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.
The following statements 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, $\map i s$ is a unit of $A_S$.
By Ring Homomorphism Preserves Invertible Elements, $\map j s = \map h {\map i s}$ 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 $j \sqbrk S \subseteq A_T^\times$ 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 $S$:
- $\ds \map {\operatorname{Sat} } S = A - \bigcup \set {\mathfrak p \in \Spec A: \mathfrak p \cap S = \O}$
Thus:
| \(\ds \map {\operatorname{Sat} } S\) | \(\subseteq\) | \(\ds \map {\operatorname{Sat} } T\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds A - \bigcup \set {\mathfrak p \in \Spec A: \mathfrak p \cap S = \O}\) | \(\subseteq\) | \(\ds A - \bigcup \set {\mathfrak p \in \Spec A: \mathfrak p \cap T = \O}\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \bigcup \set {\mathfrak p \in \Spec A: \mathfrak p \cap S = \O}\) | \(\supseteq\) | \(\ds \bigcup \set {\mathfrak p \in \Spec A: \mathfrak p \cap T = \O}\) | Subset iff Complement is Superset of Complement | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \forall \mathfrak p \in \Spec A: \, \) | \(\ds \mathfrak p \cap T = \O\) | \(\implies\) | \(\ds \mathfrak p \subseteq \bigcup \set {\mathfrak q \in \Spec A: \mathfrak q \cap S = \O}\) | Sets are Subset iff Union is Subset |
To finish, we show that the last statement is equivalent to:
- $\forall \mathfrak p \in \Spec A: \mathfrak p \cap T = \O \implies \mathfrak p \cap S = \O$
We show that, for $\mathfrak p \in \Spec A$:
- $\ds \mathfrak p \subseteq \bigcup \set {\mathfrak q \in \Spec A: \mathfrak q \cap S = \O} \iff \mathfrak p \cap S = \O$
Let $\mathfrak p \in \Spec A$.
If $\mathfrak p \cap S = \O$, then by Set is Subset of Union:
- $\ds \mathfrak p \subseteq \bigcup \set {\mathfrak q \in \Spec A: \mathfrak q \cap S = \O}$
Conversely, let $\ds \mathfrak p \subseteq \bigcup \set {\mathfrak q \in \Spec A: \mathfrak q \cap S = \O}$.
By:
we have $\mathfrak p \cap S = \O$.
We conclude that $\map {\operatorname{Sat} } S \subseteq \map {\operatorname{Sat} } T$ if and only if
- $\forall \mathfrak p \in \Spec A: \mathfrak p \cap T = \O \implies \mathfrak p \cap S = \O$
That is:
- $\forall \mathfrak p \in \Spec A: \mathfrak p \cap S \ne \O \implies \mathfrak p \cap T \ne \O$
$\blacksquare$