Éléments de Géométrie Algébrique — English

§5. Dimension and depth in preschemes

This section confines itself to restating, in the geometric language and with various complements of a technical nature, the notions and results of commutative algebra exposed in Chapter 0, §§16 and 17.

5.1. Dimension of preschemes

(5.1.1)

Let $A$ be a ring and $\mathfrak{J}$ an ideal of $A$. Recall (0, 16.1) that the definitions of the theory of dimension of a ring (0, 16.1.1) and those of the theory of combinatorial dimension of topological spaces (0, 14.1.2) give the following relations:

  dim(Spec(A)) = dim(A)                                  (5.1.1.1)

  dim(V(𝔍)) = dim(A/𝔍)                                   (5.1.1.2)

  codim(V(𝔍), Spec(A)) = ht(𝔍)                           (5.1.1.3)

(5.1.1.4) $\mathrm{Spec}(A)$ is a catenary space ⟺ $A$ is a catenary ring.

(5.1.1.5) $\mathrm{Spec}(A)$ is equidimensional ⟺ $A$ is equidimensional ⟺ the minimal prime ideals ${\mathfrak{p}}_{\alpha }$ of $A$ are such that the dimensions of the rings $A/{\mathfrak{p}}_{\alpha }$ are all equal.

(5.1.1.6) $\mathrm{Spec}(A)$ is equicodimensional ⟺ $A$ is equicodimensional ⟺ all the maximal ideals of $A$ have the same height.

Recall that a Noetherian ring $A$ is said to be biequidimensional if $\mathrm{Spec}(A)$ is biequidimensional, that is to say if $\mathrm{Spec}(A)$ is Noetherian and $A$ is equidimensional, equicodimensional, catenary, and of finite dimension.

Proposition (5.1.2).

Let $X$ be a prescheme, $Y$ an irreducible closed part of $X$, $y$ the generic point of $Y$. One has

$$codim(Y,X)=\dim ({\mathcal{O}}_{X,y})(\mathrm{5.1.2.1})$$

Indeed (I, 2.4.2), the irreducible closed parts of $X$ containing $y$ are canonically in bijective correspondence with the irreducible closed parts of $\mathrm{Spec}({\mathcal{O}}_{X,y})$, hence in bijective correspondence with the prime ideals of ${\mathcal{O}}_{X,y}$, and (5.1.2.1) follows from the definitions.

In particular, one recovers in this way the fact that the generic points of the irreducible components of $X$ are the only points $x\in X$ such that $\dim ({\mathcal{O}}_x)=0$ (I, 1.1.14).

Corollary (5.1.3).

For every closed part $Y$ of a prescheme $X$, one has

  codim(Y, X) = inf_{y ∈ Y} dim(𝒪_{X,y}).                (5.1.3.1)

Moreover, if $X$ is locally Noetherian, one has, for every $x\in Y$,

  codim_x(Y, X) = inf_{y ∈ Y, x ∈ ‾{y}} dim(𝒪_{X,y}).    (5.1.3.2)

The relation (5.1.3.2) indeed follows from (5.1.3.1) and from (0, 14.2.6).

This corollary allows us to define, for any part $Y$ of a prescheme $X$, the codimension $codim(Y,X)$ of $Y$ in $X$ as equal to the second member of (5.1.3.1).

Proposition (5.1.4).

For every prescheme $X$, one has

  dim(X) = sup_{x ∈ X} (dim(𝒪_x)).                       (5.1.4.1)

If every irreducible closed part of $X$ contains a closed point, one has also

  dim(X) = sup_{x ∈ F} (dim(𝒪_x))                        (5.1.4.2)

where $F$ is the set of closed points of $X$.

It suffices to remark (by virtue of (I, 2.4.2)) that the chains of irreducible closed parts of $X$ correspond bijectively with the chains of irreducible closed parts of a local scheme of $X$; when every irreducible closed part of $X$ contains a closed point, one can clearly restrict to the local schemes at the closed points of $X$.

We note that every irreducible closed part of $X$ contains a closed point when $X$ is quasi-compact $(0_I,\mathrm{2.1.3})$; we shall see a little further on (5.1.11) that the same holds when $X$ is locally Noetherian.

Corollary (5.1.5).

For a prescheme $X$ to be catenary it is necessary and sufficient that, for every $x\in X$, the local ring ${\mathcal{O}}_x$ be catenary. If moreover every irreducible closed part of $X$ contains a closed point, it suffices, for $X$ to be catenary, that ${\mathcal{O}}_x$ be catenary for every closed point $x$ of $X$.

The reasoning is the same as in (5.1.4), since one is comparing chains of irreducible closed parts having the same extremities.

(5.1.6)

We shall now examine the more special properties tied to Noetherian hypotheses.

Recall (0, 16.2.3) that a Noetherian local ring $A\ne 0$ is of finite dimension, equal to the minimum number of generators of an ideal of definition of $A$; for every prime ideal $\mathfrak{p}$ of $A$, the height of $\mathfrak{p}$, equal to $\dim (A_{\mathfrak{p}})$, is therefore also finite. These properties, together with (5.1.2), (5.1.3), and (5.1.4), show that:

Proposition (5.1.7).

For every non-empty closed part $Y$ of a locally Noetherian prescheme $X$, $codim(Y,X)$ is finite. If $X$ is Noetherian and affine and $Y$ an irreducible closed part of $X$, $codim(Y,X)$ is equal to the minimum number of sections $s_i$ of ${\mathcal{O}}_X$ over $X$ such that $Y$ is an irreducible component of the set of $x\in X$ such that $s_i(x)=0$ for every $i$.

Corollary (5.1.8).

Let $X$ be a locally Noetherian prescheme, $\mathcal{L}$ an invertible ${\mathcal{O}}_X$-Module, $f$ a section of $\mathcal{L}$ over $X$. Then every irreducible component of the set $Z$

of $x\in X$ such that $f(x)=0$ is of codimension $\le 1$ in $X$; it is of codimension 1 if $Z$ contains no irreducible component of $X$.

One can restrict to the case where $\mathcal{L}={\mathcal{O}}_X$. If $y$ is a generic point of an irreducible component $Y$ of $Z$, the ideal $(f_y)$ of ${\mathcal{O}}_{X,y}$ must be such that ${\mathcal{O}}_{X,y}/(f_y)$ has only one prime ideal, which means that $f_y$ generates an ideal of definition of the Noetherian local ring ${\mathcal{O}}_{X,y}$; one thus has $codim(Y,X)\le 1$ (5.1.7); if $Z$ contains no irreducible component of $X$, one cannot have $codim(Y,X)=0$ by virtue of (0, 14.2.1).

Proposition (5.1.9).

Let $X$ be a locally Noetherian prescheme, $Y$ a closed part of $X$. If $x\in Y$ is such that ${\mathcal{O}}_{X,x}$ is a catenary ring, one has

  codim_x(Y, X) = dim(𝒪_{X,x}) − codim(‾{x}, Y)          (5.1.9.1)
                = dim(𝒪_{X,x}) − dim(𝒪_{Y,x}).

Let $Y_i$ ($1\le i\le n$) be the irreducible components of $Y$ containing $x$ (which are finite in number since $Y$ is locally Noetherian), and let $y_i$ be the generic point of $Y_i$. If one sets $A={\mathcal{O}}_{X,x}$, and if ${\mathfrak{p}}_i$ is the prime ideal of $A$ corresponding to $Y_i\cap \mathrm{Spec}(A)$, the irreducible closed parts of $Y$ containing $x$ correspond bijectively with the prime ideals of $A$ containing one of the ${\mathfrak{p}}_i$, so dim(𝒪_{Y,x}) = sup_i dim(A/𝔭_i); on the other hand, one has ${\mathcal{O}}_{X,y_i}=A_{{\mathfrak{p}}_i}$, and the hypothesis on $A$ entails dim(A) = dim(A/𝔭_i) + dim(A_{𝔭_i}) (0, 16.1.4); the conclusion thus follows from the relation codim_x(Y, X) = inf_i (dim(𝒪_{X,y_i})) ((5.1.2) and (0, 14.2.6)).

Proposition (5.1.10).

(i) In every prescheme $X$, every non-empty locally constructible part $E$ contains a point $x$ such that $x$ is a locally closed part of $X$ (or, what amounts to the same, such that $x$ is isolated in $\overline{x}$).

(ii) Let $X$ be a locally Noetherian prescheme, $x$ a point of $X$ such that $x$ is locally closed in $X$; then one has $\dim (\overline{x})\le 1$; every point $y\ne x$ of $\overline{x}$ is consequently closed in $X$.

(i) The result is a particular case of the following lemma:

Lemma (5.1.10.1).

Let $X$ be a topological space having the following property: for every locally closed part $Z\ne \varnothing$ of $X$, there exists a part $Z^{\prime }$ of $Z$, locally closed in $X$ (or in $Z$, what amounts to the same), and a point $x\in Z^{\prime }$, closed in $Z^{\prime }$. Then every locally closed part $Z\ne \varnothing$ of $X$ contains a point $x$ isolated in $\overline{x}$.

Indeed, let $Z^{\prime }\subset Z$ be a locally closed part of $Z$ containing a point $x$ such that $x$ is closed in $Z^{\prime }$. There is an open neighbourhood $U$ of $x$ in $X$ such that $Z^{\prime }\cap U$ is closed in $U$, hence $x$ is also closed in $U$; this means that $U\cap \overline{x}=x$, in other words that $x$ is isolated in $\overline{x}$.

