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


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.35

Solution:

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 $A$“.


Linearity

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