I'm trying to understand the proposed solution of the following exercise:
Let $A$ be a local ring and $M,N$ finitely generated $A$-modules. Show
that $$M\otimes_A N=0$$ already implies $M = 0$ or $N=0$.
The solution we were given proposes the following:
Let $\frak m\subseteq A$ the maximal ideal in $A$. Since it holds that
$M\otimes A\big/{\frak{m}} \cong M\big/{{\frak m}M}$, as a consequence
of Nakayama's Lemma we see that $$M\otimes_A A\big/{\frak{m}}
= 0 \implies M =0$$
So far so good, I can follow everything so far. But the following part contains two parts I can't follow. They continue:
We apply that observation onto $M\otimes_A N = 0$. Thus we obtain
$$(M\otimes_A N) \otimes_A A\big/{\frak{m}} = 0.$$
My Question: If I'm not entirely mistaken, the argument we're doing here is exactly the converse of how we argued in the preceding part.
We showed that
$$M\otimes_A A\big/{\frak{m}}
= 0 \implies M =0$$
And now we basically claim that
$$ M\otimes N =0 \implies (M\otimes_A N)\otimes_A A\big/{\frak{m}}
= 0$$ which is the converse, isn't it? Am I missing something?
I'd appreciate any help, I really want to understand this.
Best Answer
Regarding the converse, the implication $$M = 0 \rightarrow M \otimes_A A/\mathfrak m = 0$$ is true (almost) by definition for any tensor product (if $M, N$ are $A$-modules and $M = 0$ then $M \otimes_A N = 0$).
Next, that $(M \otimes_A N) \otimes_A A/\mathfrak m = (M \otimes_A A/\mathfrak m) \otimes_{A/\mathfrak m}(N\otimes_A A/\mathfrak m)$ follows from the associativity of the tensor product: $$\begin{aligned} (M \otimes_A N) \otimes_A A/\mathfrak m &= M \otimes_A (N \otimes_A A/\mathfrak m) \\ &= M \otimes_A A/\mathfrak{m} \otimes_{A/\mathfrak m} \otimes (N \otimes_A A/\mathfrak m) \\ &= (M \otimes_A A/\mathfrak m)\otimes_{A/\mathfrak m}(N\otimes_A A/\mathfrak m) \end{aligned} $$
Now, we have $(M \otimes_A N) \otimes_A A/\mathfrak m = 0$, hence $(M \otimes_A N)/\mathfrak m(M \otimes_A N)$ is a zero dimensional vector space over $A/\mathfrak m$. By the above equality, this vector space is isomorphic to the tensor product of vector spaces $M/\mathfrak mM \otimes_{A/\mathfrak m} N/\mathfrak mN$. As $$0 = \dim_{A/\mathfrak m}(M/\mathfrak mM \otimes_{A/\mathfrak m} N/\mathfrak mN) = \dim_{A/\mathfrak m}(M/\mathfrak mM)\dim_{A/\mathfrak m}(N/\mathfrak mN),$$ we have that wlog $\dim_{A/\mathfrak m}(M/\mathfrak mM) = 0$, that is $M \otimes_A A/\mathfrak m = 0$. By Nakayama's lemma $M = 0$.