Existence of Homomorphism between Localizations of Ring at Elements

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

Let $f, g \in A$.


 * $(1): \quad$ 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: $\map V f \subseteq \map V g$. That is, every prime ideal containing $f$ also contains $g$.
 * $(4): \quad$ There is an inclusion of principal open subsets: $\map D f \supseteq \map D g$

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 $\set {g^n : n \in \N}$.

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

2 implies 1
Let $f$ divide some power of $g$.

Say $f d = g^n$.

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

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

3 iff 4
By definition, a principal open subset $\map D f$ is the complement in the spectrum $\Spec A$ of the vanishing set $\map V f$.

By Relative Complement inverts Subsets, $\map V f \subseteq \map V g$ $\map D f \supseteq \map D g$.

2 implies 3
Let $I$ be a prime ideal containing $f$.

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

Say $f d = g^n$.

By definition of ideal, $g^n = f d \in I$.

By definition of prime ideal, $g \in I$.

So every prime ideal containing $f$ also contains $g$.