Exposé III. Cohomological invariants and depth

1. Review

We state a few definitions and results that the reader will find, for example, in Chapter I of the course taught by J.-P. Serre at the Collège de France in 1957–58.¹

Definition.

Let $A$ be a ring (commutative with unit element, as in everything that follows) and let $M$ be an $A$-module (unitary, as in everything that follows). One calls:

• annihilator of $M$, denoted Ann M, the set of $a\in A$ such that $am=0$ for every $m\in M$.
• support of $M$, denoted Supp M, the set of prime ideals $p$ of $A$ such that the localization $M_p$ is nonzero.
• "assassin of $M$" or "set of prime ideals associated with $M$", denoted Ass M, the set of prime ideals $p$ of $A$ such that there exists a nonzero element of $M$ whose annihilator is $p$.

If $a$ is an ideal of $A$, we shall write $r(a)$ for the radical of $a$ in $A$, i.e. the set of elements of $A$ some power of which lies in $a$.

The following results hold under the assumption that $A$ is noetherian and $M$ is finitely generated.

Proposition.

1. Ass M is a finite set.
2. For an element of $A$ to annihilate a nonzero element of $M$, it is necessary and sufficient that it belong to one of the ideals associated with $M$.
3. The radical of the annihilator of $M$, $r(AnnM)$, is the intersection of the ideals associated with $M$ that are minimal (for the inclusion relation in Ass M).

Proposition.

Let $p$ be a prime ideal of $A$. The following assertions are equivalent:

1. $p\in SuppM$.
2. There exists $q\in AssM$ such that $q\subset p$.
3. $p\supset AnnM$.
4. (iii bis) $p\supset r(AnnM)$.

Proposition.

Let $N$ be a finitely generated $A$-module. One has the formula:

Ass Hom_A(N, M) = Supp N ∩ Ass M.

2. Depth

Throughout this paragraph, $A$ denotes a commutative ring, $I$ an ideal of $A$, and $M$, $N$ two $A$-modules. We write $X$ for the prime spectrum of $A$ (we shall not use its structure sheaf in this paragraph) and $Y$ for the variety of $I$, $Y=Supp(A/I)=p\in X,p\supset I$.

Lemma.

Suppose that $A$ is noetherian and that the modules $M$ and $N$ are finitely generated. Suppose moreover that $SuppN=Y$. Then the following assertions are equivalent:

1. ${\mathrm{Hom}}_A(N,M)=0$.
2. $SuppN\cap AssM=\varnothing$.
3. The ideal $I$ is not a zero divisor on $M$, meaning that for every $m\in M$, $Im=0$ implies $m=0$.
4. There exists an $M$-regular element in $I$. (An element $a$ of $A$ is said to be $M$-regular if multiplication by $a$ on $M$ is injective.)
5. For every $p\in Y$, the maximal ideal $p{A}_p$ of the local ring $A_p$ is not associated with $M_p$. In symbols: $p{A}_p\notin Ass{M}_p$.

Proof.

(i) ⇔ (ii), since $Ass{\mathrm{Hom}}_A(N,M)=\varnothing$ is equivalent to (ii) by Proposition 1.3, and to (i) by an easy consequence of Proposition 1.2.

(iii) ⇒ (ii) by contradiction: "there exists $p\in SuppN\cap AssM$" entails that $p\supset I$ and that there exists $m\in M$ whose annihilator is $p$, hence $Im\subset pm=0$, contradicting (iii).

(iv) ⇒ (iii) trivially.

(ii) ⇔ (iv), since $SuppN=Y$, so (ii) says that $I$ is contained in no ideal associated with $M$, or equivalently (since the ideals associated with $M$ are prime and finite in number) that $I$ is not contained in the union of the ideals associated with $M$. But by Proposition 1.1 (ii) this union is exactly the set of elements of $A$ that are not $M$-regular.

(i) ⇒ (v): indeed, if ${\mathrm{Hom}}_A(N,M)=0$ and if $p\in Y$, one deduces, by virtue of the formula

Hom_A(N, M)_p = Hom_{A_p}(N_p, M_p),

that ${\mathrm{Hom}}_{A_p}(N_p,M_p)=0$, hence, thanks to Proposition 1.3,

Supp N_p ∩ Ass M_p = ∅;

but $p{A}_p\in Supp{N}_p$, so $p{A}_p\notin Ass{M}_p$.

