Éléments de Géométrie Algébrique — English

§14. Combinatorial dimension of a topological space

14.1. Combinatorial dimension of a topological space

(14.1.1) Let $I$ be an ordered set; a chain of elements of $I$ is by definition a finite strictly increasing sequence $i_0<i_1<\cdots <i_n$ of elements of $I$ ($n\ge 0$); by definition the length of this chain is $n$. If $X$ is a topological space, the set of its irreducible closed subsets is ordered by inclusion, whence the notion of a chain of irreducible closed subsets of $X$.

Definition (14.1.2).

Let $X$ be a topological space. We call combinatorial dimension of $X$ (or simply dimension of $X$, if no confusion can arise) and denote ${\dim }_c(X)$, or simply $\dim (X)$, the supremum of the lengths of chains of irreducible closed subsets of $X$. For every $x\in X$, we call combinatorial dimension of $X$ at $x$ (or simply dimension of $X$ at $x$) and denote ${\dim }_x(X)$ the number ${\inf }_U(\dim (U))$, where $U$ ranges over the set of open neighbourhoods of $x$ in $X$.

It follows from this definition that one has

$$\dim (\varnothing )=-\infty$$

(the supremum in $\bar{R}$ of the empty subset being $-\infty$). If $(X_{\alpha })$ is the family of irreducible components of $X$, one has

  dim(X) = sup_α dim(X_α)                                                                  (14.1.2.1)

since every chain of irreducible closed subsets of $X$ is by definition contained in an irreducible component of $X$, and conversely these components are closed in $X$, so every irreducible closed subset of an $X_{\alpha }$ is an irreducible closed subset of $X$.

Definition (14.1.3).

We say that a topological space $X$ is equidimensional if all its irreducible components have the same dimension (then equal to $\dim (X)$ by (14.1.2.1)).

Proposition (14.1.4).

(i) For every subset $Y$ of a topological space $X$, one has $\dim (Y)\le \dim (X)$.

(ii) If a topological space $X$ is a finite union of closed subsets $X_i$, one has dim(X) = sup_i dim(X_i).

For every irreducible closed subset $Z$ of $Y$, the closure $\bar{Z}$ of $Z$ in $X$ is irreducible $(0_I,\mathrm{2.1.2})$ and $\bar{Z}\cap Y=Z$, whence (i). On the other hand, if $X={\bigcup }_{i=1}^n{X}_i$, where the $X_i$ are closed, every irreducible closed subset of $X$ is contained in one of the $X_i$ $(0_I,\mathrm{2.1.1})$, and consequently every chain of irreducible closed subsets of $X$ is contained in one of the $X_i$; whence (ii).

From (14.1.4, (i)) one deduces that for every $x\in X$, one can also write

  dim_x(X) = lim_U dim(U)                                                                  (14.1.4.1)

the limit being taken along the decreasing filtered set of open neighbourhoods of $x$ in $X$.

Corollary (14.1.5).

Let $X$ be a topological space, $x$ a point of $X$, $U$ a neighbourhood of $x$, $Y_i$ $(1\le i\le n)$ closed subsets of $U$ such that $x\in Y_i$ for every $i$ and such that $U$ is the union of the $Y_i$. Then one has

  dim_x(X) = sup_i (dim_x(Y_i)).                                                           (14.1.5.1)

It follows from (14.1.4, (ii)) that one has dim_x(X) = inf_V (sup_i (dim(Y_i ∩ V))) as $V$ ranges over the set of open neighbourhoods of $x$ contained in $U$; likewise one has dim_x(Y_i) = inf_V (dim(Y_i ∩ V)) for every $i$. The corollary is thus obvious if

$$\underset{i}{\sup }({\dim }_x(Y_i))=+\infty ;$$

in the contrary case, there is an open neighbourhood $V_0\subset U$ of $x$ such that $\dim (Y_i\cap V)={\dim }_x(Y_i)$ for $1\le i\le n$ and for every $V\subset V_0$, whence the conclusion.

Proposition (14.1.6).

For every topological space $X$, one has dim(X) = sup_{x ∈ X} dim_x(X).

