Let us prove a). If Z is an irreducible closed subset of Y, its closure $\overline{Z}$ in X is irreducible (II, § 4, No. 1, Prop. 2), and one has $\overline{Z} \cap Y = Z$. Thus every chain c of irreducible closed subsets of Y defines, by passing to the closure in X, a chain of irreducible closed subsets of X, of the same length as c. The inequality $\dim(Y) \leq \dim(X)$ follows from this. If U is an open subset of X containing a point y of Y, then therefore $\dim(U \cap Y) \leq \dim(U)$, whence $\dim_y(Y) \leq \dim_y(X)$. Let us prove b). By definition one has $\dim_x(X) \leq \dim_x(V)$, and the opposite inequality follows from a). Let us prove c). Let $Z_0 \subset \ldots \subset Z_n$ be a chain of irreducible closed subsets of X. One has $Z_n = \bigcup_{i \in I} (Z_n \cap X_i)$ and each of the sets $Z_n \cap X_i$ is closed in $Z_n$;

since I is finite, $Z_n$ is contained in one of the $X_i$. Consequently, one has $\dim(X) \leq \sup_{i} \dim(X_i)$, whence equality by a).

Now let $x$ be a point of $X$ and $n = \sup_{i \in J_x} \dim_x(X_i)$, where $J_x$ is as in the statement. One has $\dim_x(X) \geq n$ by a), and, to establish equality, one may suppose $n$ finite. For every $i \in J_x$, let $U_i$ be an open neighbourhood of $x$ in $X$, such that $\dim(U_i \cap X_i) \leq n$. Put $U = (\bigcap_{i \in J_x} U_i) \cap (\bigcap_{i \in I - J_x} \mathscr{C} X_i)$; the set $U$ is open in $X$. Moreover, one has $\dim(U) = \sup_{i \in J_x} \dim(U \cap X_i) \leq n$ by the preceding paragraph, hence $\dim_x(X) \leq n$.

b) Let $x$ be a point of $X$. Then $\dim_x(X) \geq \sup \dim_x(X_i)$, where $X_i$ runs through the family of irreducible components of $X$ that contain $x$; there is equality if $x$ has a neighbourhood $V$ which has only a finite number of irreducible components (which is the case for example if $V$ is noetherian).

The first assertion is immediate since the chains of irreducible closed subsets of $X$ are the chains of irreducible closed subsets of the irreducible components of $X$ (II, § 4, No. 1, Prop. 5). The inequality $\dim_x(X) \geq \sup_{i} \dim_x(X_i)$ follows from Prop. 1, a). Let $V$ be a neighbourhood of $x$ which has only a finite number of irreducible components, and let $(V_j)_{j \in J}$ be the family of irreducible components of $V$ which contain $x$. It follows from Prop. 1, b) and c) that one has

one concludes by remarking that each of the $V_j$ is contained in one of the $X_i$, $i \in J_x$, and that one therefore has $\sup_{j \in J} \dim_x(V_j) \leq \sup_{i \in J_x} \dim_x(X_i)$.

In fact, by definition one has $\dim(X) \geq \dim_x(X)$ for every $x \in X$. On the other hand, if $Z_0 \subset ... \subset Z_n$ is a chain of irreducible closed subsets of $X$, for every $x \in Z_0$ and every open neighbourhood $U$ of $x$, the sets $Z_0 \cap U, ..., Z_n \cap U$ form a chain of irreducible closed subsets of $U$ (II, § 4, No. 1, Prop. 7). Hence $\dim_x(X) \geq n$, whence $\dim(X) \leq \sup_{x \in X} \dim_x(X)$.

