a) Let h : A → A′ be a homomorphism of rings. Then the homomorphism ρ′ : A′ → A′ ⊗_A B deduced from ρ satisfies the condition (PM).

b) Let q be a prime ideal of B and p = ρ^{-1}(q). The canonical homomorphism ρ_q : A_p → B_q satisfies the condition (PM).

c) Let q be a prime ideal of B and p = ρ^{-1}(q). For every prime ideal p′ of A contained in p, there exists a prime ideal q′ of B lying above p′ and contained in q.

Let N be a B-module faithfully flat over A and such that N ⊗_B κ(q) ≠ 0 for every prime ideal q of B.

Prove a). The A′-module N′ = A′ ⊗_A N is faithfully flat (I, § 3, No. 3, prop. 5); let q′ be a prime ideal of B′ = A′ ⊗_A B and q its inverse image in B. We have isomorphisms

as we have N ⊗_B κ(q) ≠ 0, we also have N′ ⊗_{B′} κ(q′) ≠ 0.

Prove b). According to props. 13 and 14 of II, § 3, No. 4, the A_p-module N_q is flat. On the other hand, let b′ be a prime ideal of B_q; it is of the form bB_q where b is a prime ideal of B contained in q (II, § 3, No. 1, prop. 3); we have N ⊗_B κ(b) ≠ 0 in view of the assumption made on N, and since N_q ⊗_{B_q} κ(b′) is isomorphic to N ⊗_B κ(b′), we have N_q ⊗_{B_q} κ(b′) ≠ 0. Remark 3 allows us to conclude.

Prove c). The local homomorphism ρ_q : A_p → B_q deduced from ρ satisfies the condition (PM) according to b). The mapping Spec(B_q) → Spec(A_p) is therefore surjective (Remark 2), which was to be proved.

Let indeed a be an ideal of A such that F = V(a) and q the prime ideal of B such that Y = V(q). The inverse image by “ρ of F is the subset V(ρ(a)B) of Spec(B) and the closure of “ρ(Y) is the irreducible closed subset V(ρ^{-1}(q)) of Spec(A).

It is a matter of proving that if q is minimal among the prime ideals of B containing $\rho(a)$, then $\rho^{-1}(q)$ is minimal among the prime ideals of A containing a. Otherwise, there would exist a prime ideal $p'$ of A with $a \subset p' \subset \rho^{-1}(q)$ and $p' \neq \rho^{-1}(q)$; according to prop. 1, c), there would exist a prime ideal $q'$ of B such that $q' \subset q$ and $p' = \rho^{-1}(q')$, whence $\rho(a) \subset q' \subset q$ and $q' \neq q$ contrary to the assumption made on q.

(1)

where S is the set of maximal ideals of A.

We know that $\dim(A) = \sup_{m \in S} \dim(A_m)$ (§ 1, No. 3, prop. 8). It is therefore enough to establish the inequality

(2)

for every maximal ideal m of A. In other words, it is a matter of proving the inequality

(3)

if $p_0 \subset ... \subset p_n$ is a chain of prime ideals of A contained in m and $\overline{q}_0 \subset ... \subset \overline{q}_r$ is a chain of prime ideals of B/mB. For $0 \leq i \leq r$, there exists a prime ideal $q_{n+i}$ of B lying above m such that $\overline{q}_i = q_{n+i}/mB$, and $q_n \subset ... \subset q_{n+r}$ is a chain of prime ideals of B. Put $p'_i = p_i$ for $0 \leq i \leq n-1$ and $p'_n = m$, so that $p'_0 \subset ... \subset p'_n$ is a chain of prime ideals of A and q_n lies above $p'_n$. If $q_i$ is a prime ideal of B lying above $p'_i$ ($1 \leq i \leq n$), prop. 1, c) proves that there exists a prime ideal $q_{i-1}$ of B lying above $p'_{i-1}$ and contained in $q_i$. By descending induction, we thus construct a chain $q_0 \subset ... \subset q_n$ of prime ideals of B such that $q_i$ lies above $p_i$ for $0 \leq i \leq n$. As $q_0 \subset ... \subset q_{n+r}$ is a chain of prime ideals of B, we have proved the inequality (3).

Let q be a prime ideal of B containing $\rho(a)B$, and $p = \rho^{-1}(q)$. According to prop. 1, the local homomorphism $\rho_q : A_p \to B_q$ deduced from $\rho$ satisfies (PM), and we therefore have $\dim(A_p) \leq \dim(B_q)$ according to prop. 2. According to prop. 7 of § 7, no 3, we have $\mathrm{ht}(a) \leq \dim(A_p)$ since p contains a, whence $\mathrm{ht}(a) \leq \dim(B_q)$ for every prime ideal q of B containing $\rho(a)B$. The corollary then follows from loc. cit.

