# Does $x$ irreducible in $\Bbb{Z}[i] \implies ~N(x)$ is prime?

Does $x$ irreducible in $\Bbb{Z}[i] \implies ~N(x)$ is prime?

I tried proving it but couldn't figure it out.

• $3$ is irreducible. – Slade Oct 26 '16 at 14:18
• Okay thanks for the counter. – Ryan S Oct 26 '16 at 14:19
• @Slade How would I go about proving if $N(z)=pq$ for $p,q$ prime $p \neq q$ in $\Bbb{Z}$ then $z$ is not irreducible? – Ryan S Oct 26 '16 at 14:21

There are three kinds of irreducibles $\pi\in\mathbb{Z}[i]$. The are, up to unit:
• $\pi=1+i$, with $N(\pi)=\pi\overline{\pi}=2$.
• $\pi=a\pm bi$, with $N(\pi)=\pi\overline{\pi}=a^2+b^2=p$ an arbitrary prime that is $1\pmod{4}$.
• $\pi=p$, with $N(\pi)=\pi\overline{\pi}=p^2$, where $p$ is an arbitrary prime that is $3\pmod{4}$.
So the irreducibles that fail to have prime norm are exactly the primes in $\mathbb{Z}$ that don't "split", i.e. the ones that are still prime in $\mathbb{Z}[i]$, which are exactly the primes which are $3\pmod{4}$.