Exposé IV. Dualizing modules and dualizing functors

1. Generalities on module functors

Let

• $A$ be a commutative noetherian ring,
• $C$ the category of $A$-modules of finite type,
• $C^{\prime }$ the category of arbitrary $A$-modules,
• Ab the category of abelian groups.

The aim of this section is the study of certain properties of functors $T:C^{\circ }\to Ab$ (assumed additive). Here $C^{\circ }$ denotes the opposite category of $C$.

Note that if $M\in ObC$, then $T(M)$ may be canonically endowed with a structure of $A$-module, defined as follows: if $f_M$ is the homothety of $M$ associated to $f\in A$, then $A$ acts on $T(M)$ by $f_{T(M)}$. In other words, $T$ factors as

                         T
        C°  ───────────────────────────►  Ab
           \                            ↗
            \                        /
         T°  \                  /
              \              /
               ▼          /
                  C′

where $C^{\prime }\to Ab$ is the canonical functor. In what follows, $T(M)$ will always be considered as endowed with this $A$-module structure.

Composing with the isomorphism $M\stackrel{\sim }{\to }{\mathrm{Hom}}_A(A,M)$ the morphism ${\mathrm{Hom}}_A(A,M)\to {\mathrm{Hom}}_A(T(M),T(A))$, one obtains the following morphisms, each deduced from the other in an obvious way:

$$M⟶{\mathrm{Hom}}_A(T(M),T(A)),M\times T(M)⟶T(A),T(M)⟶{\mathrm{Hom}}_A(M,T(A)),$$

and this defines a morphism ${\varphi }_T$ of contravariant functors:

φ_T : T ⟶ Hom_A(−, T(A)).

Proposition.

The following two properties are equivalent:

1. ${\varphi }_T$ is an isomorphism of functors.
2. $T$ is left exact.

The implication (i) ⇒ (ii) is trivial.

The implication (ii) ⇒ (i) follows from the fact that, for a morphism $u:F\to F^{\prime }$ of two additive left exact functors $F$ and $F^{\prime }$ from $C^{\circ }$ to Ab, if $u(A)$ is an isomorphism, then $u$ is an isomorphism (one uses the fact that $A$ is noetherian, hence that every $A$-module of finite type is of finite presentation).

Remark.

This shows in particular that the representable functors $T:C^{\prime \circ }\to Ab$ are precisely those that commute with arbitrary inverse limits (over a preordered set, not necessarily filtered).

If $\mathrm{Hom}(C^{\circ },Ab{)}_g$ denotes the full subcategory of $\mathrm{Hom}(C^{\circ },Ab)$ whose objects are the left exact functors, one has proved the equivalence of categories

$$C^{\prime }\stackrel{\sim }{\to }\mathrm{Hom}(C^{\circ },Ab{)}_g$$

via the quasi-inverse functors

$$H\mapsto {\mathrm{Hom}}_A(-,H)$$

and

$$T\mapsto T(A).$$

Now let $J$ be an ideal of $A$, let $Y=V(J)\subset \mathrm{Spec}A$, and denote by C_Y the full subcategory of $C$ whose objects are the $A$-modules $M$ of finite type such that $SuppM\subset Y$. One has

$$C_Y=\bigcup _n{C}^{(n)},$$

where $C^{(n)}$ is the full subcategory of C_Y consisting of the modules $M$ such that $J^{n}M=0$.

Proposition.

With the same notation as above, let $T:C_Y^{\circ }\to Ab$ be a functor. To $H=\underrightarrow{\lim }T(A/J^n)$¹ is associated a natural morphism

φ_T : T ⟶ Hom_A(−, H),

and the following conditions are equivalent:

1. ${\varphi }_T$ is an isomorphism.
2. The functor $T$ is left exact.

Proof. — a) Definition of ${\varphi }_T$.

Let $M\in Ob{C}_Y$. There exists an integer $n$ such that $J^{n}M=0$. Then $M$ is an $A/J^n$-module, and if $T_n$ denotes the restriction of $T$ to $C^{(n)}$, one knows how to define the morphism

T_n ⟶ Hom_A(−, H_n),    where H_n = T(A/J^n);

whence

T(M) = T_n(M) ⟶ Hom_A(M, lim_→ H_n) = Hom_A(M, H)

and

φ_T : T ⟶ Hom_A(−, H).

b) Equivalence of (i) and (ii).

