Exposé V. Local duality and the structure of the H^i(M)

1. Complexes of homomorphisms

1.1. Let $F\bullet$ and $G\bullet$ be two graded modules; then one writes

$$\mathrm{Hom}\bullet (F\bullet ,G\bullet )$$

for the graded module of homomorphisms of graded modules from $F\bullet$ into $G\bullet$. Thus one has

Homˢ(F•, G•) = ∏ₖ Hom(Fₖ, Gₖ₊ₛ).

Let $F\bullet$ (resp. $G\bullet$) be a complex, and let $d_1$ (resp. $d_2$) be its differential; then for $h\in {\mathrm{Hom}}^s(F\bullet ,G\bullet )$ one sets¹

d(h) = h ∘ d₁ + (−1)^{s+1} d₂ ∘ h.

One verifies trivially that $d\circ d=0$, hence that $\mathrm{Hom}\bullet (F\bullet ,G\bullet )$ equipped with $d$ is a complex. The cohomology group of this complex is written

$$H\bullet (F\bullet ,G\bullet ).$$

If $G\bullet$ is injective in each degree, then

$$F\bullet \mapsto H\bullet (F\bullet ,G\bullet )$$

is an exact ∂-functor. Likewise, for arbitrary $F\bullet$,

$$G\bullet \mapsto H\bullet (F\bullet ,G\bullet )$$

is an exact δ-functor on the category of complexes $G\bullet$ that are injective in each degree.

Remark 1.2.

The cycles of $\mathrm{Hom}\bullet (F\bullet ,G\bullet )$ are the homomorphisms from $F\bullet$ into $G\bullet$ that commute or anticommute with the differentials, according to degree. The boundaries of $\mathrm{Hom}\bullet (F\bullet ,G\bullet )$ are the homomorphisms from $F\bullet$ into $G\bullet$ that are homotopic to zero.

Let $A$ be a ring, let $M$ (resp. $N$) be an $A$-module, and let $R(M)$ (resp. $R(N)$) be an injective resolution of $M$ (resp. $N$). Then there exists a canonical isomorphism²

Hˢ(R(M), R(N)) ≅ Extˢ(M, N).

Indeed, let $i:M\to R(M)$ be the canonical augmentation, and let $h\in {\mathrm{Hom}}^s(R(M),R(N))$; one writes $t_s$ for the map

$$h\mapsto h^0\circ i$$

from ${\mathrm{Hom}}^s(R(M),R(N))$ into $\mathrm{Hom}(M,R(N{)}^s)$. The family $(t_s{)}_{s⩾0}$ defines a homomorphism of (ordinary) complexes³

t: Hom•(R(M), R(N)) → Hom•(M, R(N)),

i.e. one has $(dh{)}^0\circ i=d_2\circ h^0\circ i$.

One verifies easily that, upon passing to cohomology, $t$ gives an isomorphism. In particular, it follows that

$$H\bullet (R(M),R(N))$$

does not "depend" on the chosen injective resolution $R(M)$ (resp. $R(N)$) of $M$ (resp. $N$).

To every exact sequence of $A$-modules

$$0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$$

one associates an exact sequence of injective resolutions

$$0\to R(M^{\prime })\to R(M)\to R(M^{\prime \prime })\to 0.$$

One verifies that the isomorphism (1.3) commutes with the homomorphisms

Hˢ(R(M′), R(N)) → Hˢ⁺¹(R(M″), R(N)),

Extˢ(R(M′), R(N)) → Extˢ⁺¹(R(M″), R(N)),

deduced from (6) and (5).

Let $P$ be a third $A$-module, and let $R(P)$ be an injective resolution of $P$; then composition of graded morphisms gives a pairing

Homⁱ(R(N), R(M)) × Homʲ(R(M), R(P)) → Homⁱ⁺ʲ(R(N), R(P)),

which defines a pairing

Hⁱ(R(N), R(M)) × Hʲ(R(M), R(P)) → Hⁱ⁺ʲ(R(N), R(P)),

hence a homomorphism of functors in $M$:

Hⁱ(R(N), R(M)) → Hom(Hʲ(R(M), R(P)), Hⁱ⁺ʲ(R(N), R(P))).

