# Let $\pi$ denote a prime element in $\mathbb Z[i], \pi \notin \mathbb Z, i \mathbb Z$. Prove that $N(\pi)=2$ or $N(\pi)=p$, $p \equiv 1 \pmod 4$

Let $\pi$ denote a prime element in $\mathbb Z[i], \pi \notin \mathbb Z, i \mathbb Z$. Prove that $N(\pi)=2$ or $N(\pi)=p$, $p \equiv 1 \pmod 4, p$ is a prime.

I know that $\pi$ is prime in $\mathbb Z[i]$ implies it is irreducible. Also if $\pi$ has a prime norm, that is $N(\pi) = p$ then $\pi$ is irreducible.

For any $z \in \mathbb Z[i]$ we have $N(z) = a^2 + b^2$. Since $\pi \notin \mathbb Z, i \mathbb Z$ both $a$ and $b$ must be non-zero in the case of $\pi$. By Fermat's Two Square theorem every prime number $\equiv 1 \pmod 4$ can be written as a unique sum of two squares. I have some idea I should utilize this theorem here, but no success so far.

However I've come to a dead end. I don't know how to go on from here. Could someone help me out ?

Thanks.

## 2 Answers

By Fermat's Two Square theorem every prime number $\equiv 1 \pmod{4}$ can be written as a unique sum of two squares.

So you know that rational positive primes $\equiv 1 \pmod{4}$ (and also $2$) are reducible in $\mathbb{Z}[i]$. You also know, or can easily check, that $\mathbb{Z}[i]$ is a Euclidean ring, hence a PID, hence a UFD.

Now consider $N(\pi) = \nu \in \mathbb{Z} \subset \mathbb{Z}[i]$. Let

$$\nu = \prod_{k=1}^r p_k^{\alpha_k}$$

be the prime factorisation in $\mathbb{Z}$. Let $p$ be a prime factor of $\nu$, and let $\pi_p$ be a prime element in $\mathbb{Z}[i]$ that divides $p$. Then

$$\pi_p \mid \nu = N(\pi) = \pi\overline{\pi} \Rightarrow (\pi_p \mid \pi) \lor (\overline{\pi_p} \mid \pi).$$

But that means $\pi \sim \pi_p$ or $\pi \sim \overline{\pi_p}$, in particular, $\pi \mid p$, whence $N(\pi) \mid N(p) = p^2$.

So there are two possibilities,

1. $N(\pi) = p$, and that is what we want to show (you just need to say why $p \equiv 3\pmod{4}$ is impossible under the hypothesis).
2. $N(\pi) = p^2$, but that would mean $\pi \sim p$, and $p$ itself would be prime in $\mathbb{Z}[i]$. You just need to say why that is impossible under the hypothesis.
• Shouldn't $\pi_p \in \mathbb Z[i]$ ? The only prime element dividing $p$ in $\mathbb Z$ is $\pm p$. Nov 22, 2013 at 17:24
• Sure. Typo. Thanks for noticing. Nov 22, 2013 at 17:33
• Also $\pi_p | \pi \lor \bar \pi$, and $\pi \sim \pi_p \lor \pi \sim \bar \pi_p$ Nov 22, 2013 at 17:39
• Does $\pi \sim \pi_p$ mean these elements are associated (differ by a unit) ? If so, then $\pi$ must be irreducible itself ? But $\pi$ is just arbitrary ? Nov 22, 2013 at 17:44
• $N(\pi) = a^2 + b^2 = p, a, b \neq 0 \Rightarrow p \equiv 1(\mod 4)$ since $p$ has a remainder of $1$ or $3$, and a square can have a remainder of $\{0, 1\} (\mod 4)$ we must have $a^2 + b^2 \equiv 1 \lor 2 (\mod 4)$ but $2$ is impossible hence $1$. Nov 22, 2013 at 17:56

Well...there you did all, didn't you? I mean, you say you know $\;N(\pi)=a^2+b^2\;$ is a prime in $\;\Bbb Z\;$, and thus it either is $\;2\;$ or else an odd prime that can be expressed as the sum of two squares $\;\iff p\neq 3\pmod 4\iff p=1\pmod 4\;$ ....and voila!

Now, it isn't true that $\;\alpha\in\Bbb Z[i]\;$ is a prime $\;\implies N(\alpha)\in\Bbb Z\;$ is a prime. For example, $\;N(7)=49\;$ , yet $\;7\in\Bbb Z[i]\;$ is a prime.

• Why ? I know $\pi$ is irreducible, but this doesn't necessarily mean that $N(\pi) = p$ ? I know that if $N(\pi) = p$ then $\pi$ is irreducible. I need to prove that $\pi$ is irreducible $\Rightarrow$ $N(\pi) = p$ ? Nov 22, 2013 at 14:02
• But that is false, @NicolasLykkeIversen: for example, $\;7\;$ is prime in the Gaussian Integers yet $\;N(7)=49\;$ ... Nov 22, 2013 at 14:17
• But how do I know my irreducible element $\pi$ has norm $N(\pi) = a^2 + b^2 = p \in \mathbb Z$, $p$ is a prime ? Nov 22, 2013 at 14:26
• It just doesn't, @NicolasLykkeIversen, as my example shows! Nov 22, 2013 at 14:27
• But you write $N(\pi) = a^2 + b^2$ has to be prime in $\mathbb Z$ ? Sorry if I don't understand - I appreciate your help. Nov 22, 2013 at 14:31