(v) ⇒ (i): indeed, if $p\in AssM$, there exists $m\in M$ whose annihilator is $p$, so the canonical image of $m$ in $M_p$ is nonzero, and its annihilator is an ideal containing $p$, hence containing $p{A}_p$, hence equal to $p{A}_p$. The ideal $p{A}_p$ is therefore associated with $M_p$, so $p\notin Y$ by (v), whence (i). QED

We shall now work with these conditions, replacing the functor Hom by its derived functors.

Theorem.

Let $A$ be a commutative ring, $I$ an ideal of $A$, $M$ an $A$-module. Let $n$ be an integer.

a) If there exists a sequence $f_1,\cdots ,f_{n+1}$ of elements of $I$ that is an $M$-regular sequence (i.e. $f_1$ is $M$-regular and $f_{i+1}$ is regular on $M/(f_1,\cdots ,f_i)M$ for $i⩽n$), then for every $A$-module $N$ annihilated by a power of $I$, one has:

Ext^i_A(N, M) = 0 for i ⩽ n.

b) If moreover $A$ is noetherian, $M$ is finitely generated, and there exists a finitely generated $A$-module $N$ such that $SuppN=V(I)$ and $Ex{t}_A^i(N,M)=0$ for $i⩽n$, then there exists a sequence $f_1,\cdots ,f_{n+1}$ of elements of $I$ that is $M$-regular.

Let us prove a) first, by induction. If $n<0$ the statement is empty.

Suppose $n⩾0$, and that a) has been proved for $n^{\prime }<n$. By hypothesis there exists $f_1\in I$ that is $M$-regular. Denote by $f_1^i$ multiplication by $f_1$ on $Ex{t}_A^i(N,M)$, and by $f_1^M$ multiplication by $f_1$ on $M$. The sequence

0 ⟶ M ──f_1^M──→ M ⟶ M/f_1 M ⟶ 0

is exact, hence so is the sequence:

Ext^{i−1}_A(N, M/f_1 M) ──δ──→ Ext^i_A(N, M) ──f_1^i──→ Ext^i_A(N, M).

By hypothesis $I^{r}N=0$, so $f_1^0$ is nilpotent; since $Ex{t}^i$ is a universal functor, the same holds for $f_1^i$ for every $i$. On the other hand, there is a regular sequence in $M/f_{1}M$ of length $n$, so by the induction hypothesis,

Ext^{i−1}_A(N, M) = 0 if i ⩽ n − 1.

One deduces that if $i⩽n$, then $f_1^i$ is at once nilpotent and injective, hence $Ex{t}_A^i(N,M)=0$.

Let us prove b), also by induction. If $n<0$, the statement is empty.

If $n=0$, b) follows from the implication (i) ⇒ (iv) of Lemma 2.1.

If $n>0$, by b) for $n=0$ there exists an element $f_1\in I$ that is $M$-regular; from the exact sequence (2.1) one deduces the exact sequence:

Ext^{i−1}_A(N, M) ⟶ Ext^{i−1}_A(N, M/f_1 M) ⟶ Ext^i_A(N, M).

One concludes that the hypotheses of b) are satisfied for the module $M/f_{1}M$ and the integer $n-1$. By the induction hypothesis, there exists a sequence of $n$ elements of $I$ that is $M/f_{1}M$-regular, which entails that there exists a sequence of $n+1$ elements of $I$, beginning with $f_1$, that is $M$-regular.

This theorem invites us to generalize, as follows, the classical definition of the depth of a finitely generated module over a noetherian ring:

Definition.

Let $A$ be a commutative ring with unit, $M$ an $A$-module, $I$ an ideal of $A$. One calls the $I$-depth of $M$, and denotes $pro{f}_{I}M$, the supremum in $\mathbb{N}\cup +\infty$ of the set of natural integers $n$ such that for every finitely generated $A$-module $N$ annihilated by a power of $I$, one has

Ext^i_A(N, M) = 0 for every i < n.

One deduces from the previous theorem that if $n$ is the supremum of the lengths of $M$-regular sequences of elements of $I$, one has $n⩽pro{f}_{I}M$. More precisely:

Proposition.

Let $A$ be a commutative ring, $I$ an ideal of $A$, $M$ an $A$-module, and $n\in \mathbb{N}$. Consider the assertions:

1. $n⩽pro{f}_{I}M$.

2. For every finitely generated $A$-module $N$ annihilated by a power of $I$, one has:

   Ext^i_A(N, M) = 0 for i < n.
   

