Exercise 7.18

Proof. If $p$ is a prime rational integer, with $p\equiv 3(\mathit{mod}4)$, then $p$ is a prime in $\mathbb{Z}[i]$.

Indeed, $p$ is irreducible : if $p=\mathit{uv},u,v\in \mathbb{Z}[i]$, where $u=c+\mathit{di},v$ are not units, then $p^2=N(u)N(v),N(u)>1,N(v)>1$, so $p=N(u)=u\bar{u}=c^2+d^2$.

As $c^2\equiv 0,1(\mathit{mod}4),d^2\equiv 0,1(mod4)$, so $p\equiv 0,1,2(\mathit{mod}4)$, which is in contradiction with the hypothesis.

So $p$ is irreducible in $\mathbb{Z}[i]$, and since $\mathbb{Z}[i]$ is a principal ideal domain, $p$ is prime in $\mathbb{Z}[i]$, thus $\mathbb{Z}[i]∕(p)$ is a field.

Let $z=a+\mathit{bi}\in \mathbb{Z}[i]$. The Euclidean division of $a,b$ by $p$ gives

so

Let’s verify that these $p^2$ elements are in different classes of congruences modulo $p$.

If $r+\mathit{is}\equiv r^{\prime }+i{s}^{\prime }(\mathit{mod}p)$, then $(r-r^{\prime })∕p+i(s-s^{\prime })∕p\in \mathbb{Z}[i]$, so $r\equiv r^{\prime },s\equiv s^{\prime }(\mathit{mod}p)$.

As $r,r^{\prime },s,s^{\prime }$ are between $0$ and $p-1$, $r=r^{\prime },s=s^{\prime }$.

So the cardinality of the field $\mathbb{Z}[i]∕(p)$ is $p^2$. □