Showing a finite field isomorphism

Question

Suppose $p$ is a prime number, with $p \equiv 3 \mod 4$, I need to prove that then $\mathbb{Z}[i]/(p) \cong \mathbb{F}_{p^2}$. What I have so far: $$ \mathbb{Z}[i]/(p) \cong (\mathbb{Z}[X]/(X^2+1))/(p) \cong \mathbb{Z}[X]/(X^2+1,p) \cong \mathbb{F}_5[X]/(X^2+1) \cong \mathbb{F}_{p^2}. $$ But where do we use the fact that $p \equiv 3 \mod 4$? Am I on the right track?

Answer 1

If $p\equiv 3\mod 4$, $-1$ is not a quadratic residue modulo $p$, hence $X^2+1$ remains irreducible in $\mathbf Z/p\mathbf Z$.

Actually it is a basic theorem in Algebraic Number Theory that, if $p$ is an odd prime, $-1$ is a quadratic residue mod. $p$ if and only if $p\equiv 1\mod 4$. Its proof relies on Euler's criterion:

If $p$ is an odd prime and $a$ an integer coprime to $p$, then

• If $a$ is a quadratic residue mod. $p$, $\;a^{\tfrac{p-1}2}\equiv 1\mod p$.

• If $a$ is a non-quadratic residue mod. $p$, $\;a^{\tfrac{p-1}2}\equiv -1\mod p$.

Answer 2

It's much easier to show that $p$ is irreducible in $\mathbb{Z}[i]$ (try it, without looking at this answer of mine).

Then $(p)$ is a prime, hence maximal, ideal in $\mathbb{Z}[i]$, so $\mathbb{Z}[i]/(p)$ is a field. Its elements are of the form $$ a+bi $$ with $a,b\in\mathbb{Z}/p\mathbb{Z}$.