A multiplicatively closed set $S$ is a <Left Ore set> if for every $s\in S$ and $r\in R$ there are $s'\in S,r'\in R$ with $s'r=r's$. 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 $s'\in S$ satisfies $s'r\in Rs$, say $s'r=r's$, which is precisely the Ore condition.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 139 / 4 / a / Solution

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A multiplicative subset $S$ is left Ore when, for every $s\in S$ and $r\in R$, there are $s'\in S$ and $r'\in R$ such that $s'r=r's$.


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 Incoming links: Left Ore set