Proposition 6

Let $A$ be a ring, $J$ an ideal of finite type of $A$ and $M$ an $A$-module. Let $B$ be a ring and $\rho : A \to B$ a ring homomorphism making $B$ into a flat $A$-module.

a) One has $\operatorname{prof}_A(J; M) \leq \operatorname{prof}_B(JB; B \otimes_A M)$.

b) Suppose moreover that every maximal ideal of $\operatorname{Supp}(M)$ containing $J$ belongs to the image of the canonical mapping $\operatorname{Spec}(B) \to \operatorname{Spec}(A)$. Then one has $\operatorname{prof}_A(J; M) = \operatorname{prof}_B(JB; B \otimes_A M)$. This is the case for example if the $A$-module $B$ is faithfully flat.

Assertion a) follows from Theorem 1 and Lemma 1.

Let $p$ be an integer strictly less than $\operatorname{prof}_B(JB; B \otimes_A M)$, and let $m$ be a maximal ideal of $A$ belonging to $\operatorname{Supp}(M) \cap V(J)$. Let $x$ be a finite generating family of the ideal $J$. Under the hypothesis of b), there exists a prime ideal $n$ of $B$ lying over $m$, and one has a canonical isomorphism

Now $B \otimes_A H^p(x, M)$ is isomorphic to $H^p(\rho(x), B \otimes_A M)$ (Lemma 1), hence is zero; moreover $B_n$ is faithfully flat over $A_m$ (I, § 3, No. 5, Proposition 9 and II, § 3, No. 4, Proposition 14 and Remark). Hence one has $A_m \otimes_A H^p(x, M) = 0$ and therefore $p < \operatorname{prof}_{A_m}(J_m; M_m)$ (Lemma 1 and Theorem 1). The first assertion of b) then follows from Proposition 5; the second follows from I, § 3, No. 5, Proposition 9).