It is clear that (i) implies (ii). Suppose (ii) holds and let $M\in Ob{C}^{(n)}$. We have seen that $T_n(M)\stackrel{\sim }{\to }{\mathrm{Hom}}_A(M,H_n)$; hence for every integer $n^{\prime }>n$ one has

T(M) = T_n(M) = T_{n′}(M) = lim_→ T_n(M),

and

T(M) = lim_→ Hom_A(M, H_n).

Since these are filtered direct limits, one also has the isomorphism

lim_→ Hom_A(M, H_n) ⥲ Hom_A(M, lim_→ H_n) = Hom_A(M, H).

If $C_Y^{\prime }$ denotes the category of $A$-modules with support contained in $Y$, but not necessarily of finite type, one again has the natural equivalence of categories

$$C_Y^{\prime }\stackrel{\sim }{\to }\mathrm{Hom}(C_Y^{\circ },Ab{)}_g.$$

Application. — With the same notation, let

$$T^{\bullet }:C_Y^{\circ }⟶Ab$$

be an exact $\partial$-functor. For every $i\in \mathbb{Z}$, set $H_n^i=T^i(A/J^n)$ and $H^i=\underrightarrow{\lim }{H}_n^i$.

Theorem.

Let $n\in \mathbb{Z}$. If there exists $i_0\in \mathbb{Z}$ such that $T^i=0$ for every $i<i_0$, then the following three conditions are equivalent:

1. $T^i=0$ for every $i<n$.
2. $H^i=0$ for every $i<n$.
3. There exists a module $M_0$ in C_Y such that $Supp{M}_0=Y$ and $T^i(M_0)=0$ for every $i<n$.

Proof. — It is evident that (i) implies (ii) and (iii) (take $M_0=A/J$).

We show by induction on $n$ that (ii) implies (i). It is true for $n=i_0$; suppose it has been proved up to rank $n$. Suppose then that $H^i=0$ for every $i<n+1$; by the induction hypothesis one has $T^i=0$ for $i<n$, but $T^{n-1}=0$ implies that $T^n$ is a left exact functor, and

$$T^n\stackrel{\sim }{\to }{\mathrm{Hom}}_A(-,H^n)=0.$$

We now show that (iii) implies (ii). It is again true for $n=i_0$; suppose it has been proved up to rank $n$. Let $M_0$ be an $A$-module in C_Y such that $Supp{M}_0=Y$ and $T^i(M_0)=0$ for every $i<n+1$; by the induction hypothesis one then has $H^i=0$ for every $i<n$; it remains to show that $H^n=0$. But "$H^i=0$ for every $i<n$" implies that $T^{n-1}=0$, and therefore that $T^n\stackrel{\sim }{\to }{\mathrm{Hom}}_A(-,H^n)$. One then has

Ass H^n = Ass Hom(M₀, H^n) = Supp M₀ ∩ Ass H^n = Ass H^n,

since

Ass H^n ⊂ Supp H^n ⊂ Y = Supp M₀.

Hence $T^n(M_0)=0\iff Ass{H}^n=\varnothing \iff H^n=0$; this completes the proof.

2. Characterization of exact functors

The ring $A$ is still assumed noetherian and commutative. The notation is that of Proposition 1.3:

Y = V(J),    T : C_Y° ⟶ Ab,    H = lim_→ T(A/J^n),

where we assume that $T$ is a left exact functor, whence

$$T(M)\stackrel{\sim }{\to }{\mathrm{Hom}}_A(M,H).$$

Proposition.

The following properties are equivalent:

1. The functor $T$ is exact.
2. $H$ is injective in $C^{\prime }$.

Proof. — It clearly suffices to show that (i) implies (ii), that is, to prove that if the restriction to C_Y of the functor ${\mathrm{Hom}}_A(-,H)$ is an exact functor, then $H$ is injective. But since $A$ is noetherian, in order to show that $H$ is injective it suffices to prove that every homomorphism $f:N\to H$ whose source is an $A$-module $N$ of finite type, a submodule of an $A$-module $M$ of finite type, extends to a homomorphism $\bar{f}:M\to H$.

