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

I know the converse holds but I'm not sure about this direction?

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

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    $\begingroup$ $3$ is irreducible. $\endgroup$ – Slade Oct 26 '16 at 14:18
  • $\begingroup$ Okay thanks for the counter. $\endgroup$ – Ryan S Oct 26 '16 at 14:19
  • $\begingroup$ @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? $\endgroup$ – Ryan S Oct 26 '16 at 14:21
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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 $3\pmod{4}$.

This classification also answers your question in the comments.

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  • $\begingroup$ Off topic, but did you delete your solution to this question? I'd seen it some time ago and bookmarked it for later but now it's not there. math.stackexchange.com/questions/955967/… $\endgroup$ – MathGod Jun 10 '17 at 12:39
  • $\begingroup$ @IshanSingh Yes, if I recall correctly I found a fundamental error in the solution that I was unable to repair. Basically I left out the hardest case, so all the fancy tricks weren't really doing anything. $\endgroup$ – Slade Jun 11 '17 at 4:04

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