Corollary 4

Let A be a k-algebra of finite type and let p be a prime ideal of A.

a) In order that p be maximal, it is necessary and sufficient that the field of fractions of $A/p$ be a finite extension of k.

b) Let $f \in A - p$; the ideal p is a maximal ideal of A if and only if $pA_f$ is a maximal ideal of $A_f$.

If p is a maximal ideal of A, then $A/p$ is a field, hence a ring of dimension 0; it is a finite type extension of k whose degree of transcendence is 0 (th. 3, c)), it is therefore a finite extension of k. Conversely, if the field of fractions of $A/p$ is a finite extension of k, we have $\dim(A/p) = 0$ hence p is maximal. Assertion b) follows from assertion a), taking into account that $A/p$ and $A_f/pA_f$ have the same field of fractions.

Assertion a) of Cor. 4 is a form of the theorem of zeros (V, § 3, No. 3, Prop. 1).