The definition of $H$ and the fact that $N$ is of finite type imply that there exists an integer $n$ such that $J^n\cdot f(N)=0$. Endow $M$ and $N$ with the $J$-adic topology. The $J$-adic topology of $N$ is equivalent to the topology induced by the $J$-adic topology of $M$ (Krull's theorem). There therefore exists $V=J^k\cdot M$ such that

U = V ∩ N ⊂ J^n N.

One then has the factorization

N ──────────► N/U
 \           /
  \         /
 f \       / u
    \     /
     ▼   ▼
       H

with $N/U$ and $M/V$ in C_Y. The hypothesis therefore allows one to extend $u$ to ū:

N/U ──────────► M/V
   \           /
    \         /
   u \       / ū
      \     /
       ▼   ▼
         H

and $M\to M/V\to H$ gives the desired extension $\bar{f}$.

Corollary.

Let $K$ be an injective $A$-module. Then the submodule $H_J^0(K)$ of $K$ consisting of the elements annihilated by some power of $J$ is injective.

Proof. — It suffices to verify that the restriction to C_Y of the functor ${\mathrm{Hom}}_A(-,H_J^0(K))$ is exact. Now let $M\in Ob{C}_Y$; there exists $k$ such that $J^k\cdot M=0$, and the inclusion

Hom_A(−, H⁰_J(K)) ⟶ Hom_A(M, K)

is then an isomorphism. The result follows, since ${\mathrm{Hom}}_A(-,K)$ is exact.

3. Study of the case where T is left exact and T(M) is of finite type for every M

Let, as above,

$$T:C_Y^{\circ }⟶Ab;$$

we now assume that $T$ is left exact and that one has the factorization

                         T
        C_Y°  ─────────────────────────►  Ab
            \                          ↗
             \                       /
              \                    /
               ▼                 /
                    C_Y

where, as above, $C_Y\to Ab$ is the forgetful functor. One can therefore define $T(T(M))=T\circ T(M)$, and the canonical morphism

M ⟶ Hom_A(Hom_A(M, H), H)

defines a morphism

$$M⟶T\circ T(M).$$

Proposition.

The ring $A$ being still assumed noetherian, if one makes the additional hypothesis that $A/J$ is artinian, the following conditions are equivalent:

1. $T$ is left exact and, for every $M\in Ob{C}_Y$, $T(M)$ is of finite type and $M\to T\circ T(M)$ is an isomorphism.
2. $T$ is exact and, for every residue field $k$ associated to a maximal ideal containing $J$, one has $T(k)\stackrel{\sim }{\to }k$.
3. One has $T\stackrel{\sim }{\to }{\mathrm{Hom}}_A(-,H)$ with $H$ injective, and, for every $k$ as in (ii), one has ${\mathrm{Hom}}_A(k,H)\stackrel{\sim }{\to }k$.
4. $T$ is exact and, for every $M\in Ob{C}_Y$, one has $longT(M)=longM$.

Proof. — We have already shown the equivalence of (ii) and (iii) (Prop. 2.1).

Let us show that (ii) implies (iv): first, if $M\in Ob{C}_Y$, then since $M$ is an $A/J^n$-module with $A/J^n$ artinian, long M is finite. We argue by induction on the length of $M$. Condition (iv) holds when $longM=1$, because then $M$ is a residue field falling under (ii). If $longM>1$, there exists a submodule $M^{\prime }$ of $M$ with $M^{\prime }\ne 0$ and $long{M}^{\prime }<longM$. Form the exact sequence

$$0⟶M^{\prime }⟶M⟶M^{\prime \prime }⟶0.$$

Since $T$ is exact, one has the sequence

$$0⟶T(M^{\prime })⟶T(M)⟶T(M^{\prime \prime })⟶0,$$

and long T(M) = long T(M′) + long T(M′′) = long M′ + long M′′ = long M.

(ii) ⇒ (i): Since (ii) implies (iv), let $M$ be an $A$-module in C_Y; one has $longT(M)=longM$, hence $T(M)$ is of finite length and therefore of finite type.

