Existence of Homomorphism between Localizations of Ring at Elements

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Theorem

Let $A$ be a commutative ring with unity.

Let $f, g \in A$.


The following are equivalent:

$(1): \quad$ There exists an $A$-algebra homomorphism $h : A_f \to A_g$ between localizations, the induced homomorphism.
$(2): \quad f$ divides some power of $g$.
$(3): \quad$ There is an inclusion of vanishing sets: $V(f) \subseteq V(g)$. That is, every prime ideal containing $f$ also contains $g$.
$(4): \quad$ There is an inclusion of principal open subsets: $D(f) \supseteq D(g)$


Proof

1 implies 2

Let there exist an $A$-algebra homomorphism $A_f \to A_g$.

By Existence of Homomorphism between Localizations of Ring, $f$ is contained in the saturation of the set of powers $\{ g^n : n \in \N\}$.

That is, $f$ divides some power of $g$.

$\Box$


2 implies 1

Let $f$ divide some power of $g$.

Say $fd = g^n$.

Then for all $m \in \N$, $f^m d^m = g^{nm}$.

Thus $f^m$ divides some power of $g$.

Thus the set of powers $\{ f^m : m \in \N\}$ is contained in the saturation of $\{ g^n : n \in \N\}$.

By Existence of Homomorphism between Localizations of Ring, there exists an $A$-algebra homomorphism $A_f \to A_g$.

$\Box$


3 iff 4

By definition, a principal open subset $D(f)$ is the complement in the spectrum $\operatorname{Spec} A$ of the vanishing set $V \left({f}\right)$.

By Subset iff Complement is Superset of Complement, $V(f) \subseteq V(g)$ if and only if $D(f) \supseteq D(g)$.

$\Box$