Application of Chinese Remainder Theorem to Ring Isomorphism

Question

Use Chinese Remainder Theorem to show that $\mathbb{Z}[i]/(p) \cong \mathbb{Z}_p^2$ $\ \ $with $p \in \mathbb{Z}\ \ \text{s.t. $p$ is a prime and}\ \ p = 1 \mod 4$

What I know is that this ring is finite, with order $p^2$, there is an obvious group isomorphism, but doesn't agree on the ring structure part. I am not quite sure how to set up the ideals for CRT.

Answer 1

If $p\equiv 1 \bmod 4$ because $\Bbb{Z}_p^\times$ is cyclic with $p-1$ elements there is $c^2\equiv -1 \bmod p$ so that

$$\mathbb{Z}[i]/(p)\cong \mathbb{Z}[x]/(x^2+1,p)\cong \mathbb{Z}_p[x]/(x^2+1) \cong \mathbb{Z}_p[x]/(x+c)(x-c)$$ $$\cong \mathbb{Z}_p[x]/(x+c)\times \mathbb{Z}_p[x]/(x-c)\cong \mathbb{Z}_p\times \mathbb{Z}_p$$ where the middle isomorphism is based on $\frac{p+1}{2}c (x-c)\equiv 1 \bmod (x+c),\equiv 0 \bmod (x-c)$

Answer 2

In $\Bbb Z[i]$, which is a Euclidean domain, the prime ideals are $(1+i)$, $(p)$ for primes $p\equiv3\pmod 4$ and $(a\pm bi)$ where $a^2+b^2=p$ and $p\equiv1\pmod 4$ is a prime.

So for your $p$, write $p=a^2+b^2$ and observe that $(p)=(a+bi)(a-bi)$ as ideals.