My understanding of bilinear functions is a bit sketchy so this is possibly a very dumb question. I'm reading a book which contains the following proposition (all of the tensor products in this question are over $R$).
Let $M_1, M_2$ be $R$-modules and let $N_1$ (respectively $N_2$) be an $R$-submodule of $M_1$ (respectively $M_2$). Then there is an isomorphism of $R$-modules:
\begin{equation} M_1/N_1 \otimes M_2/N_2 \cong (M_1 \otimes M_2)/(N_1 \otimes M_2 + M_1 \otimes N_2). \end{equation}
Here $R$ is a commutative ring with unity. We first note that there is an obvious $R$-bilinear map:
\begin{equation} h:M_1 \times M_2 \rightarrow M_1/N_1 \otimes M_2/N_2 \end{equation} defined by $h(a,b) = (a + N_1) \otimes (b + N_2)$. Then, by the definition of a tensor product, we have a map:
\begin{equation} \tilde{h} : M_1 \otimes M_2 \rightarrow M_1/N_1 \otimes M_2/N_2 \end{equation}
defined by $\tilde{h}(a \otimes b) = (a + N_1) \otimes (b + N_2)$.
So far, this all makes sense. The next step, which is not explained, is to notice that $N_1 \otimes_R M_2 + M_1 \otimes N_2 \subseteq \ker(\tilde{h})$. I don't see why this is the case.
For example, let $n \otimes b + a \otimes m \in N_1 \otimes M_2 + M_1 \otimes N_2$, with $n \in N_1$, $m \in N_2$, $a \in M_1$ and $b \in M_2$. Then:
\begin{align}
\tilde{h}(n \otimes b + a \otimes m) &= \tilde{h}\left( (n + a)\otimes(b + m) \right) \\
&= \left( (n+a) + N_1\right) \otimes \left((b + m) + N_2\right) \\
&= (a + N_1) \otimes (b + N_2)
\end{align}
But this is not the identity $N_1 \otimes N_2$, unless $a \in N_1$ and $b \in N_2$ which isn't the case in general.
Can anyone explain where my reasoning has gone wrong? I'm sure that I'm either misunderstanding how tensors work, but maybe there's some other aspect of the setup that I've overlooked.
Best Answer
The first line of your equality is incorrect; we do not have $n\otimes b+a\otimes m=(n+a)\otimes(b+m)$.
Instead you could use linearity of $\tilde{h}$: $$\tilde{h}(n\otimes b+a\otimes m) = \tilde{h}(n\otimes b)+\tilde{h}(a\otimes m) = (n+N_1)\otimes(b+N_2)+(a+N_1)\otimes(m+N_2).$$