The homomorphism $h$ is composed of the homomorphism of passing to the quotient of $B$ into $B/\rho(p)$ $B$ and of the canonical homomorphism of $B/\rho(p)$ $B$ into its ring of fractions $(\rho(A - p))^{-1}(B/\rho(p) \ B)$. According to the remark and the corollary to prop. 13 of II, § 4, no 3, $^a h$ therefore induces a homeomorphism of $\mathrm{Spec}(B \otimes_A \kappa(p))$ onto the subspace of $\mathrm{Spec}(B)$ formed by the prime ideals $q$ of $B$ which contain $\rho(p)$ and are disjoint from $\rho(A - p)$, that is to say which lie above $p$.

We may suppose $\dim(A) \neq -\infty$ and $n < +\infty$. Let $q_0 \subset ... \subset q_m$ be a chain of prime ideals of B; put $p_i = \rho^{-1}(q_i)$. The sequence of the $p_i$ is increasing, hence the set of its values has cardinal $\leq \dim(A) + 1$. For each $p \in \mathrm{Spec}(A)$, the set of the $q_j$ such that $p_j = p$ is a chain of the subset $^a\rho^{-1}(p)$ of $\mathrm{Spec}(B)$, hence has cardinal less than $\dim(B \otimes_A \kappa(p)) + 1$ (no 1, Lemma 1), and consequently less than or equal to $(n + 1)$. It follows that $m + 1 \leq (\dim(A) + 1)(n + 1)$, whence the proposition.

The first inequality has already been proved (§ 1, No. 3, example 4). Let us prove the second one. For every prime ideal $p$ of $A$, the ring $B \otimes_A \kappa(p)$, isomorphic to $\kappa(p)[X]$, is principal and is not a field, hence is of dimension 1 (§ 1, No. 3, example 2), and the inequality follows from prop. 3.

However, whatever the integers n and q with n + 1 ≤ q ≤ 2n + 1 may be, there exists a ring A of dimension n such that dim(A[X]) = q (see p. 84, exerc. 7).

Indeed, one deduces from cor. 2, by induction on n, that the ring A[T₁, ..., Tₙ] is of finite dimension if A is of finite dimension; fortiori, every nonzero quotient of A[T₁, ..., Tₙ] is of finite dimension (§ 1, No. 3, prop. 6).

Indeed, if q and q’ are over the same prime ideal p of A, one has dim(B ⊗ₓₐ κ(p)) ≥ 1 by Lemma 1 of No. 1, which contradicts the fact that dim(B ⊗ₓₐ κ(p)) ≤ 0 (§ 1, No. 3, example 6).

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 $N$.

Soit $\rho$ l’application canonique de $A$ dans $B$. Soient $a$ un idéal de $A$. Si $a = A$, la première égalité est claire. Supposons $a \neq A$. Comme $\rho$ est injectif, $^a\rho$ est surjectif (V, § 2, No. 1, Theorem 1). Par suite $aB \neq B$. Soit alors $q$ un idéal premier de $B$ contenant $aB$. Posons $p = q \cap A$. On a $a \subset p$, d’où $\operatorname{ht}(a) \leq \operatorname{ht}(p)$. Soit $p_0 \subset ... \subset p_n$ une chaîne d’idéaux premiers de $A$ avec $p_n = p$. D’après le deuxième théorème d’existence (V, § 2, No. 4, Theorem 3), on construit par récurrence une chaîne $q_0 \subset ... \subset q_n$ d’idéaux premiers de $B$ telle que $q_n = q$ et $q_i$ soit au-dessus de $p_i$ pour $0 \leq i \leq n$. On a $n \leq \operatorname{ht}(q)$, d’où $\operatorname{ht}(a) \leq \operatorname{ht}(q)$. En passant à la borne inférieure on obtient $\operatorname{ht}(a) \leq \operatorname{ht}(aB)$ ($§ 1$, No. 3, Prop. 7). L’inégalité $\operatorname{ht}(aB) \leq \operatorname{ht}(a)$ résulte du Theorem 1, d’où la première equality. Soit $b$ un idéal de $B$. Posons $a = \rho^{-1}(b)$. On a $aB \subset b$, d’où $\operatorname{ht}(a) = \operatorname{ht}(aB) \leq \operatorname{ht}(b)$. L’inégalité $\operatorname{ht}(b) \leq \operatorname{ht}(a)$ résulte du Theorem 1, d’où le theorem.

