Complement of maximal multiplicative set is a prime ideal Let $R$ be a commutative ring with identity. I've been trying to prove the following:

If $S \subset R$ is a maximal multiplicative set, then $R \setminus S$ is a prime ideal of $R$.

I have spent some time in it, but nothing is coming to my mind. Any idea/solution will be appreciated.
 A: We must assume $\rm 0\not\in S $ since $\rm\: 0\in S\: \Rightarrow\: 0 \not\in \overline S := R\setminus S,\: $ so $\rm\:\overline S\:$ is not an ideal. A simple Zorn lemma argument (see below) shows that, since $\rm\:S\:$ is multiplicatively closed, the ideal $\rm\{0\}\subset \overline S\:$ can be enlarged to an ideal $\rm\:P\:$ maximal w.r.t. to exclusion of $\rm\:S,\:$ and such an ideal must be prime. Therefore $\rm\: P = \overline S,\:$ else we could enlarge $\rm\:S\:$ to the monoid $\rm\:\overline P,\:$ contra maximality of $\rm\:S.\quad$ QED
Note $\: $ The prime $\rm\:P\,$ above may be alternatively constructed  as the contraction of a maximal ideal $\rm\:Q\:$ of the localization $\rm\: R_S = S^{-1} R.\: $ $\rm\:Q\:$ exists since $\rm\ 0\not\in S\ \Rightarrow\ R_S \ne \{0\}.\:$ Generally, there is a bijective order-preserving correspondence between all prime-ideals in $\rm\:R_S\:$ and all prime ideals in $\rm R$ disjoint from $\rm\:S,\:$ see Theorem 34 in Kaplansky's Commutative Rings. His Theorem $1$, p. $1$ is the above-invoked form of this result (employing no localization theory). I've appended it below.

Let $\,\overline I\,$ be the set-theoretic complement of an ideal $\,I.\,$ Then the definition of a prime ideal can be recast as follows: $\,\ I\,$ is prime $\iff$ $\, \overline I\,$ is multiplicatively closed (and nonempty). $ $ Now it goes without saying that $\,I\,$ is an ideal maximal with respect to the exclusion of $\,S = \overline I.\,$ Krull discovered a very useful converse.
Theorem $\bf 1\ \ $ Suppose that  $\,S\:$ is a nonempty multiplicatively closed set in a ring $\,R,\,$ and suppose that $\,I\:$ is an ideal in $\,R\,$  maximal with respect to the exclusion of $\,S.\,$ Then $\,I\:$ is prime.
Proof $\ \ $ Given $\,ab\in I\,$ we must show that $\,a\,$ or $\,b\,$ lies in $\,I.\:$ Suppose the contrary. Then the ideal $\,(I,a)\,$ generated by $\,I\,$ and $\,a\,$ is strictly larger than $\,I\,$ and therefore intersects $\,S.\,$ Thus there exists  $\,s\in S\,$ of the form $\,s = i + r a\,\ (i\in I,\, r\in R).\,$ Similarly  $\,\hat s = \hat i + \hat r b\in S\,\ (\hat i\in I,\, \hat r\in R).\,$ But then
$$ i,\hat i,ab\in I\ \Rightarrow\  s \hat s = (i+ ra)(\hat i + \hat r b)\in I\cap S,\ \ {\rm contradiction}\quad {\bf QED}$$
We note that, given any ideal $\,J\,$ disjoint from a nonempty multiplicatively closed set $\,S,\,$ we can by Zorn's lemma expand $\,J\,$ to an ideal $\,I\,$ maximal with respect to disjointness from $\rm\,S.\,$ Thus we have a method of constructing prime ideals.


A: Another way to proceed is to show


*

*First, that every multiplicative subset $S$ of a ring is contained in a saturated multiplicative subset $S'$, that is, one such that $$ab\in S'\implies a\in S'\wedge b\in S'.$$ In fact, there is a smallest saturated multiplicative subset containing $S$, called the saturation.

*Second, that a saturated multiplicative subset $S$ of a ring (which does not contain $0$) is the intersection of the complements of the prime ideals which are disjoint from it. (The hard part here is to show that such primes actually exist: you can construct them as the maximal elements in the family of ideals disjoint from $S$)
Once you have these two facts, your claim follows easily.
A: I assume that all multiplicative subsets discussed here are disjoint from $\{0\}$.
Let me write $S$ for the maximal multiplicative subset of $R$.  
Step 1: I claim that $S^{-1} R$ is a local ring.
Indeed, if not, there would be at least two maximal ideals in the localization, i.e., 
two prime ideals $\mathfrak{p}_1 \neq \mathfrak{p}_2$, each maximal with respect to 
being disjoint from $S$.  Thus $S$ is contained in both $R \setminus \mathfrak{p}_1$ and $R \setminus \mathfrak{p}_2$, but these are two different sets so at least one of the containments must be proper, contradicting the maximality of $S$.
Step 2: Therefore there is a unique maximal ideal in the localization, which corresponds to a prime ideal $\mathfrak{p}$ of $R$ which is disjoint from $S$.  Arguing as above, maximality of $S$ implies $S = R \setminus \mathfrak{p}$.  
Remark 1: When $R$ is a domain, the unique maximal multiplicative subset is obviously $R \setminus \{0\}$.  In this case it is tempting to construe "maximal multiplicative subset" to mean "multiplicative subset maximal with respect to being properly contained in $R \setminus \{0\}$." 
Remark 2: Conversely, the primes $\mathfrak{p}$ such that $R \setminus \mathfrak{p}$ is maximal are precisely the primes which are minimal with respect to containing $\{0\}$.  (When $R$ is a domain and the question is reconstrued as above, we want $\mathfrak{p}$ to be minimal with respect to properly containing $\{0\}$.)  
Mariano's answer makes this clear as well.
