Let $(M,\cdot,1)$ be a commutative cancellative gcd monoid. That is,
- $\forall a,b\in M\colon a\cdot b =b\cdot a$
- $\forall a,b,c\in M\colon (c\cdot a = c\cdot b)\implies (a=b)$
- $\forall a,b\in M\colon \gcd(a,b)\neq\emptyset$
Let $\Gamma=\{a_1,\dots,a_n\}$ be a finite subset of $M$. Surely, $\langle\Gamma\rangle_\cdot$ is commutative and cancellative, but not necessarily gcd-closed.
Does there exist a finitely generated gcd monoid $M'$ such that $\langle\Gamma\rangle_\cdot\le M'\le M$?
I have now crawled all of the related literature I could find on the internet and still have no clear answer. Unfortunately, I am not a frequent in this theory and can only judge from the precisely formulated statements I can find in the literature.
It seems that if $M'$ existed, it would be a unique factorization monoid, since it would be generated by primes and units (as it is gcd). My gut tells me that this is unlikely to be the case.