Isomorphism of Localization of $A_\mathfrak{p}$ and $A_\mathfrak{q}$

Question

The question is :

Let $\mathfrak{q} \subset \mathfrak{p}$ be two prime ideals of $A$. Show that the localization of $A_\mathfrak{p}$ at the prime ideal corresponding to $\mathfrak{q}$ is isomorphic to $A_\mathfrak{q}$.

I couldn't understand what is the question is asking for. Can someone elaborate as what it wants, and what is to be shown ?

Answer 1

Hint:

Let $S=A\smallsetminus\mathfrak p$, $T=A\smallsetminus\mathfrak q$. As $\mathfrak q\subset \mathfrak p$, $S\subset T$. We have a homomorphism: \begin{align*} A_\mathfrak p&\longrightarrow A_\mathfrak q\\ \Bigl(\frac as\Bigr)_\mathfrak p&\longmapsto \Bigl(\frac as\Bigr)_\mathfrak q \end{align*} by the universal property of rings of fractions (all elements of $S$ are invertible in $A_\mathfrak q$.)

On the other hand, the ideal $\mathfrak q A_\mathfrak p$ is a prime ideal in $ A_\mathfrak p$, hence we can localise this ring at this prime ideal, obtaining the ring of fractions $(A_\mathfrak p)_{\mathfrak q A_\mathfrak p}$. Typically, its elements are of the form $$\frac{\dfrac as}{\dfrac ts'}, \quad\text{where}\enspace s,s'\in S, \enspace t\in T $$ By the same universal property, you have a homomorphism from $A_\mathfrak q$ to $(A_\mathfrak p)_{\mathfrak q A_\mathfrak p}$.

All you have to do is to show this homomorphism is an isomorphism.