Proposition 5

*Let $k'$ be an extension of $k$, $A$ a k-algebra of finite type and $B$ a $k'$-algebra of finite type.

a) The $k'$-algebra $A \otimes_k B$ is of finite type and one has

b) Let $r$ be a prime ideal of $A \otimes_k B$; let $p$ (resp. $q$) denote the inverse image of $r$ in $A$ (resp. $B$). One has

Put $n = \dim(A)$ and $m = \dim(B)$. By Cor. 1 to Theorem 3 there exist injective algebra homomorphisms $\varphi : k[X_1, ..., X_n] \to A$ and $\psi : k'[Y_1, ..., Y_m] \to B$ making $A$ and $B$ respectively finite algebras over $k[X_1, ..., X_n]$ and $k'[Y_1, ..., Y_m]$. The homomorphism $\varphi \otimes \psi$ is then injective and makes $A \otimes_k B$ into a finite algebra over the $k'$-algebra $k[X_1, ..., X_n] \otimes_k k'[Y_1, ..., Y_m]$, which is identified with $k'[X_1, ..., X_n, Y_1, ..., Y_m]$. Therefore $\dim(A \otimes_k B) = n + m$ by Cor. 1 to Theorem 3, which proves a).

Let us remark that when $A$ and $B$ are integral, $A \otimes_k B$ is a torsion-free $k'[X_1, ..., X_n, Y_1, ..., Y_m]$-module by Lemma 4, and that therefore one has

for every prime ideal $r$ of $A \otimes_k B$ by Remark 1.

Prove now b). Let $r_0$ be a minimal prime ideal of $A \otimes_k B$ contained in $r$, and denote by $p_0$ (resp. $q_0$) the inverse image of $r_0$ in $A$ (resp. $B$). The ring

(A \otimes_k B)/r_0 is isomorphic to a quotient of the ring (A/p_0) \otimes_k (B/q_0). We have therefore, by $a$,

Applying cor. 5 to th. 3, we deduce the inequality

Conversely, let $p_0$ (resp. $q_0$) be a minimal prime ideal of A (resp. B) contained in $p$ (resp. q). According to the remark made above, we have

where $\overline{r}$ is the image of r by the canonical surjection $A \otimes_k B \to (A/p_0) \otimes_k (B/q_0)$. The second member of the preceding equality is obviously less than $\dim_r(A \otimes_k B)$. Applying cor. 5 to th. 3, we deduce the inequality

which completes the proof.