Indecomposable monoids

Question

Let $M$ be a commutative reduced and cancellative monoid and $K(M)$ its group of quotients.

We say that $M$ is indecomposable if for every divisor-closed submonoids $M_1$ and $M_2$, $M=M_1\oplus M_2$ implies $M_1=\{0\}$ or $M_2=\{0\}$. ($M=M_1\oplus M_2$ means that every element $x\in M$ decomposes uniquely as a sum $x=x_1+x_2$, with $x_1\in M_1$ and $x_2\in M_2$.)

It is clear that if the abelian group $K(M)$ is indecomposable, then the monoid $M$ is indecomposable. I don't know when the converse holds ?

Answer 1

I think the following is an example with $M$ indecomposable and $K(M)$ decomposable. Let $M$ be the submonoid of $\mathbb N^2$ generated by $(2,1),(3,1),(0,1)$. Then $K(M)\cong \mathbb Z^2$. Trivially $M$ is reduced since $\mathbb N^2$ is reduced.

I claim that $M$ is indecomposable. Note that $(2,1),(3,1),(0,1)$ are irreducible in $M$ (i.e., do not belong to $M\setminus \{0\}+M\setminus \{0\}$) and hence if $M=M_1\oplus M_2$, then each of these elements belongs to one of the direct summands. Note that $3(2,1)=2(3,1)+(0,1)$. Since each of $(2,1),(3,1), (1,0)$ are in one of the direct summands but they can't all be in the same summand, this gives a contradiction.