Proof of isomorphism of a field of fractions

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I'm stuck with the following exercise in my algebraic geometry lecture.

Let $A$ be a domain and $I\subset A$ be a prime ideal. Let $K$ be the field of fractions of $A$. Let $A_I$ be the subring of $K$ generated by $A$ and all $1/p$ for $p\in A\setminus I$. Let $IA_I$ be the ideal of $A_I$ generated by $I$. Show that the field of fractions of $A/I$ is isomorphic to $A_I/IA_I$.

My first thought was that this has something to do with coordinate rings, but I have no idea how to approach this problem.

Can anybody give me a hint the right direction?

Thanks in advance.

P.S.: I hope that this question doesn't violate any rules. It's the first time I'm asking something on a Stackexchange site.

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Let $R$ be an integral domain with fraction field $F$. Then a ring homomorphism $\varphi : R \to S$ extends to a homomorphism $F \to S$ if and only if $\varphi(r)$ is invertible for all $0 \neq r \in R$.

Using this criterion, try to check that the natural map $A/I \to A_I/IA_I$ induces a homomorphism $F = \operatorname{Frac}(A/I) \to A_I/IA_I$ and then check that this is an isomorphism directly.

This result is really saying that localization commutes with quotient. We can either quotient by $I$ then localize to get the fraction field, or we can localize first then quotient.