We shall see that (1.4) is a homomorphism of δ-functors in $M$. The exact sequences (5) and (6) give a commutative diagram:

Homⁱ(R(N), R(M′))   ──→   Hom(Homʲ(R(M′), R(P)), Homⁱ⁺ʲ(R(N), R(P)))
       │                                       ↑
       │                                  Hom(q, id)
       ↓                                       │
Homⁱ(R(N), R(M))    ──→   Hom(Homʲ(R(M), R(P)), Homⁱ⁺ʲ(R(N), R(P)))
       │
       │ p
       ↓                                       │
Homⁱ(R(N), R(M″))   ──→   Hom(Homʲ(R(M″), R(P)), Homⁱ⁺ʲ(R(N), R(P))).

Let $h\in {\mathrm{Hom}}^i(R(N),R(M^{\prime \prime }))$ (resp. $g\in {\mathrm{Hom}}^j(R(M^{\prime }),R(P))$) be a cycle, and let $h^{\prime }\in {\mathrm{Hom}}^i(R(N),R(M))$ (resp. $g^{\prime }\in {\mathrm{Hom}}^j(R(M),R(P))$) be such that $p(h^{\prime })=h$ (resp. $q(g^{\prime })=g$); then to say that (1.4) is a homomorphism of δ-functors in $M$ is to say that

g ∘ dh′ − dg′ ∘ h

is a coboundary in $\mathrm{Hom}\bullet (R(N),R(P))$.

Now one has

dh′ = h′ ∘ d₁ + (−1)^{i+1} d₂ ∘ h′,
dg′ = g′ ∘ d₂ + (−1)^{j+1} d₃ ∘ g′,

with the obvious notations. Hence (12) is written

g ∘ h′ ∘ d₁ + (−1)^{i+1} g ∘ d₂ ∘ h′ − g′ ∘ d₂ ∘ h − (−1)^{j+1} d₃ ∘ g′ ∘ h.

On the other hand, since $h$ and $g$ are cycles, one has

g ∘ d₂ = (−1)ʲ d₃ ∘ g,
d₂ ∘ h = (−1)ⁱ h ∘ d₁,

hence, finally, (12) is written

d(g ∘ h′ + (−1)^{i+1} g′ ∘ h),

which completes the proof.

Thus (1.3) and (1.4) give a homomorphism of δ-functors in $M$:

Extⁱ(N, M) → Hom(Extʲ(M, P), Extⁱ⁺ʲ(N, P)).

2. The local duality theorem for a regular local ring

Let $A$ be a regular local ring of dimension $r$, let $\mathfrak{m}$ be the maximal ideal of $A$, and let $M$ be a finitely generated $A$-module. One sets $H^i(M)=H_{\mathfrak{m}}^i(M)$ (hence $H^i(M)=\lim \to Ex{t}^i(A/{\mathfrak{m}}^k,M)$). One has seen (IV 5.4) that $I=H^r(A)$ is a dualizing module for $A$; denote by $D$ the associated dualizing functor. In (1.5) setting $N=A/{\mathfrak{m}}^k$, $P=A$, one obtains a homomorphism of δ-functors in $M$

φₖ: Extⁱ(A/𝔪ᵏ, M) → Hom(Extʳ⁻ⁱ(M, A), Extʳ(A/𝔪ᵏ, A)).

Passing to the direct limit over $k$, one finds a homomorphism of δ-functors

φ: Hⁱ(M) → D(Extʳ⁻ⁱ(M, A)).

Theorem 2.1 (Local duality theorem).

The functorial homomorphism $\varphi$ above is an isomorphism.

