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Suppose $(S,\cdot)$ is a semigroup with neutral element $e$ (i.e. $xe=ex=x$ for all $x\in S$) and the following properties:

  1. S is commutative: $xy=yx$.
  2. S is cancellative: $xy = xz$ implies $y = z$.
  3. 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?

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  • $\begingroup$ The question is about as broad as the question: when are irreducibles prime in a domain? It's difficult to say much of interest at that level of generality (except the obvious, e.g. when gcds exist, or equivalent properties). What is your motivation? $\endgroup$ – Bill Dubuque Jul 14 '14 at 15:20
  • $\begingroup$ Thanks. I added an example to be more specific about my motivation. $\endgroup$ – Thomas Rippl Jul 15 '14 at 5:40
  • $\begingroup$ Thomas I think you should explicitly include a $0$ element that is absorptive in your semigroup. $\endgroup$ – goblin Jul 15 '14 at 6:26
  • $\begingroup$ @user18921: thanks. I added the zero element $e$ in the description more clearly. $\endgroup$ – Thomas Rippl Jul 15 '14 at 12:00
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    $\begingroup$ You added the zero element $e$, but in 3, you say that $e$ is the neutral element of S. I strongly advise you to use $1$ for the neutral element and $0$ for the zero. $\endgroup$ – J.-E. Pin Jul 22 '14 at 8:00
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The question is not "well-posed" since it is either too general (not enough information provided) or classical.

Classical: A good overview on factorization theory of commutative semigroups is given in Chapter 15 in Clark's lecture notes in Commutative Algebra.

In some cases one may find a divisor theory for the monoid. An overview and some ideas are presented in "Realization theorems for semigroups with divisor theory" by Geroldinger and Halter-Koch

An article about the factorization (what is a prime element, etc.) in topological semigroups by Jan Snellman is here.

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