Ring of Gaussian integers modulo p=4n+3

Question

How can I prove that there are no Zero divisors in the Ring of Gaussian integers modulo p=4n+3, where p is prime, n is integer?

Thank you.

Answer 1

You have to prove that the ideal $(p)$ of $\mathbb{Z}[i]$ is prime, which amounts to saying that $p$ is prime in $\mathbb{Z}[i]$. This is generally not the same as saying that $p$ is prime in $\mathbb{Z}$: for instance, $2$ is prime in $\mathbb{Z}$, but it is not prime in $\mathbb{Z}[i]$, as $2=(1+i)(1-i)$, but $2$ divides neither $1+i$ nor $1-i$.

However, prime and irreducible elements are the same in $\mathbb{Z}[i]$, because it is a Euclidean domain, so we may as well prove that $p$ is irreducible in $\mathbb{Z}[i]$.

Suppose $p=xy$, for $x,y\in\mathbb{Z}[i]$. Then we have also $p=\bar{x}\bar{y}$ (using complex conjugation), so by multiplying term by term, $$ p^2=|x|^2|y|^2 $$ This is now a relation in the integers and we have three possibilities:

1. $|x|^2=p^2$, $|y|^2=1$;
2. $|x|^2=p$, $|y|^2=p$;
3. $|x|^2=1$, $|y|^2=p^2$.

In cases 1 and 3, we conclude that one of the factors is invertible (why?). If we can exclude case 2, we have proved that $p$ is indeed irreducible in $\mathbb{Z}[i]$.

Write $x=a+bi$, so $|x|^2=a^2+b^2$.

Can you now prove that $p=a^2+b^2$ is impossible, when $a$ and $b$ are integers?

Here you have to use the fact that $p=4n+3$. Further hint: reduce modulo $4$.