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Say $p$ is a prime number such that $p=4k+1$. I need to prove that $\mathbb{Z}[i]/p$ (quotient ring) and $\mathbb{F}^2_{p}$ are isomorphic. I know that since $p \equiv 1 \mod4$ I can say that $p=xy$ where $x,y$ are conjugate prime numbers in $\mathbb{Z}[i]$, so I tried using the CRT (Chinese remainder theorem) - so I know that $\mathbb{Z}[i]/p$ is isomorphic to $\mathbb{Z}[i]/x\times\mathbb{Z}[i]/y$, but I wasn't able to take it from there. Any hint would be helpful.

Henry
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Math101
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    The hint is to prove $\Bbb Z[i]/(x)\cong\Bbb F_p$. – Berci Nov 12 '20 at 17:52
  • as a further hint, recall that $\mathbb{Z}[i]$ is a PID, so prime ideals in it are maximal – Atticus Stonestrom Nov 12 '20 at 17:54
  • @AtticusStonestrom What is exactly the definition of $\mathbb{Z}[i]/{(x)}$? When we talk about $\mathbb{Z}(x)$ the meaning is obvious, but it's not so obvious in this notation. – Math101 Nov 12 '20 at 17:58
  • So, $(x)$ is shorthand for the ideal of $\mathbb{Z}[i]$ generated by $x$; this is also denoted $\mathbb{Z}[i]x$, and it consists of the set of elements $(a+bi)x$ for some $a,b\in\mathbb{Z}$. You can verify that this is an ideal, and hence it makes sense to quotient by it, and $\mathbb{Z}[i]/(x)$ is then just the ring consisting of all cosets $a+bi+(x)$. (I assume you've come across the definition of a general quotient rings?) – Atticus Stonestrom Nov 12 '20 at 18:06
  • I'm still not sure how to take it from here – Math101 Nov 12 '20 at 18:23

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More precisely we have $$\mathbb{Z}[i]/(p) \cong \mathbb{F}_p[X]/(X^2+1) \cong \begin{cases} \mathbb{F}_{p^2} &\text{if } p \equiv 3 \pmod 4\\ \mathbb{F}_{p} \times \mathbb{F}_{p} &\text{if } p \equiv 1 \pmod 4 \end{cases}$$

So the comment on whether it is written $\Bbb F_{p^2}$ or $\Bbb F_p^2$ was really meaningful.

Reference: Quotient ring of Gaussian integers $\mathbb{Z}[i]/(a+bi)$ when $a$ and $b$ are NOT coprime

Dietrich Burde
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