3. There exists a finitely generated $A$-module $N$ such that $SuppN=V(I)$ and $Ex{t}_A^i(N,M)=0$ for $i<n$.

4. There exists an $M$-regular sequence of length $n$ formed of elements of $I$.

The following logical implications hold:

(1) ⇐⇒ (2)
        ⇓
       (3) ⇐= (4)

Moreover, if $A$ is noetherian and $M$ is finitely generated, these conditions are equivalent.

Proof.

(1) ⇔ (2) by definition, and (2) ⇒ (3) by taking $N=A/I$. Moreover (4) ⇒ (2) by Theorem 2.2 a). Finally, if $A$ is noetherian and $M$ is finitely generated, (3) ⇒ (4) by Theorem 2.2 b).

We assume $A$ noetherian and $M$ finitely generated until the end of this paragraph.

Corollary.

Let $f\in I$ be an $M$-regular element. One has:

prof_I M = prof_I(M/f M) + 1.

Indeed, if $n⩽pro{f}_I(M/fM)$, there exists a sequence of elements of $I$, $f_1,\cdots ,f_n$, that is $(M/fM)$-regular, so the sequence $f,f_1,\cdots ,f_n$ is $M$-regular, hence $n+1⩽pro{f}_{I}M$, hence $pro{f}_{I}M⩾pro{f}_I(M/fM)+1$. On the other hand, by the exact sequence (2.2), if $i⩽pro{f}_{I}M$ one has $Ex{t}_A^{i-1}(N,M/fM)=0$, hence $pro{f}_{I}M-1⩽pro{f}_I(M/fM)$.

Corollary.

Every finite $M$-regular sequence formed of elements of $I$ may be extended to a maximal $M$-regular sequence, whose length is necessarily equal to the $I$-depth of $M$.

Remark.

One can scarcely keep oneself from saying that an $A$-module is the more beautiful the greater its depth. A module whose support does not meet $V(I)$ is among the most beautiful; indeed, one can show that for $pro{f}_{I}M$ to be finite, it is necessary and sufficient that $SuppM\cap V(I)\ne \varnothing$.

Remark.

If $A$ is a semi-local ring, let $r(A)$ be its radical and $k=A/r(A)$ its residue ring. The interesting notion of depth is obtained by taking for $I$ the radical of $A$. We shall therefore agree to write simply prof M for the $r(A)$-depth of an $A$-module $M$. One recovers in this case the notion of homological codimension (cf. Serre, op. cit. note ¹, p. 21), denoted $cod{h}_{A}M$, defined as the infimum of integers $i$ such that $Ex{t}_A^i(k,M)\ne 0$; indeed $Suppk=V(r(A))$.

Proposition.

If $A$ is noetherian and $M$ is finitely generated, one has:

prof_I M = inf_{p ∈ V(I)} prof M_p.

Corollary.

If $A$ is a noetherian semi-local ring and $M$ is a finitely generated $A$-module, one has:

prof M = inf_m prof M_m,

where $m$ ranges over the maximal ideals of $A$.

The corollary follows at once from Proposition 2.9; indeed, the prime ideals that contain the radical are precisely the maximal ideals.

Moreover, let $f\in I$. If $f$ is $M$-regular, if $p\in X$ and $p\supset I$, then the image $g$ of $f$ in $A_p$ lies in $p{A}_p$, the maximal ideal of $A_p$; and $g$ is $M_p$-regular, as follows from the exact sequence

0 ⟶ M_p ──g′──→ M_p ⟶ (M/f M)_p ⟶ 0,

where $g^{\prime }$ denotes multiplication by $g$ on $M_p$. This exact sequence also gives that $(M/fM{)}_p$ is isomorphic to $M_p/g{M}_p$; applying Corollary 2.5 to $M$ and to $M_p$, one deduces, by induction, that $pro{f}_{I}M⩽\nu (M)$, where one has set, for every $M$:

ν(M) = inf_{p ∈ V(I)} prof M_p.

More precisely, still by induction, one knows that if $f$ is $M$-regular, then $\nu (M)=\nu (M/fM)+1$; it remains therefore to show that if $\nu (M)⩾1$, there exists an $M$-regular element in $I$. But applying Lemma 2.1 to $M_p$, $A_p$, and $p{A}_p$ for each $p\in V(I)$, one sees that $p{A}_p\notin Ass{M}_p$; hence, applying Lemma 2.1 to $A$, $M$, and $I$, the conclusion follows.

