Multiplicatively Closed Subset is Saturated iff Complement is Union of Prime Ideals

Definition

Let $A$ be a commutative ring with unity.

Let $S \subseteq A$ be a multiplicatively closed subset.

Then the following are equivalent:

$(1):\quad$ $S$ is saturated

$(2):\quad$ $\relcomp A S = \bigcup \set {\mathfrak p \in \Spec A : \mathfrak p \cap S = \O }$

$(3):\quad$ $\exists \TT \subseteq \Spec A : \relcomp A S = \bigcup \TT$

where:

$\relcomp A S$ denotes the complement of $S$

$\Spec A$ denotes the prime spectrum of $A$

Proof

$(2) \implies (3)$

This is clear, choosing:

$\TT = \set {\mathfrak p \in \Spec A : \mathfrak p \cap S = \O }$.

$\Box$

$(3) \implies (1)$

Let $\TT \subseteq \Spec A$ be such that:

$\relcomp A S = \bigcup \TT$.

Let $x, y \in A$ be such that $x y \in S$.

That is:

$\forall \mathfrak p \in \TT : xy \not \in \mathfrak p$

As $\mathfrak p$s are ideals, we have:

$\forall \mathfrak p \in \TT : x \not \in \mathfrak p \land y \not \in \mathfrak p$

That is:

$x, y \in S$

$\Box$

$(1) \implies (2)$

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