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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?

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    $\begingroup$ If $\pi$ was reducible, it would not be prime. $2$ is not prime in the Gaussian integers. $\endgroup$ – tomasz Oct 14 '20 at 0:12
  • $\begingroup$ 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. $\endgroup$ – Bill Dubuque Oct 14 '20 at 0:37
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I think I got it.

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.

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