cong M$ is not valid.
Best Answer
Really $M_k=M\otimes_A A/\mathfrak{m}=M\otimes_A k$. So: $$ M_k\otimes_k N_k=(M\otimes_A k)\otimes_k(k\otimes_A N)$$ $$\cong M\otimes_A (k\otimes_k k)\otimes_A N\cong (M\otimes_A N)\otimes_A k=(M\otimes_A N)_k.$$ I think this is what you need to finish your argument. Notice that here we are only tensoring $k$ with itself over itself, and so this makes sense. It should be a good exercise to verify that the isomorphisms above are indeed isomorphisms.