Basis for $\text{Mat}_2(\mathbb{Z})$ as a $\mathbb{Z}[i]$-module Let $M=\text{Mat}_2(\mathbb{Z})$ a $\mathbb{Z}[i]$-module with scalar multiplication 
$$(a+bi)\begin{pmatrix}x&y\\z&w\end{pmatrix}\equiv\begin{pmatrix}a&-b\\b&a\end{pmatrix}\begin{pmatrix}x&y\\z&w\end{pmatrix}.$$
Let $N=\Big\{\begin{pmatrix}x&x\\z&z\end{pmatrix}:x,z\in\mathbb{Z}\Big\}$ be a submodule of $M$.

  
*
  
*What is a basis of $M$ over $\mathbb{Z}[i]$?  
  
*$N$ is a cyclic module, what is a generator of $N$?  
  
*How do I show that $M/N$ as a $\mathbb{Z}[i]$-module is isomorphic with $\mathbb{Z}[i]$?
  

What I have done:


*

*I must find linearly independent elements of $M$ such that every element of $M$ can be written as a combination of those. Are these elements the four matrices with $1$ and all zeroes?

*I'm afraid I simply don't see this one (yet). :(

*I'm guessing that $M/N=\{\begin{pmatrix}x&-y\\y&x\end{pmatrix}:x,y\in\mathbb{Z}\}$, though I can't immediately see why.
 A: For (1), the four matrices you've described are not linearly independent: for instance you have that $$\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} + i \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix} = 0.$$
However, if you write down $$z\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} + w \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix}z_1 & w_1 \\ z_2 & w_2 \end{pmatrix}$$ for arbitrary $z = z_1 + i z_2$ and $w=w_1 + iw_2$ you see that these two matrices generate all of $M$ in a unique way.
For (2), as you've already seen, the matrix $\begin{pmatrix}1 & 1 \\ 0 & 0 \end{pmatrix}$ generates $N$.
For (3), I claim that the single element $$ \mu =\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} + N $$
forms a basis for $M/N$. Indeed, we have
$$ \begin{pmatrix}x & y \\ z & w \end{pmatrix} + N = \begin{pmatrix}x-y & 0 \\ z-w & 0 \end{pmatrix} + N = ((x-y)+i(z-w))\mu$$
for any element of $M/N$. Furthermore, $\{\mu\}$ is linearly independent because for any $z = z_1 + iz_2$ we have
$$z\mu = 0 \text{ in } N \iff \begin{pmatrix}z_1 & 0 \\ z_2 & 0 \end{pmatrix} \in N \iff z_1 = z_2 = 0.$$
Thus $M/N$ is a free $1$-dimensional $\mathbb{Z}[i]$-module, hence it is isomorphic to $\mathbb{Z}[i]$.
