I know the classification of Gaussian primes: let $u$ be a unit of $\mathbb{Z}[i]$. Then the following are all Gaussian primes:

1) $u(1+i)$

2) $u(a+ib)$ where $a^2+b^2=p$ for some prime number p congruent to $1$ modulo $4$

3) $uq$ where $q$ is a prime number congruent to $3$ modulo $4$.

If i knew that $\mathbb{Z}[i]$ is a Principal ideal domain, i could say: a nonzero ideal $P$ of $\mathbb{Z}[i]$ is prime iff it is principal prime, say $P=p\mathbb{Z}[i]$ iff $p$ is a Gaussian prime, and conclude that every nonzero prime ideal $P$ of $\mathbb{Z}[i]$ belongs to one of the following families:

1) $P=(1+i)\mathbb{Z}[i]$

2) $P=(a+ib)\mathbb{Z}[i]$, where $a,b\in\mathbb{Z}$ and $a^2+b^2=p$ is an odd prime congruent to $1$ modulo $4$

3) $P=q\mathbb{Z}[i]$ where $q$ is an odd prime congruent to $3$ modulo $4$.

Now, suppose that for some reason i don't know that $\mathbb{Z}[i]$ is a PID. Can i still prove the classification above? I can't say an ideal is prime iff its generator is prime, so, how could i proceed in this case?

  • $\begingroup$ As it happens $\mathbb{Z}[i]$ is a Euclidean domain. Would using this fact be considered cheating? $\endgroup$ – awwalker May 24 '13 at 21:59

To prove this result, you must prove along the way that $\mathbb{Z}[i]$ is a PID, to guarantee that you have not missed any non-principal ideals. One cheap way to get around your restriction is to implicitly use the implication $$\text{Euclidean domain} \Longrightarrow \text{PID}.$$ In the case at hand, we obtain a Euclidean algorithm on $\mathbb{Z}[i]$ as follows. Given non-zero $x,y \in \mathbb{Z}[i]$, let $z \in \mathbb{Z}[i]$ denote the lattice point nearest $x/y$. Geometrically, it's clear that $$\vert z - x/y \vert \leq 1/\sqrt{2},$$ in which the metric above is the standard metric on the plane. Then $x-yz \in \mathbb{Z}[i]$ satisfies $\vert z-xy \vert < \vert y \vert$, so that $\vert \cdot \vert$ defines a Euclidean function on $\mathbb{Z}[i]$. As in the case with the integers, a Euclidean algorithm can be used to find GCD (up to units).

Now, suppose that $\mathfrak{p}=(a,b) \subset \mathbb{Z}[i]$ is a prime ideal. If $c$ denotes any GCD of $a$ and $b$, we have $c=(a,b)$. Induction (and the fact that integer rings are Noetherian) implies that all ideals are principal. For your problem, it would then suffice to classify all the principal ideals in $\mathbb{Z}[i]$, which is standard practice (look at norms).

Note: while this approach seems contrary to the question asked, I nevertheless believe that the best way to classify the prime ideals in $\mathbb{Z}[i]$ is to first prove that such ideals must be principal. Other techniques would simply reprove this result, without the benefit of machinery.

Edit: This edit includes a proof of the fact that $\mathbb{Z}[i]$ is Noetherian. Let $$\mathfrak{p}_1 \subsetneq \mathfrak{p}_2 \subsetneq \ldots $$ be an infinite chain of ideals in $\mathbb{Z}[i]$. We note that $a_k:=\#\mathbb{Z}[i]/\mathfrak{p}_k$ is finite for all $k$, because both $\mathbb{Z}[i]$ and $\mathfrak{p}_k$ are abelian groups of rank $2$. The sequence of positive integers $\{a_k\}$ is decreasing and does not stabilize, a contradiction. Thus no such chains exist.

  • $\begingroup$ @A Walker thank you very much for your answer, i can't still understand how does the ring of Gaussian integers inherit noetherianity from the ring of integers. If you want/have time, you could explain me that. $\endgroup$ – bateman May 25 '13 at 6:36
  • $\begingroup$ When I say "ring of integers", I refer to the ring of algebraic integers in a number field. I'll add this as an edit above. $\endgroup$ – awwalker May 25 '13 at 6:46
  • $\begingroup$ Dear A Walker, A minor comment: one doesn't need Noetherianness to conclude that $\mathbb Z[i]$ is a PID; it follows directly from being a Euclidean domain. (One that gets Noetherianness as a consequence, since PIDs are Noetherian.) Regards, $\endgroup$ – Matt E May 25 '13 at 7:00
  • $\begingroup$ @MattE I am aware. I wanted to say that arbitrary ideals were finitely generated (to proceed to show that they were generated by a single element), and could not think of an easier way. $\endgroup$ – awwalker May 25 '13 at 8:12
  • $\begingroup$ @AWalker Thank you so much for your edit, and i'm sorry if i'm annoying with so many questions, but how can i see that the quotient of two abelian groups of rank 2 is finite? $\endgroup$ – bateman May 25 '13 at 8:44

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