It remains to show that $M\to T\circ T(M)$ is an isomorphism; we again argue by induction on long M. For $M=k$ it is true. In the general case, write the commutative diagram with exact rows

$$0⟶M^{\prime }⟶M⟶M^{\prime \prime }⟶0\downarrow \downarrow \downarrow 0⟶T\circ T(M^{\prime })⟶T\circ T(M)⟶T\circ T(M^{\prime \prime })⟶0,$$

where $M^{\prime }$ is a submodule of $M$ with $M^{\prime }\ne 0$ and $long{M}^{\prime }<longM$. By the induction hypothesis the outer arrows are isomorphisms, hence

$$M⟶T\circ T(M)$$

is an isomorphism.

(i) ⇒ (ii): Let

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

be an exact sequence of $A$-modules in C_Y, and let $Q$ be the cokernel of $T(M)\to T(M^{\prime })$. Applying $T$ to the exact sequence

0 ⟶ T(M′) ⟶ T(M) ⟶ T(M′′) ⟶ Q ⟶ 0,

one obtains

$$0⟶T(Q)⟶T\circ T(M^{\prime })⟶T\circ T(M)\uparrow \uparrow \cong \cong M^{\prime }⟶M$$

hence $T(Q)=0$ and $Q\stackrel{\sim }{\to }T(T(Q))=0$.

On the other hand, let $k$ be a residue field, $k=A/\mathfrak{m}$, $J\subset \mathfrak{m}$. One must show that $T(k)\stackrel{\sim }{\to }k$. For this it suffices to note that $T(k)$ is a $k$-vector space. One then deduces

T(k) ≃ k ⊕ V,
T(T(k)) ≃ T(k) ⊕ T(V) ≃ k ⊕ V ⊕ T(V) ≃ k,

whence $V=0$.

Finally, let us show that (iv) implies (iii): it suffices to show that $T(k)\stackrel{\sim }{\to }k$. Now $longT(k)=longk=1$, so $T(k)=k^{\prime }$ is a residue field, and Supp k′ = Supp Hom_A(k, H) ⊂ Supp k. Hence $k\simeq k^{\prime }$.

Remark.

One can show that condition (iv) is equivalent to the condition

(iv)′ For every $M\in Ob{C}_Y$, one has $longT(M)=longM$.

4. Dualizing module; dualizing functor

Definition.

Let $A$ be a noetherian local ring with maximal ideal $\mathfrak{m}$. A dualizing functor for $A$ is any functor

$$T:C_{\mathfrak{m}}^{\circ }⟶Ab,$$

where we write $C_{\mathfrak{m}}$ in place of C_Y for $Y=V(\mathfrak{m})$, which satisfies the equivalent conditions of Proposition 3.1. An $A$-module $I$ is said to be dualizing for $A$ if the functor $M\mapsto {\mathrm{Hom}}_A(M,I)$ is dualizing.

Definition 4.1 can be generalized to the case where $A$ is no longer assumed to be a local ring.

Definition.

Let $A$ be a noetherian ring, and let $\bar{C}$ be the full subcategory of $C$ consisting of the $A$-modules of finite length. A dualizing functor is any $A$-linear functor $T$ from ${\bar{C}}^{\circ }$ to $\bar{C}$ which is exact and such that the morphism of functors

$$id⟶T\circ T$$

is an isomorphism.

We will prove an existence theorem and also that the module $I$ representing such a functor is locally artinian. We will likewise show that, for every maximal ideal $\mathfrak{m}$ of $A$, the $\mathfrak{m}$-primary component of the socle of $I$ is of length 1.

Proposition.

Let $A$ and $B$ be two noetherian local rings with maximal ideals ${\mathfrak{m}}_A$ and ${\mathfrak{m}}_B$, such that $B$ is a finite $A$-algebra. Then, if $I$ is a dualizing module for $A$, ${\mathrm{Hom}}_A(B,I)$ is a dualizing module for $B$.

