How do I prove $\frac{M_1 \times M_2}{N_1 \times N_2} \simeq \frac{M_1}{N_1} \times \frac{M_2}{N_2}$?

I seem to encounter this issue whenever a question involves quotient objects. In this case, I have modules $M_1$ and $M_2$ and subsets $N_1$ and $N_2$ thereof respectively. It is given that $N_1$ and $N_2$ are submodules, thus so too is their set-theoretic Cartesian product, namely $N_1 \times N_2$. Thus, all expressions in the formula below are well-defined. The task is to show that $$\frac{M_1 \times M_2}{N_1 \times N_2} \simeq \frac{M_1}{N_1} \times \frac{M_2}{N_2}.$$

As a first step, I'd like to show that there exists a function $$f : \frac{M_1 \times M_2}{N_1 \times N_2} \rightarrow \frac{M_1}{N_1} \times \frac{M_2}{N_2}$$ such that for all $(m_1,m_2) \in M_1 \times M_2$ it holds that

$$f((m_1,m_2)+N_1 \times N_2) = (m_1+N_1, m_2 + N_2),$$

but I can't work out how to do it. Since functions are generalized by relations, one approach would be to define a relation with the right properties and then prove that it is a function. But I can't even work out the appropriate relation to define.

Help, anyone?

An idea: define

$$f: M_1\times M_2\to M_1/N_1\times M_2/N_2\;\;\text{by}\;\;f(m_1,m_2):=(m_1+N_1\,,\,m_2+N_2)$$

(1) Show the above is a surjective module homomorphism;

2) Apply the first isomorphism theorem

• Thanks, I successfully completed the problem using this method. – goblin Jun 13 '13 at 5:34

You've written down a perfectly good definition of $f$ in the question:

$$f((m_1,m_2)+N_1 \times N_2) = (m_1+N_1, m_2 + N_2)$$

You do need to check that this is well-defined, but this is fairly straightforward (I can add more details on request).

You may also find it easier to define $\bar{f}\colon M_1\times M_2\to \frac{M_1}{N_1}\times\frac{M_2}{N_2}$ by:

$$\bar{f}(m_1,m_2)=(m_1+N_1,m_2+N_2)$$

and show that it has kernel $N_1\times N_2$.

Full disclosure: I haven't finished your proof, so this may not actually be an isomorphism of the kind you want (although I'd be quite surprised if it wasn't). It is a well-defined function though.