Let $q \subset p$ be a prime ideal distinct from $p$. One has $x \notin q$ in view of the minimal character of $p$. Since $A$ is an integral domain, $A_p$ can be identified with a subring of $A_q$; for every integer $n \geq 0$, let $q_n$ denote the ideal $q^n A_q \cap A_p$ of $A_p$. The minimal character of $p$ means that the local ring $A_p / xA_p$ is of dimension 0; hence it is of finite length ($§ 1$, no 3, example 1), and there exists an integer $n_0 \geq 0$ such that one has

Fix the integer $n \geq n_0$. Given $y \in q_n$, there exists $a \in A_p$ such that $y - ax \in q_{n+1}$; one then has $ax \in q_n$, whence $a \in q_n$ since $x \notin q$, and finally one has $y \in q_{n+1} + xq_n$. Hence one has

Since $x$ belongs to the maximal ideal of the noetherian local ring $A_p$, Nakayama's lemma shows that one has $q_n = q_{n+1}$. Since one has $(qA_q)^n = q_n A_q$, one concludes

Since the ring $A_q$ is local and noetherian, one has $\bigcap_{n \geq 0} (qA_q)^n = \{0\}$ (III, § 3, no 2, corollary of the prop. 5) whence $(qA_q)^{n_0} = \{0\}$ and finally the prime ideal $qA_q$ of $A_q$ is reduced to 0. Hence $q = \{0\}$, which proves that $p$ is of height 1.

The inequality $\dim(A/a) \leq \dim(A)$ follows from prop. 6 of § 1, no 3. An immediate induction on $m$ shows that it is enough to establish the inequality

for every element $x$ of the radical of $A$, that is to say, to prove that one has $\dim(A/xA) \geq n - 1$ for every chain $p_0 \subset ... \subset p_n$ of prime ideals of $A$, of length $n \geqslant 1$, and such that $x \in p_n$. It suffices to construct a chain $q_1 \subset ... \subset q_n$ of prime ideals of $A$, with $x \in q_1$ and this follows from the following lemma:

We reason by induction on $n$, the case $n = 1$ being trivial. Suppose therefore that one has $n \geqslant 2$ and that $x$ does not belong to $p_{n-1}$. Let $p'_{n-1}$ be a minimal element of the set of prime ideals of $A$ contained in $p'_n = p_n$ and containing $p_{n-2} + Ax$ (II, § 2, No. 6, lemma 2). By prop. 1, the ideal $p'_{n-1}/p_{n-2}$ of the ring $A/p_{n-2}$ is of height 1, and since $p_{n-2} \subset p_{n-1} \subset p_n$ is a chain of length 2, the same is true of $p_{n-2} \subset p'_{n-1} \subset p'_n$. One has $x \in p'_{n-1}$. The induction hypothesis applied to the chain $p_0 \subset p_1 \subset ... \subset p_{n-2} \subset p'_{n-1}$ shows that there exists a chain $p'_0 \subset p'_1 \subset ... \subset p'_{n-2} \subset p'_{n-1}$ with $x \in p'_1$ and $p'_0 = p_0$. The chain

a) Let $A$ be a nonzero semi-local noetherian ring and $a$ its radical; the quotient ring $A/a$ is artinian and nonzero, hence of dimension 0 ($§ 1$, No. 3, Example 1). There exists an integer $m \geqslant 0$ such that the ideal $a$ of $A$ is generated by $m$ elements; hence $0 \leqslant \dim(A) \leqslant m$ by Prop. 2. b) Let $a \neq A$ be an ideal of $A$, and let $m$ be a maximal ideal of $A$ containing $a$. We have $0 \leqslant \mathrm{ht}(a) \leqslant \dim(A_m)$ by Prop. 7 of $§ 1$, No. 3, and $A_m$ is a local noetherian ring. Hence $\mathrm{ht}(a)$ is finite by a). c) Every finite strictly decreasing sequence $(p_i)_{0 \leqslant i \leqslant n}$ of prime ideals of $A$ defines a chain $p_n \subset ... \subset p_0$, whence $n \leqslant \dim(A_{p_0}) < + \infty$. Therefore there cannot exist an infinite strictly decreasing sequence of prime ideals of $A$, whence c).

Let $\Phi$ be the set of minimal prime ideals of $A$, and $\Phi'$ the set of $p \in \Phi$ such that $\dim(A/p) = \dim(A)$. We know (II, § 4, No. 3, Corollary 3 of Proposition 14) that $\Phi$ is finite, hence $\Phi'$ is finite. Let $a$ be an ideal of $A$ such that $\dim(A/a) < \dim(A)$. For every $p \in \Phi'$, one has $\dim(A/a) < \dim(A/p)$, hence $a \not\subset p$. By Proposition 2 of II, § 1, No. 1, there exists therefore an element $x$ of $a$ which does not belong to any of the $p \in \Phi'$, and one then has $\dim(A/xA) = \dim(A) - 1$ by a). This proves b).