It is enough to prove that, for every point $x$ of $X$, one has $\dim_x(X) = \sup_{\alpha \in A_x} \dim_x(X_\alpha)$, where $A_x$ is the set of $\alpha \in A$ such that $x \in X_\alpha$. This is clear in the case of an open covering, and follows from Prop. 1, c), in the case of a locally finite closed covering.

a) For every nonempty closed subset $Y$ of $X$, one has

b) If $Y, Z, T$ are closed subsets of $X$ such that $Y \subset Z \subset T$, one has

It is enough to prove assertion a) in the case where $\dim(X)$ is finite. In this case, $\dim(Y)$ and $\operatorname{codim}(Y, X)$ are finite. There exists a chain $Y_0 \subset ... \subset Y_n$ of irreducible closed subsets of $Y$, of length $n = \dim(Y)$, and a chain $Y_n \subset ... \subset Y_{n+p}$ of irreducible closed subsets of $X$, of length $p \geq \operatorname{codim}(Y, X)$. It follows that $\dim(X) \geq n + p$, whence a). To establish b), one may suppose $Y$ irreducible. Since one has $\operatorname{codim}(Y, Z) \leq \operatorname{codim}(Y, T)$, the inequality is proved if $\operatorname{codim}(Y, Z) = +\infty$. Otherwise, let $Z_0$ be an irreducible component of $Z$ containing $Y$ and such that $\operatorname{codim}(Y, Z) = \operatorname{codim}(Y, Z_0)$. One has $\operatorname{codim}(Z, T) \leq \operatorname{codim}(Z_0, T)$, and one sees, as above, that $\operatorname{codim}(Y, Z_0) + \operatorname{codim}(Z_0, T) \leq \operatorname{codim}(Y, T)$, whence b).

Suppose $X$ catenary. In view of Prop. 3, b), it is enough to prove the relation when $\operatorname{codim}(Y, Z)$ and $\operatorname{codim}(Z, T)$ are finite. By joining end to end a saturated chain of irreducible closed subsets with extremities $Y$ and $Z$, of length $\operatorname{codim}(Y, Z)$, and a saturated chain of irreducible closed subsets with extremities $Z$ and $T$, of length $\operatorname{codim}(Z, T)$, one obtains a saturated chain with extremities $Y$ and $T$, of length $\operatorname{codim}(Y, Z) + \operatorname{codim}(Z, T)$. But, since $X$ is catenary, this length is necessarily equal to $\operatorname{codim}(Y, T)$.

Conversely, suppose that one has $\operatorname{codim}(Y, T) = \operatorname{codim}(Y, Z) + \operatorname{codim}(Z, T)$ for any irreducible closed subsets $Y, Z, T$ of $X$ such that $Y \subset Z \subset T$, and let us prove that $X$ is catenary. To do this, let us prove by induction on the integer $n \geq 0$ that, for every saturated chain $Z_0 \subset ... \subset Z_n$ of irreducible closed subsets of $X$, one has $\operatorname{codim}(Z_0, Z_n) = n$. If $n = 0$, this is clear. Let $n > 0$, and suppose the property holds for chains of length $\leq n - 1$. If $Z_0 \subset ... \subset Z_n$ is a saturated chain of length $n$, then $Z_0 \subset ... \subset Z_{n-1}$ is a saturated chain of length $n - 1$, hence $\operatorname{codim}(Z_0, Z_{n-1}) = n - 1$. In view of the assumption made on $X$, one has $\operatorname{codim}(Z_0, Z_n) = \operatorname{codim}(Z_0, Z_{n-1}) + \operatorname{codim}(Z_{n-1}, Z_n) = (n - 1) + 1 = n$.

The condition is necessary by Prop. 4. Conversely, suppose it satisfied, and denote by c(Z) the integer codim(Z, X) for every irreducible closed subset Z of X. If Y, Z, T are three irreducible closed subsets of X such that Y ⊂ Z ⊂ T, one has

and X is catenary by Prop. 4.

Let Y and Z be two irreducible closed subsets of X such that Y ⊂ Z. Let Y_0 ⊂ ... ⊂ Y_p be a chain such that Y_p = Y and p = \dim(Y), Z_0 ⊂ ... ⊂ Z_q a chain such that Z_0 = Z and q = \operatorname{codim}(Z, X). For every saturated chain T_0 ⊂ ... ⊂ T_r such that T_0 = Y and T_r = Z, the chain

is maximal, and of length p + q + r; by the assumption made on X, one has therefore p + q + r = \dim(X), that is r = \dim(X) - \dim(Y) - \operatorname{codim}(Z, X). It follows that X is catenary and that, for Y and Z as above, one has Taking Y = Z, one sees that the second member is equal to \dim(Z), whence the proposition.

