MORPHISMS REPRESENTED BY MONOMORPHISMS ^{1}

Kiriko Kato

Abstract. We answer a question posed by Auslander and Bridger. Every homomorphism of modules is projective-stably equivalent to an epimorphism but is not always to a monomorphism. We prove that a map is projective-stably equivalent to a monomorphism if and only if its kernel is torsionless, that is, a first syzygy. If it occurs although, there can be various monomorphisms that are projective-stably equivalent to a given map. But in this case there uniquely exists a ”perfect” monomorphism to which a given map is projective-stably equivalent.

1 Introduction

Let R be a commutative noetherian ring. Linear maps f : A → B and f′ : A′ → B′ of finite R-modules are said to be projective-stably equivalent (pse for short) if the following diagram is commutative

A ⊕ P′  \underrightarrow{({f \atop t} {s \atop u})} B ⊕ Q′
↓≅  ↓≅
A′ ⊕ P \underrightarrow{({f′ \atop t′} {s′ \atop u′} )} B′ ⊕ Q

with some projective modules P, Q, P′, Q′ and R-linear maps s, t, u, s′, t′, u′. We say a morphism f is represented by monomorphisms (”rbm” for short) if there exists a monomorphism that is pse to f.

For any homomorphism f : A → B of R-modules, ( f ρ_{B}) : A ⊕ P_{B} → B is surjective with a projective cover ρ_{B} : P_{B} → B. Thus every morphism is represented by epimorphisms. The choice of epimorphism is unique; if an epimorphism f′ is pse to f, then two sequences 0 → Ker f′ → A′ →^{f′} B′ → 0 and 0 → Ker(f ρ_{B}) → A⊕P_{B} →^{(f ρ_{B})} B → 0 becomes isomorphic after splitting off common projective summands.

The formal analogy to the representations by monomorphisms fails both in existence and in uniqueness. Every morphism is not always represented by monomorphisms (Example 1). Even if a morphism f is rbm, the choice of monomorphism is not unique; there may be two monomorphisms f′ and f′′ both pse to f and that 0 → A′ →^{f′} B′ → Cok f′ → 0 and 0 → A′′ →^{f′′} B′′ → Cok f′′ → 0 are not isomorphic by splitting off common projective summands (Example 2).

1The detailed version of this paper has been submitted for publication elsewhere.

2000 Mathematics Subjects Classification: 13D02, 13D25, 16D90

The purpose of the paper is finding a condition of a given map to be rbm. Roughly speaking, our problem is to know when an exact sequence of modules

0 → A → B →^{g} C → 0

can be modified into an exact sequence

0 → B →^{f} C → A′ → 0.

Of course the projective stabilization \underline{mod R} of mod R is not triangulated in general. So the obstruction for a given map to be rbm should be the obstruction for \underline{mod R} to be triangulated. Our first focus is an analogy to the homotopy category K(mod R) of R-complexes. In [5, Theorem 2.6], the author showed a category equivalence between \underline{mod R}

and a subcategory of K(mod R). Due to this equivalence, we describe the obstruction of being rbm with a homology of a complex associated to the given map.

The problem was originally posed by Auslander and Bridger [1]. They proved that a map is rbm if and only if it is pse to a ”perfect” monomorphism. An exact sequence of R-modules is called perfect if its R-dual is also exact. A perfect monomorphism refers to a monomorphism whose R-dual is an epimorphism. This is our next focal point. In the case that a map is rbm, the choice of a monomorphism is not unique, but then a perfect monomorphism pse to the given map is uniquely determined up to direct sum of projective modules. (Theorem 3.6.)

Looking at Theorem 3.6, we see that when a morphism is rbm, its pseudo-kernel is always the first syzygy of its pseudo-cokernel. So it is tempting to ask if the equivalent condition of rbm property is that the kernel is a submodule of a free module. This is our third point. Actually, we need to assume the total ring of fractions Q(R) is Gorenstein: the condition is satisfied for instance if R is a domain.

Theorem 4.8 : Suppose the total ring of fractions Q(R) of a ring R is Gorenstein. A morphism f is rbm if and only if Ker f is a submodule of a free module.