The assertion c) is the particular case $a = m_A$ of b).

a) If $M/SM$ is of finite length, one has $\mathrm{Card}(S) \geqslant \dim_A(M)$; if $S$ is secant for $M$, one has $\mathrm{Card}(S) \leqslant \dim_A(M)$.

b) Every secant subset for $M$ is contained in a maximal secant subset for $M$.

c) The following properties are equivalent: (i) $S$ is a maximal secant subset for $M$; (ii) $S$ is a secant subset for $M$ and $\mathrm{Card}(S) = \dim_A(M)$; (iii) $M/SM$ is of finite length and $\mathrm{Card}(S) = \dim_A(M)$; (iv) $S$ is a secant subset for $M$ and $M/SM$ is of finite length.

Since one has $S \subset m_A$, Nakayama's lemma shows that one has $M/SM \neq \{0\}$, whence $\dim_A(M/SM) \geqslant 0$ with equality if and only if $M/SM$ is of finite length. Assertion a) then results from formulas (8) and (9), as well as the equivalence of properties (ii), (iii) and (iv).

Assertion b) results from the fact that the cardinal of every subset of $m_A$ secant for $M$ is bounded above by the integer $\dim_A(M)$.

By a), every secant subset for $M$, of cardinal equal to $\dim_A(M)$, is maximal. It remains to prove that, if $S$ is secant for $M$ and if $\mathrm{Card}(S) < \dim_A(M)$, then $S$ is not maximal. Let $a$ be the annihilator of $M$, and $B$ the noetherian local ring $A/(a + SA)$. By cor. 2 of prop. 2 of No. 1, there exists an element $x$ of $m_A$ such that $\dim(B/xB) = \dim(B) - 1$ whence $x \notin S$. By remark 2, the subset $S \cup \{x\}$ of $m_A$ is secant for $A/a$, hence for $M$ by remark 1.

Since $\varnothing$ is a secant subset for $M$, Theorem 1, b) proves the existence of a maximal secant sequence for $M$, say $(x_1, ..., x_d)$. But then one has $d = \dim_A(M)$ and the $A$-module $M / \sum_{i=1}^d x_i M$ is of finite length by property (iii) of Theorem 1, c). Conversely, if $(x'_1, ..., x'_{d'})$ is a sequence of elements of $m_A$ such that the $A$-module $M / \sum_{j=1}^{d'} x'_j M$ is of finite length, one has $d' \geqslant \dim_A(M)$ by Theorem 1, a).

Recall (III, § 3, No. 2, Definition 1) that an ideal $q$ of a noetherian local ring $A$ is an ideal of definition of $A$ if the $q$-adic and $m_A$-adic topologies of $A$ coincide.

(i) $q$ is an ideal of definition of $A$; (ii) there exists an integer $n \geqslant 0$ such that $m_A^n \subset q \subset m_A$;

(iii) one has q ≠ A and A/q is an A-module of finite length ; (iv) V(q) is equal to { m_A } (in other words, m_A is the only prime ideal of A containing q).

In fact, the equivalence of (i), (ii) and (iv) was proved in III, § 2, No. 5, and the equivalence of (i) and (iii) follows from IV, § 2, No. 5, Corollary 2 to Proposition 9.

The corollary to Theorem 1 therefore makes it possible to state the following scholium:

a) For every closed subset Y of V contained in X', one has

b) One has $\operatorname{codim}(Z, X) \leq m$ for every irreducible component Z of X'. If X' is nonempty, one has $\operatorname{codim}(X', X) \leq m$.

c) If Z is an irreducible closed subset of V contained in X such that $\operatorname{codim}(Z, X) \leq m$, there exist hypersurfaces $H'_1, ..., H'_m$ such that Z is an irreducible component of $X \cap H'_1 \cap ... \cap H'_m$.

Let α be an ideal of A and $x_1, ..., x_m$ elements of A such that $X = V(\alpha)$ and $H_i = V(x_i)$ for $1 \leq i \leq m$. Let Z be an irreducible closed subset of V contained in X; there exists a prime ideal p of A containing α and such that $Z = V(p)$.

Suppose first that Z is contained in X' and denote by $\xi_i$ the image of $x_i$ in the noetherian local ring $B = A_p/\alpha A_p$. By Proposition 7, b) of § 1, No. 3, one has

By Proposition 2 of No. 1, one therefore has

(11)

If Z is an irreducible component of X', one has $\operatorname{codim}(Z, X') = 0$, whence $\operatorname{codim}(Z, X) \leq m$; this proves b). One proves a) by taking in both members of (11) the greatest lower bound over the set of irreducible components Z of Y.