Proof. If $i>r$, the right-hand side of (14) is trivially zero, and the left-hand side is zero because $H^i(M)=\lim {\to }_{k}Ex{t}^i(A/{\mathfrak{m}}^k,M)$, and this holds for each $Ex{t}^i(A/{\mathfrak{m}}^k,M)$ (syzygy theorem).

If $i=r$, by what precedes, the two functors in $M$, $H^r(M)$ and $D(\mathrm{Hom}(M,A))$, are right exact; since $A$ is noetherian and $M$ is finitely generated, it suffices to verify the isomorphism for $M=A$, which is immediate.

To show that $\varphi$ is a functorial isomorphism, it now suffices, proceeding by descending induction on $i$, to remark that every finitely generated module admits a finite presentation, and that for $i<r$ the two sides of (14) are zero when $M$ is finitely generated free. This is evident for the right-hand side, and since $H^i$ commutes with finite sums it suffices, as for the left-hand side, to show that $H^i(A)=0$ for $i<r$. But this follows, since $prof(A)=r$, from (III 3.4).

3. Application to the structure of the H^i(M)

Theorem 3.1.

Let $A$ be a noetherian local ring, $D$ a dualizing functor for $A$, and $M$ a finitely generated $A$-module with $M\ne 0$, of dimension $n$. Then one has:

(i) $H^i(M)=0$ if $i<0$ or if $i>n$.

(ii) D(Hⁱ(M))^ is a finitely generated module over Â, of dimension $⩽i$.

(iii) $H^n(M)\ne 0$, and if $A$ is complete, $D(H^n(M))$ is of dimension $n$ and

Ass(D(Hⁿ(M))) = {𝔭 ∈ Ass(M) | dim A/𝔭 = n}.

Proof. Let $I$ be the dualizing module associated to $D$. One knows that Î is a dualizing module for Â. On the other hand, one has

Hⁱ(M)^ = Hⁱ(M̂),
D(Hⁱ(M))^ = Hom(Hⁱ(M̂), Î), and
dim M̂ = dim M;

hence one may suppose $A$ complete. Now, by a theorem of Cohen, every complete local ring is a quotient of a regular local ring. To reduce to that case, one needs the following lemma:

Lemma 3.2.

Let $X$ (resp. $Y$) be a ringed space, let $X^{\prime }$ (resp. $Y^{\prime }$) be a closed subspace of $X$ (resp. $Y$), and let $f:X\to Y$ be a morphism of ringed spaces such that $f^{-1}(Y^{\prime })=X^{\prime }$. Let $F$ be an ${\mathcal{O}}_X$-Module, and denote by $A$ (resp. $B$) the ring $\Gamma ({\mathcal{O}}_X)$ (resp. $\Gamma ({\mathcal{O}}_Y)$), and by $f:B\to A$ the ring homomorphism corresponding to $f$. There exists a spectral sequence of $B$-modules, with initial term

$$E_2^{p,q}=H_{Y^{\prime }}^p(Y,R^g{f}_{\ast }(F)),$$

abutting to the $B$-module $H{\bullet }_{X^{\prime }}(X,F{)}_f$.

Proof. Let ${\mathcal{O}}_{Y,Y^{\prime }}$ be the sheaf ${\mathcal{O}}_Y|Y^{\prime }$ extended by 0 outside $Y^{\prime }$ (see Exp. I). One has an isomorphism of $B$-modules:

Hom(𝒪_{Y,Y′}, f_*(F)) ≅ Hom(f*(𝒪_{Y,Y′}), F)_[f].