Proof. — Let

$$R:C_{{\mathfrak{m}}_B}⟶C_{{\mathfrak{m}}_A}$$

be the restriction-of-scalars functor; it is exact. Let $T$ be a dualizing functor for $A$,

$$T:C_{{\mathfrak{m}}_A}⟶Ab;$$

it is exact and, for every $M\in Ob{C}_{{\mathfrak{m}}_A}$, the natural morphism $M\to T\circ T(M)$ is an isomorphism; hence $T\circ R$ is a dualizing functor for $B$. If $I$ represents $T$, then by the classical formula Hom_A(M, I) = Hom_B(M, Hom_A(B, I)), valid for every $B$-module $M$, we deduce that ${\mathrm{Hom}}_A(B,I)$ is a dualizing module for $B$.

Corollary.

Let $A$ be a noetherian local ring and $\mathfrak{a}$ an ideal of $A$; set $B=A/\mathfrak{a}$. If $I$ is a dualizing module for $A$, then the annihilator of $\mathfrak{a}$ in $I$ is a dualizing module for $B$.

Lemma.

Let $A$ be a noetherian local ring and $I$ a locally artinian $A$-module. There is a canonical isomorphism

I ⥲ Î = I ⊗_A Â.

Proof. — Let $I_n$ denote the annihilator of ${\mathfrak{m}}^n$ in $I$, where $\mathfrak{m}$ is the maximal ideal of $A$. To say that $I$ is locally artinian is to say that $I$ is the direct limit of the $I_n$ and that these are of finite length. Now the tensor product commutes with direct limits, so one is reduced to the case where $I$ is artinian. In this case $I$ is annihilated by some power of the maximal ideal, say ${\mathfrak{m}}^k$; therefore for $p⩾k$ one has $I\stackrel{\sim }{\to }I{\otimes }_{A}A/{\mathfrak{m}}^p$, and hence $I\stackrel{\sim }{\to }I{\otimes }_A\hat{A}$, since $A$ is noetherian and $I$ is of finite type.

It follows that the restriction-of-scalars functor from Â to $A$ and the extension-of-scalars functor induce quasi-inverse equivalences between the category of locally artinian Â-modules and the category of locally artinian $A$-modules.

Proposition.

Let $A$ be a noetherian local ring, Â its completion, and $I$ a dualizing module for $A$ (resp. for Â). Let $J$ be the completion² of $I$ (resp. the $A$-module obtained by restriction of scalars). Then $J$ is a dualizing module for Â (resp. for $A$). Moreover, the underlying abelian groups of $I$ and $J$ are isomorphic.

Proof. — One simply observes that the equivalence between the category of locally artinian $A$-modules and the category of locally artinian Â-modules induces an isomorphism between the bifunctors ${\mathrm{Hom}}_A(-,-)$ and ${\mathrm{Hom}}_{\hat{A}}(-,-)$, and that the characterization of a dualizing functor or a dualizing module involves only these bifunctors.

Theorem.

Let $A$ be a noetherian local ring.

a) There always exists a dualizing module $I$.

b) Two dualizing modules are isomorphic (by a non-canonical isomorphism).

c) For a module $I$ to be dualizing, it is necessary and sufficient that it be an injective envelope of the residue field $k$ of $A$.

Remark.

Proposition 4.6 reduces the proof to the case of a complete noetherian local ring. By a structure theorem of Cohen,³ such a ring is a quotient of a regular ring. Proposition 4.3 then allows one to assume $A$ regular. As we shall see later, this remark permits an explicit computation of the dualizing module;⁴ nevertheless we will prove Theorem 4.7 by other means.

Recollections. — Before proving the theorem, we make a few recollections on the notion of injective envelope. Cf. Gabriel, Thèse, Paris 1961, Des Catégories Abéliennes, ch. II § 5.

Let $\mathcal{C}$ be an abelian category in which direct limits exist and are exact⁵ (e.g. $\mathcal{C}=$ the category of modules). Every object $M$ embeds in an injective object, and one calls injective envelope of $M$ any minimal injective object containing $M$. One has the following properties:

(i) Every object $M$ has an injective envelope $I$.

(ii) If $I$ and $J$ are two injective envelopes of $M$, there exists between $I$ and $J$ an isomorphism (in general not unique) inducing the identity on $M$.

