Proving the isomorphy of two modules I am working on an old exam for linear algebra, and have been looking at this question: If $M_1,M_2$ are modules and $N_1 \subset M_1$ and $N_2 \subset M_2$ are submodules, show that $(M_1 × M_2)/(N_1 × N_2)$ is isomorph with $M_1/N_1 × M_2/N_2$. Obviously one way to do this is to define an isomorphism between the two, but because it is such a small question I was wondering whether there is a faster way to solve this? It seems like a lot of work to show well-definedness subjectivity and injectivity for such a small question.
 A: Denote by $\pi_i \colon M_i \rightarrow M_i / N_i$ and consider the map $\psi \colon M_1 \times M_2 \rightarrow M_1 / N_1 \times M_2 / N_2$ given by $\psi(m_1,m_2) = (\pi_1(m_1), \pi_2(m_2))$. This map is clearly linear and onto with kernel
$$ \ker(\psi) = \{ (m_1, m_2) \, | \, \pi_1(m_1) = \pi_2(m_2) = 0 \} = \{ (m_1, m_2) \, | \, m_1 \in N_1, m_2 \in N_2 \} = N_1 \times N_2 $$
and so by the first isomorphism theorem you get
$$ (M_1 \times M_2) / (N_1 \times N_2) = (M_1 \times M_2) / \ker(\psi) \approx \operatorname{im}(\psi) = M_1 / N_1 \times M_2 / N_2. $$
A: If you don't want to use algebra theorems that basically give your result, you have to do it the way you say. You define
$$
\phi:(x,y)+N_1\times N_2\longmapsto (x+N_1,y+N_2). 
$$
Note that 
\begin{align}
(x,y)+N_1\times N_2=(0,0)+N_1\times N_2&\iff x\in N_1\ \text{ and } y\in N_2\iff(x,y)\in N_1\times N_2\\ \ \\
&\iff (x+N_1,y+N_2)=N_1\times N_2.
\end{align}
These implication prove both that $\phi$ is well-defined and that it is injective. 
Surjectivity and linearity are trivial. 
