Show $A_{f(b)} \cong A \otimes _B B_g $

Question

Let $f: B \to A$ a ring morphism and $g \in B$.

My goal is to show that $A_{f(g)} \cong A \otimes _B B_g $

($B_g$ means localization at $g$)

Firstly using universal property of tensor product the canonical maps $A \to A_{f(g)}, B_g \to A_{f(g)}$ induce a map $h:A \otimes _B B_g \to A_{f(g)}$. Obviously it's surjective since $A$ and the multiplicative system $\{ f(g)^n \vert n \in \mathbb{N}\}$ are contained in the image.

Does anybody have an argument for injectivity?

Answer 1

This should follow from the following more general fact: Let $M$ be a $B$-module. Then, given any multiplicative system $S$ in $B$, we have $$ S^{-1}M\cong M\otimes_B S^{-1}B $$ as $B$-modules.

In your particular case we take $S=\{1,g,g^2,\dots\}$ and $M=A$ as a $B$-module. All that'd remain to show is that this map is also a(n iso)morphism in whatever category you have in mind ($B$-algebras, commutative rings?).

For a proof of this standard result see Lemma 6.19 in these notes by Gathmann: https://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c6.pdf

Answer 2

Consider the ring homomorphism $\varphi\colon A\to A\otimes_BB_g$, $a\mapsto a\otimes(1/1)$.

Suppose $\psi\colon A\to C$ is a ring homomorphism, where $\psi(f(g))$ is invertible.

By the property of $B_g$, there exists a unique homomorphism $\tau\colon B_g\to C$ such that $\tau(b/g^n)=\psi(f(b))\psi(f(g))^{-n}$.

Therefore there exists a unique homomorphism $\hat{\psi}\colon A\otimes_BB_g\to C$ such that $$ \hat{\psi}(a\otimes(b/g^n))=\psi(a)\tau(b/g^n)=\psi(a)\psi(f(b))\psi(f(g))^{-n} $$ and $$ \hat{\psi}\circ\varphi(a)=\psi(a) $$ Therefore $A\otimes_BB_g$ satisfies the property required for $A_{f(g)}$.