# Let $\pi$ denote a prime element in $\mathbb Z, \pi \notin \mathbb Z, i \mathbb Z$. 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.

#### Solutions Collecting From Web of "Let $\pi$ denote a prime element in $\mathbb Z, \pi \notin \mathbb Z, i \mathbb Z$. Prove that $N(\pi)=2$ or $N(\pi)=p$, $p \equiv 1 \pmod 4$"

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.

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.