Corollary 3

Let A be a k-algebra of finite type and let f be an element of A which belongs to no minimal prime ideal of A (for example an element of A which is not a divisor of zero, cf. IV, § 1, No. 1, Cor. 3 to Prop. 2 and No. 3, Cor. 1 to Prop. 7). We have $\dim(A) = \dim(A_f)$.

The mapping $p \mapsto pA_f$ is a bijection of the set of minimal prime ideals of A onto the set of minimal prime ideals of $A_f$. Moreover the rings $A/p$ and $A_f/pA_f = (A/p)_f$ have the same field of fractions. It is therefore sufficient to apply th. 3, b).