Proposition 6

Let A be a noetherian ring of dimension $\geq 2$, and h an integer such that $0 < h < \dim(A)$.

a) A possesses an infinity of prime ideals of height h.

b) If A is finite-dimensional, it possesses an infinity of prime ideals p such that $\mathrm{ht}(p) = h$ and $\dim(A/p) = \dim(A) - h$.

Since the dimension of A is the least upper bound of the heights of the prime ideals of A ($§ 1$, no. 3, prop. 8), since every prime ideal of A is of finite height (no. 1, cor. 1 of prop. 2), and since $h < \dim(A)$, there exists an integer $n > h$ and a prime ideal p of height n, hence a chain $p_0 \subset ... \subset p_n = p$ of prime ideals of length n. One has $\mathrm{ht}(p_i) = i$ for $0 \leq i \leq n$, whence $\mathrm{ht}(r) = h$ for every prime ideal r of A such that $p_{h-1} \subset r \subset p_{h+1}$ is a chain. The set E of these ideals is infinite by prop. 5, whence a).

If $A$ is of finite dimension, one may suppose that one has $n = \dim(A)$ in the foregoing. For every ideal $r \in E$, one has $\mathrm{ht}(r) = h$, whence $\dim(A/r) \leq n - h$, and since $r \subset p_{h+1} \subset ... \subset p_n$ is a chain of length $n - h$, one has $\dim(A/r) = n - h$.

There exist non-noetherian integral rings of dimension 2 possessing only one prime ideal of height 1, for example a valuation ring of height 2 (VI, § 4, no. 4, prop. 5).