Is a subring of an integral monoid ring an integral monoid ring?

Question

My question is basically in the title.

Given a monoid $M$, consider the integral monoid ring $R$ it defines. Any subring $S$ of it is also a subgroup of the free abelian group $(R,+)$, hence free abelian. So it is generated over the integers by some set, say $N$. I would like to be able to say that then $N$ is necessarily a submonoid of $M$ and hence $S$ is a integral monoid ring generated by $N$, but I don't think it's true. Can't think of a specific counterexample though.

Any help appreciated.

Answer 1

Take $\mathbb{Z}[x]$, so $M = \mathbb{N}$. Let $S = \mathbb{Z}[2x,x^2] \cong \mathbb{Z}[y,z]/(y^2-4z)$ be the subring.

If $S = \mathbb{Z}[N]$ for a monoid $N$, then its base change to $\mathbb{F}_2$ and to $\mathbb{Q}$ are both monoid rings and have the same number of generators. However, $S \otimes_{\mathbb{Z}} \mathbb{F}_2 \cong \mathbb{F}_2[y,z]/(y^2-4z) = \mathbb{F}_2[y,z]/(y^2)$, which has 2 generators, while $S \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{Q}[y]$, which has 1 generator.

Answer 2

The question in the title and the body are different. Let me answer the one in the body. Let $U$ be the two element semilattice $\{1,a\}$ where $aa=a$ and $1$ is identity. Then $\mathbb ZU\cong\mathbb Z^2$ via the decomposition $1=(1-a)+a$ into central idempotents. Now $U$ has only two submonoids but $\mathbb Z^2$ has more than two subrings. There is the whole ring, the diagonal subring and one can look at subrings of the form $(a,b)$ with $a-b$ divisible by some fixed $n$.

Added. The subring $R$ of $\mathbb Z\times\mathbb Z$ consisting of all elements $(a,b)$ with $a-b$ divisible by $2$ is not an integral monoid ring for any monoid. It is free abelian of rank $2$. The two element monoids are $U$ and $\mathbb Z/2\mathbb Z$. $R$ has only two idempotents and so is not isomorphic to $\mathbb ZU$ (which has four being isomorphic to $\mathbb Z^2$). But $R$ is also not isomorphic to the group ring of $\mathbb Z/2\mathbb Z$ because the only element of order $2$ in $R$ is the negative of the identity.