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Suppose $\pi = a+bi \in \mathbb{Z}[i]$ is irreducible. Is there a prime number, $p$, such that $N(\pi) \in \{p,p^2\}$? ($N(\pi) := |a+bi|^2 = a^2+b^2$)

Evan Aad
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Ryan S
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1 Answers1

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We know that $\pi|N(\pi)$, since $N(\pi)=\pi\cdot\overline{\pi}$. Since $\mathbb{Z}[i]$ is a Euclidean domain, and since an irreducible element of a Euclidean domain is prime, there is a prime number $p \in \mathbb{N}$ dividing $N(\pi)$, for which $\pi|p$.

Then $\pi|p\Rightarrow N(\pi)|N(p)$. But $N(p)=p^2$, so $N(\pi)=1, p$ or $p^2$. But $\pi$ is irreducible, hence, in particular, not a unit of $\mathbb{Z}[i]$, so we can exclude $1$.

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Bob Jones
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  • Sorry I lose you towards the end of the first line. My definition of prime in Z[i] is that is $p|ab$ then $p|a$ or $p|b$. I don't see how you know there is a prime $p$ such that $p|N(\pi)$ and $\pi |p$? – Ryan S Oct 23 '16 at 21:34
  • Take the prime factorization of $N(\pi)$, say $N(\pi)=p_1\cdot p_2\cdot\ldots$. Then $\pi$ divides this product, so it divides one of the primes. – Bob Jones Oct 23 '16 at 21:36
  • Okay Thanks I think I almost get it now. Is it correct that if $N(\pi)=1$ then $\pi $ is a unit but it can't be because $\pi $ is a irreducible so we are done. – Ryan S Oct 23 '16 at 21:41
  • @BobJones: I don't understand why, if $\pi$ divides this product, it must divide one of the primes. Could you please elaborate on this point? – Evan Aad Oct 23 '16 at 21:43
  • @EvanAad: this is the definition of prime that Ryan S uses, and in fact is very standard. That is: if $q|ab$ then $q|a$ or $q|b$. We just use the (equivalent) extended version: if $q|abcd\ldots$ then $q|a$ or $q|b$ or $q|c$ or $q|d$ or... And we use the $\mathbb{Z}[i]$ version, so we say $\pi$ instead of $q$. – Bob Jones Oct 23 '16 at 21:46
  • @BobJones: But $\pi$ is not necessarily a prime number. It is irreducible in $\mathbb{Z}[i]$, but not necessarily in $\mathbb{Z}$. In fact, in may not even belong to $\mathbb{Z}$. – Evan Aad Oct 23 '16 at 21:49
  • @EvanAad: Ryan above says that he assumes that prime and irreducible are equivalent. If you do not assume this, then my answer is incomplete, but this just comes from unique factorization in $\mathbb{Z}[i]$, or a whole host of other corollaries of the Euclidean algorithm for the Gaussian integers. Assuming unique factorization, for example, we can say that if $q|ab$ then $q$ is among the irreducible factors of $ab$, so it must come from the factors of $a$ or $b$ by uniqueness. So $q|a$ or $q|b$. – Bob Jones Oct 23 '16 at 21:52
  • @BobJones: I'm sorry, but I don't follow you. I'm sorry to be a nag, but I really want to understand your answer. You wrote: "Ryan above says that he assumes that prime and irreducible are equivalent." Where exactly does he say so? Could you please cite the phrase he uses to express this thought? – Evan Aad Oct 23 '16 at 21:57
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    @EvanAad Since $\Bbb{Z}[i]$ is a Euclidean Ring it can be shown that $p$ is irreducible $\iff$ $p$ is prime. :) – Ryan S Oct 23 '16 at 21:59
  • @RyanS: Thanks. Now I get it. – Evan Aad Oct 23 '16 at 22:05