Tensoring over $k$ and over $A$ in Exercise 3 on pg.31 of Atiyah and MacDonald

Question

Here is the question I am trying to solve:

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

And here is the hint given in Atiyah & MacDonald:

And here is an interpretation of why tensoring over $k$ or over $A$ are the same:

But I did not get the idea of what is happening in the mathematical line between the two paragraphs, and what is the idea in general, could someone explain this to me please?

EDIT Here is the solution from the beginning (I beleive that this solution was not written accurately in terms on tensoring over $A$ or over $k$ am I correct?)

EDIT 2:

In general, my question is how to prove that:

proving $$R/\mathfrak m\otimes_R (M\otimes_R N)\simeq(R/\mathfrak m\otimes_R M)\otimes_{R/\mathfrak m}(R/\mathfrak m\otimes_R N)$$

it says here https://math.stackexchange.com/posts/321382/edit use the associativity and the following property of tensor product: $L\otimes_SS\simeq L$, where in this case $S=R/\mathfrak m$ and $L=R/\mathfrak m\otimes_RM$.

But also a comment there said that this is not correct and the correct thing is that $R/ \mathfrak m$ module is bi $( R/ \mathfrak m,R)$ module.

Could anyone explain to me the justification of the comment also and which what is the correct one?

Answer 1

Let $A,B$ be rings and let $B$ be an $A$-algebra i.e. there is a ring homomorphism $\phi:A\to B$. Let $N$ be a $B$-module and $M$ an $A$-module then $B\otimes_A M$ is a $B$-module and $N$ is an $A$-module by setting $a\cdot n = \phi(a)\cdot n$.

We need to prove the identity $$ N\otimes_B(B\otimes_A M)\cong N\otimes_A M$$ as $A$-modules. (These expressions all make sense because of what I wrote in the first parahraph) So we give two maps:

1. The map $N\times M \to N\otimes_B (B \otimes_A M), \ (n,m) \mapsto n \otimes( 1 \otimes m)$ is $A$-bilinear and thus induces a map from right to left.
2. For a fixed $n\in N$ we get a map $\phi_n : M\times B \to N\otimes_A M, \ \phi_n(m,b)=nb\otimes m$ this is $A$-linear and thus induces a map $M\otimes_A B \to N\otimes_A M, m\otimes b \mapsto nb \otimes m$. Letting $n$ vary we get a map $$N\times (B \otimes_A M) \to N\otimes_A M, \ (n,\sum_i b_i \otimes m_i)\mapsto \phi_n(\sum_i b_i \otimes m_i)= \sum_i b_i\cdot n \otimes m_i.$$ This is $B$-bilinear ($B$ acts on $B\otimes_A M$ via $b\cdot (b' \otimes m)=bb' \otimes m$) so it induces a map $N\otimes_B (B\otimes_A M) \to N\otimes_A M$.

One checks that these maps are inverse to each other. Knowing this you can use associativity of tensor product to compute

$$R/\mathfrak{m} \otimes_R (M\otimes_R N) = (R/\mathfrak{m} \otimes_R M) \otimes_R N = (R/\mathfrak{m} \otimes_RM)\otimes_{R/\mathfrak{m}} (R/\mathfrak{m}\otimes_R N).$$
Btw this technique is called a base change.

Edit: In the last equation we use that we can regard $R/\mathfrak{m} \otimes_R M$ as an $R/\mathfrak{m}$-module and $N$ as an $R$-module then we use the base change formula established above. This only works because we can regard $R/\mathfrak{m}$ as an $R/\mathfrak{m}$ module.