The lemma applies to every underlying space of a prescheme $X$, for $Z$ is then also the underlying space of a prescheme (I, 5.2.1) and it suffices to take for $Z^{\prime }$ an affine open in $Z$, which is a quasi-compact Kolmogorov space, hence contains a closed point $(0_I,\mathrm{2.1.3})$.

(ii) Let $Z$ be the reduced sub-prescheme of $X$ having $\overline{x}$ as underlying space; the hypothesis entails that $x$ is open in $Z$; therefore, for every $z\in Z$, the generic point $x$ of $\mathrm{Spec}({\mathcal{O}}_{Z,z})$ is isolated in $\mathrm{Spec}({\mathcal{O}}_{Z,z})$; but the ring $A={\mathcal{O}}_{Z,z}$ is an integral Noetherian local ring, and the hypothesis entails that there exists $f\in A$ such that $A_f$ is a field; one knows (0, 16.3.3) that this entails $\dim (A)\le 1$. Since $\dim ({\mathcal{O}}_{Z,z})\le 1$ for every $z\in Z$, one indeed has $\dim (Z)\le 1$.

Corollary (5.1.11).

If $X$ is a locally Noetherian prescheme, every non-empty closed part of $X$ contains a closed point.

Indeed, every closed part of $X$ is constructible $(0_{III},\mathrm{9.1.1}and\mathrm{9.1.5})$ and it suffices to apply (5.1.10).

(5.1.12)

Let $X$ be a prescheme, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module of finite type, $S=Supp(\mathcal{F})$ its support, which is closed in $X$ $(0_I,\mathrm{5.2.2})$. If, for every $x\in X$, one considers $Supp({\mathcal{F}}_x)$ as a closed part of the local scheme $\mathrm{Spec}({\mathcal{O}}_x)$, one has, by definition (0, 16.1.7) dim(ℱ_x) = dim(Supp(ℱ_x)); but one has

  Supp(ℱ_x) = S ∩ Spec(𝒪_{X,x}) = Spec(𝒪_{S,x})

where, in the last term, $S$ denotes any closed sub-prescheme of $X$ having $S$ as underlying space. If one remarks that the irreducible components of $\mathrm{Spec}({\mathcal{O}}_{S,x})$ are the intersections of $\mathrm{Spec}({\mathcal{O}}_{X,x})$ with the irreducible components of $S$ containing $x$, and correspond bijectively with the latter, one sees that

$$\dim ({\mathcal{F}}_x)=\dim ({\mathcal{O}}_{S,x})(\mathrm{5.1.12.1})$$

by virtue of (5.1.1.1); it also follows from (5.1.2) that one has

$$\dim ({\mathcal{F}}_x)=codim(\overline{x},S)(\mathrm{5.1.12.2})$$

if $X$ is locally Noetherian.

One says that $\mathcal{F}$ is equidimensional at the point $x\in X$ if ${\mathcal{F}}_x$ is an equidimensional ${\mathcal{O}}_{X,x}$-module, that is to say (0, 16.1.7) if $Supp({\mathcal{F}}_x)$ is equidimensional as a closed part of $\mathrm{Spec}({\mathcal{O}}_{X,x})$; this amounts to saying that the ring ${\mathcal{O}}_{S,x}$ is equidimensional.

One calls dimension of $\mathcal{F}$ and denotes by $\dim (\mathcal{F})$ the dimension of the support $Supp(\mathcal{F})$; it follows from (5.1.4) and (5.1.12.1) that one has

  dim(ℱ) = sup_{x ∈ X} dim(ℱ_x).                         (5.1.12.3)

If $X=\mathrm{Spec}(A)$ is an affine scheme and if $\mathcal{F}=\tilde{M}$, where $M$ is an $A$-module of finite type, one has $\dim (\mathcal{F})=\dim (M)$ by virtue of (0, 16.1.7) and (5.1.4).

Proposition (5.1.13).

Let $X$ be a prescheme, $\mathcal{F}$ a quasi-coherent ${\mathcal{O}}_X$-Module of finite type, $x$ a point of $X$, $x^{\prime }$ a generization of $x$ in $X$; one then has

$$\dim ({\mathcal{F}}_{x^{\prime }})\le \dim ({\mathcal{F}}_x).(\mathrm{5.1.13.1})$$

This follows at once from (5.1.12.1) and the definitions.

5.2. Dimension of an algebraic prescheme

Proposition (5.2.1).

Let $k$ be a field, $X$ an irreducible prescheme locally of finite type over $k$, $\xi$ its generic point. Then $X$ is biequidimensional and one has $\dim (X)=deg.t{r}_{k}k(\xi )$.

Every local ring ${\mathcal{O}}_x$ is the local ring of a $k$-algebra of finite type, hence a quotient of a regular local ring (0, 17.3.9), and one knows (0, 16.5.12) that such rings are catenary; consequently $X$ is a catenary space (5.1.5), and as $X$ is irreducible, it suffices (5.1.1) to show that for every closed point $x\in X$, one has

$$\dim ({\mathcal{O}}_x)=deg.t{r}_{k}k(x).(\mathrm{5.2.1.1})$$

One may evidently suppose $X$ reduced and affine, hence integral with ring $A$, an algebra of finite type over $k$. Let $n=deg.t{r}_{k}k(\xi )$, with $k(\xi )=K$ the field of fractions of $A$. One knows (Bourbaki, Alg. comm., chap. V, §3, n° 1, th. 1) that there exists a sub-$k$-algebra $B=k[t_1,\cdots ,t_n]$ of $A$, where the $t_i$ are algebraically independent over $k$, such that $A$ be a finite $B$-algebra. Let $\mathfrak{m}=j_x$, which by hypothesis is a maximal ideal of $A$; $\mathfrak{n}=B\cap \mathfrak{m}$ is therefore a maximal ideal of $B$ (Bourbaki, Alg. comm., chap. V, §2, n° 1, prop. 1), and $A_{\mathfrak{m}}$ is a local ring of the finite $B_{\mathfrak{n}}$-algebra $S^{-1}A$, where $S=B-\mathfrak{n}$; as $B_{\mathfrak{n}}$ is integrally closed and $S^{-1}A$ integral, one has $\dim (A_{\mathfrak{m}})=\dim (B_{\mathfrak{n}})$ (0, 16.1.6). One may therefore restrict to the case where $A=k[t_1,\cdots ,t_n]$; one knows then that $k^{\prime }=k(x)$ is a finite extension of $k$ (I, 6.4.2); let $A^{\prime }=k^{\prime }[t_1,\cdots ,t_n]$; taking into account (I, 2.4.6 and 3.3.14), there exists a maximal ideal ${\mathfrak{m}}^{\prime }$ of $A^{\prime }$ above $\mathfrak{m}$ and such that $k^{\prime }$ be the residue field of $A_{{\mathfrak{m}}^{\prime }}^{\prime }$. Since $A^{\prime }$ is integral and is a finite $A$-algebra, the same reasoning as above shows that $\dim (A_{{\mathfrak{m}}^{\prime }}^{\prime })=\dim (A_{\mathfrak{m}})$, so that one is reduced to the case where $A/\mathfrak{m}=k$. Now in this case $\mathfrak{m}$ is generated by polynomials $t_i-a_i$ (where $a_i\in k$, $1\le i\le n$, the $a_i$ being the canonical images of the $t_i$ in $A/\mathfrak{m}$). Replacing $t_i$ by $t_i-a_i$, one sees finally that one is reduced to the case where $\mathfrak{m}={\sum }_{i=1}^{n}A{t}_i$. The completion of the local ring $A_{\mathfrak{m}}$ is then the formal series ring $k[[t_1,\cdots ,t_n]]$; one knows that it has the same dimension as $A_{\mathfrak{m}}$ (0, 16.2.4), and on the other hand, the dimension of $k[[t_1,\cdots ,t_n]]$ is equal to $n$ (0, 17.1.4); whence the conclusion.

Corollary (5.2.2).

For a prescheme $X$ locally of finite type over a field $k$, $\dim (X)$ coincides with the number defined in (4.1.1).

Indeed, if $X_{\alpha }$ are the reduced sub-preschemes having as underlying spaces the irreducible components of $X$, one has dim(X) = sup_α (dim(X_α)) (0, 14.1.2.1) and it suffices to apply (5.2.1) to the $X_{\alpha }$.

Corollary (5.2.3).

Let $X$ be a prescheme locally of finite type over a field $k$, $x$ a point of $X$. One has

  dim_x(X) = dim(𝒪_x) + deg.tr_k k(x).                   (5.2.3.1)

One knows that there exists an open neighbourhood $U$ of $x$ in $X$ such that ${\dim }_x(X)=\dim (U)$ (0, 14.1.4.1), and one may suppose that the irreducible components of $U$ are the $X_i\cap U$, where the $X_i$ are the irreducible components of $X$ containing $x$; as $U\cap X_i$

is dense in $X_i$, it follows from (4.1.1.3) that $\dim (X_i)=\dim (U\cap X_i)$, so one has dim_x(X) = sup_i (dim(X_i)). Moreover, the minimal prime ideals of ${\mathcal{O}}_x$ correspond to the generic points of the $X_i$, hence (0, 16.1.1.1), one has dim(𝒪_{X,x}) = sup_i (dim(𝒪_{X_i, x})). One is thus reduced to the case where $X$ is irreducible; as $X$ is biequidimensional by (5.2.1), one has dim(X) = dim(‾{x}) + codim(‾{x}, X) (0, 14.3.5.1), and one knows that $\dim (\overline{x})=deg.t{r}_{k}k(x)$ by (5.2.1) and $codim(\overline{x},X)=\dim ({\mathcal{O}}_x)$ by (5.1.2).

Corollary (5.2.4).

