Ring Homomorphism from $M_{2}(\mathbb{Z}_{p}[i])$ to $M_{2}(\mathbb{Z}_{p})$

Let $M_{2}(\mathbb{Z}_{p}[i])$ be a 2 x 2 matrix having element over the ring Gaussian Integers modulo $p$, $M_{2}(\mathbb{Z}_{p})$ be a 2 x 2 matrix over the ring of Intergers modulo $p$. And

I want to find a nontrivial surjective ring homomorphism

$$\varphi \colon M_{2}(\mathbb{Z}_{p}[i]) \longrightarrow M_{2}(\mathbb{Z}_{p})$$

between these two but I can't seem to find a definition for $\varphi$.

What could be a possible $\varphi$ for it to be a surjective ring homomorphism?

Any help would be greatly appreciated.

• What do you mean precisely by the Gaussian integers mod $p$? – Tobias Kildetoft Mar 14 '18 at 10:13
• Map $i$ to $\pmatrix{0&1\\-1&0}$. – Berci Mar 14 '18 at 10:21
• @Berci What about when $i$ is not invertible? – Tobias Kildetoft Mar 14 '18 at 10:30
• @TobiasKildetoft As I presume $i^4=1$ then $i$ is surely invertible. I take $\Bbb Z_p[i]=\Bbb Z_p[X]/\left<X^2+1\right>$ to be here. – Lord Shark the Unknown Mar 14 '18 at 10:31
• @LordSharktheUnknown Ahh, right, it is not $i$ itself that becomes a zero-divisor. – Tobias Kildetoft Mar 14 '18 at 10:35

If $p\equiv1\pmod 4$, then $u^2+1\equiv 0\pmod p$ is soluble, so one can define $\varphi(A+iB)=A+uB$ where $A$, $B\in M_2(\Bbb Z_p)$.
If $p\equiv3\pmod 4$, then $\Bbb Z_p[i]\cong\Bbb F_{p^2}=k$, the finite field of order $p^2$. The centre of $M_2(k)$ is isomorphic to $k$, and in a surjective homomorphism $\varphi:M_2(k)\to M_2(\Bbb Z_p)$ it must be mapped into the centre of $M_2(\Bbb Z_p)$, which is isomorphic $\Bbb Z_p$. But there is no unital ring homomorphism from $k$ to $\Bbb Z_p$.
I'll leave the case $p=2$ for you to consider.