It follows from definition (14.1.2) and proposition (14.1.4) that ${\dim }_x(X)\le \dim (X)$ for every $x\in X$. On the other hand, let $Z_0\subset Z_1\subset \cdots \subset Z_n$ be a chain of irreducible closed subsets of $X$, and let $x\in Z_0$; for every open set $U\subset X$ containing $x$, $U\cap Z_i$ is irreducible $(0_I,\mathrm{2.1.6})$ and closed in $U$, and since $U\cap Z_i\overline{=}{Z}_i$ in $X$, the $U\cap Z_i$ are pairwise distinct; hence $\dim (U)\ge n$, which completes the proof.

Corollary (14.1.7).

If $(X_{\alpha })$ is an open cover of $X$, or a locally finite closed cover of $X$, one has dim(X) = sup_α (dim(X_α)).

If $X_{\alpha }$ is a neighbourhood of $x\in X$, one has ${\dim }_x(X)\le \dim (X_{\alpha })$, whence the first assertion. On the other hand, if the $X_{\alpha }$ are closed and if $U$ is a neighbourhood of $x\in X$ meeting only a finite number of the sets $X_{\alpha }$, one has dim_x(X) ≤ dim(U) = sup_α (dim(U ∩ X_α)) ≤ sup_α (dim(X_α)) by (14.1.4), whence the second assertion.

Corollary (14.1.8).

Let $X$ be a Noetherian Kolmogorov space $(0_I,\mathrm{2.1.3})$, and let $F$ be the set of closed points of $X$. Then dim(X) = sup_{x ∈ F} dim_x(X).

With the notation of the proof of (14.1.6), it suffices to remark that there exists in Z_0 a closed point $(0_I,\mathrm{2.1.3})$.

Proposition (14.1.9).

Let $X$ be a non-empty Noetherian Kolmogorov space. For $\dim (X)=0$, it is necessary and sufficient that $X$ be finite and discrete.

If a space $X$ is Hausdorff (and a fortiori if $X$ is a discrete space), the only irreducible closed subsets of $X$ are reduced to a point, so $\dim (X)=0$. Conversely, suppose that $X$ is a Noetherian Kolmogorov space such that $\dim (X)=0$; since every irreducible component of $X$ contains a closed point $(0_I,\mathrm{2.1.3})$, it must be reduced to that point. Since $X$ has only a finite number of irreducible components, it is finite and discrete.

Corollary (14.1.10).

Let $X$ be a Noetherian Kolmogorov space. For a point $x\in X$ to be isolated, it is necessary and sufficient that ${\dim }_x(X)=0$.

The condition is obviously necessary (without hypothesis on $X$). It is sufficient,

since it follows that $\dim (U)=0$ for an open neighbourhood $U$ of $x$, and since $U$ is a Noetherian Kolmogorov space, $U$ is finite and discrete.

Proposition (14.1.11).

The function $x\mapsto {\dim }_x(X)$ is upper semi-continuous on $X$.

It is clear that this function is upper semi-continuous at every point where it equals $+\infty$. So suppose ${\dim }_x(X)=n<+\infty$; then formula (14.1.4.1) shows that there exists an open neighbourhood U_0 of $x$ such that $\dim (U)=n$ for every open neighbourhood $U\subset U_0$ of $x$. Granting that, for every $y\in U_0$ and every open neighbourhood $V\subset U_0$ of $y$, one has $\dim (V)\le \dim (U_0)=n$ (14.1.4); one then deduces from (14.1.4.1) that ${\dim }_y(X)\le n$.

Remark (14.1.12).

If $X$, $Y$ are two topological spaces, and $f:X\to Y$ a continuous map, one can have $\dim (f(X))>\dim (X)$; an example is obtained by taking for $X$ a discrete space with two elements $a$, $b$, for $Y$ the set {a, b} equipped with the topology for which the closed sets are $\varnothing$, $a$ and {a, b}; if $f:X\to Y$ is the identity map, one has $\dim (Y)=1$ and $\dim (X)=0$. One will note that $Y$ is the spectrum of a discrete valuation ring $A$, of which $a$ is the unique closed point and $b$ the generic point; if $K$ and $k$ are the field of fractions and the residue field of $A$, $X$ is the spectrum of the ring $k\times K$ and $f$ the continuous map corresponding to the homomorphism $(\varphi ,\psi ):A\to k\times K$, where $\varphi :A\to k$ and $\psi :A\to K$ are the canonical homomorphisms (cf. (IV, 5.4.3)).

14.2. Codimension of a closed subset

Definition (14.2.1).