Let us give easy examples: Example 1 Set R = k[[X, Y]]/(XY) with any field k, g : R^{2}/\binom{X}{Y}R → R/(X) ⊕ R/(Y) with g(\binom{a}{b} mod \binom{X}{Y}R) = (a mod(X), b mod(Y)). Since Ker g ≅ R/(X, Y) is not a first syzygy, g is not rbm due to Theorem 4.8. Example 2 Set R = k[[X, Y]]/(XY) with any field k, f : R^{2}/\binom{X}{Y}R → R^{2}/\binom{X^{2}}{Y^{2}}R with f(\binom{a}{b} mod \binom{X}{Y}R) = \binom{Xa}{Y b} mod \binom{X^{2}}{Y^{2}}R. The map f is not a monomorphism; Ker f ≅ R/(X)⊕R/(Y) is a first syzygy. By Theorem 4.8, f is rbm. In fact, let f′ : R^{2}/\binom{X}{Y}R → R^{2}/\binom{X^{2}}{Y^{2}}R ⊕ R^{2} be defined as f′(\binom{a}{b} mod \binom{X}{Y}R) = (\binom{Xa}{Y b} mod \binom{X^{2}}{Y^{2}}R, \binom{Y a}{Xb}). Obviously f′ is a monomorphism that is pse to f. On the other hand, f′′ : R^{2}/\binom{X}{Y}R → R^{2}/\binom{X^{2}}{Y^{2}}R⊕R^{2} with f′′(\binom{a}{b} mod \binom{X}{Y}R) = (\binom{Xa}{Y b} mod \binom{X^{2}}{Y^{2}}R, \binom{Y^{2}a}{X^{2}b}) is also a monomorphism and pse to f. We have two exact sequences

θ_{f} : 0 → R^{2}/\binom{X}{Y}R →^{f′} R^{2}/\binom{X^{2}}{Y^{2}}R ⊕ R^{2} → R^{2}/\binom{X}{Y}R ⊕ R^{2}/\binom{Y}{X}R → 0,

and

σ : 0 → R^{2}/\binom{X}{Y}R →^{f′′} R^{2}/\binom{X^{2}}{Y^{2}}R ⊕ R^{2} → R^{2}/\binom{X}{Y^{2}}R ⊕ R^{2}/\binom{Y}{X^{2}}R → 0,

that are not isomorphic. We see θ_{f} is perfect but σ is not.

2 Stable module category and homotopy category

Throughout the paper, R is a commutative noetherian ring, By an ”R-module” we mean a finitely generated R-module. For an R-module M, ρ_{M} : P_{M} → M denotes a projective cover of M.

Definition 2.1 The projective stabilization \underline{mod R} is defined as follows.

• Each object of \underline{mod R} is an object of mod R.

• For objects A, B of \underline{mod R}, a set of morphisms from A to B is \underline{Hom}_{R}(A, B) = Hom_{R}(A, B)/P(A, B) where P(A, B) := \{f ∈ Hom_{R}(A, B) | f factors through some projective module\}. Each element is denoted as \underline{f} = f mod P(A, B).

A morphism mod R is called a stable isomorphism if \underline{f} is an isomorphism in \underline{mod R}. If two R-modules A and A′ are isomorphic in \underline{mod R}, we say A and A′ are stably

isomorphic and write A ≅^{st} A′.

Definition 2.2 Morphisms f : A → B and f′ : A′ → B′ in mod R are said to be

projective-stably equivalent (pse for short) and denoted as f ≅^{st} f′ if if there exist stable isomorphisms α : A → A′ and β : B → B′ such that \underline{β ∘ f} = \underline{f′ ∘ α}.

Let L be a full subcategory of K(mod R) defined as

L = {F^{•} ∈ K(proj R) | H^{i}(F^{•}) = 0 (i < 0), H_{j}(F^{∗}_{•}) = 0 (j ≥ 0)}.

Lemma 2.3 ( [5] Proposition 2.3, Proposition 2.4 )

1) For A ∈ \underline{mod R}, there exists F_{A}^{•} ∈ L that satisfies

H^{0}(τ_{≤0}F_{A}^{•}) ≅^{st} A.

Such an F_{A}^{•} is uniquely determined by A up to isomorphisms. We fix the notation F_{A}^{•} and call this a standard resolution of A.