Let $X$ be a prescheme locally of finite type over a field $k$, $\mathcal{L}$ an invertible ${\mathcal{O}}_X$-Module, $f$ a section of $\mathcal{L}$ over $X$, such that the set $Y$ of $x\in X$ for which $f(x)=0$ is rare in $X$. Then one has $\dim (Y)\le \dim (X)-1$; the two sides of this inequality are equal if $Y$ meets every irreducible component of maximum dimension of $X$.

If $(X_{\alpha })$ is the family of irreducible components of $X$, one has

  dim(Y) = sup_α (dim(Y ∩ X_α))

and one is therefore reduced to the case where $X$ is irreducible ($Y\cap X_{\alpha }$ being rare in $X_{\alpha }$ since each $X_{\alpha }$ has a non-empty interior in $X$). One may restrict to the case where $Y\ne \varnothing$; then, for every maximal point $x$ of $Y$, one has (since $X$ is biequidimensional) dim(‾{x}) = dim(X) − codim(‾{x}, X), and since $Y$ is rare in $X$, one has

$$codim(\overline{x},X)=1$$

by (5.1.8); whence the corollary.

Remarks (5.2.5).

(i) Contrary to what happens for algebraic $k$-preschemes, a locally Noetherian prescheme $X$ is not necessarily catenary (cf. (5.6.11)); it is, however, if all its local rings ${\mathcal{O}}_x$ are quotients of regular rings (0, 16.5.12) and in particular if $X$ is regular (I, 4.1.4). Nevertheless, even an (integral) scheme $X=\mathrm{Spec}(A)$, where $A$ is a regular ring, is not necessarily biequidimensional; in other words (0, 14.3.3) the codimensions in $X$ of the various closed points of $X$ are not necessarily the same. For example, let $B$ be a discrete valuation ring, $\mathfrak{m}=B\pi$ its maximal ideal, $k=B/\mathfrak{m}$ its residue field, $K$ the field of fractions of $B$; let $A$ be the polynomial ring B[T]. In $A$ there are maximal ideals of height 2, for example $(\pi )+(T)$; but there are also maximal ideals of height 1, for example the principal ideal $(\pi T-1)$: indeed, a principal prime ideal $\ne 0$ is of height 1 (5.1.8); on the other hand $A/(\pi T-1)$ is isomorphic to the ring of fractions $B_{\pi }$ $(0_I,\mathrm{1.2.3})$, which is none other than $K$ here, which proves that the ideal $(\pi T-1)$ is maximal and of height 1.

(ii) When $X$ is a prescheme locally of finite type over a field, one has seen (4.1.1.3) that $\dim (U)=\dim (X)$ for every dense open $U$ in $X$. This result does not extend to regular Noetherian preschemes, even if they are biequidimensional. For example, let $B$ be a discrete valuation ring, $\mathfrak{m}$ its maximal ideal; $X=\mathrm{Spec}(A)$ has two points, the ideal (0) and $\mathfrak{m}$, the latter being the only closed point; one has $\dim (X)=1$, but the open set $U=(0)$ is of dimension 0 (cf. §10).

5.3. Dimension of the support of a Module and Hilbert polynomial

This number uses the results of Chapter III; it will not be used in the sequel of this chapter.

Proposition (5.3.1).

Let $A$ be an Artinian local ring, $X$ a projective scheme over $\mathrm{Spec}(A)$, $\mathcal{L}$ an invertible ${\mathcal{O}}_X$-Module very ample relative to $A$, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module $\ne 0$; set $\mathcal{F}(n)=\mathcal{F}{\otimes }_{{\mathcal{O}}_X}{\mathcal{L}}^{\otimes n}$ for $n\in \mathbb{Z}$. Then the degree of the Hilbert polynomial $P(n)={\chi }_A(\mathcal{F}(n))$ of $\mathcal{F}$ relative to $A$ (III, 2.5.3) is equal to the dimension of $Supp(\mathcal{F})$.

We reason by induction on $d=\dim (Supp(\mathcal{F}))$. One knows that there exists a closed sub-prescheme $Y$ of $X$ whose $Supp(\mathcal{F})$ is the underlying space, and an ${\mathcal{O}}_Y$-Module coherent $\mathcal{G}$ such that $\mathcal{F}=j_{\ast }(\mathcal{G})$, where $j:Y\to X$ is the canonical injection $(Er{r}_{III},30)$. It is immediate that the Hilbert polynomials of $\mathcal{F}$ and of $\mathcal{G}$ are the same, so one may restrict to the case where $X=Supp(\mathcal{F})$. Suppose first $d=0$; all the points of $X$ being closed, $X$ is an Artinian scheme (I, 6.2.2), hence $\mathcal{F}(n)=\mathcal{F}$ for every integer $n$, and one has consequently (III, 2.5.3)

$${\chi }_A(\mathcal{F}(n))=lon{g}_A(\Gamma (X,\mathcal{F}))$$

for $n$ large enough, and the degree of the Hilbert polynomial is therefore 0. Suppose now $d>0$, and set $Z=Ass(\mathcal{F})$, which is a finite set (3.1.6); there exist an integer $m>0$ and a section $f\in \Gamma (X,{\mathcal{L}}^{\otimes m})$ such that $X_f$ be a neighbourhood of $Z$ (II, 4.5.4). Multiplication by $f$ is a homomorphism

$${\mu }_f:\mathcal{F}\to \mathcal{F}(m)$$

which is injective (3.1.8); one therefore has an exact sequence

  0 → ℱ ─μ_f→ ℱ(m) → 𝒢 → 0                              (5.3.1.1)

where $\mathcal{G}$ is coherent. By virtue of Nakayama's lemma, the points $x\in Supp(\mathcal{G})$ are exactly those for which $f(x)=0$. We shall deduce from this that one has

$$\dim (Supp(\mathcal{G}))=d-1.(\mathrm{5.3.1.2})$$

This will follow from the following lemma:

Lemma (5.3.1.3).

Let $A$ be an Artinian local ring, $X$ a projective scheme over $\mathrm{Spec}(A)$, $\mathcal{L}$ an invertible ${\mathcal{O}}_X$-Module ample relative to $A$; then, for every section $g$ of $\mathcal{L}$ over $X$, the set $X_g$ of $x\in X$ such that $g(x)=0$ meets every irreducible component of $X$ of dimension $\ne 0$.

Indeed (II, 5.5.7), the set $X_g$ is an affine open of $X$, and if it contains an irreducible component $X^{\prime }$ of $X$, the reduced closed sub-prescheme of $X$ having $X^{\prime }$ as underlying space is at once projective and affine over $A$, hence finite over $A$ (III, 4.4.2), and consequently an Artinian scheme, hence of dimension 0.

This lemma being established, note that since $Z$ contains the maximal points of $X$, $X_f$ is dense, hence $Supp(\mathcal{G})$ is rare in $X$, and the relation (5.3.1.2) follows from the lemma and from (5.2.4).

This being so, from the exact sequence (5.3.1.1) one deduces, for every $n$, the exact sequence

  0 → ℱ(n) → ℱ(n + m) → 𝒢(n) → 0

and for $n$ large enough, one therefore also has the exact sequence (III, 2.2.3)

  0 → Γ(X, ℱ(n)) → Γ(X, ℱ(n + m)) → Γ(X, 𝒢(n)) → 0

whence, taking (III, 2.5.3) into account, for $n$ large enough,

  χ_A(𝒢(n)) = χ_A(ℱ(n + m)) − χ_A(ℱ(n)).

As, by virtue of (5.3.1.2) and the induction hypothesis, the degree of the polynomial ${\chi }_A(\mathcal{G}(n))$ is $d-1$, the preceding relation entails that the degree of ${\chi }_A(\mathcal{F}(n))$ is $d$. Q.E.D.

5.4. Dimension of the image of a morphism

Proposition (5.4.1).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a morphism.

(i) If $f$ is quasi-finite, one has dim(X) ≤ dim(‾{f(X)}) ≤ dim(Y).

(ii) If $f$ is surjective and open (resp. surjective and closed), one has $\dim (X)\ge \dim (Y)$.

(i) One can replace $f$ by $f_{red}$ (II, 6.2.4), hence suppose $X$ and $Y$ reduced; if $Z$ is the reduced closed sub-prescheme of $Y$ having as underlying space $\overline{f(X)}$, one has then $f=j\circ g$, where $j:Z\to Y$ is the canonical injection and $g:X\to Z$ is a quasi-finite morphism (I, 5.2.2 and II, 6.2.4). One may therefore restrict to the case where $f(X)$ is dense in $Y$. For every $x\in X$, ${\mathcal{O}}_x$ is then a quasi-finite ${\mathcal{O}}_{f(x)}$-module (II, 6.2.2), and consequently ${\mathfrak{m}}_{f(x)}{\mathcal{O}}_x$ is an ideal of definition of ${\mathcal{O}}_x$ $(0_I,\mathrm{7.4.4})$; but one knows (0, 16.3.10) that if $A\to B$ is a local homomorphism of Noetherian local rings such that, if $\mathfrak{m}$ is the maximal ideal of $A$, $\mathfrak{m}B$ is an ideal of definition of $B$, then one has $\dim (B)\le \dim (A)$; this completes the proof of (i) by virtue of (5.1.4).

