Existence of Homomorphism between Localizations of Ring

Theorem
Let $A$ be a commutative ring with unity.

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


 * $(1): \quad$ 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 $T$.

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 $h : A_S \to A_T$.

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 $T$.

3 implies 2
By definition, $S$ is a subset of its saturation.

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:

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

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

Also see

 * Uniqueness of Homomorphism between Localizations of Ring