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I have shown that for $q \in \mathbb Z[i]$, if $N(q)=p$ ($p$ prime), or $N(q)=p^2$ ($p$ prime, $p \equiv 3 \pmod 4)$, then $q$ is irreducible ($(N(q)$ denotes the norm of $q$). But how can I prove that these are the only irreducible elements in $\mathbb Z[i]$?

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  • $\begingroup$ No, I just saw it. $\endgroup$ Mar 1 '15 at 18:16
  • $\begingroup$ Okay upvoted. I'm a bit new so I don't know how all these things work :) $\endgroup$ Mar 1 '15 at 18:23
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Let $\pi\in\mathbb Z[i]$ irreducible. As you know $N(\pi)\ne 1$, so $N(\pi)$ is a product of primes in $\mathbb Z_{>0}$.
If $N(\pi)$ is prime, you are done.
Suppose $N(\pi)$ is not prime.
If there is a prime $p\equiv 3\pmod 4$ such that $p\mid N(\pi)=\pi\bar{\pi}$ then $p\mid\pi$, so $\pi$ and $p$ are associates in $\mathbb Z[i]$ and therefore $N(\pi)=N(p)=p^2$. (Here I've used that any prime $p\equiv 3\pmod 4$ is prime in $\mathbb Z[i]$.)
Otherwise, $N(\pi)$ is a product of primes $p\equiv1\pmod 4$. Let $p$ be such a prime. We also know that $p=z\bar z$ where $z\in\mathbb Z[i]$ is a prime element. Then $z\mid \pi\bar{\pi}$ and therefore $\pi$ (or $\bar{\pi}$) and $z$ are associates, so $N(\pi)=N(z)=p$, a contradiction.

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    $\begingroup$ This is old post but I have a question. If there is a prime 3 mod 4, or prime 1 mod 4, we're good. What if $N(\pi)=2^n$ for $n>1$? $\endgroup$
    – name
    Aug 2 '18 at 21:25
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    $\begingroup$ If $2\mid N(\pi)$ then $1\pm i\mid\pi$, and since $\pi$ is irreducible... $\endgroup$
    – user26857
    Aug 3 '18 at 6:05
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    $\begingroup$ And this show that $N(\pi)=2^n$ with $n>1$ is not possible. $\endgroup$
    – user26857
    Aug 4 '18 at 19:10
  • $\begingroup$ Yes I see thank you $\endgroup$
    – name
    Aug 4 '18 at 19:44
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If $q \in \mathbb Z[i]$ is irreducible, it generates a maximal ideal, since we work in an euclidean domain. So $\mathbb Z[i]/(q)$ is a finite field, which is - as an abelian group - generated by at most two elements, so it is isomorphic to $\mathbb F_p$ or $\mathbb F_{p^2}$ for some prime number $p$. This shows $N(q)=p$ or $N(q)=p^2$.

In the latter case we have to show $p \equiv 3 \mod 4$. To that account, we note that we have $p \in (q)$, hence $q | p$ and thus $q = p$ (up to a unit) due to $N(q)=p^2=N(p)$. So $p$ is prime in $\mathbb Z[i]$ and I think you already know that this implies $p \equiv 3 \pmod 4$.

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  • $\begingroup$ A finite field isn't supposed to be generated by just one element though? $\endgroup$ Feb 28 '15 at 14:41
  • $\begingroup$ Not as an abelian group. $\endgroup$
    – MooS
    Feb 28 '15 at 14:42
  • $\begingroup$ Can you clarify why? $\endgroup$ Feb 28 '15 at 14:43
  • $\begingroup$ As an abelian group $\mathbb F_{p^n}$ is isomorphic to $(\mathbb Z/p\mathbb Z)^n$ $\endgroup$
    – MooS
    Feb 28 '15 at 14:44
  • $\begingroup$ Still, I do not understand where the "at most two elements" part comes from. $\endgroup$ Feb 28 '15 at 15:25

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