Krull's principal ideal theorem

Proofs

Proof of the principal ideal theorem

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

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

Proof of the height theorem

Krull's height theorem can be proved as a consequence of the principal ideal theorem by induction on the number of elements. Let x_1, \dots, x_n be elements in A, \mathfrak{p} a minimal prime over (x_1, \dots, x_n) and \mathfrak{q} \subsetneq \mathfrak{p} a prime ideal such that there is no prime strictly between them. Replacing A by the localization A_{\mathfrak{p}} we can assume (A, \mathfrak{p}) is a local ring; note we then have \mathfrak{p} = \sqrt{(x_1, \dots, x_n)}. By minimality of \mathfrak{p}, it follows that \mathfrak{q} cannot contain all the x_i; relabeling the subscripts, say, x_1 \not\in \mathfrak{q}. Since every prime ideal containing \mathfrak{q} + (x_1) is between \mathfrak{q} and \mathfrak{p}, \sqrt{\mathfrak{q} + (x_1)} = \mathfrak{p} and thus we can write for each i \ge 2, :x_i^{r_i} = y_i + a_i x_1 with y_i \in \mathfrak{q} and a_i \in A. Now we consider the ring \overline{A} = A/(y_2, \dots, y_n) and the corresponding chain \overline{\mathfrak{q}} \subset \overline{\mathfrak{p}} in it. If \overline{\mathfrak{r}} is a minimal prime over \overline{x_1}, then \mathfrak{r} contains x_1, x_2^{r_2}, \dots, x_n^{r_n} and thus \mathfrak{r} = \mathfrak{p}; that is to say, \overline{\mathfrak{p}} is a minimal prime over \overline{x_1} and so, by Krull's principal ideal theorem, \overline{\mathfrak{q}} is a minimal prime (over zero); \mathfrak{q} is a minimal prime over (y_2, \dots, y_n). By inductive hypothesis, \operatorname{ht}(\mathfrak{q}) \le n-1 and thus \operatorname{ht}(\mathfrak{p}) \le n. \square

References

• , see in particular section (12.I), p. 77
• http://www.math.lsa.umich.edu/~hochster/615W10/supDim.pdf

References

1. {{harvnb. Eisenbud. 1995