Proposition.

Let $u:A\to B$ be a homomorphism of noetherian rings. Let $I$ be an ideal of $A$, $M$ a finitely generated $A$-module. Set $IB=I{\otimes }_{A}B$ and $M_B=M{\otimes }_{A}B$. If $B$ is $A$-flat, one has:

prof_{IB} M_B ⩾ prof_I M;

moreover, if $B$ is faithfully flat over $A$, one has equality.

Indeed, let $N=A/I$. By flatness, $N{\otimes }_{A}B=B/IB$; set $N_B=N{\otimes }_{A}B$. Again by flatness and the noetherian hypotheses, one has:

Ext^i_B(N_B, M_B) = Ext^i_A(N, M) ⊗_A B,

so $Ex{t}_A^i(N,M)=0$ implies $Ex{t}_B^i(N_B,M_B)=0$, and the converse holds if $B$ is faithfully flat over $A$.

3. Depth and topological properties

Lemma.

Let $X$ be a topological space, $Y$ a closed subspace, $F$ a sheaf of abelian groups on $X$. Set $U=X-Y$. Let $n$ be an integer. The following conditions are equivalent:

1. $H_Y^i(X,F)=0$ for $i<n$.

2. For every open $V$ of $X$, the homomorphism of groups

   H^i(V, F) ⟶ H^i(V ∩ U, F)
   

   is bijective for $i<n-1$ and injective for $i=n-1$.

3. For every open $V$ of $X$,

   H^i_{Y ∩ V}(V, F|V) = 0 for i < n.
   

Proof.

(ii) ⇔ (iii): indeed, let $V$ be an open of $X$, set $X^{\prime }=V$, $Y^{\prime }=Y\cap V$, $F^{\prime }=F|V$, $U^{\prime }=X^{\prime }-Y^{\prime }$. Then $Y^{\prime }$ is closed in $X^{\prime }$, so one has an exact sequence:

H^i_{Y′}(X′, F′) ⟶ H^i(X′, F′) ──ρ^i──→ H^i(U′, F′) ⟶ H^{i+1}_{Y′}(X′, F′).

If the outer terms vanish, the homomorphism ${\rho }^i$ is bijective; and if the left-hand term vanishes, ${\rho }^i$ is injective. So (iii) ⇒ (ii). Conversely, if $i<n$, then $H_{Y^{\prime }}^i(X^{\prime },F^{\prime })$ vanishes because ${\rho }^i$ is injective and ${\rho }^{i-1}$ is surjective.

(i) ⇒ (iii): indeed, the "local-to-global" spectral sequence gives:

H^p(X, ℋ^q_Y(X, F)) ⟹ H^*_Y(X, F).

Now, by hypothesis ${\mathcal{H}}_Y^q(X,F)=0$ for $q<n$, hence $H_Y^{p+q}(X,F)=0$ for $p+q<n$.

(iii) ⇒ (i): indeed, (iii) says that the presheaf

V ⟼ H^i_{Y ∩ V}(V, F|V)

is zero, hence so is the associated sheaf, which is ${\mathcal{H}}_Y^i(X,F)$, since $Y$ is closed.

Remark.

The equivalence of (i) and (ii) was proved in Proposition I 2.13. As remarked there, if $n⩾2$, one may omit the condition that ${\rho }^{n-1}$ be injective.²

Proposition.

Let $X$ be a locally noetherian prescheme, $Y$ a closed subprescheme of $X$, and $F$ a coherent ${\mathcal{O}}_X$-module. The conditions of Lemma 3.1 are equivalent to each of the following:

1. (iv) For every $x\in Y$, one has $prof{F}_x⩾n$.

2. (v) For every coherent ${\mathcal{O}}_X$-module $G$ on $X$ whose support is contained in $Y$, one has

   Ext^i_{𝒪_X}(G, F) = 0 for i < n;
   

3. (vi) There exists a coherent ${\mathcal{O}}_X$-module $G$ whose support is equal to $Y$ such that

   Ext^i_{𝒪_X}(G, F) = 0 for i < n.
   

If $X$ is affine, all the work has been done (cf. Proposition 2.4) to establish the equivalence of the three conditions of Proposition 3.3. These conditions are local, except for the implication (v) ⇒ (vi); but in that case one may take $G={\mathcal{O}}_Y$ and invoke Proposition 2.4 again. It therefore suffices to prove (i) ⇒ (vi) and (v) ⇒ (i).

