Corollary 6

Let A be a finitely generated k-algebra not reduced to 0, and let n be an integer $\geqslant 0$. The following conditions are equivalent:

a) For every $p \in \operatorname{Ass}(A)$, one has $\dim(A/p) = n$.

b) Every associated prime ideal of A is minimal and all the irreducible components of Spec(A) have dimension n.

c) There exists an injective k-homomorphism $\varphi : k[X_1, ..., X_n] \to A$ making A a $k[X_1, ..., X_n]$-module of finite type without torsion.

The equivalence of a) and b) is immediate. Let us show that a) implies c). According to b), the ring A has dimension n and there therefore exists (cor. 1) an injective k-homomorphism $\varphi : k[X_1, ..., X_n] \to A$ making A a $k[X_1, ..., X_n]$-module of finite type. For every prime ideal $p \in \operatorname{Ass}(A)$, the ring $A/p$ is then integral over $k[X_1, ..., X_n]$, and we therefore have $n = \dim(A/p) = \dim(k[X_1, ..., X_n]/\varphi^{-1}(p))$ by th. 1, a) of no. 3, whence $\varphi^{-1}(p) = 0$. The image by the injective homomorphism $\varphi$ of a nonzero element of $k[X_1, ..., X_n]$ is not a zero divisor in A (IV, § 1, no. 1, cor. 3 to prop. 2), whence c).

Conversely, suppose that condition c) is satisfied. For every prime ideal $p \in \operatorname{Ass}(A)$, the homomorphism $k[X_1, ..., X_n] \to A/p$ deduced from $\varphi$ is injective (loc. cit.). We therefore have $\dim(A/p) = n$ by cor. 1.