Now one has

$$f\ast ({\mathcal{O}}_{Y,Y^{\prime }})={\mathcal{O}}_{X,X^{\prime }},$$

and moreover if $G$ is an injective ${\mathcal{O}}_X$-Module, then $f_{\ast }(G)$ is an injective ${\mathcal{O}}_Y$-Module, at least if $f$ is flat — a case to which one easily reduces by replacing ${\mathcal{O}}_X$, etc., by the constant sheaves of rings $\mathbb{Z}$. Hence the spectral sequence of the composite functor

$$F\mapsto \mathrm{Hom}({\mathcal{O}}_{Y,Y^{\prime }},f_{\ast }(F)),$$

with initial term

E₂^{p,q} = Extᵖ(Y; 𝒪_{Y,Y′}, Rᵍf_*(F)),

abuts, taking into account (16) and (17), to

$$Ext\bullet (X;{\mathcal{O}}_{X,X^{\prime }},F{)}_f.$$

The lemma then follows from (I 13 bis). QED

Let now $f:B\to A$ be a surjective homomorphism of local rings. Let

$$f:\mathrm{Spec}(A)\to \mathrm{Spec}(B)$$

be the corresponding morphism of affine schemes. Set $X=\mathrm{Spec}(A)$ (resp. $X^{\prime }={\mathfrak{m}}_A$), $Y=\mathrm{Spec}(B)$ (resp. $Y^{\prime }={\mathfrak{m}}_B$), and let $M$ be an $A$-module and $M$ the corresponding ${\mathcal{O}}_X$-Module. Since $R^g{f}_{\ast }(M)=0$ for $q>0$, the spectral sequence (15) degenerates, and by (3.2) one obtains an isomorphism of $B$-modules:

Hⁿ_{{𝔪_B}}(Y, f_*(M̃)) ≅ Hⁿ_{{𝔪_A}}(X, M̃)_[f],

hence an isomorphism of $B$-modules:

$$H_{{\mathfrak{m}}_B}^n(M_f)\cong H_{{\mathfrak{m}}_A}^n(M{)}_f.$$

On the other hand, if D_A (resp. D_B) is the dualizing functor for $A$ (resp. $B$), one has

$$D_A(M{)}_f\cong D_B(M_f).$$

Finally, since one has a ring isomorphism

B/Ann M_[f] ≅ A/Ann M,

one sees that the change of base rings under consideration changes nothing. So suppose $A$ is regular of dimension $r$.

By (2.1) one has

$$D(H^i(M))=Ex{t}^{r-i}(M,A).$$

We shall prove the equivalence of the following properties:

(a) $\dim Ex{t}^j(M,A)⩽r-j$;

(b) for every $\mathfrak{p}\in X=\mathrm{Spec}(A)$ such that $\dim A_{\mathfrak{p}}<j$, one has $Ex{t}^j(M,A{)}_{\mathfrak{p}}=0$;

(c) $codim(Supp(Ex{t}^j(M,A)),X)⩾j$.

To prove (a) ⇒ (b), let $\mathfrak{p}\in X$ with $\dim A_{\mathfrak{p}}<j$; then $\dim A/\mathfrak{p}>r-j$, hence by (a) $Ann(Ex{t}^j(M,A))\not\subset \mathfrak{p}$, which entails $Ex{t}^j(M,A{)}_{\mathfrak{p}}=0$. Let $\mathfrak{p}\in Supp(Ex{t}^j(M,A))$; then $Ex{t}^j(M,A{)}_{\mathfrak{p}}\ne 0$, so by (b) $\dim A_{\mathfrak{p}}⩾j$. Hence codim(Supp(Extʲ(M, A)), X) = inf{dim A_𝔭 | 𝔭 ∈ Supp(Extʲ(M, A))} ⩾ j, that is, (b) ⇒ (c). Finally (c) implies (a) trivially.

Let us now prove the theorem.

(i) Let $x=(x_1,\cdots ,x_r)$ be a system of parameters for $A$ such that $x_i\in AnnM$ for $i=1,\cdots ,r-n$. Let $K\bullet ((x^k),M)$ be the Koszul complex. One sees easily that the map $K^i((x^k),M)\to K^i(({x^k}^{\prime }),M)$ for $k<k^{\prime }$ is zero, if $i>n$. It follows that $H^i(M)=\lim \to H^i((x^k),M)=0$ if $i>n$. On the other hand, it is trivial that $H^i(M)=0$ if $i<0$, so (i) is proved.

