Corollary 5

Let A be a k-algebra of finite type, p a prime ideal of A and $(\mathfrak{p}_i)_{i \in I}$ the family of minimal prime ideals of A contained in p. We have:

We have $\dim_p(A) = \sup_{i \in I} \dim_{p/p_i}(A/p_i)$ (§ 1, No. 3, Prop. 6). But, according to Cor. 3, we have $\dim_{p/p_i}(A/p_i) = \dim(A/p_i)$, whence the first equality. According to Cor. 2, we have $\dim(A/p_i) = \dim((A/p_i)_p) + \dim(A/p)$. The second equality of the corollary therefore follows from the fact that $\dim(A_p) = \sup_{i \in I} \dim((A/p_i)_p)$ and the third from th. 3.

Thus $\dim_p(A)$ is the supremum of the lengths of the chains of prime ideals of $A$ of which $p$ is an element.