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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


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.