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 $J\subset I$.
A maximal ideal among those disjoint from a multiplicative set is maximum
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ring-theory
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0A similar statement which possibly you want to prove that *is* true: Let $S$ be a multiplicative set and $I$ an ideal disjoint from $S$. Then there is a prime ideal $P$ containing $I$ that is disjoint from $S$. – 2012-05-19
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1As @Benjamin Lim points out, there seems to be a unique standard result of minimal distance from your statement: **multiplicative avoidance**. See e.g. $\S 4.1$ of http://math.uga.edu/~pete/integral.pdf. (But anyway, where is all this coming from?) – 2012-05-20
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0@PeteL.Clark This is from an example question for second year math undergraduates – 2012-05-20
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0[link](http://math.stackexchange.com/questions/147199/if-i-is-an-ideal-in-r-and-s-subset-r-is-multipicatively-closed-disjoint) is the complete question. @PeteL.Clark – 2012-05-20
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3Please don't start your post in the title and continue in the body. The title should be an informative general description of what you are posting about, and the body should be self-contained. – 2012-05-20
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0Closing 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