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