A multiplicatively closed subset $S$ of a ring $ A$ is said to be saturated if $xy\in S$ $\iff$ $x\in S$ and $y\in S$. Show that if $S$ is saturated if and only if $A-S$ is a union of a primes.
First of all I assume they mean the union where there are no redundant ideals in the union. For example in $\mathbb Z$, $(2)\cup (3)\cup (6)=(2)\cup (3)$ and this statement cannot be both true and false.
Lets work assuming nonredundant union is used. If $A-S$ is a union of primes it follows easily that $S$ is saturated.
Let $S$ be saturated. Then certainly $A-S$ is a union of primes as it it is the union of each $(x)$ with $x\notin S$. We would like to know that we can write this as a union of prime ideals.
Since we know we can write $A-S$ as a union of ideals, I think we can assume that we have written as a union of ideals in a non-redundant way. We do not know that these are prime. Suppose that one of the ideals $I$ in the union is not prime. Then there is $xy\in I$ such that neither $x$ or $y$ are in $I$. We cannot have $x$ and $y$ in $S$ since $xy\notin S$. Therefore $x$ must be in some other ideal $J$ in the union. This does not seem to give us enough to contradict the union being non-redundant.
Is it possible to continue with this approach or how should I tackle this?