(ii) Since $A$ is regular, $\dim A_{\mathfrak{p}}<j$ entails that the global homological dimension of $A_{\mathfrak{p}}$ is strictly less than $j$, and hence $Ex{t}^j(M,A{)}_{\mathfrak{p}}=Ex{t}_{A_{\mathfrak{p}}}^j(M_{\mathfrak{p}},A_{\mathfrak{p}})=0$; so one has proved (b) and consequently (a). (ii) then follows from (22) and from (a).

(iii) There exists a $\mathfrak{p}\in Supp(M)$ such that $\dim A_{\mathfrak{p}}=r-n$ and such that $Supp(M_{\mathfrak{p}})={\mathfrak{m}}_{A_{\mathfrak{p}}}$. Since $A_{\mathfrak{p}}$ is regular if $A$ is, one finds $prof{A}_{\mathfrak{p}}=r-n$, hence

Extʳ⁻ⁿ_A(M, A)_𝔭 = Extʳ⁻ⁿ_{A_𝔭}(M_𝔭, A_𝔭) ≠ 0.

This implies, taking (22) into account, that on the one hand

$$H^n(M)\ne 0,$$

and on the other hand

$$\dim D(H^n(M))⩾n,$$

hence by (ii)

$$\dim D(H^n(M))=n.$$

Let now $Y=Supp(M)$. By (i) one knows that $D(H^n(M^{\prime }))=Ex{t}^{r-n}(M^{\prime },A)$ is a functor in $M^{\prime }$, left exact, on the category $({\mathcal{C}}_Y{)}^{\circ }$. Hence there exists an $A$-module $H$ and an isomorphism of functors in $M^{\prime }$:

Extʳ⁻ⁿ(M′, A) = Hom(M′, H).

Let Yᵢ, $i=1,\cdots ,k$, be the irreducible components of $Y$ of maximum dimension. We shall see that the assertion $Ex{t}^{r-n}(M^{\prime },A)\ne 0$ is equivalent to the assertion: there exists an $i$ such that $Supp{M}^{\prime }\supset Y_i$. Indeed, if $Supp{M}^{\prime }\supset Y_i$, then $\dim (M^{\prime })=n$, hence $Ex{t}^{r-n}(M^{\prime },A)\ne 0$.

If $Supp{M}^{\prime }\not\supset Y_i$ for every $i=1,\cdots ,k$, then $\dim M^{\prime }<n$ and

$$D(H^n(M^{\prime }))=Ex{t}^{r-n}(M^{\prime },A)=0.$$

Since Ass(Extʳ⁻ⁿ(M, A)) = Supp M ∩ Ass(H), one sees that the last assertion of (iii) follows from the following lemma:

Lemma 3.3.

Let $X=\mathrm{Spec}(A)$, let $Y$ be a closed subset of $X$, let $T:({\mathcal{C}}_Y{)}^{\circ }\to Ab$ be a left exact functor, and let Yᵢ, $i=1,\cdots ,k$, be a family of irreducible components of $Y$ such that the assertion: $T(M)=0$ is equivalent to the assertion: $\forall i,SuppM\not\supset Y_i$. Then $T$ is representable by a module $H$ such that $Ass(H)={\bigcup }_{i=1}^k{y}_i$, where yᵢ is the generic point of Yᵢ, $i=1,\cdots ,k$.

Proof. Let $y\in Y$; one constructs an $A$-module $M(y)$ such that $Supp(M(y))=y$. Suppose that $y\ne y_i$ for every $i=1,\cdots ,k$; then $Y_i\not\subset Supp(M(y))$ for every $i=1,\cdots ,k$, so $T(M(y))=0$. It follows that

Ass(T(M(y))) = Supp(M(y)) ∩ Ass(H) = ∅,

hence $y\notin Ass(H)$. If $y=y_i$, then $Y_i\subset Supp(M(y))$, so $T(M(y))\ne 0$, whence

Ass(T(M(y))) = Supp(M(y)) ∩ Ass(H) ≠ ∅.

