6
$\begingroup$

Let $H$ be a multiplicatively written monoid with identity $1_H$. We write $H^\times$ for the set of units (or invertible elements) of $H$. We say that an element $a \in H$ is an atom if $a \notin H^\times$ and there do not exist $x, y \in H \setminus H^\times$ such that $a = xy$, and a prime if $a \notin H^\times$ and $a \mid_H xy$ implies $a \mid_H x$ or $a \mid_H y$. Here, $\mid_H$ is the divisibility preorder on $H$ (that is, $x \mid_H y$ iff $y \in HxH$).

Q. What is known about the class, $\mathcal M_{\sf p}$, of monoids for which every prime is an atom? Have they ever been studied? Do they have a special name?

It is seen that $\mathcal M_{\sf p}$ includes all commutative, unit-cancellative monoids ($H$ is unit-cancellative if $xy = x$ or $yx=x$ for some $x, y \in H$ implies $y \in H^\times$), while a free monoid with basis a set containing at least two elements is a non-commutative, cancellative example.

As for a non-example, it is enough to consider the case when $H$ is a non-trivial monoid with an absorbing element $0_H$, but no zero divisors (e.g., the multiplicative monoid of a domain): Here, $0_H = 0_H \cdot 0_H$ and $0_H \notin H^\times$, so $0_H$ is neither a unit nor an atom. However, $0_H$ is a prime, because $0_H \mid_H xy$, for some $x, y \in H$, only if $x = 0_H$ or $y = 0_H$.

$\endgroup$
7
$\begingroup$

Fact 1: If $M$ is a monoid where primes are atoms, then $M$ is Dedekind-finite.

Proof. Working contrapositively, assume $ab=1$ with $a,b\in M\setminus M^{\times}$. Now $a$ is prime since it divides every element $x\in M$ (because $x=xab$). On the other hand $a=a(ba)$ is a product of two non-units [if $ba\in M^{\times}$ then $a$ is left invertible, and $ab=1$ implies it is right invertible, a contradiction], so $a$ is not an atom. $\boxed{\,}$

Fact 2: If $M$ is a monoid satifying $$ (\dagger)\qquad \forall r,s,t\in M,\ (rst=s)\implies r,t\in M^{\times}, $$ then primes are atoms.

Proof. Assume contrapositively $p\in M$ is prime but not an atom. If $M$ is not Dedekind-finite, then $(\dagger)$ fails, and we are done. So we hereafter assume $M$ is Dedekind-finite.

Since $p$ is not an atom write $p=xy$ with $x,y\in M\setminus M^{\times}$. As $p|xy$, without loss of generality we may assume $p|x$. Thus $x=apb$ for some $a,b\in M$, and we then have $p=ap(by)$. As $y$ is not a unit, Dedekind-finiteness implies $by\notin M^{\times}$. Taking $r=a$, $s=p$, and $t=by$ shows that $(\dagger)$ fails. $\boxed{\,}$

Note that this class (the $(\dagger)$ monoids) encompasses both the commutative, unit-cancellative monoids, and the free monoids.

$\endgroup$
  • 1
    $\begingroup$ Nice! The non-example in the OP shows, on the other hand, that Dedekind-finiteness is not sufficient: In that example, $xy = 1_H$ iff $x=y=1_H$, but condition ($\dagger$) is not satisfied, since $x\cdot 0_X \cdot y = 0_X \ne 1_H$ for all $x, y \in X$. $\endgroup$ – Salvo Tringali Apr 3 '17 at 4:54
  • $\begingroup$ By the way, I was asking for some information about condition ($\dagger$) in a related topic: mathoverflow.net/questions/253337. Does it have a standard name? $\endgroup$ – Salvo Tringali Apr 3 '17 at 8:39
  • $\begingroup$ I don't know if $(\dagger)$ has a standard name. From the ring theory side of things, I've never seen it arise. Regarding your question of whether the property "every prime is an atom" is studied-- again from a ring theory perspective, primes come up all the time, but atoms (in my limited experience) mainly arise when studying commutative domains, and there the property is true. $\endgroup$ – Pace Nielsen Apr 3 '17 at 14:08
  • $\begingroup$ As for the 2nd part of your comment, you may want to have a look at a recent volume edited by S.T. Chapman, Fontana, Geroldinger & Olberding, Multiplicative Ideal Theory and Factorization Theory: Commutative & Non-Commutative Perspectives (Springer, 2016). In the meanwhile, your answer and the remarks made so far raise an obvious question: Is ($\dagger$) a necessary condition for a (Dedekind-finite) monoid $H$ to have the property that every prime is an atom? $\endgroup$ – Salvo Tringali Apr 3 '17 at 14:45
1
$\begingroup$

This should actually be a comment, but it's too long for that, so I'm posting it as an answer.


Pace Nielsen proved in this thread that, if $H \in \mathcal M_{\sf p}$ (the class of all monoids with the property that every prime is an atom, as specified in the OP), then $H$ is Dedekind-finite (i.e., $xy = 1_H$, for some $x, y \in H$, only if $yx = 1_H$). Moreover, he observed that every monoid $H$ such that $xyz=y$, for some $x,y,z \in H$, implies $x, z \in H^\times$ (condition ($\dagger$) in his answer), belongs to $\mathcal M_{\sf p}$. In a comment, I asked, as an obvious follow-up, if the converse of this latter fact is true, and I've just realized that the answer is no.

Indeed, let $X$ be a set with $|X| \ge 3$, pick two distinct elements $0_X$ and $1_X$ in $X$, and define a multiplication $\ast$ on $X$ by taking, for all $x \in X$ and $y, z \in X \setminus \{1_X\}$, $x \ast 1_X :=1_X \ast x := x$ and $y\ast z:= 0_X$. Then $H=(X,\ast)$ is a reduced, commutative monoid with zero ("reduced" means that the only unit is the identity), where $1_X$ is the identity and $0_X$ the absorbing element.

Clearly, $0_X$ is not an atom of $H$, because $0_X \ast 0_X = 0_X \ne 1_X$. Moreover, if $a \in X \setminus \{1_X, 0_X\}$ and $a = x \ast y$ for some $x, y \in X$, then one of $x$ and $y$ must be equal to $1_X$, otherwise $x \ast y = 0_X \ne a$. Therefore, the set of atoms of $H$ is given by $X \setminus \{1_X, 0_X\}$.

On the other hand, $0_X$ is not a prime of $H$, because having assumed $|X| \ge 3$ yields the existence of an element $x \in X \setminus \{0_X\}$ with $x \ast x = 0_X$, so that $0_X \mid x \ast x$, yet $0_X \nmid x$. Therefore, the primes of $H$, if any, are necessarily contained in $H \setminus \{1_X, 0_X\}$, that is, are atoms.

Finally, it is clear that $H$ doesn't satisfy condition ($\dagger$), because $x \ast 0_X \ast y = 0_X$ for all $x, y \in X$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.