By the remark of II, § 4, No. 3, if a is an ideal of A, the topological space Spec(A/a) is homeomorphic to the closed subspace V(a) of Spec(A). Assertion a) results from this and from prop. 1, a) of No. 1. Assertion b) results from loc. cit., corollary to prop. 13. By loc. cit., cor. 2 to prop. 14, the irreducible components of Spec(A) are homeomorphic to the spaces Spec(A/p), where p is a minimal prime ideal of A, and assertion c) results from the corollary of prop. 1 of No. 1. Under the assumption of d), the space Spec(A) has only a finite number of irreducible components; the irreducible components of Spec(A) containing p are the sets V(q), where q is a minimal prime ideal contained in p. Assertion d) then results from cor., b) of prop. 1 of No. 1.

Let us prove e) finally. It is a question of proving that, for every chain $p_0 \subset ... \subset p_n$ of prime ideals of A such that $a \subset p_0$, one has dim(A) $\geq n + 1$. In view of the assumption made on a, there exists a prime ideal $p_{-1}$ of A contained in $p_0$, distinct from $p_0$, and $p_{-1} \subset p_0 \subset ... \subset p_n$ is a chain of prime ideals of A, of length $n + 1$.

b) Let p be a prime ideal of A and a an ideal of A. Then one has $\dim(A_p/aA_p) = -\infty$ if a is not contained in p, and $\dim(A_p/aA_p) = \operatorname{codim}(V(p), V(a))$ if a is contained in p. In particular, if p is a prime ideal of A, one has $\dim(A_p) = \operatorname{ht}(p)$.

c) If a is an ideal of A, one has $\operatorname{ht}(a) = \inf_{p} \operatorname{ht}(p) = \inf_{p} \dim(A_p)$ where p runs through the set of prime ideals of A containing a.

Assertion a) is the translation of Definition 3, a) of No. 2. Assertion b) follows from the fact that the mapping $q \mapsto q(A_p/aA_p)$ is an increasing isomorphism of the set of prime ideals q of A such that $a \subset q \subset p$ onto the set of prime ideals of the local ring $A_p/aA_p$ (II, § 2, No. 5, Proposition 11). Let a be an ideal of A; the irreducible closed subsets of $V(a)$ are the sets $V(p)$, where p is a prime ideal of A containing a. Assertion c) therefore follows from Remark 1 of No. 2.

This follows from Proposition 7, a), and from II, § 2, No. 5, Proposition 11.

a) One has $\dim(A) = \sup_m \dim(A_m) = \sup_m \operatorname{ht}(m)$, where m runs through the set of maximal (resp. prime) ideals of A.

b) Let b be an ideal of A distinct from A, and a an ideal of A contained in b. Then one has $\operatorname{codim}(V(b), V(a)) + \dim(A/b) \leq \dim(A/a)$. In particular, for every ideal b of A distinct from A, one has the inequality $\operatorname{ht}(b) + \dim(A/b) \leq \dim(A)$.

The first assertion follows from Remark 2 of No. 2 and from Proposition 7, b). The second follows from Proposition 3, a) of No. 2 and from the relations $\dim(A/b) = \dim(V(b))$, $\dim(A/a) = \dim(V(a))$.

a) For every $p \in \mathrm{Supp}(M)$, one has $\dim_{A_p}(M_p) = \mathrm{codim}(V(p), \mathrm{Supp}(M))$.

b) $\dim_A(M)$ is the least upper bound of the $\dim_{A_p}(M_p)$, where $p$ ranges over $\mathrm{Spec}(A)$ (resp. where $p$ ranges over the set of maximal ideals of $A$ belonging to $\mathrm{Supp}(M)$).

