When is $A \setminus S$ a prime ideal? One of the definitions of a prime ideal $P$ of a commutative ring $A$ is

$P \subset A$ is prime if $A \setminus P$ is a multiplicative set.

My question is: what conditions do I have to impose on a set $\emptyset \neq S \subset A$ so that $A \setminus S$ was a prime ideal?
By definition above $A \setminus(A \setminus S) = S$ must be a multiplicative set. Of course $S \neq A$. What are other conditions?
 A: Another necessary condition is that $S$ is saturated meaning that if $ab\in S$, both $a,b\in S$. Of course $0$ cannot be in $S$ either if you want the complement to be an ideal.
Unfortunately, saturated multiplicative sets still don't characterize complements of prime ideals: it turns out that they are exactly the complements of unions of prime ideals. So another ingredient is needed (and I'm not sure if there is a concise one.)
One obvious sufficient condition to add, of course, is to ask for $R\setminus S$ to be an additive subgroup of $R$. Then the multiplicative ideal requirements work out automatically and you have a prime ideal. In fact this translates to a fairly natural condition on $S$: $a+b\in S\implies a\in S$ or $b\in S$. The fact that $S$ is saturated already implies that $R\setminus S$ already contains additive inverses for its elements.
Actually, reviewing everything here again, it looks like we've found a complete set of conditions. $R\setminus S$ is a prime ideal of $R$ iff $S$ is a saturated, multiplicatively closed, "additively prime" set not containing $0$.
