Let \(R\) be a commutative ring and let \(M\) be an \(R\)-module. Recall that for \(\mu \in M\), the annihilator of \(\mu\) is the set: \[\operatorname{Ann}(\mu) = \{ r \in R \mid r \mu = 0 \}.\]
Suppose that \(I\) is an ideal in \(R\) which is maximal among the set of annihilators of non-zero elements of \(M\) (i.e. \(I = \operatorname{Ann}(\mu)\) for some \(\mu \in M \setminus \{0\}\), and if \(\operatorname{Ann}(\nu) \supsetneq I\) for some \(\nu \ne 0\), no such \(\nu\) exists).
Prove that \(I\) is a prime ideal in \(R\).
Solution
Write \[ I=\operatorname{Ann}(\mu), \qquad \mu\ne0, \] and assume \(I\) is maximal among annihilators of nonzero elements of \(M\). Since \(\mu\ne0\), we have \(1\notin I\), so \(I\) is proper.
Suppose \[ ab\in I \] and \(b\notin I\). Then \[ b\mu\ne0. \] Moreover, for every \(r\in I\), \[ r(b\mu)=b(r\mu)=0, \] so \[ I\subseteq\operatorname{Ann}(b\mu). \] By maximality and \(b\mu\ne0\), \[ I=\operatorname{Ann}(b\mu). \] But \[ a(b\mu)=(ab)\mu=0, \] so \(a\in\operatorname{Ann}(b\mu)=I\). Therefore \[ ab\in I,\ b\notin I\implies a\in I, \] and hence \[ \boxed{I\text{ is prime}}. \]