Existence of Homomorphism between Localizations of Ring

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Theorem

Let $A$ be a commutative ring with unity.

Let $S, T \subseteq A$ be multiplicatively closed subsets.


The following are equivalent:

$(1): \quad$ There exists an $A$-algebra homomorphism $h : A_S \to A_T$ between localizations, the induced homomorphism.
$(2): \quad S$ is a subset of the saturation of $T$.
$(3): \quad$ The saturation of $S$ is a subset of the saturation of $T$.
$(4): \quad$ Every prime ideal meeting $S$ also meets $T$.


Proof

Let $i : A \to A_S$ and $j : A \to A_T$ be the localization homomorphisms.


1 implies 2

Let $h : A_S \to A_T$ be an $A$-algebra homomorphism.

Then by definition, $j = h \circ i$:

$\xymatrix{ A \ar[d]_i \ar[r]^{j} & A_T\\ A_S \ar[ru]_{h} }$

Let $s \in S$.

By definition of localization, $i(s)$ is a unit of $A_S$.

By Ring Homomorphism Preserves Invertible Elements, $j(s) = h(i(s))$ is a unit of $A_T$.

Thus $s$ is an element of the saturation of $T$.

$\Box$


2 implies 1

Let $S$ be a subset of the saturation of $T$.

Then its image $j(S) \subseteq A_T^\times$ consists of units of $A_T$.

By definition of localization at $S$, there exists a unique $A$-algebra homomorphism $h : A_S \to A_T$.

$\Box$


2 implies 3

Let $S$ be a subset of the saturation of $T$.

By definition, its saturation is the smallest saturated multiplicatively closed subset of $A$ containing $S$.

Thus the saturation of $S$ is a subset of the saturation of $T$.

$\Box$


3 implies 2

By definition, $S$ is a subset of its saturation.

$\Box$


3 iff 4

By definition, the saturation of $S$ is the complement of the union of prime ideals that are disjoint from $S$:

$\operatorname{Sat}(S) = A - \displaystyle \bigcup \left\{ \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap S = \varnothing \right\}$

Thus:

\(\displaystyle \operatorname{Sat}(S) \subseteq \operatorname{Sat}(T)\) \(\iff\) \(\displaystyle A - \bigcup \left\{ \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap S = \varnothing \right\} \subseteq A - \bigcup \left\{ \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap T = \varnothing \right\}\)
\(\displaystyle \) \(\iff\) \(\displaystyle \bigcup \left\{ \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap S = \varnothing \right\} \supseteq \bigcup \left\{ \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap T = \varnothing \right\}\) Subset iff Complement is Superset of Complement
\(\displaystyle \) \(\iff\) \(\displaystyle \forall \mathfrak p \in \operatorname{Spec} A : \left( \mathfrak p \cap T = \varnothing \implies \mathfrak p \subseteq \bigcup \left\{ \mathfrak q \in \operatorname{Spec} A : \mathfrak q \cap S = \varnothing \right\} \right)\) Sets are Subset iff Union is Subset

To finish, we show that the last statement is equivalent to:

$\forall \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap T = \varnothing \implies \mathfrak p \cap S = \varnothing$

We show that, for $\mathfrak p \in \operatorname{Spec} A$:

$\mathfrak p \subseteq \displaystyle \bigcup \left\{ \mathfrak q \in \operatorname{Spec} A : \mathfrak q \cap S = \varnothing \right\} \iff \mathfrak p \cap S = \varnothing$

Let $\mathfrak p \in \operatorname{Spec} A$.

If $\mathfrak p \cap S = \varnothing$, then by Set is Subset of Union:

$\mathfrak p \subseteq \displaystyle \bigcup \left\{ \mathfrak q \in \operatorname{Spec} A : \mathfrak q \cap S = \varnothing \right\}$

Conversely, let $\mathfrak p \subseteq \displaystyle \bigcup \left\{ \mathfrak q \in \operatorname{Spec} A : \mathfrak q \cap S = \varnothing \right\}$.

By:

Union of Sets Disjoint with Set
Subset of Disjoint Set

we have $\mathfrak p \cap S = \varnothing$.

We conclude that $ \operatorname{Sat}(S) \subseteq \operatorname{Sat}(T)$ if and only if

$\forall \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap T = \varnothing \implies \mathfrak p \cap S = \varnothing$

That is:

$\forall \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap S \neq \varnothing \implies \mathfrak p \cap T \neq \varnothing$

$\blacksquare$


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