Editing Krull's principal ideal theorem (section)

=== Proof of the principal ideal theorem ===
Let <math>A</math> be a Noetherian ring, ''x'' an element of it and <math>\mathfrak{p}</math> a minimal prime over ''x''. Replacing ''A'' by the [[Localization (commutative algebra)#Localization at primes|localization]] <math>A_\mathfrak{p}</math>, we can assume <math>A</math> is local with the maximal ideal <math>\mathfrak{p}</math>. Let <math>\mathfrak{q} \subsetneq \mathfrak{p}</math> be a strictly smaller prime ideal and let <math>\mathfrak{q}^{(n)} = \mathfrak{q}^n A_{\mathfrak{q}} \cap A</math>, which is a <math>\mathfrak{q}</math>-[[primary ideal]] called the ''n''-th [[symbolic power (mathematics)|symbolic power]] of <math>\mathfrak{q}</math>. It forms a descending chain of ideals <math>A \supset \mathfrak{q} \supset \mathfrak{q}^{(2)} \supset \mathfrak{q}^{(3)} \supset \cdots</math>. Thus, there is the descending chain of ideals <math>\mathfrak{q}^{(n)} + (x)/(x)</math> in the ring <math>\overline{A} = A/(x)</math>. Now, the [[radical of an ideal|radical]] <math>\sqrt{(x)}</math> is the intersection of all minimal prime ideals containing <math>x</math>; <math>\mathfrak{p}</math> is among them. But <math>\mathfrak{p}</math> is a unique maximal ideal and thus <math>\sqrt{(x)} = \mathfrak{p}</math>. Since <math>(x)</math> contains some power of its radical, it follows that <math>\overline{A}</math> is an Artinian ring and thus the chain <math>\mathfrak{q}^{(n)} + (x)/(x)</math> stabilizes and so there is some ''n'' such that <math>\mathfrak{q}^{(n)} + (x) = \mathfrak{q}^{(n+1)} + (x)</math>. It implies:
:<math>\mathfrak{q}^{(n)} = \mathfrak{q}^{(n+1)} + x \, \mathfrak{q}^{(n)}</math>,

from the fact <math>\mathfrak{q}^{(n)}</math> is <math>\mathfrak{q}</math>-primary (if <math>y</math> is in <math>\mathfrak{q}^{(n)}</math>, then <math>y = z + ax</math> with <math>z \in \mathfrak{q}^{(n+1)}</math> and <math>a \in A</math>. Since <math>\mathfrak{p}</math> is minimal over <math>x</math>, <math>x \not\in \mathfrak{q}</math> and so <math>ax \in \mathfrak{q}^{(n)}</math> implies <math>a</math> is in <math>\mathfrak{q}^{(n)}</math>.) Now, quotienting out both sides by   <math>\mathfrak{q}^{(n+1)}</math>  yields <math>\mathfrak{q}^{(n)}/\mathfrak{q}^{(n+1)} = (x)\mathfrak{q}^{(n)}/\mathfrak{q}^{(n+1)}</math>. Then, by [[Nakayama's lemma]] (which says a finitely generated module ''M'' is zero if <math>M = IM</math> for some ideal ''I'' contained in the radical), we get <math>M = \mathfrak{q}^{(n)}/\mathfrak{q}^{(n+1)} = 0</math>; i.e., <math>\mathfrak{q}^{(n)} = \mathfrak{q}^{(n+1)}</math> and thus <math>\mathfrak{q}^{n} A_{\mathfrak{q}} = \mathfrak{q}^{n+1} A_{\mathfrak{q}}</math>. Using Nakayama's lemma again, <math>\mathfrak{q}^{n} A_{\mathfrak{q}} = 0</math> and <math>A_{\mathfrak{q}}</math> is an Artinian ring; thus, the height of <math>\mathfrak{q}</math> is zero. <math>\square</math>