By the first part of the proof, this implies $y\in Ass(H)$, whence the lemma. QED

Example 3.4.

Let $A$ be a noetherian ring, let $X=\mathrm{Spec}(A)$, and let $Y$ be a closed subset of $X$ such that $X-Y$ is affine; then for every irreducible component $Y_{\alpha }$ of $Y$ one has $codim(Y_{\alpha },X)⩽1$.

Indeed, consider $X$ as a prescheme over $X$. Let $y_{\alpha }\in Y_{\alpha }$ be a generic point, and consider the morphism $\mathrm{Spec}({\mathcal{O}}_{X,y_{\alpha }})\to X$. The affine scheme obtained by base extension of $X$ to $\mathrm{Spec}({\mathcal{O}}_{X,y_{\alpha }})$ is canonically isomorphic to $\mathrm{Spec}({\mathcal{O}}_{X,y_{\alpha }})$.

By (EGA I 3.2.7) one sees that if $y_0$ is the unique closed point of $Y_0=\mathrm{Spec}({\mathcal{O}}_{X,y_{\alpha }})$, then $Y_0-y_0$ is affine. By (EGA III 1.3.1) one finds

Hⁱ(Y₀ − y₀, 𝒪_{Y₀}) = 0   if i > 0,

hence by (I 2.9)

Hⁱ⁻¹(𝒪_{X,y_α}) = Hⁱ_{{y₀}}(Y₀, 𝒪_{Y₀}) = 0   if i ⩾ 2.

Taking 3.1 (iii) into account, it follows that

$$\dim {\mathcal{O}}_{X,y_{\alpha }}⩽1,$$

hence codim(Y_α, X) = inf_{y ∈ Y_α} dim 𝒪_{X,y} ⩽ 1. QED

Let $A$ be a noetherian local ring, $\mathfrak{m}$ its maximal ideal, and $M$ a finitely generated $A$-module. Suppose that $A$ is a quotient of a regular local ring. Set $X=\mathrm{Spec}(A)$, and for every $x\in X$, ${\mathfrak{m}}_x=\mathfrak{m}{A}_x$.

Proposition 3.5.

The following two conditions are equivalent:

a) $H^i(M)$ is of finite length;

b) $\forall x\in X-\mathfrak{m},H^{i-dim}{x}_{{\mathfrak{m}}_x}(M_x)=0$.

Proof. Taking (3.2) into account, we may suppose $A$ regular. By (2.1) we have

$$H^i(M)=D(Ex{t}^{r-i}(M,A)),$$

where $r=\dim A$. By (IV 4.7), a) is equivalent⁴ to

Extʳ⁻ⁱ(M, A) is of finite length.

Now (24) is equivalent to

∀ x ∈ X − {𝔪}, one has Extʳ⁻ⁱ(M, A)_x = 0.

On the other hand $A_x$ is regular of dimension $r-\dim x$, so by (2.1)

Hⁱ⁻ᵈⁱᵐ {x}_{𝔪_x}(M_x) = D(Ext^{(r − dim {x}) − (i − dim {x})}_{A_x}(M_x, A_x)) = D(Extʳ⁻ⁱ_{A_x}(M_x, A_x)).

Since $M$ is finitely generated, one has

Extʳ⁻ⁱ_A(M, A)_x = Extʳ⁻ⁱ_{A_x}(M_x, A_x),

whence the proposition.

Corollary 3.6.

In order that $H^i(M)$ be of finite length for $i⩽n$, it is necessary and sufficient that

prof(M_x) > n − dim {x}

for every $x\in X-\mathfrak{m}$.

Proof. Follows from (3.5) and (III 3.1).