PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 128, Number 5, Pages 1307–1312 S 0002-9939(99)05130-8 Article electronically published on October 18, 1999

GENERALIZED MATLIS DUALITY

RICHARD G. BELSHOFF, EDGAR E. ENOCHS, AND JUAN RAMON GARCÍA ROZAS

(Communicated by Wolmer V. Vasconcelos)

Abstract. Let R be a commutative noetherian ring and let E be the minimal injective cogenerator of the category of R-modules. A module M is said to be reflexive with respect to E if the natural evaluation map from M to Hom_{R}(Hom_{R}(M, E), E) is an isomorphism. We give a classification of modules which are reflexive with respect to E. A module M is reflexive with respect to E if and only if M has a finitely generated submodule S such that M/S is artinian and R/ ann(M) is a complete semi-local ring.

Matlis and Gabriel in [7] and [5] considered modules over a complete local ring R. They showed that if the dual of an R-module is taken with respect to E_{R}(k) (the injective envelope of the residue field k of R), then finitely generated and artinian modules are reflexive.

Various authors have considered related questions. For example, dropping the condition that R be complete or weakening local to semilocal ([1], [2], [6], [8], [9]).

In this paper we let R be any commutative noetherian ring and let E be the minimal injective cogenerator of the category of R-modules. We give a classification of modules which are reflexive with respect to E. The result is that a module M is reflexive with respect to E if and only if M has a finitely generated submodule S such that M/S is artinian and if R/I is a complete semilocal ring where I = ann(M).

We denote by Ω the maximal spectrum of R, and we let E = ⨁_{m∈Ω} E_{R}(R/m) be the minimal injective cogenerator in the category of R-modules. For an R-module M we let M^{∨} = Hom_{R}(M, E) and call M^{∨} the Matlis dual of M. If the canonical map M → M^{∨∨} is an isomorphism we say that M is (Matlis) reflexive. We note that for any M, the map M → M^{∨∨} is an injection. From this it is easy to conclude that ann(M) = ann(M^{∨}).

If S ⊂ R is a multiplicative set and the canonical map M → S^{−1}M is an isomorphism, we write M = S^{−1}M. If M = S^{−1}M, then also M^{∨} = S^{−1}(M^{∨}). When S = R − P is the complement of a prime ideal P of R we use the usual notation M_{P}.

If R is a local ring we let R̂ denote its completion. If M is finitely generated we note that R̂ ⊗_{R} M ≅ M̂ (the completion of M). We write R̂ ⊗_{R} M ≅ M̂ = M to mean that M → R̂ ⊗_{R} M ≅ M̂ is an isomorphism.

We note that if m ∈ Ω and M is a finitely generated R-module M, then Hom_{R}(M, E(R/m)) ≠ 0 if and only if ann(M) ⊂ m.

\begin{matrix}c \\ ◯ \end{matrix}2000 American Mathematical Society

Received by the editors January 28, 1998 and, in revised form, July 1, 1998.

1991 Mathematics Subject Classification. Primary 13C05; Secondary 13H99.

Key words and phrases. Matlis, duality.

If R is a complete local ring, then all finitely generated and all artinian R-modules are reflexive. If R is a complete semilocal ring, then since R is the product of a finite number of complete local rings, we still have finitely generated and artinian modules over R are reflexive.

Lemma 1. If I ⊂ R is an ideal and IM = 0 for an R-module M, then M is reflexive as an R-module if and only if M is reflexive as an R/I-module.

Proof. This follows from the fact that Hom_{R}(R/I, E) is a minimal injective cogenerator over R/I and that, since M ⊗_{R} R/I ≅ M, we have

Hom_{R/I}(M, Hom_{R}(R/I, E)) ≅ Hom_{R}(M, E).

Lemma 2. If S ⊂ R is multiplicative and S^{−1}M = M for an R-module M, then M is reflexive as an R-module if and only if M is reflexive as an S^{−1}R-module.

Proof. As in Lemma 1, Hom_{R}(S^{−1}R, E) is a minimal injective cogenerator over S^{−1}R and

Hom_{S^{−1}R}(M, Hom_{R}(S^{−1}R, E)) ≅ Hom_{R}(M, E).

The following result strengthens Theorem 2(i) in [1].

