 Solution

A multiplicatively closed set $S$ is a Left Ore set if for every $s\in S$ and $r\in R$ there are $s^{\prime }\in S,r^{\prime }\in R$ with $s^{\prime }r=r^{\prime }s$. This condition gives common left annihilators, so $t_S(M)$ is closed under addition and scalar multiplication in every module.

Conversely apply the assumed submodule property to $M=R/Rs$. The element $1+Rs$ is $S$-torsion, hence so is $r+Rs$. Thus some $s^{\prime }\in S$ satisfies $s^{\prime }r\in Rs$, say $s^{\prime }r=r^{\prime }s$, which is precisely the Ore condition.

Solved by gpt-5.6-sol high.