Let $\mathcal{M}$ be a commutative monoid and $x, y \in \mathcal{M}$.

In general, is there something one can say about $x$ and $y$ if for any $n \in \mathbb{N}$ the element $y^n$ divides $x$, i.e., there is a $z_n$ such that $x = y^n \cdot z_n$?

In certain concrete monoids the following possibilities appear.

  • $x$ is a zero element, that is, $x \cdot z = x$ for all $z$
  • $y$ must be idempotent (e.g., natural numbers, non-negative rationals under addition, non-negative reals under addition ...)
  • $y$ is invertible (any group)

What other possibilities are there, if any, and can one say something meaningful about commutative monoids where if $x$ and $y$ satisfy the condition above then either $x$ is a zero element or $y$ is idempotent?

Of course such a monoid cannot have non-idempotent invertible elements.

  • $\begingroup$ Then $y^n$ is a ''repeated prefix'' of $x$ $\endgroup$ – Wuestenfux May 26 '17 at 9:12
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    $\begingroup$ Take $\mathcal{M} = \mathbb{N}_{>0} +X\mathbb{Q}_{>0}[X]$ as a submonoid of $(\mathbb{Q}_{>0}[X],\times)$. In this one, $X \neq 0$, and $2$ is neither invertible nor idempotent, but its powers divide $X$. $\endgroup$ – nombre May 26 '17 at 9:17
  • $\begingroup$ I am a bit confused. Why are you saying "Of course such a monoid cannot have invertible elements". Your definition makes perfect sense in this case and furthermore, you consider yourself the case "y is invertible (any group)". $\endgroup$ – J.-E. Pin May 27 '17 at 9:47
  • $\begingroup$ @J.-E.Pin The last sentence is in reference to the previous one. Namely, if for every pair of elements $x$ and $y$ satisfying the condition it must be that either $x$ is a zero element or $y$ is idempotent, then it follows that the monoid does not have non-idempotent invertible elements. Does make it clearer? Did I make a mistake somewhere? $\endgroup$ – Aleš Bizjak May 28 '17 at 6:22

Recall that an epigroup (also called group-bound semigroup) is a semigroup in which every element has a power that belongs to a subgroup. Note that the identity of the subgroup is an idempotent which is not required to be an identity for the semigroup.

Let $S$ be an epigroup and let $x \in S$. Then some power of $x$, say $x^n$, belongs to group $G$ with identity $e_x$. Note that $e_x$ is entirely determined by $x$. Indeed, suppose that $x^m$ belongs to some group $H$ with identity $e$. Then $x^{nm} \in G \cap H$ and thus $e_x = e$.

Now, I claim that $y^n$ divides $e_y$ for all $n \geqslant 0$. Indeed, since $S$ is an epigroup, there exists $k > 0$ such that $y^k$ belongs to a group $G$ with identity $e_y$. Let us choose $q$ and $r$ such that $n = kq - r$ with $q > 0$ and $r \geqslant 0$. Then $y^ny^r = y^{kq} = (y^k)^q \in G$ and thus $y^n$ divides $y^ny^r$ which divides $e_y$.

It follows that $y^n$ divides $x$ for all $n \geqslant 0$ if and only if $e_y$ divides $x$.

  • $\begingroup$ This is nice! But I don't think it quite answers the question. You show that in epigroups if $y^n$ divides $n$ for all $n$ there is some idempotent $e_y$ which divides $x$. The requirement I have is that in such a case $y$ must be idempotent (or $x$ is a zero element). I don't think that follows, does it? But, I think what you wrote is still very useful for the application I have in mind, because you have shown that $y$ divides $e_y$ which divides $x$, so every element $x$ has an idempotent below it and above $y$ (in divisibility preorder). $\endgroup$ – Aleš Bizjak May 28 '17 at 6:53

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