For $A=\begin{pmatrix} a\_1 & b\_1 \\ c\_1&d\_1 \end{pmatrix}, B=\begin{pmatrix} a\_2 & b\_2 \\ c\_2&d\_2 \end{pmatrix}\in M\_2(\mathbb Z)$, define 


$A\*B:=a\_1L\_1BR\_1+b\_1L\_1BR\_2+c\_1L\_2BR\_1+d\_1L\_2BR\_2$, where 


$L\_1=I\_2=\begin{pmatrix} 1 & 0 \\ 0&1 \end{pmatrix}, L\_2=\begin{pmatrix} 0 & 1 \\ 1&0 \end{pmatrix}, R\_1=\begin{pmatrix} 0 & 1 \\ -1&-1 \end{pmatrix}, R\_2=-I\_2-R\_1$.


Is there an elegant/clever way to see that $(M\_2(\mathbb Z),+,\*)$ forms a commutative ring with unity, where $+$ is the usual matrix addition, and $\*$ acts as the multiplication? 



A peculiar operation on $M_2(\mathbb Z)$ which along with the usual matrix addition, makes $M_2(\mathbb Z)$ into a commutative ring with unity
There is an isomorphism $f:\mathbb{C\times C}\rightarrow (M\_2(\mathbb{R}),+,\*)$ given by


$$f((1,0))=\left(\begin{array}{cc}
-1 & -1\\-1 & -1
\end{array}\right)\!/2,\ \ \
f((i,0))=\left(\begin{array}{cc}
1 & -1\\1 & -1
\end{array}\right)\!/2\sqrt{3},
$$


$$f((0,1))=\left(\begin{array}{cc}
-1 & -1\\1 & 1
\end{array}\right)\!/2,\ \ \
f((0,i))=\left(\begin{array}{cc}
1 & -1\\-1 & 1
\end{array}\right)\!/2\sqrt{3}
$$


which I found by finding the identity for \*, then finding two idempotents which add up to the identity, then finding two elements whose squares are the negatives of those idempotents.


So there is an isomorphism 
$g:(M\_2(\mathbb{Z}),+,\*) \rightarrow \mathbb{Z[\omega]\times Z[\omega]}$ given by


$$g\left(\left(\begin{array}{cc}
1 & 0\\0 & 0
\end{array}\right)\right)
= \left(\omega,\ \omega\right)
,\ \ \
g\left(\left(\begin{array}{cc}
0 & 1\\0 & 0
\end{array}\right)\right)
= \left(\omega^2,\omega^2\right)
,\\
g\left(\left(\begin{array}{cc}
0 & 0\\1 & 0
\end{array}\right)\right)
= \left(\omega, -\omega\right)
,\ \ \ 
g\left(\left(\begin{array}{cc}
0 & 0\\0 & 1
\end{array}\right)\right)
= \left(\omega^2,-\omega^2\right)
$$
Here $\omega=(-1+\sqrt{3i})/2$, and there are copies of $\mathbb{Z}$ in the image because $1=-(\omega+\omega^2)$.


