$\phi: M_2(\mathbb{Z})\to \mathbb{Z}$ is necessarily trivial Let $\varphi: M_2(\mathbb{Z})\to \mathbb{Z}$ be a ring homomorphism. Then $\varphi$ is necessarily trivial.
My reasoning:
Suppose not. Then $\varphi(I)=1\implies \forall A\in M_2(\mathbb{Z}),\exists A^{-1}\in M_2(\mathbb{Z})$ such that $\varphi(I)=\phi(A)\phi(A^{-1})=\phi(A)\phi(A)^{-1}=aa^{-1}$. However, only $1$ and $-1$ have multiplicative inverses in $\mathbb{Z}$. Hence, if $A\ne I$, then $\phi(A)^{-1}$ may not exist. This implies that $\phi$ must be trivial (that is, map everything to $1$). [For if $\phi(A)=0$ for all $A\ne I$, then $\phi(A)^{-1}$ admits division by $0$].
Do you think this is correct?
 A: Let $\phi$ be any such morphism. Then $Ker\phi$ is an ideal in $M_2(\mathbb{Z})$. But it's a two-sided ideal. 
What do ideals in $M_2(\mathbb{Z})$ look like ?
Assume $Ker\phi$ is not $\{0\}$, and let $M\in Ker\phi$, $M\neq 0$. But then since $Ker\phi$ is an ideal, the matrix consisting of $rk(M)$ times the same integer $k$ (for some $k$) in the top left-hand corner and zeroes elsewhere is also in $Ker\phi$, since the one where $k=1$ is equivalent to $M$ in $M_2(\mathbb{Q})$, and then multiplying by sufficiently large integers does the trick.Then composing with projectors, one gets the matrix with only one $k$ in the top left hand corner, and zeroes elsewhere.
Composing with some pivot matrices gives all the $kE_{i,j}$ (with the usual notations), so that $kM_2(\mathbb{Z})\subset Ker\phi$, which implies $\phi(kI_n)=0$, which in turn gives $k\phi(I_n) =0$. Since $k\neq 0$, this gives $\phi(I_n) =0$, $Ker\phi = M_2(\mathbb{Z})$.  Thus $\phi$ is trivial.
Therefore, we can assume that $Ker\phi = \{0\}$. But then $M_2(\mathbb{Z})$ is isomorphic with its image, which is a subring of the commutative ring $\mathbb{Z}$. However $M_2(\mathbb{Z})$ is not commutative, which is a contradiction. 
Thus $\phi$ is trivial.