(iii) $I$ is an essential extension of $M$, that is to say $P\subset I$ and $P\cap M=0$ imply $P=0$. Moreover, if $I$ is injective and an essential extension of $M$, then $I$ is an injective envelope of $M$.

Granting these results, to prove Theorem 4.7 it suffices to prove c).

Proof. — Let $I$ be a dualizing module for $A$. Then $I$ is injective and ${\mathrm{Hom}}_A(k,I)$ is isomorphic to $k$. Composing the isomorphism $k\simeq {\mathrm{Hom}}_A(k,I)$ with the inclusion

Hom_A(k, I) ↪ Hom_A(A, I) ≃ I,

one obtains the inclusion

$$k\hookrightarrow I.$$

Let us show that $I$ is an injective envelope of $k$. Let $J$ be an injective module such that

$$k\subset J\subset I.$$

Since $J$ is injective, there exists an injective $A$-submodule $J^{\prime }$ of $I$ such that $I=J\oplus J^{\prime }$. We show that ${\mathrm{Hom}}_A(k,J^{\prime })=0$. One has

Hom_A(k, I) ≃ Hom_A(k, J) ⊕ Hom_A(k, J′);

${\mathrm{Hom}}_A(k,J)$ is a vector subspace of ${\mathrm{Hom}}_A(k,I)\simeq k$ not reduced to zero (since it contains the inclusion $k\subset J$), so ${\mathrm{Hom}}_A(k,J)\simeq k$, and consequently ${\mathrm{Hom}}_A(k,J^{\prime })=0$.

Arguing by induction on the length, one deduces that ${\mathrm{Hom}}_A(M,J^{\prime })=0$ for every $A$-module $M$ of finite length; since $I$ is the direct limit of the modules $\mathrm{Hom}(A/{\mathfrak{m}}^n,I)$ (cf. Proposition 1.3), which are of finite length by hypothesis, the projection $I\to J^{\prime }$ is zero, and consequently $J^{\prime }=0$.

Conversely, let $I$ be an injective envelope of $k$. To see that $I$ is a dualizing module, it suffices, by 2.1 and 3.1 (ii), to show that $V={\mathrm{Hom}}_A(k,I)$ is isomorphic to $k$. Now one has the double inclusion

$$k\subset V\subset I;$$

$V$ is a vector space over $k$ that decomposes as the direct sum of $k$ and a vector subspace $V^{\prime }$ of $I$ such that $V^{\prime }\cap k=0$. Now $I$ is an essential extension of $k$, hence $V^{\prime }=0$ and $V=k$.

Corollary.

Let $A$ be a noetherian local ring; every dualizing module for $A$ is locally artinian.

Proof. — Let $I$ be a dualizing module; it is an injective envelope of $k$. Using the notation and the result of Corollary 2.2, one has

$$k\subset H_{\mathfrak{m}}^0(I)\subset I,$$

and $H_{\mathfrak{m}}^0(I)$ is injective. One deduces that $I=H_{\mathfrak{m}}^0(I)$, and hence that $I$ is locally artinian.⁶

5. Consequences of the theory of dualizing modules

The functor

T = Hom_A(−, I) : C_𝔪 ⟶ C_𝔪

is an anti-equivalence. Indeed, $T\circ T$ is isomorphic to the identity functor, and the argument is formal from there.

One deduces the usual properties of the notion of orthogonality:

Let $M\ast ={\mathrm{Hom}}_A(M,I)=T(M)$, and let $N\subset M$ be a submodule. Define the orthogonal of $N$ to be the submodule $N^{\prime }$ of $M\ast$ consisting of the elements of $M\ast$ that vanish on $N$. One thereby obtains a bijection between the set of submodules of $M$ and the set of submodules of $M\ast$, which reverses the order.

In particular:

• $lon{g}_{M}N=colon{g}_{M\ast }{N}^{\prime }$.
• The monogenic modules, i.e. those such that $M/\mathfrak{m}M$ is of dimension 0 or 1, correspond under duality to the modules whose socle is of length 0 or 1.
• If $A$ is artinian, the ideals of $A$ correspond to the submodules of $I$.

and so on.

