# For which abelian groups $G$ does the monoid of zero-sum sequences over $G$ embed into a ring as a divisor-closed subsemigroup?

Let $K$ be a multiplicatively written semigroup (either commutative or not) and $H$ a subsemigroup of $K$. We say that $H$ is divisor-closed (in $K$) if $x \in H$ for all $x, y \in K$ such that $x \mid_K y$ (i.e., $y = uxv$ for some $u, v \in K$) and $y \in H$.

Accordingly, we say that a semigroup $S$ is annular (bear with me, I don't have a better word for the moment) if it embeds as a divisor-closed subsemigroup into the multiplicative monoid of a ring. So here is my question:

Q. Given an abelian group $G$, denote by $\mathscr B(G)$ the monoid of zero-sum sequences over $G$, that is, the submonoid of $\mathscr F(G)$, the free abelian monoid with basis $G$, given by the inverse image of $0_G$ under the canonical (monoid) epimorphism $\mathscr F(G) \to G$. For which $G$ is $\mathscr B(G)$ annular?

This is a very special case of a more general question (namely, when does a semigroup embeds into a ring etc.?), which, however, is also much harder and beyond the scope of this post. Indeed, my motivation is simply the following: Factorization theory (that is, the theory of non-unique factorization) grew up out of algebraic number theory, and a turning point in its history was when the theory, until then focused on integral domains, was reforged in the language of monoids, based at least in part on the consideration that the latter provide "models" (for studying various phenomena of interest) that wouldn't have been available otherwise, with monoids of zero-sum sequences being probably the most representative of these models. So, the whole point of my question is that I'd like to understand to which extent the statement in bold is well-grounded.

Update #1. A fruitful exchange with Benjamin Steinberg in the comments below has eventually shown that if $H$ is a non-trivial monoid with trivial group of units (as in the case of interest here) embedding as a divisor-closed subsemigroup into the multiplicative monoid of a unital ring $R$, then the characteristic of $R$ is $2$, simply because the condition of divisor-closedness implies $R^\times \cong H^\times = \{1_H\}$.

• Does the semigroup ring work? – Benjamin Steinberg Jul 25 '17 at 9:32
• @BenjaminSteinberg Let me answer with another question: Let $K$ be a field and $H$ a commutative, cancellative, torsion-free monoid. Pick $f, g \in K[H]$, and denote by $\ast$ the multiplication in the monoid ring $K[H]$. Is it true that $|{\rm supp}(f \ast g)| \ge |{\rm supp}(f)| + |{\rm supp}(g)|-1$? Assume, if necessary, that $K$ has characteristic zero. – Salvo Tringali Jul 25 '17 at 17:52
• I am not sure the support can be controlled that way but H is divisor free in the monoid ring ZH – Benjamin Steinberg Jul 26 '17 at 11:00
• Do you not want factor closed up to units? – Benjamin Steinberg Jul 26 '17 at 11:36
• As for 2nd to last of yours comments: Divisor-free = divisor-closed? Sorry, I'm not familiar with the former term, while the use of the latter is widespread (at least in factorization theory). As for your last comment: No, I don't want divisor-closedness to be defined up to associates. Therefore, if $H$ and $K$ are monoids and $H$ is divisor-closed (in $K$), then $H^\times = K^\times$ and $\mathcal A(H) = \mathcal A(K) \cap H$, where $\mathcal A(M)$ denotes the set of atoms (aka irreducible elements) of a given monoid $M$. – Salvo Tringali Jul 26 '17 at 12:04

Figured it out (sorry for answering my own question). I'll prove the following:

Lemma. Let $H$ be a linearly orderable monoid and $R$ a domain whose group of units is trivial. Then $H$ embeds as a divisor-closed submonoid into the multiplicative monoid of the monoid ring $R[H]$.

This will show that the answer to the question in the OP is "For all $G$", since $\mathscr B(G)$ is a torsion-free, cancellative, commutative monoid, and hence is linearly orderable.

Proof of the lemma. Let $\preceq$ be a total order on $H$ for which $(H, \preceq)$ is a linearly ordered monoid, that is, $xz \prec yz$ and $zx \prec zy$ for all $x, y, z \in H$ with $x \prec y$ (as usual, $u \prec v$ means $u \preceq v$ and $u \ne v$). Moreover, let $\delta$ be the canonical embedding $H \to R[H]$, so that $\delta(x)$ is, for every $x \in H$, a Kronecker delta $H \to R$ centered at $x$, which I'll rather denote by $\delta_x$.

Now pick $x \in H$, and assume $\delta_x = f\ast g$ for some $f, g \in R[H]$, with $\ast$ being the multiplication in $R[H]$. I claim that $|{\rm supp}(f)| = |{\rm supp}(g)| = 1$. Indeed, it is clear that $f$ and $g$ cannot be identically zero, so the support of each of them is non-empty. Accordingly, let $y_f$ and $z_f$ be, respectively, the minimum and the maximum of $S_f := {\rm supp}(f)$ relative to the order $\preceq$ (which exist because $\preceq$ is total and $S_f$ is not only non-empty, but also finite); define $y_g$ and $z_g$ in a similar way (only with $g$ in lieu of $f$). Then set $y := y_f y_g$ and $z := z_f z_g$. We have $$\delta_x(y) = \sum_{uv=y} f(u) g(v) = f(y_f) g(y_g) \ne 0_R.$$ To see this, note that, if $u \prec y_f$ or $v \prec y_g$, then $f(u) g(v) = 0_R$, and on the other hand, if $y_f \preceq u$ and $y_g \preceq v$, then $y = y_f y_g \preceq uv$, with equality iff $u = y_f$ and $v = y_g$ (by the assumption that $H$ is linearly ordered by $\preceq$). Then the rest is trivial, because $y_f \in {\rm supp}(f)$ and $y_g \in {\rm supp}(g)$ give $f(y_f) g(y_g) \ne 0_R$ (by the fact that $R$ is a domain).

To wit, we have shown that $y$ is in the support of $\delta_x$, and so is $z$ by an analogous argument. It follows $x = y = z$, which is, however, possible only if $y_f = z_f$ and $y_g = z_g$ (by construction, $y_f \preceq z_f$ and $y_g \preceq z_g$, and we can't have $y_f \neq z_f$ or $y_g \ne z_g$, otherwise $H$ being linearly ordered by $\preceq$ would yield a contradiction).

In other words, $f = a\delta_u$ and $g = b \delta_v$ for some $a, b \in R$ and $u, v \in H$ such that $ab= 1_R$. But $R^\times$ is trivial (by hypothesis), and therefore $a = b = 1_R$. So $f, g \in \delta(H)$, and we are done. []

In the last line of the proof, we use that domains are, of course, Dedekind-finite rings, so that a product is a unit iff so are all the factors (see here).