Theorem 3. Let R be a local ring and M a finitely generated R-module. Then M is reflexive if and only if R̂ ⊗_{R} M = M.

Proof. For any such M we have the commutative diagram

M  ⟶ Hom_{R}(Hom_{R}(M, E_{R}(k)), E_{R}(k))
↓  ↓
M̂ = R̂ ⊗_{R} M ⟶ HomR̂(HomR̂(M̂, ER̂(k)), ER̂(k))

where k is the residue field of R and of R̂. The bottom horizontal arrow is an isomorphism since M̂ is a reflexive R̂-module.

But we claim that the right vertical arrow is an isomorphism. For E_{R}(k) = ER̂(k) and

HomR̂(M̂, E_{R}(k)) = HomR̂(R̂ ⊗_{R} M, E_{R}(k))
= Hom_{R}(M, HomR̂(R̂, E_{R}(k))) = Hom_{R}(M, E_{R}(k)).

Since A = Hom_{R}(M, E_{R}(k)) and E_{R}(k) are artinian R and R̂-modules, we have HomR̂(A, E_{R}(k)) = Hom_{R}(A, E_{R}(k)). It follows that M → R̂ ⊗_{R} M is an isomorphism if and only if M is reflexive.

Corollary 4. If I is an ideal in a local ring R, then R/I is reflexive as an R-module (or as an R/I-module) if and only if R/I is a complete local ring.

Proof. This follows from the fact that R̂ ⊗_{R} R/I = R̂/I.

Lemma 5. If M is an R-module and S ⊂ M is a submodule, then M is reflexive if and only if S and M/S are reflexive.

Proof. The commutative diagram

0 ⟶ S ⟶ M ⟶ M/S  ⟶ 0
↓ ↓ ↓
0 ⟶ S^{∨∨} ⟶ M^{∨∨} ⟶ (M/S)^{∨∨} ⟶ 0

has exact rows and the vertical maps are all injections. So the result follows.

Lemma 6. No infinite direct sum of nonzero R-modules is reflexive.

Proof. Let ⨁_{i∈I} M_{i} be reflexive where I is infinite and M_{i} ≠ 0 for all i ∈ I. We have

(⨁_{i∈I} M_{i})^{∨} = ∏_{i∈I}(M_{i}^{∨}), and the canonical map

⨁_{i∈I} M_{i} → (∏_{i∈I}(M_{i}^{∨}))^{∨} ≅ (⨁_{i∈I} M_{i})^{∨∨}

sends (x_{i})_{i∈I} to the map (ϕ_{i})_{i∈I} ↦ ∑_{i∈I} ϕ_{i}(x_{i}). So the image of (x_{i})_{i∈I} is 0 on

⨁_{i∈I}(M_{i}^{∨}) ⊂ ∏_{i∈I}(M_{i}^{∨}) if and only if x_{i} = 0 for all i, i.e., if and only if (x_{i})_{i∈I} = 0.

But ⨁_{i∈I}(M_{i}^{∨}) ≠ ∏_{i∈I}(M_{i}^{∨}) so there is a nonzero linear f : ∏_{i∈I}(M_{i}^{∨}) → E which is 0

on ⨁_{i∈I}(M_{i}^{∨}). By the above no such f can be the image of an (x_{i})_{i∈I} ∈ ⨁_{i∈I} M_{i} and

so ⨁_{i∈I} M_{i} is not reflexive.

Corollary 7. If M is reflexive there is a finitely generated submodule S such that M/S is artinian.

Proof. If M = 0 this is trivial. If M ≠ 0 there is a finitely generated S_{1} ⊂ M such that Soc(M/S_{1}) ≠ 0. If Soc(M/S_{1}) is essential in M/S_{1}, it is well known that M/S_{1} (and in fact E(M/S_{1})) is artinian. If it is not essential, let N/S_{1} ∩ Soc(M/S_{1}) = 0 with S_{1} ⊂ N, S_{1} ≠ N. Then there is a finitely generated S_{2} with S_{1} ⊂ S_{2} ⊂ N and Soc(N/S_{2}) ≠ 0. But then Soc(M/S_{1}) → Soc(M/S_{2}) is injective but not surjective. We repeat the procedure and see that it must stop, for otherwise if T = ⋃ S_{n}, then Soc(M/T) is an infinite direct sum. This is impossible by Lemma 6.

