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Let $a \in \mathbb{Z}[i]$ be a prime element. Show that $N(a)$ is a prime or a square of a prime.

I know that the converse is true. That is, if the norm of $a$ is prime then $a$ is a Gaussian prime. But how do I prove the claim above? Thanks!

user112358
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1 Answers1

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Note that $N(a)=a\overline{a}$ is an integer. Since $a$ is prime (in particular not a unit and nonzero), it follows that $a$ divides some prime $p \in \mathbb{Z}$, $p=ab$ say. And therefore, $p^2=N(p)=N(a)N(b)$. By uniqueness of the factorization, $N(a)=p$ or $N(a)=p^2$.

Plankton
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  • Can you explain why $a$ divides some prime $p \in \mathbb{Z}$? From my understanding $a=x+yi$ and being a Gaussian prime means there is no non-unit Gaussian integer dividing $a$. Thank you! – user112358 Feb 04 '16 at 20:00
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    You can factorize $N(a)$ over the integers and there will be non trivial factors since $a$ is not a unit and nonzero. But the equation $N(a)=a\overline{a}$ says that $a$ divides $N(a)$ and thus, by the definition of being a prime element, it must divide some prime $p\in \mathbb{Z}$ appearing in the factorization of $N(a)$. – Plankton Feb 04 '16 at 20:02