Existence of Homomorphism between Localizations of Ring at Elements

From ProofWiki
Jump to navigation Jump to search

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


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

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

$\Box$


2 implies 1

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

Say $f d = g^n$.

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 $A_f \to A_g$.

$\Box$


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, $\map V f \subseteq \map V g$ if and only if $\map D f \supseteq \map D g$.

$\Box$


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 $g$.

$\Box$


3 implies 2