Krull's height theorem

Remarks

The proof is by induction on $n$, with Krull's principal ideal theorem as the case $n = 1$. For the step, take a prime $Q \subsetneq P$ with no prime strictly between them. Since $P$ is minimal over $I$, some $x_i$ lies outside $Q$, and one shows that $Q$ is minimal over an ideal with $n - 1$ generators. For example, in $k[x, y, z]$ over a field $k$, the prime $(x, y)$ is minimal over the ideal generated by $x$ and $y$, and its height is $2$. A converse holds: every prime ideal of height $n$ in a Noetherian ring is minimal over some ideal generated by $n$ elements. As a consequence, every prime ideal of a Noetherian ring has finite height. A Noetherian local ring has finite Krull dimension, at most the number of generators of its maximal ideal.

Going deeper

For a student

The theorem is the base of dimension theory of Noetherian local rings. It gives $\dim R \le$ the minimal number of generators of the maximal ideal, which leads to regular local rings and to systems of parameters. It also shows that the prime ideals of a Noetherian ring satisfy the descending chain condition.