Let $A$ be a ring, $S$ a multiplicative set, and $I$ an ideal of $A$ disjoint from $S$; then there exists a prime ideal $P$ of $A$ containing $I$ and disjoint from $S$.
The author of the book I'm reading, proves this by saying:
Suppose that I can find an ideal $P\supset I$ maximal subject to the condition $P\cap S=\emptyset$. I claim that $P$ is a prime ideal.
He then shows that $P$ is indeed a prime ideal and then uses Zorn's lemma to show we can find such a $P$. (You might call this the standard proof.)
But why can't we just define $P:=A\backslash S$. It is prime, contains $I$ and is disjoint from $S$. Where am I off, when I think this will do?
I think that there's probably something I've overlooked, or else the author would have just used this argument.
To remove this from the unanswered questions; the problem is that $A\backslash S$ is not necessarily closed under addition and thus there's no reason to expect that it's an ideal. A counterexample would be $S=\{1\}$, where $A$ has other invertible elements.
With thanks to Nate, mt_ and brick.