According to the normalization lemma (V, § 3, No. 1, th. 1), there exists an integer $n \geq 0$, a finite sequence $(x_1, ..., x_n)$ of elements of $A$ algebraically free and an increasing sequence $(h(j))_{0 \leq j \leq m}$ of integers $\leq n$ such that the ideal $p_j \cap B$ is equal to the prime ideal $q_j$ of $B$ generated by the $x_k$ with $1 \leq k \leq h(j)$, and that $A$ is entire over the ring $B$. Let $j$ be an integer such that $0 \leq j < m$. By passing to quotients, one deduces from the canonical injection of $B$ into $A$ an injective homomorphism of $B/q_j$ into $A/p_j$ which makes $A/p_j$ a finite $(B/q_j)$-algebra. Since the ring $B/q_j$ is isomorphic to an algebra of polynomials in $n - h(j)$ indeterminates over $k$, it is integrally closed (V, § 1, No. 3, Cor. 3 of Prop. 13). According to th. 2 of No. 3, one therefore has

It follows that one has $h(j+1) \leq h(j) + 1$ and $q_{j+1} \neq q_j$, whence $h(j+1) = h(j) + 1$. But one has $h(m) = n$ since $q_m$ is maximal (V, § 2, No. 1, Prop. 1), whence finally $h(j) = n - m + j$.

Suppose $A$ is integral and consider a maximal chain $p_0 \subset ... \subset p_m$ of prime ideals of $A$. We have $p_0 = 0$. We deduce then from Lemma 3 the existence of an injective homomorphism $\varphi : k[X_1, ..., X_m] \to A$ of $k$-algebras which makes $A$ a finite $k[X_1, ..., X_m]$-algebra. Consequently, the degree of transcendence over $k$ of the field of fractions of $A$ is equal to $m$, whence c). Assertion a) follows from assertion c) applied to the ring $A/p$ and assertion b) is an immediate consequence of a) and Prop. 5 of § 1, No. 2.

For a finite type k-algebra A to have dimension n, it is necessary and sufficient that there exist an injective k-homomorphism $\varphi : k[X_1, ..., X_n] \to A$ making A a finite algebra over $k[X_1, ..., X_n]$.

This follows from th. 3, from the normalization lemma (V, § 3, No. 1, th. 1) and from th. 1, a) of No. 3.

In particular, we have $\operatorname{ht}(m) = \dim(A_m) = \dim(A)$ for every maximal ideal m of A.

This follows from th. 3 and Remark 4 of § 1, No. 3.

The mapping $p \mapsto pA_f$ is a bijection of the set of minimal prime ideals of A onto the set of minimal prime ideals of $A_f$. Moreover the rings $A/p$ and $A_f/pA_f = (A/p)_f$ have the same field of fractions. It is therefore sufficient to apply th. 3, b).

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).

We have $\dim_p(A) = \sup_{i \in I} \dim_{p/p_i}(A/p_i)$ (§ 1, No. 3, Prop. 6). But, according to Cor. 3, we have $\dim_{p/p_i}(A/p_i) = \dim(A/p_i)$, whence the first equality. According to Cor. 2, we have $\dim(A/p_i) = \dim((A/p_i)_p) + \dim(A/p)$. The second equality of the corollary therefore follows from the fact that $\dim(A_p) = \sup_{i \in I} \dim((A/p_i)_p)$ and the third from th. 3.

Thus $\dim_p(A)$ is the supremum of the lengths of the chains of prime ideals of $A$ of which $p$ is an element.

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.

Suppose first that B is integral. The algebra $B \otimes_A K$ is then a ring of fractions of B defined by a multiplicative subset not containing 0. It therefore has as its field of fractions the field of fractions L of B. By th. 3, one has

Now, by the corollary of A, V, p. 106

whence the result in this case.

Let us pass to the general case. Every minimal prime ideal $p$ of $B$ is formed of zero divisors in $B$, and is therefore above the ideal 0 of $A$. It follows that the mapping $p \mapsto p . (B \otimes_A K)$ is a bijection from the set of minimal prime ideals of $B$ onto the set of minimal prime ideals of $B \otimes_A K$. The proposition therefore follows from the first part of the proof and from prop. 6, c) of § 1, no 3.

Let $K$ (resp. $L$) be the field of fractions of $A$ (resp. $B$). There exists a set $I$ (resp. $J$) such that $M$ (resp. $N$) is isomorphic to a submodule of $K^{(I)}$ (resp. $L^{(J)}$). The $(A \otimes_k B)$-module $M \otimes_k N$ is therefore isomorphic to a submodule of $K^{(I)} \otimes_k L^{(J)}$, which is isomorphic to $(K \otimes_k L)^{(I \times J)}$. Since $K \otimes_k L$ is a ring of fractions of the integral ring $A \otimes_k B$, it is a torsion-free module over $A \otimes_k B$, whence the lemma.

*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.

a) We have $\dim(A') = \dim(A)$.

b) Let p' be a prime ideal of A' and p its inverse image in A; we have $\dim_{p'}(A') = \dim_p(A)$.

c) Let p' be a minimal prime ideal of A' and p its inverse image in A. Then p is minimal and we have $\dim(A'/p') = \dim(A/p)$.

The assertions a) and b) follow from prop. 5 by taking $B = k'$. Let us prove c). The ideal p is minimal (No. 1, prop. 1) and we have