# A maximal ideal among those disjoint from a multiplicative set is maximum [closed]

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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## closed as not a real question by mixedmath♦, Benjamin Lim, Dylan Moreland, Willie WongMay 21 '12 at 15:07

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A 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 $S$. – user38268 May 19 '12 at 23:55
As @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 math.uga.edu/~pete/integral.pdf. (But anyway, where is all this coming from?) – Pete L. Clark May 20 '12 at 0:04
@PeteL.Clark This is from an example question for second year math undergraduates – Montez May 20 '12 at 0:12
link is the complete question. @PeteL.Clark – Montez May 20 '12 at 0:14
Please 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. – Arturo Magidin May 20 '12 at 0:57

The statement seems to be false: e.g. take $R = \mathbb{Z}$, $S = \{1\}$, $I = 2\mathbb{Z}$ and $J = 3 \mathbb{Z}$.
You are right, but given $J\in X$, is it possible to find $I\in X$ maximal such that $J\subset I$? – Montez May 19 '12 at 23:57
@Zhang: Assuming by "maximal" you mean "maximal with respect to being an element of $X$": yes, this is just Zorn's Lemma. – Pete L. Clark May 20 '12 at 0:00
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.