How to show two rings $(A_f)_{g'}$ and $A_{f g''}$, where $g' = g''/f^n$, are isomorphic?

Question

Let $A$ be a commutative ring with unity. I would like to show the two rings $(A_f)_{g'}$ and $A_{f g''}$, where $g' = g''/f^n$, are isomorphic. First is the $A$ localized at $f$, then localized at$g'$. What is the map that gives the isomorphism? Thank you.

Answer 1

$f$ is invertible in $A_{fg''}$ since $f \cdot g''/(fg'') = 1$. Thus, the canonical homomorphism $A \to A_{fg''}$ factors through $A_f$ via $$ x/f^a \mapsto \frac{xg''^a}{(fg'')^a}. $$

The image of $g'$ is invertible in $A_{fg''}$ since $$ g''/f^n \mapsto \frac{g''^{n+1}}{(fg'')^n} $$ and $$ \frac{g''^{n+1}}{(fg'')^n} \cdot \frac{f^{2n+1}g''^n}{(fg'')^{n+1}} = 1. $$

This gives us a homomorphism $(A_f)_{g'} \to A_{fg''}$ via $$ \frac{x/f^a}{g'^b} \mapsto \frac{xg''^a}{(fg'')^a} \cdot \left(\frac{f^{2n+1}g''^n}{(fg'')^{n+1}}\right)^b. $$

Similarly, we can go in the opposite direction and define a homomorphism $A_{fg''} \to (A_f)_{g'}$ then verify that the resulting homomorphisms are inverses to each other.

Alternatively, one can verify that both rings satisfy the same universal property, thus avoiding the cumbersome calculations above.