A slick proof of "The ring of integers of a number field has infinitely many non-associated atoms"?

Question

Let $\mathbf Z_K$ be the ring of integers of an algebraic number field $K$. It is well known that $\mathbf Z_K$ has infinitely many non-associated atoms (and hence is not a Cohen-Kaplansky domain).

Q. Is there a slick proof of this result? If so, could you provide a reference?

I feel the answer should be yes, but the only proof I knew until a few days ago is based on proving something much stronger and is sensibly more complicated than what I'm keen to consider a ``slick proof'': that $\mathbf Z_K$ has infinitely many prime ideals, and each of these contains a prime element.

Answer 1

Fact 1. The ring $\textbf{Z}_K$ has infinitely many prime ideals and has Krull dimension $1$.

Proof. We have a surjective morphism $\mathrm{Spec}\,\textbf{Z}_K\rightarrow \mathrm{Spec}\mathbb{Z}$. This follows from the fact that the inclusion $\mathbb{Z}\subseteq \textbf{Z}_K$ is an integral extension of rings. So it suffices to invoke the going-up theorem of Cohen-Seidenberg. There is at least one prime ideal of $\textbf{Z}_K$ in each fiber of the morphism.

Fact 2. Let $A$ be a Noetherian domain. Then every nonzero and non-invertible element $a\in A$ can be decomposed as

$$a = f_1\cdot f_2\cdot ...\cdot f_n$$

where $f_1,...,f_n\in A$ are irreducible.

Proof. Fix $a\in A$ as in the statement. Let $\mathcal{F}$ be a family of proper principial ideals of $A$ that contain $a$. Since $A$ is Noetherian, in $\mathcal{F}$ there exists a maximal element with respect to inclusion. Say that this maximal element is generated by some $f_1\in A$. Then $a = f_1\cdot a_1$. If $a_1$ is non-invertible apply this argument to $a_1$ and obtain $f_2$ and $a_2$. In other words, continue by induction. It is clear that

1. All $f_1,f_2,...$ are irreducible (otherwise ideals generate by them won't be maximal among principial ideals).

2. This procedure must stop at some point. Indeed, $a_{n+1}$ must be invertible for some $n\in \mathbb{N}$, since otherwise you will obtain infinite and increasing chain of ideals $Aa_1\subseteq Aa_2\subseteq ...\subseteq Aa_n\subseteq ...$ in the Noetherian ring $A$.

This implies that $a = f_1\cdot f_2\cdot ...\cdot f_n$.

Remark. The decomposition in Fact 2 is by no means unique.

Corollary. All nonzero prime ideals of $\textbf{Z}_K$ are maximal.

Now you can construct infinite sequence of irreducible elements of $\textbf{Z}_K$ as follows. Pick $\{\mathfrak{p}_n\}_{n\in \mathbb{N}}$ a sequence of pairwise distinct maximal ideals in $\textbf{Z}_K$. This is possible by Fact 1 and Corollary. Prime avoidance shows that

$$\mathfrak{p}_{n+1}\subsetneq \bigcup_{k=0}^n\mathfrak{p}_k$$

Hence for every $n\in \mathbb{N}$ there exists $$r_{n+1}\in \mathfrak{p}_{n + 1}\setminus \left(\bigcup_{k=0}^n\mathfrak{p}_k\right)$$

and by virtue of Fact 2 you may pick irreducible divisor $f_{n+1}$ of $r_{n+1}$ in $\mathfrak{p}_{n+1}$. You still have that $$f_{n+1}\in \mathfrak{p}_{n + 1}\setminus \left(\bigcup_{k=0}^n\mathfrak{p}_k\right)$$ and hence $\{f_n\}_{n\in \mathbb{N}}$ are irreducible elements of $\textbf{Z}_K$ such that no two differ by unit.

This actually proves the following.

Proposition. Let $A$ be a Noetherian domain with infinitely many maximal ideals, then $A$ contains an infinite set of irreducible elements such that no two differ by unit.

Remark. If $A$ is a Dedekind ring with finitely many prime ideals, then one can prove that $A$ is a UFD and hence the proposition above does not hold.