Use Zorn’s Lemma to construct an ideal which maximally does not contain a given finitely generated ideal

Let R be a ring with 1 \neq 0, and let A = (a_1,\ldots,a_n) be a nonzero finitely generated (proper) ideal of R. Prove that there is an ideal in R which is maximal with respect to the property “does not contain A“.

Let \mathcal{C} denote the set of all ideals in R which do not contain I; this set is partially ordered by inclusion. Moreover \mathcal{C} is nonempty since 0 does not contain A. Let \{C_i\}_\mathbb{N} be a chain in \mathcal{C}. Now \bigcup C_i is an ideal in R by this previous exercise. Suppose A \subseteq \bigcup C_i. Then for each generator a_k of A, we have a_k \in C_{i_k} for some i_k. Since \{C_i\}_\mathbb{N} is totally ordered and there are only finitely many a_k, there exists an inclusion-maximal element C_t such that a_k \in C_t for each k; thus A \subseteq C_t, a contradiction. Thus A \nsubseteq \bigcup C_i, and we have \bigcup C_i \in \mathcal{C}. So \bigcap C_i is an upper bound of the chain \{C_i\}_\mathbb{N} in \mathcal{C}. Thus every chain in \mathcal{C} has an upper bound, and by Zorn’s lemma there exists a maximal ideal B with respect to the property “does not contain A“.

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  • Barum  On October 12, 2011 at 11:57 pm

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

    • nbloomf  On October 13, 2011 at 10:02 am

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

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