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In a ring $R$, if $S$ is a multiplicatively closed set excluding $0$... letting $X$ be the collection of ideals disjoint from $S$, if $I\in X$ maximal, $J\in X$, prove that $J\subset I$.

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    Closing as "not a real question" partly by request of OP. Please see [the related question here](http://math.stackexchange.com/questions/147199/if-i-is-an-ideal-in-r-and-s-subset-r-is-multipicatively-closed-disjoint) instead.2012-05-21

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The statement seems to be false: e.g. take $R = \mathbb{Z}$, $S = \{1\}$, $I = 2\mathbb{Z}$ and $J = 3 \mathbb{Z}$.

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    @Zhang: Assuming by "maximal" you mean "maximal with respect to being an element of $X$": yes, this is just Zorn's Lemma.2012-05-20
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For a multiplicative set $S$, there is a one-to-one, inclusion preserving correspondence between the prime ideals of $S^{-1}R$, the localization of $R$ at $S$, and the prime ideals of $R$ that are disjoint from $S$. In particular, there is a maximum ideal of $R$ that is disjoint from $S$ if and only if $S^{-1}R$ is a local ring (a ring with a unique maximal ideal).

This will not happen in general. For example, with $R=\mathbb{Z}$, for any integer $m$ you can let $S=\{1,m,m^2,m^3,\ldots\}$ to obtain a multiplicative set; the prime ideals of $S^{-1}R$ are precisely the images of prime ideals of $R$ generated by primes that do not divide $m$. (In Pete Clark's example above, all primes). For instance, any odd prime $p$ will give you an ideal $p\mathbb{Z}$ that is disjoint from $S=\{1,2,4,8,16,\ldots\} = \{2^n\mid n\in\mathbb{N}\}$), and these ideals are all maximal and not contained in one another.