(ii) The definition of dimension (0, 14.1.2) shows that it suffices to prove that for every sequence $(y_i{)}_{0\le i\le n}$ of distinct elements of $Y$ such that $y_i\in \overline{y_{i+1}}$ for $0\le i\le n-1$, there exists a sequence $(x_i{)}_{0\le i\le n}$ of points of $X$ such that $x_i\in \overline{x_{i+1}}$ for $0\le i\le n-1$ and $f(x_i)=y_i$ for every $i$. Suppose first $f$ surjective and open and let us prove the existence of the $x_i$ by induction on $i$; the existence of $x_0\in X$ such that $f(x_0)=y_0$ follows from the fact that $f$ is surjective. If the $x_i$ have been determined for $i\le m$ so as to satisfy $f(x_i)=y_i$ for $i\le m$ and $x_i\in \overline{x_{i+1}}$ for $i<m$, one notes that since $f$ is open and $y_{m+1}$ a generization of $y_m$, there exists $x_{m+1}\in f^{-1}(y_{m+1})$ which is a generization of $x_m$ (1.10.3) and the induction can proceed.

Suppose now $f$ surjective and closed and let us prove the existence of the $x_i$ by descending induction on $i$, the existence of $x_n$ such that $f(x_n)=y_n$ still resulting from the fact that $f$ is surjective. If the $x_i$ have been determined so as to satisfy the desired conditions for $i>m$, one notes that $f(\overline{x_{m+1}})$ is the closure of $f(x_{m+1})=y_{m+1}$ since $f$

is closed (Bourbaki, Top. gén., chap. I, 3rd ed., §5, n° 4, prop. 9); there therefore exists $x_m\in \overline{x_{m+1}}$ such that $f(x_m)=y_m$ and the descending induction can continue.

Corollary (5.4.2).

If $X$, $Y$ are two locally Noetherian preschemes, $f:X\to Y$ a finite morphism (hence closed), one has $\dim (X)=\dim (\overline{f(X)})$. If moreover $f$ is surjective, one has $\dim (X)=\dim (Y)$.

Remarks (5.4.3).

(i) One has seen in (4.1.2) that if $X$ and $Y$ are preschemes locally of finite type over a field $k$ and $f$ a $k$-morphism, the inequality $\dim (Y)\le \dim (X)$ is already verified when $f$ is quasi-compact and dominant. On the contrary, one can have $\dim (Y)>\dim (X)$ even when $f$ is of finite type, bijective, and a local immersion, if one only supposes $X$ and $Y$ locally Noetherian. For example, let $A$ be a discrete valuation ring, $K$ its field of fractions, $k$ its residue field; if $Y=\mathrm{Spec}(A)$, and if $a$ is the generic point of $Y$ and $b$ its closed point, one has therefore $k(a)=K$, $k(b)=k$. Let then $X$ be the sum prescheme of $\mathrm{Spec}(K)$ and $\mathrm{Spec}(k)$, $f:X\to Y$ the morphism which coincides on $\mathrm{Spec}(K)$ and $\mathrm{Spec}(k)$ with the respective canonical morphisms (I, 2.4.5); it is clear that $f$ is a bijective local immersion, $a$ being open in $Y$; on the other hand $f$ is of finite type, for $K=A[{\pi }^{-1}]$, where $\pi$ is a uniformizer of $A$, so $K$ is an $A$-algebra of finite type. However one has $\dim (X)=0$ and $\dim (Y)=1$.

(ii) If $A$ and $B$ are two Noetherian rings such that $A\subset B$ and that $B$ is a finite $A$-algebra, the corollary (5.4.2) shows again that $\dim (B)=\dim (A)$ (0, 16.1.5). Suppose moreover that $A$ is a Noetherian local ring; then $B$ is a Noetherian semi-local ring (Bourbaki, Alg. comm., chap. IV, §2, n° 5, cor. 3 of prop. 9); if ${\mathfrak{n}}_i$ ($1\le i\le r$) are the maximal ideals of $B$, one has therefore

  dim(A) = sup_i dim(B_{𝔫_i}).                           (5.4.3.1)

One will observe that, under these conditions, one does not necessarily have $\dim (B_{{\mathfrak{n}}_i})=\dim (A)$ for every $i$: it suffices to see this to replace $B$ by the direct product of $B$ and the residue field $k$ of $A$. But one even has examples where $A$ and $B$ are in addition integral rings of dimension 2 and where certain of the $B_{{\mathfrak{n}}_i}$ have dimension $<\dim (A)$ (5.6.11). We shall, however, see further on ((5.6.4) and (5.6.10)) that this last phenomenon cannot present itself when one supposes that $A$ is a quotient of a regular local ring.

5.5. Dimension formula for a morphism of finite type

(5.5.1)

Recall (0, 16.3.9) that if $A$, $B$ are two Noetherian local rings, $k$ the residue field of $A$, $\varphi :A\to B$ a local homomorphism, one has

  dim(B) ≤ dim(A) + dim(B ⊗_A k).                        (5.5.1.1)

One deduces:

Proposition (5.5.2).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a morphism, $x$ a point of $X$, $y=f(x)$. Then one has

  dim(𝒪_x) ≤ dim(𝒪_y) + dim(𝒪_x ⊗_{𝒪_y} k(y)).           (5.5.2.1)

In particular, if $x$ is a maximal point of the fibre $f^{-1}(y)$, one has

$$\dim ({\mathcal{O}}_x)\le \dim ({\mathcal{O}}_y)(\mathrm{5.5.2.2})$$

since ${\mathcal{O}}_x{\otimes }_{{\mathcal{O}}_y}k(y)={\mathcal{O}}_x/{\mathfrak{m}}_y{\mathcal{O}}_x$ is the local ring of $x$ in the prescheme $f^{-1}(y)$, and is therefore of dimension 0 by hypothesis.

We shall obtain a more precise formula than (5.5.2.1) when one supposes that $f$ is a morphism of finite type.

Proposition (5.5.3).

Let $A$ be a Noetherian ring, $T$ an indeterminate, $\mathfrak{p}$ a prime ideal of $A$; then ${\mathfrak{p}}^{\prime }=\mathfrak{p}A[T]$ is prime in $B=A[T]$ and ${\mathfrak{p}}^{\prime }\cap A=\mathfrak{p}$. There exists an infinity of prime ideals of $B$ distinct from ${\mathfrak{p}}^{\prime }$ whose intersection with $A$ is $\mathfrak{p}$; these ideals are pairwise without inclusion relation. Moreover, if $\mathfrak{q}$ is such an ideal, one has

  dim(B_𝔮) = dim(B_{𝔭'}) + 1 = dim(A_𝔭) + 1.             (5.5.3.1)

In the first assertions, one reduces at once, by replacing $A$ by $A/\mathfrak{p}$ and observing that $A[T]/\mathfrak{p}A[T]=(A/\mathfrak{p})[T]$, to the case $\mathfrak{p}=0$; they then follow from the fact that the prime ideals of $B=A[T]$ whose intersection with $A$ reduces to 0 are exactly those which do not meet the multiplicative part $S=A-0$ of the integral ring $A$; now one knows that there is an increasing bijection of the set of these ideals onto the set of prime ideals of $S^{-1}A[T]=K[T]$, where $K$ is the field of fractions of $A$ (Bourbaki, Alg. comm., chap. II, §2, n° 5, prop. 11). Moreover, one has, according to (5.5.1.2), dim(B_𝔮) ≤ dim(A_𝔭) + dim(B_𝔮/𝔭 B_𝔮), and if $k$ is the field of fractions of $A/\mathfrak{p}$, $B_{\mathfrak{q}}/\mathfrak{p}{B}_{\mathfrak{q}}$ is canonically identified with $(k[T]{)}_{\mathfrak{q}}$, hence is a discrete valuation ring, so of dimension 1. Finally, if $\mathfrak{p}={\mathfrak{p}}_0\supset {\mathfrak{p}}_1\supset \cdots \supset {\mathfrak{p}}_m$ is a chain of prime ideals of $A$ of maximum length, the ideals ${\mathfrak{p}}_{j}B$ ($0\le j\le m$) are prime in $B$, pairwise distinct, and contained in $\mathfrak{q}$; hence $\dim (B_{\mathfrak{q}})\ge m+1=\dim (A_{\mathfrak{p}})+1$ and consequently $\dim (B_{\mathfrak{q}})=\dim (A_{\mathfrak{p}})+1$. This relation can also be written $ht(\mathfrak{q})=ht(\mathfrak{p})+1$; as $\mathfrak{q}\ne {\mathfrak{p}}^{\prime }$, one has moreover $ht(\mathfrak{q})\ge ht({\mathfrak{p}}^{\prime })+1\ge ht(\mathfrak{p})+1$ by definition of the height of a prime ideal; this completes the proof of (5.5.3.1).

Corollary (5.5.4).

For every Noetherian ring $A$, one has

  dim(A[T_1, …, T_r]) = dim(A) + r                       (5.5.4.1)

($T_i$ indeterminates).

One will note, on the other hand, that there are examples of non-Noetherian local rings $A$ such that $\dim (A)=1$ and $\dim (A[T])=3$ [30].

Corollary (5.5.5).

For every locally Noetherian prescheme $X$, the dimension of $X[T_1,\cdots ,T_r]=X{\otimes }_{\mathbb{Z}}\mathbb{Z}[T_1,\cdots ,T_r]$ ($T_i$ indeterminates) is $\dim (X)+r$.

This follows from (5.5.4) and (0, 14.1.7).

Corollary (5.5.6).

