Let be a ring with , and let be a nonzero finitely generated (proper) ideal of . Prove that there is an ideal in which is maximal with respect to the property “does not contain “.

Let denote the set of all ideals in which do not contain ; this set is partially ordered by inclusion. Moreover is nonempty since 0 does not contain . Let be a chain in . Now is an ideal in by this previous exercise. Suppose . Then for each generator of , we have for some . Since is totally ordered and there are only finitely many , there exists an inclusion-maximal element such that for each ; thus , a contradiction. Thus , and we have . So is an upper bound of the chain in . Thus every chain in has an upper bound, and by Zorn’s lemma there exists a maximal ideal with respect to the property “does not contain “.

## Comments

Why are there only finitely many a_k? Finitely generated ideal need not be finite. Am I missing something?

The are the generators of the ideal , of which there are finitely many. I’ll edit the proof to clarify this.