A maximal ideal among those disjoint from a multiplicative set is maximum 
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: The statement seems to be false: e.g. take $R = \mathbb{Z}$, $S = \{1\}$, $I = 2\mathbb{Z}$ and $J = 3 \mathbb{Z}$.  
A: 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.