1 Recall that $V(x)$ consists of the prime ideals of A containing x.

Conversely, suppose that one has codim(Z, X) $\leq m$, that is to say dim(B) $\leq m$. Since every element of $A_p$ is the product of an invertible element of $A_p$ by the image of an element of A, the scholium of No. 2 proves the existence of elements $x'_1, ..., x'_m$ of A whose images in B generate an ideal of definition of B. Put $H'_i = V(x'_i)$ for $1 \leq i \leq m$. It is clear that Z is an irreducible component of $X \cap H'_1 \cap ... \cap H'_m$.

One has codim(H, V) = 1 if and only if H contains no irreducible component of V. If this is so, all the irreducible components of H are of codimension 1 in V.

For every irreducible component Z of H, one has

by Proposition 4, b), and one has codim(Z, V) = 0 if and only if Z is an irreducible component of V. One has by definition

where Z ranges over the set of irreducible components of H. Corollary 1 follows at once from these remarks.

1) one has $X \subset H$; 2) the set $X \cap H$ is nonempty and each of its irreducible components Z satisfies $\operatorname{codim}(Z, X) = 1$; 3) the set $X \cap H$ is empty.

Suppose $X' = X \cap H$ nonempty and distinct from X; every irreducible component Z of $X'$ is distinct from X and satisfies $\operatorname{codim}(Z, X) \leq 1$ by Prop. 4, b). Corollary 2 follows at once from this.

Let x be an extremal element of A. Then Ax is a prime ideal because x is extremal, of height 1 because A is an integral domain (No. 1, Prop. 1). Let p be a prime ideal of height 1 of A. Then $V(p)$ is an irreducible component of a hypersurface $V(Ax)$ for a suitable x (Prop. 4, c)). Let $x = \prod_i y_i^{n_i}$ be a decomposition of x into a product of extremal elements such that $y_i$ and $y_j$ are relatively prime if $i \neq j$. The irreducible components of $V(Ax)$ are the $V(Ay_i)$. Hence $p = Ay_i$ for a suitable i. The last assertion follows from Corollary 2 to Prop. 2 of No. 1.

Replacing A by $A/p$ if necessary, one is reduced to the case where $p = \{0\}$.

By Lemma 1 of No. 1, one has $q = \bigcup_{r \in E} r$, and Prop. 2 of II, $§ 1$, No. 1 shows that E is infinite.

Let $y \neq 0$ be an element of $\bigcap_{r \in E} r$. The height of q is finite (No. 1, Cor. 1 of Prop. 2), and one has $\mathrm{ht}(q) \geq 2$ by hypothesis. There therefore exists a prime ideal $q' \subset q$ of height 2. The first part of the proof applied to $q'$ shows that the set E’ of prime ideals of height 1 contained in $q'$ is infinite; each of these ideals contains y by hypothesis. Now the local noetherian ring $B = A_{q'}/yA_{q'}$ is of dimension 1 by Corollary 2 of Prop. 2 of No. 1. For every $r \in E'$, the prime ideal $r/yA_{q'}$ of B is therefore minimal; consequently, the noetherian ring B has an infinity of minimal prime ideals, which is absurd (II, $§ 4$, No. 3, Cor. 3 of Prop. 14). Hence $\bigcap_{r \in E} r = \{0\}$.

a) A possesses an infinity of prime ideals of height h.

b) If A is finite-dimensional, it possesses an infinity of prime ideals p such that $\mathrm{ht}(p) = h$ and $\dim(A/p) = \dim(A) - h$.

Since the dimension of A is the least upper bound of the heights of the prime ideals of A ($§ 1$, no. 3, prop. 8), since every prime ideal of A is of finite height (no. 1, cor. 1 of prop. 2), and since $h < \dim(A)$, there exists an integer $n > h$ and a prime ideal p of height n, hence a chain $p_0 \subset ... \subset p_n = p$ of prime ideals of length n. One has $\mathrm{ht}(p_i) = i$ for $0 \leq i \leq n$, whence $\mathrm{ht}(r) = h$ for every prime ideal r of A such that $p_{h-1} \subset r \subset p_{h+1}$ is a chain. The set E of these ideals is infinite by prop. 5, whence a).

If $A$ is of finite dimension, one may suppose that one has $n = \dim(A)$ in the foregoing. For every ideal $r \in E$, one has $\mathrm{ht}(r) = h$, whence $\dim(A/r) \leq n - h$, and since $r \subset p_{h+1} \subset ... \subset p_n$ is a chain of length $n - h$, one has $\dim(A/r) = n - h$.