Let $A$ be a noetherian local ring, let ${\mathcal{D}}_A$ be the category of $A$-modules $M$ such that, for every $n\in \mathbb{N}$, $M_n=M/{\mathfrak{m}}^{n+1}M$ is of finite length and such that $M=\underleftarrow{\lim }{M}_n$, and let Â be the completion of $A$. The restriction-of-scalars functor and the completion functor are quasi-inverse equivalences between ${\mathcal{D}}_A$ and ${\mathcal{D}}_{\hat{A}}$, which commute up to isomorphism with the formation of the underlying abelian groups of the modules considered. Let ${\mathcal{C}}_A$ denote the category of locally artinian $A$-modules with socle of finite dimension.

Proposition.

Let $A$ be a noetherian local ring and let $I$ be a dualizing module for $A$. The functors

Hom_A(−, I) : (𝒞_A)° ⟶ 𝒟_A

and

Hom_{Â}(−, I) : 𝒟_A ⟶ (𝒞_A)°

are equivalences of categories, quasi-inverse to one another.

Moreover, if one transports these functors via the equivalences of categories between ${\mathcal{D}}_A$ and ${\mathcal{D}}_{\hat{A}}$ on the one hand, and ${\mathcal{C}}_A$ and ${\mathcal{C}}_{\hat{A}}$ on the other, one finds the functor ${\mathrm{Hom}}_{\hat{A}}(-,I)$.

Proof. — Let $X\in Ob{\mathcal{C}}_A$. By definition,

X = lim_→ X_k,    X_k = Hom_A(A/𝔪^{k+1}, X),
                  k ∈ ℕ

so

Hom_A(X, I) = lim_← Hom_A(X_k, I).

Therefore $Y=\underleftarrow{\lim }{X}_k$ is an Â-module of finite type, as follows from EGA 0_I 7.2.9. We note in this connection that ${\mathcal{D}}_A$ is also the category of Â-modules of finite type, or, if one prefers, that ${\mathcal{D}}_A$ is the category of complete $A$-modules of finite type over Â. Let then $Y$ be such a module, and let $f:Y\to I$ be an Â-homomorphism. The image of $f$ is a submodule of finite type, hence is annihilated by ${\mathfrak{m}}^k$ for some $k$; indeed every $x\in I$ is annihilated by a power of $\mathfrak{m}$. So $f$ factors through $Y/{\mathfrak{m}}^{k}Y$, whence it follows that

Hom_{Â}(Y, I) = lim_→ Hom_{Â}(Y_(k), I)    with Y_(k) = Y/𝔪^{k+1}Y
                  k

              = lim_→ (Y_(k))*
                  k

belongs to $Ob{\mathcal{C}}_A$. It then follows immediately that the two functors of the statement are quasi-inverse to one another.

It follows from the foregoing that neither the categories nor the functors under consideration, nor the underlying abelian groups of the modules considered, are changed by replacing $A$ by Â; Proposition 5.1 then states as follows:

The restriction of the functor ${\mathrm{Hom}}_{\hat{A}}(-,I)$ to the category of Â-modules of finite type takes its values in the category of locally artinian Â-modules with socle of finite dimension, and admits a quasi-inverse functor, which is the restriction of the functor ${\mathrm{Hom}}_{\hat{A}}(-,I)$. On the intersection of these two categories these two functors coincide (obviously!) and establish an auto-duality of the category of Â-modules of finite length.

Example (Macaulay).

Let $A$ be a local ring with residue field $k$. Let $k_0$ be a subfield of $A$ such that $k$ is finite over $k_0$, with $[k:k_0]=d$. Every $A$-module of finite length can be viewed as a $k_0$-vector space of finite dimension equal to $d\cdot long(M)$. The functor $T$:

$$M⟶{\mathrm{Hom}}_{k_0}(M,k_0)$$

is then exact and preserves length, hence is dualizing for $A$. The associated dualizing module is therefore

A′ = lim_→ Hom_{k₀}(A/𝔪^n, k₀),
       n

the topological dual of $A$ endowed with the $\mathfrak{m}$-adic topology.

Example.

