TITLE: Is there a ring homomorphism $M_2(\mathbb Z)\to \mathbb Z$?
QUESTION [16 upvotes]: I have the following problem:

Is it possible to construct ring homomorphism from $M_2(\mathbb Z)\to \mathbb Z$, or in other words, a homomorphism from ring of all $2\times2$ matrices over the integers into integers?

I tried determinant, trace and mapping that maps matrix to it's element in the position $(1,1)$ but none of that obviously works, which led me to believe there might not be such a homomorphism.
Determinant doesn't work because it is obviously not a linear map.  Trace doesn't work because it doesn't respect multiplication.  And the mapping to the position $(1,1)$ also doesn't work because
$$\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}1&1\\1&1\end{pmatrix}=\begin{pmatrix}2&2\\2&2\end{pmatrix}.$$

REPLY [3 votes]: Let $i$ be the image of the identity matrix. Then $i^2=i$ and so $i=0$ or $i=1$.
If $i=0$, then the map is the zero homomorphism because $A=AI$. 
If $i=1$, let $J=\pmatrix{0&-1\\1&0}$. Then $J^2=-I$ translates to $j^2=-1$, which cannot happen in $\mathbb Z$.
Therefore, the only ring homomorphism $M_2(\mathbb Z)\to \mathbb Z$ is the zero map. 
If you require that a ring homomorphism must preserve the multiplicative identity, then there is none.