Given an irreducible closed subset $Y$ of a topological space $X$, we call combinatorial codimension (or simply codimension) of $Y$ in $X$, and denote $codim(Y,X)$, the supremum of the lengths of chains of irreducible closed subsets of $X$ of which $Y$ is the smallest element. If $Y$ is an arbitrary closed subset of $X$, we call codimension of $Y$ in $X$, and again denote $codim(Y,X)$, the infimum of the codimensions in $X$ of the irreducible components of $Y$. We say that $X$ is equicodimensional if all the minimal irreducible closed subsets of $X$ have the same codimension in $X$.

It follows from this definition that $codim(\varnothing ,X)=+\infty$, the infimum in $\bar{R}$ of the empty subset being $+\infty$. If $Y$ is closed in $X$ and if $(X_{\alpha })$ (resp. $(Y_{\beta })$) is the family of irreducible components of $X$ (resp. $Y$), every $Y_{\beta }$ is contained in some $X_{\alpha }$, and more generally every chain of irreducible closed subsets of $X$ of which $Y_{\beta }$ is the smallest element consists of subsets of some $X_{\alpha }$; one therefore has

  codim(Y, X) = inf_β (sup_α (codim(Y_β, X_α)))                                            (14.2.1.1)

where for each $\beta$, $\alpha$ ranges over the set of indices such that $Y_{\beta }\subset X_{\alpha }$.

Proposition (14.2.2).

Let $X$ be a topological space.

(i) If $\Phi$ is the set of irreducible closed subsets of $X$, one has

  dim(X) = sup_{Y ∈ Φ} (codim(Y, X)).                                                      (14.2.2.1)

(ii) For every non-empty closed subset $Y$ of $X$, one has

  dim(Y) + codim(Y, X) ≤ dim(X).                                                           (14.2.2.2)

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

  codim(Y, Z) + codim(Z, T) ≤ codim(Y, T).                                                 (14.2.2.3)

(iv) For a closed subset $Y$ of $X$ to be such that $codim(Y,X)=0$, it is necessary and sufficient that $Y$ contain an irreducible component of $X$.

Assertions (i) and (iv) are immediate consequences of the definition (14.2.1). To prove (ii), one can restrict to the case where $Y$ is irreducible, and then the formula follows from the definitions (14.1.1) and (14.2.1). Finally, to prove (iii), one can, by virtue of definition (14.2.1), first restrict to the case where $Y$ is irreducible; then codim(Y, Z) = sup_α (codim(Y, Z_α)) for the irreducible components $Z_{\alpha }$ of $Z$ containing $Y$; it is clear that $codim(Y,T)\ge codim(Y,Z)$, so the inequality is true if $codim(Y,Z)=+\infty$; otherwise, there exists an $\alpha$ such that $codim(Y,Z)=codim(Y,Z_{\alpha })$ and by virtue of (14.2.1), one can restrict to the case where $Z$ is also irreducible; but then the inequality (14.2.2.3) is an obvious consequence of definition (14.2.1).

Proposition (14.2.3).

Let $X$ be a topological space, $Y$ a closed subset of $X$. For every open set $U$ in $X$, one has

  codim(Y ∩ U, U) ≥ codim(Y, X).                                                           (14.2.3.1)

Moreover, for the two members of (14.2.3.1) to be equal, it is necessary and sufficient that, if $(Y_{\alpha })$ is the family of irreducible components of $Y$ meeting $U$, one have codim(Y, X) = inf_α (codim(Y_α, X)).

We know $(0_I,\mathrm{2.1.6})$ that $Z\mapsto \bar{Z}$ is a bijection from the set of irreducible closed subsets of $U$ onto the set of irreducible closed subsets of $X$ meeting $U$, and in particular makes the irreducible components of $Y\cap U$ correspond to the irreducible components of $Y$ meeting $U$; if $Y_{\alpha }$ is one of the latter, one therefore has $codim(Y_{\alpha },X)=codim(Y_{\alpha }\cap U,U)$, and the proposition then follows from definition (14.2.1).

Definition (14.2.4).

Let $X$ be a topological space, $Y$ a closed subset of $X$, $x$ a point of $X$. We call codimension of $Y$ in $X$ at the point $x$ and denote $codi{m}_x(Y,X)$ the number ${\sup }_U(codim(Y\cap U,U))$, where $U$ ranges over the set of open neighbourhoods of $x$ in $X$.