c) Let $M'$ be a finitely generated submodule of $M$; then

a) Let $a$ be the annihilator of $M$; then the annihilator of the $A_p$-module $M_p$ is $aA_p$ (II, § 2, n° 4, formule (9)), whence $\dim_{A_p}(M_p) = \dim(A_p/aA_p)$. One concludes by prop. 7, b) of n° 3.

b) This follows at once from a) and from the fact that $\dim_{A_p}(M_p) = -\infty$ if $p$ does not belong to $\mathrm{Supp}(M)$.

c) One has $\mathrm{Supp}(M) = \mathrm{Supp}(M') \cup \mathrm{Supp}(M/M')$ (II, § 4, n° 4, prop. 16), and one applies prop. 1 of n° 1.

In order that $M$ be of type $C_{\leq d}$, that is to say of dimension $\leq d$, it is necessary and sufficient that for every minimal prime ideal $p$ of $\mathrm{Supp}(M)$, one have $\dim(A/p) \leq d$, which means that $z(M) \in Z_{\leq d}$. Suppose that one has $\dim(M) \leq d$, and let $p \in \mathrm{Spec}(A)$ be such that $\dim(A/p) = d$; then either $p \notin \mathrm{Supp}(M)$ and hence $M_p = 0$, or else $p \in \mathrm{Supp}(M)$, and $p$ is a minimal element of $\mathrm{Supp}(M)$; in both cases the coefficient of $[V(p)]$ in $z(M)$ is $\mathrm{long}_{A_p}(M_p)$, whence a). Part b) is proved analogously; it should be noted that a finitely generated module $M$ is of type $C^{>d}$ if and only if one has $M_p = 0$ for every prime ideal $p$ of height $< d$.

According to Prop. 9, c) and Remark 3 of No. 4, the sets $C_{\leq d}$ and $C^{>d}$ are hereditary (A, VIII, § 10, No. 1, Def. 1), and one may consider the corresponding Grothendieck groups $K(C_{\leq d})$ and $K(C^{>d})$ (loc. cit., No. 2); for every $A$-module $M$ of type $C_{\leq d}$ (resp. $C^{>d}$), let $[M]_{\leq d}$ (resp. $[M]^{>d}$) denote the associated element in $K(C_{\leq d})$ (resp. $K(C^{>d})$). By Lemma 1, the mappings $z_d$ and $z^d$ are additive; there therefore exist (loc. cit., Prop. 3) homomorphisms

such that $\zeta_d([M]_{\leq d}) = z_d(M)$ for every $A$-module $M$ of type $C_{\leq d}$ and $\zeta^d([N]^{>d}) = z^d(N)$ for every $A$-module $N$ of type $C^{>d}$. Moreover, since $C_{\leq d-1} \subset C_{\leq d}$ and $C^{>d+1} \subset C^{>d}$, one has canonical homomorphisms With these notations:

One has $\zeta_d \circ i_d = 0$ by Lemma 1. For every $p \in \mathrm{Spec}(A)$ such that $\dim(A/p) = d$, one has $\zeta_d([A/p]_{\leq d}) = z_d(A/p) = [V(p)]$, hence the homomorphism $\zeta_d$ is surjective. By IV, § 1, No. 4, Theorem 1, $K(\mathscr{C}_{\leq d})$ is generated by the $[A/p]_{\leq d}$, where $p \in \mathrm{Spec}(A)$ and $\dim(A/p) \leq d$; consequently, every element $\xi$ of $K(\mathscr{C}_{\leq d})$ can be written $\xi = i_d(\eta) + \sum_{i=1}^k n_i[A/p_i]_{\leq d}$, with $\eta \in K(\mathscr{C}_{\leq d-1})$, $n_i \in \mathbf{Z}$ and $\dim(A/p_i) = d$ for $1 \leq i \leq k$; one has $\zeta_d(\xi) = \sum_{i=1}^k n_i[V(p_i)]$ and consequently $\zeta_d(\xi) = 0$ implies $\xi = i_d(\eta) \in \mathrm{Im}(i_d)$, whence $\mathrm{Ker}(\zeta_d) = \mathrm{Im}(i_d)$.

One reasons analogously for the second sequence.