Let $J$ be the ideal of $Y$: it is a coherent sheaf of ideals. Set ${\mathcal{O}}_Y^m={\mathcal{O}}_X/J^{m+1}$: this is a coherent ${\mathcal{O}}_X$-module whose support is equal to $Y$, and one knows (Theorem II 6.b) that

ℋ^i_Y(X, F) = lim_{→ m} Ext^i_{𝒪_X}(𝒪_Y^m, F),

so (v) ⇒ (i). Moreover, the transition morphisms in the projective system of the ${\mathcal{O}}_Y^m$ are epimorphisms.

If the functor $Ex{t}^i$ is left exact in its first argument — at least when this argument is taken in the category of coherent ${\mathcal{O}}_X$-modules with support contained in $Y$ — then the transition morphisms of the inductive system obtained by applying $Ex{t}^i$ to the ${\mathcal{O}}_Y^m$ will be injective; but (i) entails that the limit is zero, so (i) will entail that the modules $Ex{t}_{{\mathcal{O}}_X}^i({\mathcal{O}}_Y^m,F)$ are zero for every $m$. We argue by induction. The statement is trivial for $n<0$. Suppose (i) ⇒ (vi) for $n<q$, then (i) ⇒ (v), so, by the long exact sequence of Ext, $Ex{t}^q$ is left exact in its first argument, hence the modules $Ex{t}_{{\mathcal{O}}_X}^q({\mathcal{O}}_Y^m,F)$ are zero for every $m$. So (i) ⇒ (vi) for $n⩽q$. QED

Example.

Let $A$ be a noetherian local ring, $m$ its maximal ideal, $M$ a finitely generated $A$-module, and $n$ an integer. Set $X=\mathrm{Spec}(A)$, $Y=m$, $U=X-Y$. Let $F$ be the sheaf associated with $M$. The following conditions are equivalent:

1. $profM⩾n$.

2. The natural homomorphism

   H^i(X, F) ⟶ H^i(U, F)
   

   is injective for $i=n-1$ and bijective for $i<n-1$.

3. $Ex{t}_A^i(k,M)=0$ for $i<n$, where $k=A/m$.

4. $H_Y^i(X,F)=0$ for $i<n$.

Taking Remark 3.2 into account, one obtains:

Corollary.

Let $X$ be a locally noetherian prescheme, $Y$ a closed subprescheme of $X$, $F$ a coherent ${\mathcal{O}}_X$-module. The following conditions are equivalent:

1. For every $x\in Y$, $prof{F}_x⩾2$.

2. For every open $V$ of $X$, the natural homomorphism

   H^0(V, F) ⟶ H^0(V ∩ (X − Y), F)
   

   is bijective.

Theorem (Hartshorne).

Let $X$ be a locally noetherian prescheme, $Y$ a closed subprescheme of $X$. Suppose that, for every $x\in Y$, $prof{\mathcal{O}}_{X,x}⩾2$; then the natural map

$${\pi }_0(X)⟶{\pi }_0(X-Y)$$

is bijective.

Proof. Since $X$ is locally noetherian, $X$ is locally connected; it therefore suffices to prove that for $X$ to be connected it is necessary and sufficient that $X-Y$ be. Now, for a ringed space in local rings $(X,{\mathcal{O}}_X)$ to be connected, it is necessary and sufficient that $H^0(X,{\mathcal{O}}_X)$ not be a direct product of two nonzero rings. But the hypothesis implies, by Corollary 3.5 applied to $F={\mathcal{O}}_X$, that the homomorphism

H^0(X, 𝒪_X) ⟶ H^0(X − Y, 𝒪_X)

is an isomorphism, whence the conclusion.

Corollary.

Let $X$ be a locally noetherian prescheme. Let $d$ be an integer such that $\dim {\mathcal{O}}_{X,x}⩾d$ implies $prof{\mathcal{O}}_{X,x}⩾2$. Then, if $X$ is connected, $X$ is connected in codimension $d-1$, i.e. if $X^{\prime }$ and $X^{\prime \prime }$ are two irreducible components of $X$, there exists a sequence of irreducible components of $X$:

X′ = X_0, X_1, …, X_n = X″

such that for every $i$ with $0⩽i<n$, the codimension of $X_i\cap X_{i+1}$ in $X$ is at most $d-1$.

