TITLE: Example of an application of a theorem about ideals in rings of fractions in Atiyah-MacDonald
QUESTION [7 upvotes]: In Atiyah-MacDonald, we have the following theorem (p. 41): 
Proposition 3.11.
i) Every ideal in $S^{-1}R$ is an extended ideal.
ii) If $I$ is an ideal in $R$ then $I^{ec} = \bigcup_{s \in S} (I : \langle s \rangle )$. Hence $I^e = (1) = S^{-1}R$ if and only if $I$ meets $S$.
iii) $I = I^{ec}$ if and only if no element of $S$ is a zero-divisor in $R/I$.
iv) The prime ideals of $S^{-1}R$ are in one-to-one correspondence with the prime ideals of $R$ which don't meet $S$.
(I'm omitting point v) of the theorem since my question is about ii),iii) and iv). )
While I am able prove this theorem I'm wondering about how I'll be able to remember it, in particular, statements ii)-iv). Can anyone give me an example of where I'll be using either one of these three statements or all of them?
Thanks.

REPLY [5 votes]: Of all of the statements above, number 4 is the one that I've used the most. Let me tell you about what number 4 can be used for:

Proving the lying over theorem: Given a finite extension $A \subset B$, for every prime ideal $P \subset A$ there exists a prime ideal $Q$ of $B$ lying over $P$, i.e. $Q^{c} = P$. You can prove this by noticing that $S^{-1}B$ is a finitely generated $S^{-1}A$ - module (To prove this bit, I remember using some tensor products iirc) and then if you draw an appropriate diagram you can apply (iv). I believe you have Nakayama's Lemma available to you as $S^{-1}B$ is a finitely generated $S^{-1}A$ - module.
Proving that a ring $A$ is absolutely flat iff $A_\mathfrak{m}$ for each maximal ideal $\mathfrak{m}$.
Proving that  every prime ideal of $A$ is maximal iff $A/\mathfrak{R}$ is absolutely flat. $\mathfrak{R}$ is the nilradical of $A$.
Atiyah Macdonald problem 3.6 - For this problem one of the localisations iirc reduces to the case of a local ring, and that is very powerful!
Proving that if ring $A$ has no nilpotent elements and $\mathfrak{p}$ a minimal prime ideal of $A$, then $A_\mathfrak{p}$ is a field. You can see a proof here.
This isn't really about the ideal correspondence in (iv) but about the following result: If you have an ideal $I$ disjoint from $S$, then you can always find a prime ideal $P$ such that $P \supset I$ and is maximal with respect to the property that $P \cap S = \emptyset$. This is Krull's lemma, and you can use it to prove problems 1.14 and 3.7(i) of Atiyah - Macdonald.
Proving that the nilradical of $S^{-1}A$ is  $S^{-1}\mathfrak{R}$. This is used in the first step in the "if" direction of #3 above.