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](https://www.sciencedirect.com/science/article/pii/002186939290234D)). 



> 
> **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.



A slick proof of "The ring of integers of a number field has infinitely many non-associated atoms"?
**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**](https://en.wikipedia.org/wiki/Prime_avoidance_lemma) 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.