Note first that if $X$ is Cohen-Macaulay, then $d=2$ enjoys the property invoked above. In this connection, recall that one defines the codimension of $Y$ in $X$ as the infimum of the dimensions of the local rings in $X$ at the points of $Y$.

Proof. Let $\mathcal{F}$ be the collection of closed subsets of $X$ whose codimension in $X$ is at least $d$. One notes that $\mathcal{F}$ is an antifilter of closed subsets of $X$. Moreover, for a closed $Y\subset X$ to belong to $\mathcal{F}$, it is necessary and sufficient that, for every $y\in Y$, there exist an open neighborhood $V$ of $y$ in $X$ such that $Y\cap V$ has codimension $⩾d$ in $V$. Finally, if $X$ is connected and $Y\in \mathcal{F}$, then $X-Y$ is connected by Hartshorne's theorem. The corollary thus follows from the next lemma, which is of a purely topological nature.

Lemma.

Let $X$ be a connected, locally noetherian topological space, and let $\mathcal{F}$ be an antifilter of closed subsets of $X$. Suppose that every closed $Y\subset X$ that locally belongs to $\mathcal{F}$ (i.e. for every point $x\in X$ there exist an open neighborhood $V$ of $x$ in $X$ and a $Y^{\prime }\in \mathcal{F}$ such that $V\cap Y=V\cap Y^{\prime }$) belongs to $\mathcal{F}$. The following conditions are equivalent:

1. For every $Y\in \mathcal{F}$, $X-Y$ is connected.
2. If $X^{\prime }$ and $X^{\prime \prime }$ are two distinct irreducible components of $X$, there exists a sequence of irreducible components of $X$, $X_0,X_1,\cdots ,X_n$, such that $X^{\prime }=X_0$, $X^{\prime \prime }=X_n$ and, for each $i$ with $1⩽i<n$, $X_i\cap X_{i+1}\notin \mathcal{F}$.

(ii) ⇒ (i). Let $Y\in \mathcal{F}$; we must show that the open set $U=X-Y$ is connected. Now, if $U^{\prime }$ and $U^{\prime \prime }$ are two irreducible components of $U$, there exist two irreducible components $X^{\prime }$ and $X^{\prime \prime }$ of $X$ such that $X^{\prime }\cap U=U^{\prime }$ and $X^{\prime \prime }\cap U=U^{\prime \prime }$. Let $X_0,\cdots ,X_n$ be a sequence of irreducible components of $X$ having the property invoked above; if one sets $U_i=X_i\cap U$ for $0⩽i⩽n$, the $U_i$ are irreducible components of $U$, and moreover $U_i\cap U_{i+1}$ is nonempty for $0⩽i<n$, since otherwise $X_i\cap X_{i+1}\subset Y$ would be an element of $\mathcal{F}$, contrary to the choice of the sequence of the $X_i$. This entails that $U$ is connected.

(i) ⇒ (ii). Let $Y={\bigcup }_{X^{\prime },X^{\prime \prime }}{X}^{\prime }\cap X^{\prime \prime }$, where one requires $X^{\prime }$ and $X^{\prime \prime }$ to be two distinct irreducible components of $X$ such that $X^{\prime }\cap X^{\prime \prime }\in \mathcal{F}$. The family of the $X^{\prime }\cap X^{\prime \prime }$ is locally finite since $X$ is locally noetherian; moreover the $X^{\prime }\cap X^{\prime \prime }$ are closed, so $Y$ is closed. Also, $Y$ belongs locally to $\mathcal{F}$, so $Y\in \mathcal{F}$. Hence $U=X-Y$ is connected. Let $X^{\prime }$ and $X^{\prime \prime }$ be two distinct irreducible components of $X$, and let $U^{\prime }$ and $U^{\prime \prime }$ be their traces on $U$, which are nonempty by construction of $Y$. These are irreducible components of $U$; but $U$ is connected, so, $U$ being locally noetherian, there exists a sequence of irreducible components $U_0,\cdots ,U_n$ of $U$ such that $U_0=U^{\prime }$, $U_n=U^{\prime \prime }$, $U_i\cap U_{i+1}\ne \varnothing$, and $U_i\cap U_{i+1}\ne U_i$ for $0⩽i<n$. Let $X_0,\cdots ,X_n$ be the sequence of irreducible components of $X$ such that $X_i\cap U=U_i$; if $X_i\cap X_{i+1}\in \mathcal{F}$, then by the construction of $\mathcal{F}$ one would have $U_i\cap U_{i+1}=\varnothing$ or $U_i=U_{i+1}$, which is impossible by the choice of the $U_i$. QED

Corollary.