The argument above is taken from Enochs [3], Proposition 1.3, and is included here for completeness.

Proposition 8. Let M be an R-module and suppose that for some m ∈ Ω, Hom_{R}(M, E(R/n)) = 0 for n ∈ Ω, n ≠ m. Then if M is reflexive, M_{m} = M and Hom_{R}(M^{∨}, E(R/n)) = 0 for n ≠ m.

Proof. Let M be reflexive and m ∈ Ω be such that Hom_{R}(M, E(R/n)) = 0 for n ≠ m. If M ≠ 0, we have a natural nonzero homomorphism M → M^{∨∨} = Hom_{R}(M^{∨}, E) (n ∈ Ω). So if Hom_{R}(M^{∨}, E(R/n)) ≠ 0, then the projection E → E(R/n) induces Hom_{R}(M^{∨}, E) → Hom_{R}(M^{∨}, E(R/n)) and so we have a nonzero homomorphism M → Hom_{R}(M^{∨}, E(R/n)). So then

Hom_{R}(M, Hom_{R}(M^{∨}, E(R/n))) ≠ 0.

But

Hom_{R}(M, Hom_{R}(M^{∨}, E(R/n))) ≅ Hom_{R}(M^{∨}, Hom_{R}(M, E(R/n)))

so Hom_{R}(M, E(R/n)) ≠ 0. Hence n = m.

Since M^{∨} = Hom_{R}(M, E(R/m)), it follows easily using properties of E(R/m) that (M^{∨})_{m} = M^{∨}. But then (M^{∨∨})_{m} = M^{∨∨} and so M_{m} = M. It now follows by Lemma 2 that M is a reflexive R_{m}-module as well as a reflexive R-module.

In the next result we use the fact that if A is an artinian R-module, then there are distinct maximal ideals n_{1}, n_{2}, …, n_{t} ∈ Ω and a decomposition

A = A_{1} ⊕ A_{2} ⊕ ⋯ ⊕ A_{t}

such that (A_{i})_{n_{i}} = A_{i}. The n_{1}, …, n_{t} and the decomposition are unique in the obvious sense. If B ⊂ A is a submodule, then

B = (B ∩ A_{1}) ⊕ (B ∩ A_{2}) ⊕ ⋯ ⊕ (B ∩ A_{t})

gives the corresponding decomposition of B. We note that Hom_{R}(A_{i}, E(R/m)) = 0 for m ∈ Ω, m ∉ {n_{1}, …, n_{t}}.

Theorem 9. Let M be a finitely generated R-module and let I = ann(M). Then M is reflexive if and only if R/I is a complete semilocal ring.

Proof. We have M^{∨} = Hom_{R}(M, ⨁ E(R/m)) ≅ ⨁ Hom_{R}(M, E(R/m)) (the direct sum over all m ∈ Ω) since M is finitely generated. Since M^{∨} is also reflexive, we see by Proposition 8 that Hom_{R}(M, E(R/m)) = 0 except for a finite number of m ∈ Ω. Let n_{1}, n_{2}, …, n_{t} ∈ Ω be distinct elements of Ω such that Hom_{R}(M, E(R/m)) =

0 for m ∉ {n_{1}, n_{2}, …, n_{t}}, so M^{∨} = ⨁_{i=1}^{t} Hom_{R}(M, E(R/n_{i})). We can assume

Hom_{R}(M, E(R/n_{i})) ≠ 0 for i = 1, 2, …, t. If M_{i} denotes Hom_{R}(M, E(R/n_{i})), then since Hom_{R}(M_{i}, E(R/m)) = 0 for m ≠ n_{i}, we see Hom_{R}(R/I_{i}, E(R/m)) = 0 for m ≠ n_{i} where I_{i} = ann(M_{i}). Hence I_{i} is contained in only one maximal ideal, namely n_{i}. Hence R/I_{i} is a local ring and (R/I_{i})_{n_{i}} = R/I_{i}.