2) For \underline{f} ∈ \underline{Hom}_{R}(A, B), there exists f^{•} ∈ Hom_{K(proj R)}(F_{A}^{•}, F_{B}^{•}) that satisfies

\underline{H^{0}(τ_{≤0}f^{•})} ≅ \underline{f}.

Such an f^{•} is uniquely determined by \underline{f} up to isomorphisms, so we use the notation f^{•} to describe a chain map with this property for given \underline{f}.

Theorem 2.4 ( [5] Theorem 2.6) The mapping A ↦ F_{A}^{•} gives a functor from \underline{mod R} to K(mod R), and this gives a category equivalence between \underline{mod R} and L.

For f ∈ Hom_{R}(A, B), there exists a triangle

C(f)^{•−1} →^{n_{f}^{•}} F_{A}^{•} →^{f^{•}} F_{B}^{•} →^{c_{f}^{•}} C(f)^{•}.  (2.1)

In general, C(f)^{•} does not belong to L but it satisfies the following:

H^{i}(C(f)^{•}) = 0 (i < −1), H_{j}(C(f)^{∗}_{•}) = 0 (j > −1).

Definition and Lemma 2.5 ([5], Definition and Lemma 3.1) As objects of \underline{mod R}, \underline{Ker} \underline{f} := H^{−1}(τ_{≤−1}C(f)^{•}) and \underline{Cok} \underline{f} := H^{0}(τ_{≤0}C(f)^{•}) are uniquely determined by \underline{f}, up to isomorphisms. We call these the pseudo-kernel and the pseudo-cokernel of \underline{f}.

For a given map f : A → B, from (2.1), we have an exact sequence

0 → \underline{Ker} \underline{f} → A ⊕ P →^{(f ρ)} B → 0  (2.2)

with some projective module P. This characterizes the pseudo-kernel.

Lemma 2.6 For a given f ∈ Hom_{R}(A, B), suppose A ⊕ P′ →^{(f p′)} B is epimorphism

with projective module P′. Then Ker (f p′) ≅^{st} \underline{Ker}\underline{f} and the sequence

0 → Ker (f p′) → A ⊕ P′ →^{(f p′)} B → 0

is isomorphic to 2.2 after splitting off some aplit exact sequence of projective modules.

Lemma 2.7 ([5] Lemma 3.6)

1) There is an exact sequence

0 → Ker f → \underline{Ker} f → Ω_{R}^{1}(Cok f) → 0.

2) There is an exact sequence

0 → L → \underline{Cok} \underline{f} → Cok f → 0

such that Ω_{R}^{1}(L) is the surjective image of Ker f.

3 Representation by monomorphisms and perfect exact sequences

Definition 3.1 A morphism f : A → B in mod R is said to be represented by monomorphisms (rbm for short) if some monomorphism f′ : A′ → B′ in mod R is pse to f, that is, there exist stable isomorphisms α : A → A′ and β : B → B′ such that \underline{β ∘ f} = \underline{f′ ∘ α}.

Each morphism is not always rbm. It was Auslander and Bridger who first defined and studied ”represented by monomorphisms” property.

Theorem 3.2 (Auslander-Bridger) The following are equivalent for a morphism f : A → B in mod R.

1) There exists a monomorphism f′ : A → B ⊕ P with a projective module P such that f = s ∘ f′ via some split epimorphism s : B ⊕ P → B.

2) There exists a monomorphism f′ : A → B ⊕ P with a projective module P such that f = s ∘ f′ via some split epimorphism s : B ⊕ P → B, and f′^{∗} is an epimorphism.

3) Hom_{R}(B, I) → Hom_{R}(A, I) is surjective if I is an injective module.

The condition 1) of Theorem 3.2 turns out to be equivalent to the rbm condition.

Lemma 3.3 For a morphism f : A → B in mod R, f is rbm if and only if there exists a monomorphism f′ : A → B ⊕ P with a projective module P such that f = s ∘ f′ via some split epimorphism s : B ⊕ P → B.

The most remarkable point in Auslander-Bridger’s Theorem is that being rbm is equivalent to being represented by ”perfect monomorphisms” whose R-dual is an epimorphism.

Definition 3.4 An exact sequence 0 → A → B → C → 0 of R-modules is called a perfect exact sequence or to be perfectly exact if its R-dual 0 → Hom_{R}(C, R) → Hom_{R}(B, R) → Hom_{R}(A, R) → 0 is also exact. A monomorphism f is called a perfect monomorphism if Hom_{R}(f, R) is an epimorphism.

