Theorem 1

Let ρ : A → B be a homomorphism of rings making B an integral A-algebra.

a) Let M be a finitely generated A-module. Then one has dim_B(M ⊗ₓₐ B) ≤ dim_A(M). In particular, one has dim(B) ≤ dim(A). If the mapping “ρ : Spec(B) → Spec(A) is surjective, for example (V, § 2, No. 1, th. 1) if ρ is injective, one has dim_B(M ⊗ₓₐ B) = dim_A(M), and in particular dim(B) = dim(A).

b) Let b be an ideal of B and a = ρ⁻¹(b) its inverse image in A. One has ht(b) ≤ ht(a) and dim(B/b) = dim(A/a). If “ρ : Spec B → Spec A is surjective, one has ht(aB) ≤ ht(a) for every ideal a of A.

c) Suppose that B is finite over A and let N be a finitely generated B-module. Then one has dim_B(N) = dim_A(N). In particular, one has dim(B) = dim_A(B).

Let us prove a). By prop. 5 of V, § 1, No. 1, the κ(p)-algebra B ⊗ₓₐ κ(p) is integral for every prime ideal p of A. Let q₀ ⊂ ... ⊂ qₘ be a chain of prime ideals of B; by Lemma 2, the ideals pᵢ = ρ⁻¹(qᵢ) are pairwise distinct, hence p₀ ⊂ ... ⊂ pₘ is a chain of prime ideals of A, whence m ≤ dim(A). Hence dim(B) ≤ dim(A).

Suppose now that the mapping “ρ is surjective. Let p₀ ⊂ ... ⊂ pₙ be a chain of prime ideals of A; there exists therefore a prime ideal q₀ of B over p₀. By cor. 2 to the first existence theorem (V, § 2, No. 1, th. 1), one can construct, by induction on n, a chain q₀ ⊂ ... ⊂ qₙ of prime ideals of B such that qᵢ is over pᵢ for 0 ≤ i ≤ n. Hence n ≤ dim(B) and consequently dim(A) ≤ dim(B).

This proves $a$ in the case where $M = A$. In the general case, let $a$ be the annihilator of $M$, so that the support of $M$ is identified with $\operatorname{Spec}(A/a)$, and one has $\dim_A(M) = \dim(A/a)$. By II, § 4, No. 4, prop. 19, the support of $M \otimes_A B$ is the inverse image by $^a\rho$ of the support of $M$, hence is identified with $\operatorname{Spec}(B/\rho(a) B)$, and one has $\dim_B(M \otimes_A B) = \dim(B/\rho(a) B)$. It remains to observe that the homomorphism $\rho': A/a \to B/\rho(a) B$ deduced from $\rho$ makes $B/\rho(a) B$ an $(A/a)$-algebra integral, and that “$\rho'$ is surjective when $^a\rho$ is so.

Let us prove $b$. By prop. 7 of § 1, No. 3, it suffices to prove that $\operatorname{ht}(b) \leq \dim(A_p)$ for every prime ideal $p$ of $A$ containing $a$; let $p$ be such an ideal. By V, § 2, No. 1, cor. 2 to th. 1, there exists a prime ideal $q$ of $B$ over $p$ and containing $b$, and one has $\operatorname{ht}(b) \leq \dim(B_q)$ by prop. 7 of § 1, No. 3.

Now $B_q$ identifies with a ring of fractions of the $A_p$-algebra $B \otimes_A A_p$, whence

by Prop. 6 of § 1, No. 3 and the assertion $a$ above. We have thus proved the inequality $\operatorname{ht}(b) \leq \operatorname{ht}(a)$. Moreover, the homomorphism from $A/a$ into $B/b$ deduced from $\rho$ is injective and makes $B/b$ an $(A/a)$-algebra which is integral; we therefore have $\dim(B/b) = \dim(A/a)$ by $a$. Suppose “$\rho$ is surjective and let $a$ be an ideal of $A$ and $p$ a prime ideal of $A$ containing $a$. There exists by assumption a prime ideal $q$ of $B$ lying above $p$. We have $aB \subset q$, whence $\operatorname{ht}(aB) \leq \operatorname{ht}(q) \leq \operatorname{ht}(p)$ by what precedes. Passing to the greatest lower bound, we obtain $\operatorname{ht}(aB) \leq \operatorname{ht}(a)$.

Finally, $c$ follows from $b$ applied to the annihilator $b$ of $N$.