(January 2010) Let \(R\) be a commutative ring and \(S\subseteq R\) a subset of \(R\). Define the annihilator of \(S\) in \(R\) to be \[ \operatorname{Ann}(S)=\{r\in R\mid rs=0\text{ for every }s\in S\}. \] Prove that \(\operatorname{Ann}(S)\) is an ideal of \(R\).
Solution
Proof. (Nonempty) We have 0∈ Ann(S), since 0s = 0∀s∈S. So Ann(S)⁄= ∅. (Closed under−) Given a,b ∈ Ann(S) and s∈ S, we have (a−b)s = as−bs = 0− 0 = 0, so that a−b∈ Ann(S). (Sticky) Given r∈ R and x∈ Ann(S), and given any s∈ S, we have ( rx)s = r(xs) = r(0) = 0. Thus,rx∈ Ann(S). Since R is commutative, xr =rx∈ Ann(S). QED