Form equivalence class Howard-Rubin Number: 14
Statement: Suppose \(R\) is a commutative ring, \(A\) is aproper ideal in \(R\), and \(S\) is a multiplicative semigroup in \(R\)not meeting \(A\). Then there is a prime ideal \(p\) in \(R\) such that\(A \subseteq p\) and \(p\cap S=\emptyset\). Rav [1977] and Note 80.
Howard-Rubin number: 14 AO
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