3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.