By virtue of (14.2.3), one can also write

  codim_x(Y, X) = lim_U (codim(Y ∩ U, U))                                                  (14.2.4.1)

the limit being taken along the decreasing filtered set of open neighbourhoods of $x$ in $X$. One will note that one has

  codim_x(Y, X) = +∞    if    x ∈ X − Y.

Proposition (14.2.5).

If $(Y_i{)}_{1\le i\le n}$ is a finite family of closed subsets of a topological space $X$, and $Y$ the union of this family, one has

  codim(Y, X) = inf_i (codim(Y_i, X)).                                                     (14.2.5.1)

Indeed, every irreducible component of one of the $Y_i$ is contained in an irreducible component of $Y$, and conversely every irreducible component of $Y$ is also an irreducible component of one of the $Y_i$ $(0_I,\mathrm{2.1.1})$; the conclusion thus follows from definition (14.2.1) and inequality (14.2.2.3).

Corollary (14.2.6).

Let $X$ be a topological space, $Y$ a locally Noetherian closed subspace of $X$.

(i) For every $x\in X$, there exist only a finite number of irreducible components $Y_i$ $(1\le i\le n)$ of $Y$ containing $x$, and one has codim_x(Y, X) = inf_i (codim(Y_i, X)).

(ii) The function $x\mapsto codi{m}_x(Y,X)$ is lower semi-continuous on $X$.

Indeed, by hypothesis there is an open neighbourhood U_0 of $x$ in $X$ such that $Y\cap U_0$ is Noetherian, hence has only a finite number of irreducible components, which are traces on U_0 of irreducible components of $Y$; a fortiori, there are only a finite number of irreducible components $Y_i$ $(1\le i\le n)$ of $Y$ containing $x$, and one can, by replacing U_0 by an open neighbourhood $U\subset U_0$ of $x$ meeting none of the $Y_i$ that do not contain $x$, assume that the $Y_i\cap U$ are the irreducible components of $Y\cap U$; for every open neighbourhood $V\subset U$ of $x$ in $X$, the $Y_i\cap V$ are then the irreducible components of $Y\cap V$, and (14.2.3) then shows that $codim(Y_i,X)=codim(Y_i\cap V,V)$, which proves (i). Moreover, for every $x^{\prime }\in U$, the irreducible components of $Y$ containing $x^{\prime }$ are some of the $Y_i$, so $codi{m}_{x^{\prime }}(Y,X)\ge codi{m}_x(Y,X)$, which proves (ii).

14.3. The chain condition

(14.3.1) In a topological space $X$, we shall say that a chain $Z_0\subset Z_1\subset \cdots \subset Z_n$ of irreducible closed subsets is saturated if there is no irreducible closed subset $Z^{\prime }$ distinct from the $Z_i$ and such that $Z_k\subset Z^{\prime }\subset Z_{k+1}$ for some index $k$.

Proposition (14.3.2).

Let $X$ be a topological space such that, for any two irreducible closed subsets $Y$, $Z$ of $X$ with $Y\subset Z$, one has $codim(Y,Z)<+\infty$. The two following conditions are equivalent:

a) Two saturated chains of irreducible closed subsets of $X$, having the same extremities, have the same length.

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

  codim(Y, T) = codim(Y, Z) + codim(Z, T).                                                 (14.3.2.1)

It is immediate that a) entails b). Conversely, suppose b) verified, and let us show that if two saturated chains with the same extremities have lengths $m$ and $n\le m$, one necessarily has $m=n$. We argue by induction on $n$, the proposition being obvious for $n=1$. So suppose $n>1$, $n<m$, and let $Z_0\subset Z_1\subset \cdots \subset Z_n$ be a

saturated chain such that there exists another saturated chain with extremities Z_0, $Z_n$ and of length $m$. Since $codim(Z_0,Z_n)\ge m>n$ and $codim(Z_0,Z_1)=1$, it follows from b) that $codim(Z_1,Z_n)=codim(Z_0,Z_n)-1>n-1$, which contradicts the induction hypothesis.

When the conditions of (14.3.2) are satisfied, one says that $X$ satisfies the chain condition, or also is a catenary space. It is clear that every closed subspace of a catenary space is catenary.

Proposition (14.3.3).