Proposition 3.5 ( [5] Lemma 2.7) The following are equivalent for an exact sequence

θ : 0 → A →^{f} B →^{g} C → 0.

1) θ is perfectly exact.

2) 0 → F_{A}^{•} →^{f^{•}} F_{B}^{•} →^{g^{•}} F_{C}^{•} → 0 is exact.

3) F_{C}^{•−1} → F_{A}^{•} →^{f^{•}} F_{B}^{•} →^{g^{•}} F_{C}^{•} is a distinguished triangle in K (mod R).

For a morphism f : A → B, A ⊕ P_{B} ⟶^{(f ρ_{B})} B is an epimorphism with a projective cover ρ_{B} : P_{B} → B. Thus each morphism is represented by epimorphisms. And the choice of the representing epimorphism is unique up to direct sum of projective modules, as we have seen in Lemma 2.6.

Unlikely, we already know an example of a morphism that is not rbm. And moreover, even if a given map is represented by a monomorphism, there would be another representing monomorphism. (Example 1 and Example 2.)

However, uniqueness theorem is obtained in this way. Due to Theorem 3.2, a morphism is rbm if and only if it is represented by a perfect monomorphism. And if this is the case, the representing perfect monomorphism is uniquely determined up to direct sum of projective modules.

Theorem 3.6 Let f : A → B be a morphism in mod R. Then f is rbm if and only if H^{−1}(C(f)^{•}) vanishes. If this is the case, we have the following:

1) We have a perfect exact sequence

θ_{f} : 0 → A \underrightarrow{\binom{f}{ϵ}} B ⊕ F_{A}^{1} →^{(c^{f} π)} \underline{Cok}\underline{f} → 0.

2) For any exact sequence of the form

σ : 0 → A \underrightarrow{\binom{f}{q}} B ⊕ P′ →^{(g p)} C → 0

with some projective module P′, there is a commutative diagram

θ_{f} : 0 → A \underrightarrow{\binom{f}{ϵ}} B ⊕ F_{A}^{1} →^{(c^{f} π)} \underline{Cok}\underline{f} → 0
↓^{α}  ↓^{β}  ↓^{γ}
σ : 0 → A \underrightarrow{\binom{f}{q}} B ⊕ P′  →^{(g p)}  C → 0

where α and β are stable isomorphisms.

3) There is an exact sequence with some projective module Q and Q′

0 → Q′ → \underline{Cok}\underline{f} ⊕ Q →^{(γ ρ)} C → 0.

In other words, \underline{Ker}\underline{γ} is projective.

4) If σ is also perfectly exact, then σ is isomorphic to θ_{f} up to direct sum of split exact sequences of projective modules. proof. We have a triangle

F_{A}^{•} →^{f^{•}} F_{B}^{•} → C(f)^{•} →^{n_{f}^{•+1}} F_{A}^{•+1}  (3.3)

which induces a term-wise exact sequence of complexes in C(proj R)

0 → F_{A}^{•} → C(n_{f})^{•} → C(f)^{•} → 0  (3.4)

Applying τ_{≤0} to the diagram above and taking homology, we get the following exact sequence of modules:

θ_{f} : 0 → H^{−1}(C(f)^{•}) → A \underrightarrow{\binom{f}{ϵ}} B ⊕ F_{A}^{1} →^{(c^{f} π)} \underline{Cok}\underline{f} → 0  (3.5)

Suppose that H^{−1}(C(f)^{•}) = 0. Then C(f)^{•} ≅ F_{\underline{Cok}\underline{f}}, and the exact sequence 3.4 shows that θ_{f} is perfectly exact.

Conversely, suppose that f is rbm; there is an exact sequence

σ : 0 → A \underrightarrow{\binom{f}{q}} B ⊕ P′ →^{(g p)} C → 0.