There exist non-noetherian integral rings of dimension 2 possessing only one prime ideal of height 1, for example a valuation ring of height 2 (VI, § 4, no. 4, prop. 5).

One may suppose $M$ and $N$ non-zero. By the corollary of th. 1 (no. 2), there exists a subset $S$ (resp. $T$) of $m_A$ (resp. $m_B$) such that $\mathrm{Card}(S) = \dim_A(M)$ (resp. $\mathrm{Card}(T) = \dim_{\overline{B}}(\overline{N})$) and that $M/SM$ is an $A$-module of finite length (resp. $\overline{N}/T\overline{N}$ is a $B$-module of finite length). One has $\rho(S) \subset m_B$. Let $E$ be the $B$-module $M \otimes_A N$; the $B$-module $E/(\rho(S).E + T.E)$ is isomorphic to $M/SM \otimes_A N/TN$, hence is of finite length: by props. 18 and 19 of II, § 4, no. 4, one has

It follows, by th. 1, that one has the inequality

Now suppose $N$ flat over $A$, and let us show that there is equality in formula (12). Let $a$ (resp. $b$) be the annihilator of $M$ (resp. $N$). By props. 18 and 19 of II, § 4, no. 4, one has

Hence

and also

and since the ring $B'/\rho'(\mathfrak{m}_{A'}).B'$ is isomorphic to $B/(b + \rho(\mathfrak{m}_A).B)$, our assertion follows from formulas (13), (14) and (15).

a) One has

with equality if $\rho$ makes $B$ into a flat $A$-module.

b) Suppose that $\rho$ makes $B$ into a flat $A$-module, and let $S$ be a subset of $\mathfrak{m}_A$. Then $S$ is secant for $A$ if and only if $\rho(S)$ is secant for $B$.

Assertion a) is the particular case $M = A, N = B$ of Prop. 7. Let us prove b). Under the hypotheses made, we have

with $\overline{B} = B \otimes_A \kappa_A = B/\rho(\mathfrak{m}_A).B$. Since $\rho$ is injective (I, § 3, No. 5, Prop. 9), we have

Finally $B' = B/\rho(S).B$ is a flat module over $A' = A/SA$, whence

since $B'/\rho(\mathfrak{m}_{A'}).B'$ is isomorphic to $\overline{B}$. Our assertion follows at once from formulas (17), (18) and (19).

Let $q$ be a prime ideal of $B$ and $p = \rho^{-1}(q)$. By Cor. 1, we have $\dim(B_q) \leq \dim(A_p) + \dim(B_q \otimes_A \kappa(p))$. It follows (Prop. 6, b) of § 1, No. 3) that we have the inequality $\dim(B_q) \leq \dim(A) + \dim(B \otimes_A \kappa(p))$, and one concludes by Prop. 8 of § 1, No. 3.

Put $B = A[X_1, ..., X_n]$. For every prime ideal $p$ of $A$, the ring $B \otimes_A \kappa(p)$ is a polynomial ring in $n$ variables over a field, hence is of dimension $n$ ($§ 2$, No. 4, Cor. 1 to Theorem 3). By Cor. 2, we have $\dim(B) \leqslant \dim(A) + n$; the reverse inequality follows from Example 4 of $§ 1$, No. 3.

If $B$ is faithfully flat over $A$, then $^a\rho$ is surjective (II, $§ 2$, No. 5, Cor. 4 to Prop. 11) and we have $\text{ht}(a) \leqslant \text{ht}(\rho(a).B)$ ($§ 2$, No. 1, corollary of Prop. 2).

It remains therefore to prove inequality (20) under the assumption that $^a\rho$ is surjective. Let $p$ be a prime ideal of $A$ such that $a \subset p$ and $\text{ht}(a) = \text{ht}(p)$ ($§ 1$, no 3, prop. 7). Put $\overline{B} = B \otimes_A \kappa(p)$ and denote by $h$ the canonical homomorphism of $B$ into $\overline{B}$. If $X = {}^a\rho^{-1}(p)$ is the nonempty set of prime ideals of $B$ lying over $p$, it is known ($§ 2$, no 1, lemme 1) that the mapping $^ah$ is a homeomorphism of $\mathrm{Spec}(\overline{B})$ onto the subspace $X$ of $\mathrm{Spec}(B)$. Let $q$ be the image under $^ah$ of a minimal prime ideal of $\overline{B}$; one has $\dim(B_q \otimes_A \kappa(p)) = 0$ and Cor. 1 yields the inequality $\dim(B_q) \leqslant \dim(A_p) = \text{ht}(p) = \text{ht}(a)$, finally