Let $X$ be a Noetherian Kolmogorov space of finite dimension. The following conditions are equivalent:

a) Two maximal chains of irreducible closed subsets of $X$ have the same length.

b) $X$ is equidimensional, equicodimensional, and catenary.

c) $X$ is equidimensional, and for any two irreducible closed subsets $Y$, $Z$ of $X$ such that $Y\subset Z$, one has

  dim(Z) = dim(Y) + codim(Y, Z).                                                           (14.3.3.1)

d) $X$ is equicodimensional and for any two irreducible closed subsets $Y$, $Z$ of $X$ such that $Y\subset Z$, one has

  codim(Y, X) = codim(Y, Z) + codim(Z, X).                                                 (14.3.3.2)

The hypotheses on $X$ entail that the extremities of a maximal chain of irreducible closed subsets of $X$ are necessarily a closed point and an irreducible component of $X$ $(0_I,\mathrm{2.1.3})$; moreover, every saturated chain with extremities $Y$, $Z$ (with $Y\subset Z$) is contained in a maximal chain whose elements other than those of the given chain are either contained in $Y$ or contain $Z$. These remarks at once establish the equivalence of a) and b), and also prove that if a) is verified, one has, for every irreducible closed subset $Y$ of $X$,

  dim(Y) + codim(Y, X) = dim(X);                                                           (14.3.3.3)

since (14.3.2.1) holds, one deduces (14.3.3.1) and (14.3.3.2) at once from (14.3.3.3). Conversely, (14.3.3.1) entails (14.3.2.1), hence (14.3.3.1) entails the chain condition by virtue of (14.3.2); moreover, applying (14.3.3.1) to the case where $Y$ is reduced to a closed point $x$ of $X$ and $Z$ to an irreducible component of $X$, one obtains $codim(x,X)=\dim (Z)$; one concludes that c) entails b). Similarly, (14.3.3.2) entails (14.3.2.1), hence the chain condition; moreover, with the same choice of $Y$ and $Z$ as above, (14.3.3.2) again entails $codim(x,X)=\dim (Z)$, so (since every irreducible component of $X$ contains a closed point by virtue of $(0_I,\mathrm{2.1.3})$), d) entails b).

One says that a Noetherian Kolmogorov space is biequidimensional if it is of finite dimension and if it verifies the equivalent conditions of (14.3.3).

Corollary (14.3.4).

Let $X$ be a biequidimensional Noetherian Kolmogorov space; then, for every closed point $x$ of $X$ and every irreducible component $Z$ of $X$, one has

  dim(X) = dim(Z) = codim({x}, X) = dim_x(X).                                              (14.3.4.1)

The last equality follows from the fact that if $Y_0=x\subset Y_1\subset \cdots \subset Y_m$ is a maximal chain of irreducible closed subsets of $X$ and $U$ an open neighbourhood of $x$, the $U\cap Y_i$ are pairwise distinct irreducible closed subsets of $U$ (since $U\cap Y_i\overline{=}{Y}_i$), whence $\dim (U)=\dim (X)$ by virtue of (14.1.4).

Corollary (14.3.5).

Let $X$ be a Noetherian Kolmogorov space; if $X$ is biequidimensional, so is every union of irreducible components of $X$ and every irreducible closed subset of $X$. Moreover, for every closed subset $Y$ of $X$, one then has

  dim(Y) + codim(Y, X) = dim(X).                                                           (14.3.5.1)

Every chain of irreducible closed subsets of $X$ being contained in an irreducible component of $X$, the first assertion follows at once from (14.3.3). Moreover, if $X^{\prime }$ is an irreducible closed subset of $X$, $X^{\prime }$ trivially verifies conditions (14.3.3, c)), whence the second assertion.

Finally, to prove (14.3.5.1), we remark that we have seen in the proof of (14.3.3) that this relation is verified when $Y$ is irreducible; if $Y_i$ $(1\le i\le m)$ are the irreducible components of $Y$, the one among the $Y_i$ for which $\dim (Y_i)$ is greatest is also the one for which $codim(Y_i,X)$ is smallest; hence (14.3.5.1) follows from the definitions of $\dim (Y)$ and $codim(Y,X)$.

Remark (14.3.6).

The reader will note that the proof of (14.3.2) applies to an arbitrary ordered set, the fact that one is dealing with irreducible closed subsets of a topological space not intervening in that proof. The same holds for the proof of (14.3.3) in an ordered set $E$ such that for every $x\in E$, there exists $z\le x$ which is minimal in $E$, and in which the length of chains of elements of $E$ is bounded.