Why is $\bigcap_{\mathfrak{p} \in D(f)} A_\mathfrak{p} \subset A_f$?

Question

I have some problems regarding exercise 5 in Bosch's book, chapter 6.3. Well, actually it's just the problem stated in the title. $A$ is an integral domain so the natural localization maps are injective. But I feel like I'm running in circles trying to show that inclusion. Any hint would be great.

Answer 1

The standard trick is the following:

Consider $x \in \bigcap_{\mathfrak{p} \in D(f)} A_{\mathfrak{p}},$ and let $$\mathfrak{a}=(A:_{A}x):=\{g \in A\;|\; gx \in A\}.$$ Clearly $\mathfrak{a}$ is an ideal of $A$, and we may assume proper (otherwise $x\in A$, so there is nothing to prove).

Now let $\mathfrak{q}$ be a prime containing $\mathfrak{a}.$ Then $\mathfrak{q}$ cannot be in $D(f)$, since for any $\mathfrak{p} \in D(f)$ there is $s \in \mathfrak{a} \setminus \mathfrak{p}$ by the fact that $x \in A_{\mathfrak{p}}$ (this yoga might be best to think through on your own). So we may conclude that $f \in \mathfrak{q}.$

Altogether, we have $f \in \bigcap_{\mathfrak{q} \supseteq \mathfrak{a}} \mathfrak{q}= \sqrt{\mathfrak{a}} $. Thus, there is a power of $f$, $f^N$, which belongs to $\mathfrak{a}$, and so $f^Nx=a \in A,$ hence $x = \frac{a}{f^N}\in A_f$.