Let $A$ be a noetherian local ring. Suppose that for every prime ideal $p$ of $A$ one has:

(dim A_p ⩾ 2) ⟹ (prof A_p ⩾ 2).

Suppose moreover that $A$ satisfies the chain condition.³ Then, for every minimal prime ideal $p$ of $A$, $\dim A/p=\dim A$, or equivalently, all the irreducible components of $\mathrm{Spec}A$ have the same dimension: that of $A$.

If $X^{\prime }$ and $X^{\prime \prime }$ are two irreducible components of $X$, one joins them by a chain having the properties enumerated in Corollary 3.7; it then suffices to show that two successive components have the same dimension, which follows from the second hypothesis.

Example.

Let $X$ be the union of two complementary linear subspaces, of respective dimensions 2 and 3, in a vector space of dimension 5; more precisely, let $X=\mathrm{Spec}A$, with $A=B/p\cap q$, where $B=k[X_1,\cdots ,X_5]$, $p$ is the ideal generated by $X_1,X_2,X_3$ and $q$ the ideal generated by X_4 and X_5. Then $X$ can be disconnected by the intersection point $x$ of the two linear subspaces, so the depth of ${\mathcal{O}}_{X,x}$ is equal to 1, since it cannot be $⩾2$ by Theorem 3.6. Another reason: the equidimensionality conclusion of the previous corollary fails.

More generally, taking a union $X$ of two linear subspaces of dimensions $p,q⩾2$ in a vector space of dimension $p+q$, for no embedding of $X$ in a regular scheme is $X$ even set-theoretically a complete intersection at the origin: for (possibly modifying it without changing the underlying topological space in a neighborhood of the origin), $X$ would be Cohen-Macaulay, hence of depth $⩾2$ at the origin, which is not the case.

Remark.

Let $X$ be a locally noetherian prescheme, $Y$ a closed subprescheme of $X$, $F$ an ${\mathcal{O}}_X$-module. Depth is a purely topological notion, expressed in terms of the vanishing of the $H_Y^i(X,F)$ for $i<n$. One also wishes to study these sheaves for a given $i$, or for $i>n$. In this connection one proves the following result:

Lemma.

Let $m$ be an integer. For $H_Y^i(X,F)=0$, $i>m$, to hold for every coherent $F$, it is necessary and sufficient that it hold for $F={\mathcal{O}}_X$.

By inductive limit, the property then holds for every quasi-coherent sheaf. For instance, if $Y$ can be described locally by $m$ equations, or, as one says, if $Y$ is locally set-theoretically a complete intersection (which occurs for example if $X$ and $Y$ are non-singular), it follows from the computation of the $H_Y^i(X,F)$ by the Koszul complex that these sheaves are zero for $i>m$. We have, moreover, used this fact implicitly in Example 3.10.

This cohomological condition is, however, not sufficient, as the next example shows.

Example.

Let $X=\mathrm{Spec}(A)$, where $A$ is a normal noetherian local ring of dimension 2. Let $Y$ be a curve in $X$. One can show that the complement of the curve is an affine open, so⁴ $H_Y^i({\mathcal{O}}_X)\cong H_{X-Y}^{i-1}({\mathcal{O}}_X)=0$ for $i>1$, since $H^{i-1}(X-Y,{\mathcal{O}}_X)=0$. Nevertheless, one can construct a curve that is not described by a single equation.

We shall seek⁵ conditions for the $H_Y^i(X,F)$ to be coherent for a given $i$, which is not the case in general, as obvious examples show — for instance $H_m^n(A)$ for $A$ a noetherian local ring of dimension $n>0$; for example when $A$ is a discrete valuation ring with field of fractions $K$, one finds $H_m^1(A)\simeq K/A$, which is not a finitely generated module over $A$. To enlighten the reader, let us say that the problem posed is equivalent to the following: let $f:U\to X$ be an open immersion, let $G$ be a coherent sheaf on $U$; find criteria for the higher direct images $R^i{f}_{\ast }G$ to be coherent sheaves on $X$ for a given $i$. These conditions are necessary for the use of formal geometry that we saw in Exposé IX and the following ones.