The maps f̃ = \binom{f}{q} and g̃ = (g p) produce the similar diagram as (3.3) :

\begin{aligned}F_{A}^{•} & →^{f̃^{•}} & F_{B⊕P′}^{•} & → & C(f̃)^{•} & → & F_{A}^{•+1} \\ ↓^{}^{α̃^{•}} & & ∥ & & ↓^{γ̃^{•}} & & ↓_{}_{α̃^{•+1}} \\ C(g̃)^{•−1} & → & F_{B⊕P′}^{•} & →^{g̃^{•}} & F_{C}^{•} & → & C(g̃)^{•} \end{aligned}  (3.6)

Since A ≅^{st} \underline{Ker}\underline{g̃}, τ_{≤0}α̃^{•} = 0 is an isomorphism, equivalently τ_{≤−1}C(α̃)^{•} = 0 hence τ_{≤−2}C(γ̃)^{•} = 0. From the long exact sequence of homology groups H^{−2}(C(γ̃)^{•}) → H^{−1}(C(f̃)^{•}) → H^{−1}(F_{C}^{•}), we get H^{−1}(C(f̃)^{•}) = 0. Obviously, H^{−1}(C(f̃)^{•}) ≅ H^{−1}(C(f)^{•}) hence H^{−1}(C(f)^{•}) = 0. Now it remains to prove 2) - 4) in the case H^{−1}(C(f)^{•}) = 0.

2) Applying τ_{≤0} to the diagram (3.6) and taking homology, we get the following diagram with exact rows:

θf̃: 0 →  A  \underrightarrow{\binom{f_{q}}{ϵ}} B ⊕ P′ ⊕ F_{A}^{1} → \underline{Cok}\underline{\binom{f}{q}} → 0
↓_{}_{α̃′}  ↓β̃′  ↓^{γ̃′}
σg̃ : 0 → \underline{Ker}\underline{(g p)} → B ⊕ P′ ⊕ P_{C} →^{(g p ρ_{C})}  C  → 0.

Notice that α̃′ and β̃′ are stable isomorphisms. The upper row is a direct sum of θ_{f} and a trivial complex, and the lower row is that of σ and a trivial complex. Splitting off trivial complexes we get a desired diagram:

0 → A \underrightarrow{\binom{f}{ϵ}} B ⊕ F_{A}^{1} →^{(c^{f} π)} \underline{Cok}\underline{f} → 0
↓_{α′}  ↓^{β′}  ↓^{γ}
0 → A \underrightarrow{\binom{f}{q}} B ⊕ P′  →^{(g p)}  C → 0

3) As we see above, γ̃^{•} : C(f̃)^{•} = F_{\underline{Cok}\underline{f̃}}^{•} → F_{C}^{•} has τ_{≤−2}C(γ̃)^{•} = 0. We may consider γ̃^{•} as γ̃_{i} = id (i ≤ −1) hence Q′ = \underline{Ker}\underline{γ̃} is projective;

0 → Q′ → \underline{Cok}\underline{f̃}⊕ P_{C} →^{(γ̃′ ρ_{C})} C → 0.

Since \underline{Cok}\underline{f̃} ≅^{st} \underline{Cok}\underline{f} and γ̃′ ≅^{st} γ, the above sequence is the desired sequence 0 → Q′ → \underline{Cok}\underline{f} ⊕ Q → C → 0 with some projective module Q.

4) Suppose σ is perfect. From Proposition 3.5, F_{C}^{•−1} → F_{A}^{•} →^{f^{•}} F_{B}^{•} →^{g^{•}} F_{C}^{•} is a distinguished triangle, and F_{C}^{•} ≅ C(f)^{•}, hence the induced sequence σ is isomorphic to θ_{f}. (q.e.d.)

4 Representation by monomorphisms and torsionless modules.

In the previous section, we see that a given map f is represented by monomorphisms if and only if H^{−1}(C(f)^{•}) = 0. If this is the case, \underline{Ker}\underline{f} = Cok d_{C(f)}^{−2} is the first syzygy of \underline{Cok}\underline{f} = Cok d_{C(f)}^{−1}. So it is natural to ask the converse: Is a given map f represented by monomorphisms if \underline{Ker}\underline{f} is a first sygyzy? This section deals with the problem. As a conclusion, the answer is yes if the total ring of fractions Q(R) of R is Gorenstein. What is more, if Q(R) is Gorenstein, instead of a pseudo-kernel, we can use a (usual) kernel to describe rbm condition.

The next is well known. See [1] and [4] for the proof.

Definition and Lemma 4.1 The following are equivalent for an R-module M.

1) The natural map ϕ : M → M^{∗∗} is a monomorphism.