Since R/I_{i} is also reflexive and finitely generated, we get that R/I_{i} is a reflexive R_{n_{i}}-module by Lemma 2. But then by Theorem 3, R̂_{n_{i}} ⊗_{R_{n_{i}}} R/I_{i} = R/I_{i}. This means that R/I_{i} is a complete local ring.

But now if I = ann(M), we have I = ⋂_{i=1}^{t} I_{i}. The I_{1}, …, I_{t} are pairwise comaxi-

mal so by the Chinese remainder theorem R/I ≅ R/I_{1} × R/I_{2} × ⋯ × R/I_{t}. Hence R/I is a complete semilocal ring.

Now assume that M is a finitely generated R-module and that R/I is a complete semilocal ring where I = ann(M). Then M is reflexive as an R/I-module and so by Lemma 1 is reflexive as an R-module.

Corollary 10. If I ⊂ R is an ideal, then R/I is reflexive if and only if R/I is a complete semilocal ring.

Proof. Immediate.

Corollary 11. If I, J ⊂ R are ideals such that R/I and R/J are complete semilocal rings, then R/IJ is a complete semilocal ring.

Proof. We only need argue that R/IJ is a reflexive R-module. We have the exact sequence

0 → I/IJ → R/IJ → R/I → 0.

But R/I is a reflexive R-module, and since I/IJ is a quotient of (R/J)^{n} for some n ≥ 1 it is a reflexive R-module. Hence R/IJ is a reflexive R-module. (These claims all use Lemma 5.)

We now give the complete classification of reflexive modules.

Theorem 12. An R-module M is reflexive if and only if it has a finitely generated submodule S such that M/S is artinian and if R/I is a complete semilocal ring where I = ann(M).

Proof. Assume that M is reflexive. By Corollary 7 we have the finitely generated S with M/S artinian. Since S is reflexive and finitely generated, R/J is a complete semilocal ring where J = ann(S) by Theorem 9. Since M/S is artinian and reflexive, (M/S)^{∨} is noetherian, i.e., finitely generated. Hence, if K = ann(M/S)^{∨}, then R/K is complete semilocal. But ann(M/S)^{∨} = ann(M/S). Since I = ann(M) ⊃ JK and since R/JK is complete semilocal by Corollary 11, we see that R/I is complete semilocal.

Conversely assume that M has a finitely generated submodule S ⊂ M with M/S artinian and that R/I is a complete semilocal ring where I = ann(M).

Then, since J = ann(S) ⊃ ann(M) = I, we see that R/J is complete and semilocal. Hence S is reflexive.

Since M/S is artinian and K = ann(M/S) ⊃ ann(M) = I, we see that R/K is complete and semilocal. Then the artinian R/K-module is reflexive as an R/Kmodule. By Lemma 1, this implies that M/S is a reflexive R-module. But then by an appeal to Lemma 5 it follows that M is a reflexive R-module.

Examples. 1. If k is an algebraically closed field and I ⊂ k[x_{1}, …, x_{n}] is such that k[x_{1}, …, x_{n}]/I is semilocal, then in fact k[x_{1}, …, x_{n}]/I is artinian. Hence a finitely generated reflexive module M over k[x_{1}, …, x_{n}] has finite length. Since the dual A^{∨} of an artinian reflexive k[x_{1}, …, x_{n}]-module A is finitely generated, we see that A also has finite length. So by Theorem 12 the reflexive k[x_{1}, … x_{n}]-modules are exactly those of finite length.

2. If k is any field and R = k[[x]][y], then R/(y) is a complete local ring and so is a reflexive R-module which is not of finite length.

Remark. We recall that a module M is said to be cotorsion if xt^{1}(F, M) = 0 for all flat R-modules F. In [4], finitely generated cotorsion modules were characterized over a commutative noetherian ring R of finite Krull dimension. In effect, over such a ring R, a finitely generated module M is cotorsion if and only if it is reflexive. We do not know if this holds if we allow R to have infinite Krull dimension.

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Department of Mathematics, Southwest Missouri State University, Springfield, Missouri 65804

E-mail address: belshoff@math.smsu.edu

Department of Mathematics, University of Kentucky, Lexington, Kentucky 40506

E-mail address: enochs@ms.uky.edu

Department of Algebra and Analysis, University of Almería 04120 Almería, Spain

E-mail address: jrgrozas@ualm.es