Under the hypotheses of (5.5.3), let $\mathfrak{q}$ be a prime ideal of $B=A[T]$ such that $\mathfrak{q}\cap A=\mathfrak{p}$; if $k$ and $k^{\prime }$ are the residue fields of $A_{\mathfrak{p}}$ and $B_{\mathfrak{q}}$ respectively, one has

  dim(A_𝔭) + 1 = dim(B_𝔮) + deg.tr_k k'.                 (5.5.6.1)

If $\mathfrak{q}=\mathfrak{p}B$, one has, according to (5.5.3.1), $\dim (B_{\mathfrak{q}})=\dim (A_{\mathfrak{p}})$, and in this case $k^{\prime }=k(T)$, hence the formula (5.5.6.1) is indeed verified. In the contrary case, $\dim (B_{\mathfrak{q}})=\dim (A_{\mathfrak{p}})+1$, and as $\mathfrak{q}$ corresponds to a prime ideal ${\mathfrak{q}}^{\prime }\ne 0$ of k[T], $k^{\prime }$ is an algebraic extension of $k$, so one still has the formula (5.5.6.1).

Lemma (5.5.7).

Let $A$ be an integral Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $k$ its residue field.

(i) For every integral ring $B$ containing $A$, such that $B=A[x]$ (for some $x\in B$) and every prime ideal $\mathfrak{q}$ of $B$ such that $\mathfrak{q}\cap A=\mathfrak{m}$, one has

  dim(A) + deg.tr_A B ≥ dim(B_𝔮) + deg.tr_k k'           (5.5.7.1)

denoting by $k^{\prime }$ the residue field of $B_{\mathfrak{q}}$ and by $deg.t{r}_{A}B$ the transcendence degree of the field of fractions of $B$ over that of $A$.

(ii) Suppose that for every maximal ideal $\mathfrak{n}$ of A[T] such that $\mathfrak{n}\cap A=\mathfrak{m}$, the ring $(A[T]{)}_{\mathfrak{n}}$ is catenary; then the two sides of (5.5.7.1) are equal.

(i) If $x$ is transcendent over the field of fractions of $A$, one has $deg.t{r}_{A}B=1$ and the two sides of (5.5.7) are equal by virtue of (5.5.6). In the contrary case, one has $B=A[T]/\mathfrak{p}$, where $\mathfrak{p}$ is a prime ideal $\ne 0$ of A[T], such that $\mathfrak{p}\cap A=0$ since $B$ contains $A$; one therefore has $ht(\mathfrak{p})=1$ by virtue of (5.5.3). The ideal $\mathfrak{q}$ of $B$ is of the form $\mathfrak{n}/\mathfrak{p}$, where $\mathfrak{n}\supset \mathfrak{p}$ is a prime ideal of A[T] such that $\mathfrak{n}\cap A=\mathfrak{m}$, and one has $B_{\mathfrak{q}}=(A[T]{)}_{\mathfrak{n}}/\mathfrak{p}(A[T]{)}_{\mathfrak{n}}$; the formula (0, 16.1.4.1), applied to $X=\mathrm{Spec}((A[T]{)}_{\mathfrak{n}})$ and to $Y=\mathrm{Spec}(B_{\mathfrak{q}})$, gives

  dim((A[T])_𝔫) ≥ ht(𝔭 (A[T])_𝔫) + dim(B_𝔮) = 1 + dim(B_𝔮)     (5.5.7.2)

since $ht(\mathfrak{p}(A[T]{)}_{\mathfrak{n}})=ht(\mathfrak{p})=1$ by virtue of the bijective correspondence between prime ideals of A[T] contained in $\mathfrak{n}$ and prime ideals of $(A[T]{)}_{\mathfrak{n}}$. Finally, the formula (5.5.6.1) gives

  dim((A[T])_𝔫) = dim(A) + 1 − deg.tr_k k'               (5.5.7.3)

since the residue fields of $B_{\mathfrak{q}}$ and of $(A[T]{)}_{\mathfrak{n}}$ are the same; on the other hand, one has then $deg.t{r}_{A}B=0$, which completes the proof of (5.5.7.1).

(ii) If $(A[T]{)}_{\mathfrak{n}}$ is catenary, the two sides of (5.5.7.2) are equal (0, 16.1.4), hence also the two sides of (5.5.7.1).

Theorem (5.5.8) (dimension formula).

Let $A$ be an integral Noetherian local ring, $B$ an integral ring containing $A$ which is an $A$-algebra of finite type, $\mathfrak{q}$ a prime ideal of $B$ such that $\mathfrak{q}\cap A$ is the maximal ideal $\mathfrak{m}$ of $A$, $k$ and $k^{\prime }$ the residue fields of $A$ and $B_{\mathfrak{q}}$ respectively. One then has the inequality

  dim(A) + deg.tr_A B ≥ dim(B_𝔮) + deg.tr_k k'.          (5.5.8.1)

Moreover the two sides are equal if, for every sub-$A$-algebra $A^{\prime }$ of finite type of B[T] and every maximal ideal ${\mathfrak{m}}^{\prime }$ of $A^{\prime }$ such that ${\mathfrak{m}}^{\prime }\cap A=\mathfrak{m}$, $A_{{\mathfrak{m}}^{\prime }}^{\prime }$ is catenary.

Let $B=A[x_1,\cdots ,x_n]$ and let us reason by induction on $n$. Set $C=A[x_1,\cdots ,x_{n-1}]$ and $\mathfrak{r}=\mathfrak{q}\cap C$; $C_{\mathfrak{r}}$ is an integral Noetherian local ring, and if one sets $S=C-\mathfrak{r}$, $B^{\prime }=S^{-1}B$, ${\mathfrak{q}}^{\prime }=S^{-1}\mathfrak{q}$, one has $B_{\mathfrak{q}}=B_{{\mathfrak{q}}^{\prime }}^{\prime }$, $B^{\prime }=C_{\mathfrak{r}}[x_n]$ and ${\mathfrak{q}}^{\prime }\cap C_{\mathfrak{r}}=\mathfrak{r}{C}_{\mathfrak{r}}$; in addition the fields of fractions

of $B^{\prime }$ and $C_{\mathfrak{r}}$ are those of $B$ and $C$ respectively. If $k_1$ is the residue field of $C_{\mathfrak{r}}$, one has therefore, according to (5.5.7.1),

  dim(C_𝔯) + deg.tr_C B ≥ dim(B_𝔮) + deg.tr_{k_1} k'.    (5.5.8.2)

On the other hand, the induction hypothesis gives

  dim(A) + deg.tr_A C ≥ dim(C_𝔯) + deg.tr_k k_1          (5.5.8.3)

whence (5.5.8.1) by adding (5.5.8.2) and (5.5.8.3) term by term. To prove the second assertion, note first that every sub-$A$-algebra of finite type of C[T] is also a sub-$A$-algebra of finite type of B[T]; by induction on $n$, one may therefore suppose that the two sides of (5.5.8.3) are equal. On the other hand, to see that the two sides of (5.5.8.2) are equal, it suffices, by virtue of (5.5.7), to verify that if $\mathfrak{n}$ is a maximal ideal of $C_{\mathfrak{r}}[T]$ such that $\mathfrak{n}\cap C_{\mathfrak{r}}=\mathfrak{r}{C}_{\mathfrak{r}}$, then $(C_{\mathfrak{r}}[T]{)}_{\mathfrak{n}}$ is catenary; but one has $C_{\mathfrak{r}}[T]=S^{\prime -1}C[T]$ where $S^{\prime }=C-\mathfrak{r}$; the ideal $\mathfrak{n}$ is therefore of the form $S^{\prime -1}{\mathfrak{n}}^{\prime }$, where ${\mathfrak{n}}^{\prime }$ is a prime ideal of C[T] such that ${\mathfrak{n}}^{\prime }\cap C=\mathfrak{r}$, whence $(C_{\mathfrak{r}}[T]{)}_{\mathfrak{n}}=(C[T]{)}_{{\mathfrak{n}}^{\prime }}$; if ${\mathfrak{m}}^{\prime }$ is a maximal ideal of C[T] containing ${\mathfrak{n}}^{\prime }$, $(C[T]{)}_{{\mathfrak{n}}^{\prime }}$ is therefore a local ring of the ring $(C[T]{)}_{{\mathfrak{m}}^{\prime }}$, and as by hypothesis this latter is catenary, so is $(C[T]{)}_{{\mathfrak{n}}^{\prime }}$ (0, 16.1.4).

5.6. Dimension formula and universally catenary rings

Proposition (5.6.1).

Let $A$ be a Noetherian ring. The following properties are equivalent:

a) Every polynomial ring $A[T_1,\cdots ,T_n]$ is catenary.

b) Every algebra of finite type over $A$ is catenary.

c) $A$ is catenary, and for every integral local $A$-algebra essentially of finite type (1.3.8) $B^{\prime }$, one has, denoting by $\mathfrak{p}$ the inverse image in $A$ of the maximal ideal $\mathfrak{q}$ of $B^{\prime }$, by $A^{\prime }$ the image of $A_{\mathfrak{p}}$ in $B^{\prime }$, by $k$ and $k^{\prime }$ the residue fields of $A^{\prime }$ and $B^{\prime }$ respectively,

  dim(A') + deg.tr_{A'}(B') = dim(B') + deg.tr_k k'.     (5.6.1.1)

The fact that b) entails c) follows from (5.5.8); indeed, $B^{\prime }$ is a local $A^{\prime }$-algebra essentially of finite type (1.3.10), hence of the form $B_{{\mathfrak{q}}^{\prime \prime }}^{\prime \prime }$, where B'' is a sub-$A^{\prime }$-algebra of finite type of $B^{\prime }$, ${\mathfrak{q}}^{\prime \prime }$ a prime ideal of B'' above the maximal ideal ${\mathfrak{p}}^{\prime }$ of $A^{\prime }$; moreover the fields of fractions of $B^{\prime }$ and B'' are the same, hence $deg.t{r}_{A^{\prime }}(B^{\prime })=deg.t{r}_{A^{\prime }}(B^{\prime \prime })$. To prove (5.6.1.1), it suffices to show that (under hypothesis b)) every sub-$A^{\prime }$-algebra A_1 of finite type of B'[T] is catenary; indeed the two sides of (5.5.8.1), where one replaces $A$, $B$, and $\mathfrak{q}$ by $A^{\prime }$, B'', and ${\mathfrak{q}}^{\prime \prime }$, will then be equal, whence the equality (5.6.1.1). Now the hypothesis b) entails that every $A$-algebra essentially of finite type is catenary (0, 16.1.4), and A_1 is such an $A$-algebra (1.3.9).

