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Let $R$ be a commutative ring with $1 \neq 0$. Assume that $a \in R$ is such that $a^n \neq 0$ for each positive integer $n$ and let $\mathcal S = \{a^n\}_{n \geq 0}$.

  1. Prove that there exists an ideal $I$ of $R$ such that $I$ is maximal among ideals of $R$ with $I \cap S = \emptyset$.

  2. Prove that an ideal $I$ as in (a) is a prime ideal.

My thoughts...

  1. Let $\mathcal C = \{J_\alpha\}$ denote the collection of ideals of $R$ not intersecting $\mathcal S$. Note that $\mathcal C$ is nonempty since $\{0\} \in \mathcal C$ and $\mathcal C$ is partially ordered under inclusion. For any ascending chain $\{J_n\}$ the ideal $\bigcup_n J_n$ is an upper bound. Hence, by Zorn's Lemma, $\mathcal C$ has a maximal element, say $I$.

  2. Suppose $A$ and $B$ are ideals of $R$ with $AB \subset I$. If $A \not\subset I$ and $B \not\subset I$ then, by maximality of $I$ with respect to empty $\mathcal S$-intersection, $A \cap \mathcal S$ and $B \cap \mathcal S$ are nonempty. In this case, $AB \cap \mathcal S$ is nonempty, contradicting $AB \subset I$.

Is my solution for [2.] okay? I'm concerned that if $I$ is not the unique ideal with this maximality condition then it's possible that $A, B \in \mathcal C$ but are included in different maximal ideals $I_A$ and $I_B$.

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You are right to worry. To make a correct argument consider for example the ideals $I+A$ and $I+B$, these are clearly bigger than $I$ and hence meet $S$.

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