Let $H$ be a multiplicatively written monoid with identity $1\_H$. An *atom* of $H$ is a non-unit element $a \in H$ that doesn't split into the product of two non-unit elements. 


Given $x \in H$, we take $\mathsf L\_H(x) := \{0\} \subseteq \mathbf N$ if $x = 1\_H$; otherwise, $\mathsf L\_H(x)$ is the set of all $k \in \mathbf N^+$ for which there exist atoms $a\_1, \ldots, a\_k \in H$ such that $x = a\_1 \cdots a\_k$. Then, we say that an element $x \in H$ is *strongly irreducible* if $\mathsf L\_H(x^n) = \{n\}$ for all $n \in \mathbf N$ (in particular, every strongly irreducible element is an atom). 



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> **Q1.** Assume $H$ is the multiplicative monoid of the ring of integers, $\mathbf Z\_K$, of a number field $K$. Does there exist $v \in \mathbf N^+$ (depending only on $K$) such that, for every atom $a \in H$, the following holds: Either $a$ is strongly irreducible, or $|\mathsf L\_H(a^v)| \ge 2$?
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Of course, the second clause is equivalent to $|\mathsf L\_H(a^n)| \ge 2$ for every $n \ge v$. Moreover, it is obvious that, if $a \in H$ is an atom, but is not absolutely irreducible, then $|\mathsf L\_H(a^n)| \ge 2$ for all large $n$: The point of the question is that the exponent $v$ is required to be uniform with respect to the choice of $a$.


*Update 1.* The question can be shifted from $\mathbf Z\_K$ to the monoid of zero-sum sequences over a finite abelian group $G$, and from there to the subcase when $G$ is cyclic. 


Indeed, it is (well?) known, see Propositions 8.3 and 12 in [A. Geroldinger, *Sets of lengths*, Amer. Math. Monthly **123** (2016), No. 10, 960-988], that the multiplicative monoid of the non-zero elements of $\mathbf Z\_K$ is [essentially equimorphic](https://mathoverflow.net/questions/266082/) to the monoid of zero-sum sequences over the (ideal) class group of $K$, which is a finite abelian group: This proves the first of the reductions alluded to in the previous paragraph, because if $\varphi: M \to N$ is an essentially surjective equimorphism, then 


* (i) $\mathsf L\_M(x) = \mathsf L\_N(\varphi(x))$ for all $x \in M$, and
* (ii) for every $y \in N$ there is $x\_y \in M$ such that $\mathsf
 L\_M(x\_y) = \mathsf L\_N(y)$.


As for the second reduction, it is more or less straightforward from the [primary decomposition of finite abelian groups](https://en.wikipedia.org/wiki/Finitely_generated_abelian_group#Primary_decomposition).


All in all, the above considerations show that the *real* question I'm asking is:



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> **Q2.** Let $G$ be a finite cyclic group, and let $\mathscr B(G)$ be the monoid of zero-sum sequences over $G$ (namely, the kernel of the canonical epimorphism from the free abelian monoid with basis $G$ to $G$). Is it true that there exists $v \in \mathbf N^+$ (depending only on $G$) such that, for every *minimal* zero-sum sequences $\mathfrak a \in \mathscr B(G)$, the following holds: Either $\mathfrak a$ is strongly irreducible, or $|\mathsf L\_{\mathscr B(G)}(\mathfrak a^v)| \ge 2$?
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In this way, the question has a more combinatorial flavor, and my *guess* is that the answer is obviously yes (though I'm still missing some details).


*Update 2.* As a matter of fact, I was missing the obvious. As pointed out by Fedor Pedrov in the comments, reducing **Q1** to the monoid $\mathscr B(G)$ of zero-sum sequences over a *finite* group $G$ is (more than) enough to conclude, because $\mathscr B(G)$ has finitely many atoms: The length of a minimal zero-sum sequence over $G$ is bounded above by the Davenport constant of $G$, which is, in turn, $\le |G|$.



On the factorization of powers of atoms in the ring of integers of a number field
For given $G$, $H$ contains finitely many minimal zero-sum sequences. Indeed, any sequence containing the same element more than $|G|$ times is not minimal. Thus the result. 


