TITLE: Show $R \setminus S$ is a union of prime ideals
QUESTION [5 upvotes]: I'm stuck on the following question:

Let $R$ be a commutative ring with $1$, and $S \subseteq R$ a saturated multiplicative set (that is, $1 \in S$ and $x, y \in S$ if and only if $xy \in S$).  Show that $R \setminus S$ is a union of prime ideals.  

Here's my attempt so far.  If $a \in R \setminus S$, then the ideal $Ra$ does not intersect $S$, because if $ra$ were in $S$ for some $r \in R$, then $a$ would have to be in $S$.  Thus, it follows by Zorn's lemma that there exist ideals $I$ of $R$ which are maximal with respect to the property that $I \cap S = \emptyset$.
Let $I$ be any such ideal.  If I can show that $I$ is prime, the conclusion will follow.  Let $xy \in I$.  I want to show that $x$ or $y$ is in $I$.  Suppose $x \not\in I$.  Then $I + Rx$ intersects $S$, so there exists $s \in S, a \in I, r \in R$ such that $s = a + rx$.  Then $sy = ay + rxy \in I$, so $sy$, and hence $y$, cannot be in $S$.
Similarly if $y$ is not in $I$, then $x \not\in S$.  So we are reduced to proving the following is impossible: $xy \in I$, $x$ and $y$ are not in $S$, but they aren't in $I$ either.

REPLY [4 votes]: You need to show that if $x,y \notin I$, then $xy \notin I$.  Consider the product $(I  + (x))(I + (y)) \subseteq I + (xy)$.
Edit:  Further details...  (Spoiler alert.)  See the original question for what $I$ is and why we need to show that $I$ is prime.  Suppose to the contrary that $xy \in I$ for some $x,y \notin I$.  Then $(I + (x))(I + (y)) = I^2 + xI + yI + (xy) \subseteq I$.  Since $I + (x)$ and $I + (y)$ properly contain $I$, they must each contain an element of $S$, and hence their product contains an element of $S$.  (Here we use the fact that $S$ is multiplicatively closed.)  Thus $I$ contains an element of $S$, a contradiction.