It is trivial that b) entails a); conversely, a) entails b), every $A$-algebra of finite type being a quotient of a polynomial algebra (0, 16.1.4).

It remains to prove that c) entails b). Since every quotient by a prime ideal

of an $A$-algebra of finite type is an integral $A$-algebra of finite type, it amounts to seeing that, if $B$ is an integral $A$-algebra of finite type, $\mathfrak{q}$ and ${\mathfrak{q}}^{\prime }$ two prime ideals of $B$ such that ${\mathfrak{q}}^{\prime }\subset \mathfrak{q}$, one has (0, 16.1.4.2)

  dim(B_𝔮 / 𝔮' B_𝔮) + dim(B_{𝔮'}) = dim(B_𝔮).            (5.6.1.2)

Let $\mathfrak{p}$, ${\mathfrak{p}}^{\prime }$ be the inverse images of $\mathfrak{q}$, ${\mathfrak{q}}^{\prime }$ respectively in $A$, $\mathfrak{n}$ the kernel of the homomorphism $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$.

The image $A^{\prime }$ of $A_{\mathfrak{p}}$ in $B_{\mathfrak{q}}$ being isomorphic to $A_{\mathfrak{p}}/\mathfrak{n}$, the formula (5.6.1.1) applied to $A^{\prime }$ and to $B^{\prime }=B_{\mathfrak{q}}$ is written

  dim(A_𝔭/𝔫) + deg.tr_{A'}(B_𝔮) = dim(B_𝔮) + deg.tr_{k(𝔭)} k(𝔮).  (5.6.1.3)

On the other hand, the kernel of $A_{{\mathfrak{p}}^{\prime }}\to B_{{\mathfrak{q}}^{\prime }}$ is $\mathfrak{n}{A}_{{\mathfrak{p}}^{\prime }}$, hence, applying the formula (5.6.1.1) where one replaces $A^{\prime }$ by $A_{{\mathfrak{p}}^{\prime }}/\mathfrak{n}{A}_{{\mathfrak{p}}^{\prime }}$ and $B^{\prime }$ by $B_{{\mathfrak{q}}^{\prime }}$, one has

  dim(A_{𝔭'} / 𝔫 A_{𝔭'}) + deg.tr_{A_{𝔭'} / 𝔫 A_{𝔭'}} (B_{𝔮'})
        = dim(B_{𝔮'}) + deg.tr_{k(𝔭')} k(𝔮').            (5.6.1.4)

Finally, $B/{\mathfrak{q}}^{\prime }$ is an integral $A$-algebra of finite type, the inverse image of $\mathfrak{q}/{\mathfrak{q}}^{\prime }$ in $A$ is $\mathfrak{p}$, and the kernel of the homomorphism $A_{\mathfrak{p}}\to B_{\mathfrak{q}}/{\mathfrak{q}}^{\prime }{B}_{\mathfrak{q}}$ is ${\mathfrak{p}}^{\prime }{A}_{\mathfrak{p}}$, hence, applying (5.6.1.1) where one replaces $A^{\prime }$ by $A_{\mathfrak{p}}/{\mathfrak{p}}^{\prime }{A}_{\mathfrak{p}}$ and $B^{\prime }$ by $B_{\mathfrak{q}}/{\mathfrak{q}}^{\prime }{B}_{\mathfrak{q}}$, one has

  dim(A_𝔭/𝔭' A_𝔭) + deg.tr_{k(𝔭')}(B_𝔮 / 𝔮' B_𝔮)
        = dim(B_𝔮 / 𝔮' B_𝔮) + deg.tr_{k(𝔭)} k(𝔮).        (5.6.1.5)

Let us now add (5.6.1.4) and (5.6.1.5) term by term, and note that $k({\mathfrak{p}}^{\prime })$ (resp. $k({\mathfrak{q}}^{\prime })$) is the field of fractions of $A_{\mathfrak{p}}/{\mathfrak{p}}^{\prime }{A}_{\mathfrak{p}}$ (resp. $B_{\mathfrak{q}}/{\mathfrak{q}}^{\prime }{B}_{\mathfrak{q}}$); on the other hand $A_{{\mathfrak{p}}^{\prime }}/\mathfrak{n}{A}_{{\mathfrak{p}}^{\prime }}$ and $A_{\mathfrak{p}}/\mathfrak{n}$ have the same field of fractions, and $B_{{\mathfrak{q}}^{\prime }}$ and $B_{\mathfrak{q}}$ have the same field of fractions; finally, since $A$ is supposed catenary, one has (0, 16.1.4.2)

  dim(A_𝔭 / 𝔭' A_𝔭) + dim(A_{𝔭'} / 𝔫 A_{𝔭'}) = dim(A_𝔭/𝔫).

Taking these remarks and (5.6.1.3) into account, the relation (5.6.1.2) follows. Q.E.D.

Definition (5.6.2).

One says that a Noetherian ring $A$ is universally catenary if it verifies the equivalent conditions a), b), c) of (5.6.1).

Remarks (5.6.3).

(i) If $A$ is universally catenary, so is $S^{-1}A$ for every multiplicative part $S$ of $A$, as follows at once from (5.6.1, a)) and from the fact that every ring of fractions of a catenary ring is catenary. Conversely, if, for every prime ideal $\mathfrak{p}$ of $A$, the ring $A_{\mathfrak{p}}$ is universally catenary, then $A$ is universally catenary: this follows from (5.6.1, c)), if one notes that setting $S=A-\mathfrak{p}$, $S^{-1}B$ is an $A_{\mathfrak{p}}$-algebra of finite type, and $B_{\mathfrak{q}}=(S^{-1}B{)}_{S^{-1}\mathfrak{q}}$.

(ii) One says that a locally Noetherian prescheme $X$ is universally catenary if, for every $x\in X$, the ring ${\mathcal{O}}_x$ is universally catenary. It follows from (i) that for $A$ to be a universally catenary ring, it is necessary and sufficient that the scheme $\mathrm{Spec}(A)$ be so.

(iii) The criterion (5.6.1, c)) involves only the images of $A$ in integral $A$-algebras of finite type, hence integral quotient rings of $A$. One concludes that if

${\mathfrak{p}}_i$ ($1\le i\le n$) are the minimal prime ideals of $A$, it amounts to the same to say that $A$ is universally catenary or that each of the $A/{\mathfrak{p}}_i$ is so. This also shows that it amounts to the same to say that a locally Noetherian prescheme $X$ is universally catenary, or that $X_{red}$ is so.

(iv) The criterion (5.6.1, b)) and the remark (i) show that, if $A$ is a universally catenary ring, so is every $A$-algebra essentially of finite type (1.3.8).

Proposition (5.6.4).

Every quotient ring of a regular ring is universally catenary.

It all comes down to seeing that a regular ring $A$ is universally catenary (5.6.3, (iv)); now, one knows that $A[T_1,\cdots ,T_n]$ is regular (0, 17.3.7), hence catenary (0, 16.5.12), and one concludes by (5.6.1, a)).

In particular, it follows from Cohen's theorem (0, 19.8.8) that every complete Noetherian local ring is universally catenary. Likewise, every algebra of finite type over a field being a quotient of a regular ring, every prescheme locally of finite type over a field is universally catenary.

We shall see further on (6.3.7) that in (5.6.4), one can replace "regular ring" by "Cohen-Macaulay ring".

Proposition (5.6.5).

Let $Y$ be an irreducible locally Noetherian prescheme, $X$ an irreducible prescheme, $f:X\to Y$ a dominant morphism locally of finite type. Let $\xi$ (resp. $\eta$) be the generic point of $X$ (resp. $Y$), and let $e=\dim (f^{-1}(\eta ))=deg.t{r}_{k(\eta )}k(\xi )$ ("dimension of the generic fibre", cf. $(0_I,\mathrm{2.1.8})$ and (4.1.1)). For every $x\in X$, one has then, setting $y=f(x)$,

  e + dim(𝒪_y) ≥ deg.tr_k(y) k(x) + dim(𝒪_x)             (5.6.5.1)

  dim(𝒪_x) ≤ dim(𝒪_y) + dim(𝒪_x ⊗_{𝒪_y} k(y)) − δ(x)    (5.6.5.2)

setting $\delta (x)={\dim }_x(f^{-1}(y))-e$.

