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I need help with the following question:

Let $R$ be a commutative ring, $S \subset R \ \ $ closed under multiplication.

define $\Omega $ as the set of all ideal $I$ in $R$ such that $I \cap S = \phi$

proof that if P is maximal under containing in $\Omega\ $ then P is a prime ideal.

so far I've done the following:

assume $xy \in P$. we need to proof that either $x \in P$ or $y \in P$.

If $xy \in P$ then $xy \notin S$ therefore either $x \notin S$ or $y \notin S$.

what I want to do is to proof that one of the ideals $P + \langle x \rangle$ or $P + \langle y \rangle$ is in $\Omega$, and then because P is maximal in $\Omega$, the ideal is in P, and therefore $x$ or $y$ is in p.

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Shai Alkoby
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1 Answers1

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You are on the right track : If $xy \in P$, suppose that $x\notin P$ and $y\notin P$, then $P+\langle x\rangle \neq P$ and $P+\langle y\rangle \neq P$, whence $$ (P+\langle x\rangle) \cap S \neq \emptyset, \text{ and } (P+\langle y\rangle )\cap S \neq\emptyset $$ Hence, $\exists z_1\in P, r_1\in R$ such that $$ z_1+r_1x \in S $$ And $\exists z_2\in P, r_2\in R$ such that $$ z_2+r_2y \in S $$ Hence, $$ r := z_1z_2 + z_1r_2y + z_2r_1x + r_1r_2xy \in S $$ But by hypothesis, $r \in P$, which contradicts the fact that $P\cap S =\emptyset$.

Hence, either $x\in P$ or $y\in P$.

Prahlad Vaidyanathan
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