# Why is $\pi$ irreducible in $\mathbb{Z}[i]$? [duplicate]

I'm reading Dummit&Foot and got stuck at page 290.

Let O be a quadratic integer ring and $$\pi$$ be prime in O. Then $$(\pi)\cap \mathbb{Z} =pZ$$ for some prime integer. Since $$p\in (\pi)$$ we have $$p=\pi \pi'$$. So $$p^2=N(\pi)N(\pi')$$. Assume that $$N(\pi)=\pm p^2$$. Then $$N(\pi')=\pm 1$$ so $$\pi'$$ is a unit and $$p=\pi$$ (up to assoicates) is irreducible in $$\mathbb{Z}[i]$$.

Why $$\pi$$ is irreducible?

If we consider the case where $$p=\pi$$, then $$p$$ is not always irreducible. For example, $$p=2=(1+i)(1-i)$$ and none of these factors are units, because their norms are $$1^2+1^2=2$$. So what am I missing?

• If $\pi$ was reducible, it would not be prime. $2$ is not prime in the Gaussian integers. Oct 14 '20 at 0:12
• Primes are always irreducible in domains (as you claim in you answer). This is well-known and proved here many times, e.g. the linked dupe. It's best for site health to delete questions that are dupes of FAQs. Oct 14 '20 at 0:37

So we have shown that $$p$$ is irreducible, because in the domains prime means irreducible, and the fact that $$\pi$$ is irreducible implies that $$p$$ is irreducible.