Moreover, if $Y$ is universally catenary, the two sides of (5.6.5.1) (resp. (5.6.5.2)) are equal. If moreover $x$ is closed in $f^{-1}(y)$, one has

$$\dim ({\mathcal{O}}_x)=\dim ({\mathcal{O}}_y)+e.(\mathrm{5.6.5.3})$$

One may evidently restrict to the case where $X$ and $Y$ are affine (hence $f$ of finite type) and reduced (I, 5.4), hence integral. As $f$ is dominant, ${\mathcal{O}}_y$ is then identified with a sub-ring of ${\mathcal{O}}_x$, and the respective fields of fractions of ${\mathcal{O}}_y$ and ${\mathcal{O}}_x$ with $k(\eta )$ and $k(\xi )$ (I, 8.2.7); moreover, ${\mathcal{O}}_x$ is a local ring of an integral ${\mathcal{O}}_y$-algebra of finite type $B$ at a prime ideal $\mathfrak{q}$ (I, 3.6.5); applying (5.5.8.1) to $A={\mathcal{O}}_y$, $B$, and $\mathfrak{q}$, one obtains (5.6.5.1). Moreover, ${\mathcal{O}}_x{\otimes }_{{\mathcal{O}}_y}k(y)={\mathcal{O}}_x/{\mathfrak{m}}_y{\mathcal{O}}_x$ is none other than the local ring of the fibre $f^{-1}(y)$ at the point $x$ (I, 3.6.1); as $f^{-1}(y)$ is a prescheme of finite type over $k(y)$, it follows from (5.2.3) that one has

  dim(𝒪_x ⊗_{𝒪_y} k(y)) = dim_x(f⁻¹(y)) − deg.tr_{k(y)} k(x);

the inequality (5.6.5.2) is therefore only another form of (5.6.5.1).

The fact that the two sides of (5.6.5.1) are equal when $Y$ is universally

catenary is none other than the equality (5.6.1.1), applied to $A={\mathcal{O}}_y$ and $B^{\prime }={\mathcal{O}}_x$. To prove that one moreover has (5.6.5.3) when $x$ is closed in $f^{-1}(y)$, it suffices to note that, in general, $deg.t{r}_{k(y)}k(x)$ is the dimension of $\overline{x}\cap f^{-1}(y)$, as follows from (5.2.1) applied to the reduced closed sub-prescheme of $f^{-1}(y)$ having this sub-space as underlying space.

We shall prove further on (13.1.1) that, under the conditions of (5.6.5), one always has $\delta (x)\ge 0$, and (5.6.5.2) therefore in this case makes (5.5.2.1) more precise.

Corollary (5.6.6).

Under the general hypotheses of (5.6.5), one has

$$\dim (X)\le \dim (Y)+e.(\mathrm{5.6.6.1})$$

If one supposes moreover $Y$ universally catenary, then, for the two sides of (5.6.6.1) to be equal, it is necessary and sufficient that one have

  dim(Y) = sup_{y ∈ f(X)} dim(𝒪_y).                      (5.6.6.2)

In particular, the two sides of (5.6.6.1) are equal when $Y$ is locally of finite type over a field.

The inequality (5.6.6.1) indeed follows from (5.6.5.1) applied at a closed point $x$ of $X$, taking (5.1.4.2) and (5.1.11) into account. On the other hand, every non-empty fibre $f^{-1}(y)$ contains a point $x$ closed in this fibre (5.1.11), and if one supposes $Y$ universally catenary, one has at this point the relation (5.6.5.3). Taking the upper bounds of the two sides of (5.6.5.3) as $x$ runs through $X$, and noting that every point $x$ closed in $X$ is a fortiori closed in $f^{-1}(f(x))$, the second assertion of the corollary follows from (5.1.4.1) and (5.1.4.2).

To prove the last assertion, note that there is an affine open neighbourhood $U$ of the generic point of $X$ such that $f|U$ be of finite type, hence $f(U)$ is a constructible part of $Y$, dense in $Y$ (1.8.5), and consequently contains a non-empty open part (hence everywhere dense) of $Y$ $(0_{III},\mathrm{9.2.2})$. The hypothesis that $Y$ is locally of finite type over a field then ensures, on the one hand, that $Y$ is universally catenary (5.6.4), and on the other hand that the two sides of (5.6.6.2) are equal (5.2.2 and 4.1.1.3).

One will note that the condition (5.6.6.2) is trivially verified when $f$ is surjective.

Corollary (5.6.7).

Let $Y$ be a locally Noetherian prescheme, $f:X\to Y$ a morphism locally of finite type, $n$ an integer. If, for every $y\in Y$, $\dim (f^{-1}(y))\le n$, then one has

$$\dim (X)\le \dim (Y)+n.(\mathrm{5.6.7.1})$$

By virtue of (0, 14.1.7) and (1.5.4), one may restrict to the case where $X$ and $Y$ are affine, hence $f$ of finite type, $X$ and $Y$ reduced; let $X_i$ ($1\le i\le m$) be the closed (integral) sub-preschemes of $X$ having as underlying spaces the irreducible components of $X$; one has dim(X) = sup_i (dim(X_i)). If $Z_i$ is the reduced closed sub-prescheme of $Y$ having as underlying space $\overline{f(X_i)}$, $Z_i$ is integral $(0_I,\mathrm{2.1.5})$; the restriction $X_i\to Y$ of $f$ factors as $X_i\stackrel{g_i}{\to }{Z}_i\stackrel{j_i}{\to }Y$, where $j_i$ is the canonical injection (I, 5.2.2), and $g_i$ is dominant

and of finite type (1.5.4). One has $\dim (Z_i)\le \dim (Y)$ for every $i$; on the other hand, if ${\zeta }_i$ is the generic point of $Z_i$, one has $\dim (g_i^{-1}({\zeta }_i))\le n$ for every $i$ by hypothesis; one sees thus that to prove (5.6.7.1), it suffices to show that $\dim (X_i)\le \dim (Z_i)+n$ for every $i$, which follows from (5.6.6.1).

Corollary (5.6.8).

Let $Y$ be a locally Noetherian prescheme, $X$ an irreducible prescheme, $f:X\to Y$ a morphism locally of finite type, $n$ an integer $\ge 0$. Suppose that $Y$ is universally catenary, and that for every $y\in Y$, one has $\dim (f^{-1}(y))\ge n$ (resp. $\dim (f^{-1}(y))=n$). Then one has

$$\dim (X)\ge \dim (Y)+n(\mathrm{5.6.8.1})$$

(resp.

  dim(X) = dim(Y) + n).                                  (5.6.8.2)

As $n\ge 0$, the hypothesis entails that $f$ is surjective, hence $Y$ irreducible; moreover, if $\eta$ is the generic point of $Y$, one has $\dim (f^{-1}(\eta ))\ge n$ (resp. $\dim (f^{-1}(\eta ))=n$). The conclusion therefore follows from (5.6.6).

Remarks (5.6.9).

(i) Even if $Y$ is regular, $X$ irreducible, $f:X\to Y$ dominant and of finite type, the two sides of (5.6.6.1) are not necessarily equal, as shown by the example where $Y=\mathrm{Spec}(A)$ where $A$ is a discrete valuation ring, $X=\mathrm{Spec}(K)$ where $K$ is the field of fractions of $A$, $f$ being the canonical morphism.

(ii) The example (5.4.3, (i)) shows that in (5.6.6) and (5.6.8), one cannot suppress the hypothesis that $X$ is irreducible, the other hypotheses being verified. We shall, however, see (10.6.1) that with supplementary hypotheses on $Y$, verified for example when $Y$ is a prescheme locally of finite type over a field, or over a Dedekind ring having an infinity of prime ideals, such phenomena cannot present themselves.

Proposition (5.6.10).

Let $A$ be an integral universally catenary Noetherian local ring, $B$ an integral ring containing $A$ which is a finite $A$-algebra. Then, for every maximal ideal $\mathfrak{n}$ of $B$, one has $\dim (B_{\mathfrak{n}})=\dim (A)$.

Indeed, one has $deg.t{r}_{A}B=0$ and the residue field $k^{\prime }$ of $B_{\mathfrak{n}}$ is an algebraic extension of the field of fractions of $A$. The conclusion follows from the formula (5.6.1.1), every maximal ideal of $B$ being above that of $A$.

Example (5.6.11).

Let $A$ be an integral Noetherian local ring and integrally closed; then one has seen (0, 16.1.6) that the conclusion of (5.6.10) is valid for every finite integral $A$-algebra $B$ containing $A$. On the contrary, we shall construct an example of an integral catenary Noetherian local ring $A$ for which the conclusion of (5.6.10) will be in default. We shall use the construction of (5.2.5, (i)). Let $k_0$ be a field, $k$ a pure transcendental extension $k_0(S_i{)}_{i\in \mathbb{N}}$ of infinite transcendence degree, $V$ the discrete valuation ring $k[S{]}_{(S)}$, local ring of the polynomial ring k[S] at the prime ideal (S); finally let $E$ be the polynomial ring V[T]; one has seen that in $E$ the maximal ideal $\mathfrak{m}=(S)+(T)$ is of height 2 and the maximal ideal ${\mathfrak{m}}^{\prime }=(ST-1)$ of height 1;

the corresponding residue fields are $k(\mathfrak{m})=k$ and $k({\mathfrak{m}}^{\prime })=k(S)$, field of fractions of $V$; by virtue of the choice of $k$, these fields are isomorphic. Denote by $ϵ$ and ${ϵ}^{\prime }$ the canonical homomorphisms of $E$ onto $E/\mathfrak{m}=k(\mathfrak{m})$ and $E/{\mathfrak{m}}^{\prime }=k({\mathfrak{m}}^{\prime })$ respectively; let on the other hand $\sigma$ be an isomorphism of $k(\mathfrak{m})$ onto $k({\mathfrak{m}}^{\prime })$, and consider the sub-ring $C$ of $E$ formed by $x\in E$ such that ${ϵ}^{\prime }(x)=\sigma (ϵ(x))$ (this construction is a particular case of the "gluing procedures" which will be studied systematically in Chap. V). It is immediate that $\mathfrak{n}=\mathfrak{m}{\mathfrak{m}}^{\prime }=\mathfrak{m}\cap {\mathfrak{m}}^{\prime }$ is a maximal ideal of $C$, $C/\mathfrak{m}{\mathfrak{m}}^{\prime }$ being identified with the sub-field of $(E/\mathfrak{m})\times (E/{\mathfrak{m}}^{\prime })$ formed by pairs $(z,\sigma (z))$. One has $E=C+C(ST-1)$, in other words $E$ is a finite $C$-algebra and is evidently the integral closure of $C$; we shall see in addition that $C$ is Noetherian: this will follow from the following lemma:

Lemma (5.6.11.1).