Let $A$ be a regular noetherian local ring of dimension $n$. Let $\mathfrak{m}$ be its maximal ideal and $k$ its residue field. There exists a regular system of parameters $(x_1,x_2,\cdots ,x_n)$ that generates $\mathfrak{m}$ and that is an $A$-regular sequence. One can therefore compute the $Ex{t}_A^i(k,A)$ by the Koszul complex; one finds

Ext^i_A(k, A) = 0    if i ≠ n,
Ext^n_A(k, A) ≃ k.

The depth of $A$ being $n$, for every $M$ annihilated by a power of $\mathfrak{m}$, $Ex{t}_A^i(M,A)=0$ if $i<n$; furthermore $Ex{t}_A^i(M,A)=0$ if $i>n$, since the global cohomological dimension of $A$ is equal to $n$. Hence $Ex{t}_A^n(-,A)$ is exact, and moreover $Ex{t}_A^n(k,A)\simeq k$; it follows that:

Proposition.

If $A$ is a regular noetherian local ring of dimension $n$, the functor

$$M⟶Ex{t}_A^n(M,A)$$

is dualizing. The associated dualizing module is

I = lim_→ Ext^n_A(A/𝔪^r, A);
      r

it is isomorphic to $H_{\mathfrak{m}}^n(A)$ (Exposé II, Th. 6).⁷

Remark.

If $A$ satisfies the hypotheses of both preceding examples, the two dualizing modules so obtained are isomorphic. Suppose for example that $A$ is regular of dimension $n$, complete, and of equal characteristic. There then exists a field of representatives, say $K$. If one chooses a system of parameters $(x_1,\cdots ,x_n)$ of $A$, one can construct an isomorphism between $A$ and the ring of formal power series $K[[T_1,\cdots ,T_n]]$; whence, as we shall now see, an explicit isomorphism between the two dualizing modules

$$v:H_{\mathfrak{m}}^n(A)⟶A^{\prime }.$$

One can find an intrinsic interpretation of this isomorphism using the module ${\Omega }^n={\Omega }^n(A/K)$ of completed relative differentials of maximal degree. Indeed, it is known that ${\Omega }^n$ admits a basis consisting of the element $d{x}_1\wedge d{x}_2\cdots \wedge d{x}_n$. Whence an isomorphism

$$u:H_{\mathfrak{m}}^n({\Omega }^n)⟶H_{\mathfrak{m}}^n(A).$$

A remarkable fact is then that the composite

vu = w : H^n_𝔪(Ω^n) ⟶ A′

does not depend on the choice of system of parameters and commutes with change of base field.

To construct $v$, one computes $H_{\mathfrak{m}}^n(A)$ using the Koszul complex associated to the $x_i$; one finds

H^n_𝔪(A) = lim_→ A/(x₁^r, …, x_n^r);
            r

where the transition morphisms are defined as follows: set $I_r=A/(x_1^r,\cdots ,x_n^r)$; let $e_{a_1,\cdots ,a_n}^r$ denote the image of $x_1^{a_1}{x}_2^{a_2}\cdots x_n^{a_n}$ in $I_r$. The $e_{a_1,\cdots ,a_n}^r$, for $0⩽a_i<r$, form a basis of $I_r$.

That said, if $s$ is an integer, the transition morphism

t_{r, r+s} : I_r ⟶ I_{r+s}

is multiplication by $x_1^s{x}_2^s\cdots x_n^s$, so

u_{r, r+s}(e^r_{a₁,…,a_n}) = e^{r+s}_{a₁+s, …, a_n+s}.

Note that giving an $A$-homomorphism $w$ from an $A$-module $M$ to $A^{\prime }$ is equivalent to giving a $K$-linear form $w^{\prime }:M\to K$ that is continuous on submodules of finite type. In the case $M=H_{\mathfrak{m}}^n({\Omega }^n)$, the definition of $w$ is therefore equivalent to that of a linear form

$$\rho :H_{\mathfrak{m}}^n({\Omega }^n)⟶K,$$

called the residue form.⁸ To construct $\rho$, it suffices to define forms ${\rho }_r:I_r\to K$ that fit together, and one will take

ρ_r(e^r_{a₁,…,a_n}) =
    1   if a_i = r − 1 for 1 ⩽ i ⩽ n,
    0   otherwise.

Translation ledger (Exposé IV-specific)