# On the factorization of powers of atoms in the ring of integers of a number field

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).

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 $|\mathsf L_H(a^v)| \ge 2$?

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 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.

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

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 $|\mathsf L_{\mathscr B(G)}(\mathfrak a^v)| \ge 2$?

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, $\le |G|$.

• For given $G$, $H$ contains finitely many minimal zero-sum sequences, or where I am wrong? – Fedor Petrov Jun 12 '17 at 18:27
• @FedorPetrov Nowhere: you're right, I was missing the obvious! So this finishes the proof. Actually, much more is true: The answer to Q1 is yes not only when $H$ is the multiplicative monoid of $\mathbf Z_K$, but more in general for any transfer Krull monoid of finite type (and there is a wealth of them out there), or even more in general for any monoid $H$ that is essentially equimorphic to a commutative monoid $M$ with finitely many non-associate atoms. Why don't you post your comment/remark as an answer (in such a way that we can mark the thread as "answered")? – Salvo Tringali Jun 12 '17 at 19:39

## 1 Answer

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.