Let $R$ be a ring, $S$ a sub-ring, $\mathfrak{K}=An{n}_S(R/S)$ the conductor of $S$ in $R$ (largest ideal of $S$ which is also an ideal of $R$).

(i) For every ideal $\mathfrak{J}\subset \mathfrak{K}$ of $R$, there exists a strictly increasing map of the set of ideals $\mathfrak{a}$ of $S$ such that $R\cdot \mathfrak{a}=\mathfrak{J}$ to the set of sub-$(S/\mathfrak{K})$-modules of $\mathfrak{J}/\mathfrak{KJ}$.

(ii) If $S/\mathfrak{K}$ and $R$ are Noetherian and if $R$ is an $S$-module of finite type, then every increasing sequence of ideals of $S$ contained in $\mathfrak{K}$ is stationary.

(i) One has indeed, $\mathfrak{KJ}=\mathfrak{K}(R\cdot \mathfrak{a})=\mathfrak{Ka}\subset \mathfrak{a}$ ($\mathfrak{a}$ being an ideal of $S$), whence the conclusion.

(ii) Let $({\mathfrak{L}}_n)$ be an increasing sequence of ideals of $S$ contained in $\mathfrak{K}$. The sequence of ideals $R\cdot {\mathfrak{L}}_n$ of $R$ is stationary since $R$ is Noetherian; one may therefore suppose that all the ideals $R\cdot {\mathfrak{L}}_n$ are equal to the same ideal $\mathfrak{J}$ of $R$. As $R$ is Noetherian, $\mathfrak{J}/\mathfrak{KJ}$ is an $R$-module of finite type, hence also an $S$-module of finite type, and consequently an $(S/\mathfrak{K})$-module of finite type. But since $S/\mathfrak{K}$ is Noetherian, $\mathfrak{J}/\mathfrak{KJ}$ is an $(S/\mathfrak{K})$-Noetherian module, and the conclusion follows from (i).

We shall apply this lemma taking $R=E$, $S=C$; it is clear that $\mathfrak{n}$ is the conductor of $C$ in $E$; moreover, for every ideal $\mathfrak{a}$ of $C$, $\mathfrak{a}/(\mathfrak{n}\cap \mathfrak{a})$ is isomorphic to $(\mathfrak{a}+\mathfrak{n})/\mathfrak{n}$, hence a sub-$C$-module of $C/\mathfrak{n}$, which is a simple $C$-module; it therefore suffices to show that every ideal $\mathfrak{a}\subset \mathfrak{n}$ of $C$ is of finite type, and this follows from (5.6.11.1).

It follows from (0, 16.1.5) that one has $\dim (C)=\dim (E)=2$. Take $A=C_{\mathfrak{n}}$; if $U=C-\mathfrak{n}$, the ring of fractions $B=U^{-1}E$ is therefore the integral closure of $A$, and is an $A$-module of finite type; moreover, as $\mathfrak{m}$ and ${\mathfrak{m}}^{\prime }$ are the only prime ideals of $E$ containing $\mathfrak{n}$, $B$ is a semi-local ring whose local rings at the two maximal ideals are isomorphic to $E_{\mathfrak{m}}$ and $E_{{\mathfrak{m}}^{\prime }}$ respectively, hence are respectively of dimension 2 and 1. This shows that $\dim (B)=2$, so also $\dim (A)=2$ (0, 16.1.5). As $A$ is an integral local ring, it is necessarily catenary (every prime ideal distinct from 0 and from the maximal ideal being necessarily of height 1); but it does not verify the conclusion of (5.6.10), and a fortiori is not universally catenary.

Note further that $\mathfrak{n}$ is an ideal of height 2 in $C$, and that for every prime ideal $\mathfrak{p}$ of height 1 in $C$, there exists a unique prime ideal ${\mathfrak{p}}^{\prime }$ of $E$ above $\mathfrak{p}$, necessarily of height 1, and such that $C_{\mathfrak{p}}=E_{{\mathfrak{p}}^{\prime }}$. Indeed, there is at least one prime ideal ${\mathfrak{p}}^{\prime }$ of $E$ above $\mathfrak{p}$ and it follows from the Cohen-Seidenberg theorem that such

an ideal is necessarily of height 1; moreover, as $E$ is a finite $C$-algebra and $\mathfrak{p}\ne \mathfrak{n}$, $C_{\mathfrak{p}}$ is integrally closed (Bourbaki, Alg. comm., chap. V, §1, n° 5, cor. 5 of prop. 16), hence $E_{\mathfrak{p}}=C_{\mathfrak{p}}$, which proves our assertions.

It would be interesting to know whether every integral Noetherian local ring verifying the conclusion of (5.6.10) is universally catenary; this is what Nagata [33] affirmed, but his proof does not seem to be complete.

5.7. Depth and property $(S_n)$

Definition (5.7.1).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module. One calls depth (resp. codepth) of $\mathcal{F}$ at a point $x\in X$ the number $prof({\mathcal{F}}_x)$ (resp. $coprof({\mathcal{F}}_x)$) (0, 16.4.5 and 16.4.9). One calls codepth of $\mathcal{F}$ the number

  coprof(ℱ) = sup_{x ∈ X} coprof(ℱ_x).                   (5.7.1.1)

One says that $\mathcal{F}$ is a Cohen-Macaulay ${\mathcal{O}}_X$-Module at $x$ if ${\mathcal{F}}_x$ is a Cohen-Macaulay ${\mathcal{O}}_x$-module, that is to say (0, 16.5.1) if $coprof({\mathcal{F}}_x)=0$. One says that $\mathcal{F}$ is a Cohen-Macaulay ${\mathcal{O}}_X$-Module if it is so at every point, in other words if $coprof(\mathcal{F})=0$. A point $x\in X$ such that ${\mathcal{O}}_x$ is a Cohen-Macaulay ring is also called a Cohen-Macaulay point of $X$.

One calls codepth of $X$ and denotes by $coprof(X)$ the number $coprof({\mathcal{O}}_X)$. One says that $X$ is a Cohen-Macaulay prescheme if ${\mathcal{O}}_X$ is a Cohen-Macaulay ${\mathcal{O}}_X$-Module, in other words if $coprof(X)=0$. Every locally Noetherian prescheme of dimension 0 is evidently a Cohen-Macaulay prescheme. To say that $\mathrm{Spec}(A)$ is a Cohen-Macaulay scheme means that $A$ is a Cohen-Macaulay ring (0, 16.5.13).

Definition (5.7.2).

Let $X$ be a locally Noetherian prescheme, $\mathcal{F}$ a coherent ${\mathcal{O}}_X$-Module, $k$ a positive or negative integer. One says that $\mathcal{F}$ possesses the property $(S_k)$ if, for every $x\in X$, one has

  prof(ℱ_x) ≥ inf(k, dim(ℱ_x)).                          (5.7.2.1)

One says that $\mathcal{F}$ possesses the property $(S_k)$ at a point $x\in X$ if, for every generization $x^{\prime }$ of $x$ in $X$, one has

  prof(ℱ_{x'}) ≥ inf(k, dim(ℱ_{x'})).                    (5.7.2.2)

One says that $X$ verifies the property $(S_k)$ (resp. verifies the property $(S_k)$ at a point $x$) if ${\mathcal{O}}_X$ verifies the property $(S_k)$ (resp. verifies the property $(S_k)$ at the point $x$).

For $\mathcal{F}$ to verify the property $(S_k)$, it is evidently necessary and sufficient that it verify it at every point of $X$. If $U$ is an open of $X$ and if $\mathcal{F}$ verifies $(S_k)$, so does $\mathcal{F}|U$; conversely, if $(U_{\alpha })$ is an open cover of $X$ and if $\mathcal{F}|U_{\alpha }$ verifies $(S_k)$ for every $\alpha$, $\mathcal{F}$ verifies $(S_k)$.

Remarks (5.7.3).

(i) Recall that one always has $prof({\mathcal{F}}_x)\le \dim ({\mathcal{F}}_x)$ (0, 16.4.5.1) if ${\mathcal{F}}_x\ne 0$. To say that $\mathcal{F}$ possesses the property $(S_k)$ therefore means that one has $prof({\mathcal{F}}_x)\ge k$ except at points $x\in X$ such that $\dim ({\mathcal{F}}_x)<k$ and that at these latter points one has $\dim ({\mathcal{F}}_x)=prof({\mathcal{F}}_x)$, that is to say (0, 16.5.1) that $\mathcal{F}$ is a Cohen-Macaulay ${\mathcal{O}}_x$-Module at these points; one will note that at points where $\dim ({\mathcal{F}}_x)=k$, one has $prof({\mathcal{F}}_x)=k$,

hence $\mathcal{F}$ is again a Cohen-Macaulay ${\mathcal{O}}_x$-Module at these points. To say that $\mathcal{F}$ possesses the property $(S_k)$ for every $k$ therefore means that $\mathcal{F}$ is a Cohen-Macaulay ${\mathcal{O}}_X$-Module. It is clear that for $k^{\prime }\ge k$, the property $(S_{k^{\prime }})$ implies $(S_k)$; for $k\le 0$, every coherent ${\mathcal{O}}_X$-Module has the property $(S_k)$.