Tensor product of two quotient modules

Solution 1:

One way is to show that $M \otimes N$ modulo the the submodule $P$ generated by $m' \otimes n$ and $m \otimes n'$ (with $m' \in M', m \in M, n' \in N', n \in N$) directly satisfies the same universal property as $M/M' \otimes N/N'$.

You have a well defined $A$-bilinear map $j: M/M' \times N/N' \rightarrow (M \otimes N)/P$ given by

$$(m+M',n+N') \mapsto m \otimes n + P$$

Suppose $Q$ is an $A$-module, and $f: M/M' \times N/N' \rightarrow Q$ is $A$-bilinear. I claim that there exists a unique $A$-module homomorphism $\phi: (M\otimes N)/P \rightarrow Q$ such that $\phi \circ j = f$. The uniqueness of the tensor product will then guarantee you an isomorphism $M/M' \otimes N/N' \rightarrow (M \otimes N)/P$ such that $(m +M') \otimes (n+N') \mapsto m \otimes n + P$.

The map $M \times N \rightarrow Q, (m,n) \mapsto f(m + M',n + N')$ is certainly $A$-bilinear, so there exists a unique $A$-linear map $\psi: M \otimes N \rightarrow Q$ given on generators by $\psi(m\otimes n) = f(m+M',n+N')$. Also, for any $m' \in M', n' \in N$ you have

$$\psi(m' \otimes n) = f(m' + M',n + N') = f(0 + M', n+ N') = 0$$

$$\psi(m \otimes n') = f(m + M', n' + N') = f(m+M',0+N') = 0$$

so $P$ is contained in the kernel of $\psi$. Therefore the $A$-module homomorphism $\phi: (M \otimes N)/P \rightarrow Q$ defined by $\phi(x + P) = \psi(x)$ is well defined.

Now $\phi$ does what is required:

$$\phi \circ j(m+M',n+N') = \phi(m\otimes n + P) = \psi(m \otimes n) = j(m+M',n+N')$$

and the uniqueness of $\phi$ is easily seen from the uniqueness of $\psi$.