## 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 $A$“.

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