# Integral monoid rings and Ore conditions

Consider a cancellative monoid $$S$$ satisfying the left Ore condition, so it embeds in a group $$G=S^{-1}S$$. Consider also the integral monoid rings $$\mathbb Z[S]$$ and $$\mathbb Z[G]$$.

I have two, probably trivial, questions:

1. Can one prove that $$S$$ satisfies the left Ore condition (for rings) as a multiplicative set in $$\mathbb Z[S]$$? I have tried both to find references and to prove it myself, but without success.

2. Even if the answer to (1) is negative, one can still consider the map $$\mathbb Z[S]\to \mathbb Z[G]$$ and use it to make a left $$\mathbb Z[S]$$-module $$M$$ into a left $$\mathbb Z[G]$$-module $$\bar M:=\mathbb Z[G]\otimes_{\mathbb Z[S]}M.$$ Furthermore, there is a canonical map $$\varphi \colon M\to \bar M$$, defined by $$\varphi(m):=1\otimes m$$, for all $$m\in M$$. Is there an easy description for $$\ker(\varphi)$$? For example, is $$\ker(\varphi)=\{m\in M:\exists s\in S, sm=0\}$$?

Let $$s\in S$$ and $$r\in \mathbb{Z}[S]$$. We can then write $$r=\sum_{i=1}^{n}\alpha_i t_i$$ for some $$\alpha_{i}\in \mathbb{Z}$$ and some $$t_i\in S$$.

The left Ore condition on $$S$$ implies that there exists some $$u_1$$ and $$v_1$$ (in $$S$$) with $$u_1t_1 = v_1 s$$. Similarly, the left Ore condition gives us $$u_2$$ and $$v_2$$ with $$u_2(u_1t_2) = v_2s$$. Repeating, we see that $$(u_n u_{n-1}\cdots u_2 u_1)r\in \mathbb{Z}[S]s.$$ This shows that $$S$$ is a left Ore set inside the domain $$\mathbb{Z}[S]$$.