Krull's principal ideal theorem

Symbolic statement
Let $$R$$ be a Noetherian and $$x \in R$$. Let $$P$$ be a minimal element among prime ideals containing $$x$$. Then, the codimension of $$P$$ is at most 1.

Property-theoretic statement
The property of commutative unital rings of being a Noetherian ring is stronger than the property of being a ring satisfying PIT.

Generalizations

 * Krull's height theorem: This is often also called the final version of the principal ideal theorem.
 * Determinantal ideal theorem: This generalizes the principal ideal theorem to the ideal generated by the determinants of minors of a matrix

Starting assumptions
In the above setup, we show that if $$Q$$ is a prime ideal in $$R$$ contained inside $$P$$, then the codimension of $$Q$$, which is the same as the dimension of $$R_Q$$ is zero. This will show that the codimension of $$P$$ is at most 1. The crucial thing we shall use is that $$x \notin Q$$.

First note that we can replace $$R$$ by $$R_P$$, so we may assume that $$P$$ is a maximal ideal in $$R$$. We now begin the proof.

Argument setup
Since $$P$$ is minimal over $$x$$, we see that in the ring $$R/(x)$$, the ideal $$P/(x)$$ is the unique maximal ideal of a local ring, and is also a minimal prime ideal. Thus, $$R/(x)$$ is a local Artinian ring with unique maximal ideal $$P/(x)$$.

Hence, in $$R$$, consider the descending chain $$x + Q^{(n)}$$, where $$Q^{(n)}$$ denotes the $$n^{th}$$ symbolic power of $$Q$$. This descending chain stabilizes, so we get:

$$Q^{(n)} \subset (x) + Q^{(n+1)}$$

In particular, we can find $$f \in Q^{(n)}$$ and $$a \in R, g \in Q^{(n+1)}$$ such that:

$$f = ax + g$$

This yields $$ax \in Q^{(n)}$$, so since $$x \notin Q$$, we get $$a \in Q^{(n)}$$

Nakayama's lemma
The above reasoning shows that:

$$Q^{(n)} = (x)Q^{(n)} + Q^{(n+1)}$$

Now consider the module $$M = Q^{(n)}/Q^{(n+1)}$$. The above equation yields that:

$$M = (x)M$$

But since $$x \in P$$, we see that $$x$$ is in the Jacobson radical of $$R$$, so Nakayama's lemma yields that $$M = 0$$. Thus $$Q^{(n)} = Q^{(n+1)}$$.

We now apply Nakayama's lemma in the localization at $$Q$$, to conclude that:

$$(Q_Q)^n = 0$$

This yields that $$R_Q$$ isa zero-dimensional ring.