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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 $\ker(\varphi)=\{m\in M:\exists s\in S, sm=0\}$?

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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 $\mathbb{Z}[S]$.

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