Torsion of $\mathbb Z ^{2 \times 2}$ I am trying to find the torsion of the ring $R= \mathbb Z ^{2 \times 2}$ seen as an $R-$ left module. So I want to see for which matrices $B$, there exists $A \neq 0$ such that $AB=0$. So if $B=\begin{pmatrix} a & b \\  c & d  \\  \end{pmatrix}$, then I want to find $A=\begin{pmatrix} x & y \\  z & w  \\  \end{pmatrix}$ with $$\begin{pmatrix} x & y \\  z & w  \\  \end{pmatrix}\begin{pmatrix} a & b \\  c & d  \\  \end{pmatrix}=\begin{pmatrix} 0 & 0 \\  0 & 0  \\  \end{pmatrix}$$
From this condition I get these equations:
$$ax+cy=0$$$$bx+dy=0$$$$az+cw=0$$$$bz+dw=0$$
If $B$ is an invertible matrix in $R$, then it is clear that $B$ is not in the torsion. This led me to think that the torsion may be the set of non invertible matrices, or at least is contained in it. I've tried to find the invertible matrices in this ring but I got stuck.
Any suggestions would be appreciated.

It's easier to find the right zero divisors, transposition will give the left zero divisors.
If $A=\left[\begin{smallmatrix}a&\smash{b}\\c&d\end{smallmatrix}\right]$ is a left zero divisor, then there is a vector $v=\left[\begin{smallmatrix}x\\y\end{smallmatrix}\right]\ne0$ such that $Av=0$ and conversely, because we can use the matrix having $v$ on both columns.
In particular, the rank of $A$ (as a matrix in $\mathbb{Q}^{2\times2}$) must be $<2$. Conversely, if $A$ has rank $<2$, we can find $v\ne0$ with rational coefficients such that $Av=0$. But upon multiplying $v$ by a common multiple of the denominators of the coefficients, we get a vector with integer coefficients and the same property.
So a matrix is a left zero divisor if and only if its rank is $<2$, that is, its determinant is $0$. Since transposition doesn't change the determinant, we have also characterized the right zero divisors.
Note that this set is properly contained in the set of noninvertible matrices: a matrix can have rank $2$ but be noninvertible in $\mathbb{Z}^{2\times2}$.
