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.

  • 1
    $\begingroup$ What do you mean precisely by the Gaussian integers mod $p$? $\endgroup$ – Tobias Kildetoft Mar 14 '18 at 10:13
  • $\begingroup$ Map $i$ to $\pmatrix{0&1\\-1&0}$. $\endgroup$ – Berci Mar 14 '18 at 10:21
  • $\begingroup$ @Berci What about when $i$ is not invertible? $\endgroup$ – Tobias Kildetoft Mar 14 '18 at 10:30
  • $\begingroup$ @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. $\endgroup$ – Lord Shark the Unknown Mar 14 '18 at 10:31
  • $\begingroup$ @LordSharktheUnknown Ahh, right, it is not $i$ itself that becomes a zero-divisor. $\endgroup$ – 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.


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