Suppose $(S,\cdot)$ is a semigroup with neutral element $e$ (i.e. $xe=ex=x$ for all $x\in S$) and the following properties:
- S is commutative: $xy=yx$.
- S is cancellative: $xy = xz$ implies $y = z$.
- For $n \geq 2$ it is never true that $x^n= x$ except for $x = e$, the neutral element of $S$.
Question: Are there elementary additional conditions ensuring that an irreducible element is prime?
$x\in S \setminus \{e\}$ is called irreducible, if $x = yz$ implies $y = x$ or $y = e$.
$x \in S \setminus \{e\}$ is called prime, if for any $y,y' \in S$ with $x z = yy'$ for a $z \in S$ it is true that there exists a $w \in S$ such that $x w = y$ or $x w =y'$ (one can also formulate: if $x$ divides $yz$ then it already divides $y$ or $z$).
I would be happy if someone could provide a result not restricted to finite semigroups. An example I have in mind is $S = \mathcal{N}_f(\mathbb{R})$ the set of finite point measures $m$ on $\mathbb{R}$, i.e. measures of the form $m=\sum_{i=1}^k n_i \delta_{x_i}$ for some $x_i \in \mathbb{R}$, $n_i \in \mathbb{N}$ for $i = 1, \dotsc, k\in \mathbb{N}_0$. A binary operation on $S$ is the addition of measures and the neutral element is $e=0$, the null measure. One can check that irreducible and prime elements coincide: $\{\delta_x: x \in \mathbb{R}\}$. Are there additional "features" of this particular semigroup responsible for the coincidence of the two sets?