2) Ext_{R}^{1}(Tr M, R) = 0

3) M is a first syzygy; there exists a monomorphism from M to a projective module.

If M satisfies these conditions, M is said to be torsionless. ^{1}

To solve our problem, the special kind of maps is a key. For M ∈ mod R, consider a module J^{2}M = Tr Ω_{R}^{1} Tr Ω_{R}^{1}M. Since Tr J^{2}M is a first sygyzy, we have Ext_{R}^{1}(J^{2}M, R) = 0, which means H_{−1}(F_{J^{2}M}^{∗}_{•}) = 0 and τ_{≥−2}F_{J^{2}M}^{∗}_{•}is a projective resolution of Tr Ω_{R}^{1}M = Cok (d_{F_{J^{2}M}} ^{−2})^{∗} = Cok (d_{F_{M}} ^{−2})^{∗}. The identity map on Tr Ω_{R}^{1}M induces a chain map (F_{M})_{•}^{∗} → (F_{J^{2}M})_{•}^{∗} and its R-dual ψ_{M}^{•} : F_{J^{2}M}^{•} → F_{M}^{•} subsequently.

Lemma 4.2 The map ψ_{M} : J^{2}M → M is rbm if and only if an R-module M has (Ext_{R}^{1}(M, R))^{∗} = 0. proof. From Theorem 3.6, ψ_{M} is rbm if and only if H^{−1}(C(ψ_{M})^{•}) = 0. By definition, ψ_{M}^{−1} and ψ_{M}^{−2} are identity maps hence ψ_{M}^{i} are identity maps for i ≤ −1. We may assume τ_{≤−2}C(ψ_{M})^{•} = 0, which implies H^{−1}(C(ψ_{M})^{•}) = Ker d_{C(ψ_{M})}^{−1}. As τ_{≥−1}C(ψ_{M})_{•}^{∗} is a projective resolution of Cok (d_{ψ_{M}} ^{−1})^{∗} ≅ H_{−1}(C(ψ_{M})_{•}^{∗}), we get H^{−1}(C(ψ_{M})^{•}) ≅

1In [1], Auslander and Bridger use the term ”1-torsion free” for ”torsionless”. Usually a module M is called torsion-free if the natural map M → M ⊗ Q(R) is injective.

(H_{−1}(C(ψ_{M})_{•}^{∗}))^{∗}. A triangle F_{J^{2}M}^{•} →^{ψ_{M}^{•}} F_{M}^{•} → C(ψ_{M})^{•} → F_{J^{2}M}^{•+1} induces an R-dual

triangle F_{J^{2}M} _{•+1}^{∗} → C(ψ_{M})_{•}^{∗} → F_{M} _{•}^{∗} → F_{J^{2}M} _{•}^{∗} which produces an exact sequence of modules

0 → H_{−1}(C(ψ_{M})_{•}^{∗}) → H_{−1}(F_{M} _{•}^{∗}) → H_{−1}(F_{J^{2}M} _{•}^{∗}) → 0.

As we see in the discussion above, H_{−1}(F_{J^{2}M} _{•}^{∗}) = 0. Hence H_{−1}(C(ψ_{M})_{•}^{∗}) ≅ H_{−1}(F_{M} _{•}^{∗}) = Ext_{R}^{1}(M, R), and we get H^{−1}(C(ψ_{M})^{•}) ≅ (Ext_{R}^{1}(M, R))^{∗}. (q.e.d.)

The above result is generalized as follows:

Lemma 4.3 Let f : A → B be a morphism in mod R. Suppose (Ext_{R}^{1}(B, R))^{∗} = 0. If \underline{Ker}\underline{f} is projective, then f is rbm.

proof. We may assume τ_{≤−2}C(f)^{•} = 0. Similarly as in the proof of Lemma 4.2, we have H^{−1}(C(f)^{•}) ≅ (H_{−1}(C(f)_{•}^{∗}))^{∗}. Since \underline{Ker}\underline{f} is projective, f induces a stable

isomorphism J^{2}A ≅^{st} J^{2}B, and via this stable isomorphism, ψ_{B} is projective stably equivalent to f ∘ ψ_{A}, equivalently ψ_{B}^{•} ≅ f^{•} ∘ ψ_{A}^{•} in K(mod R). We have a triangle

C(ψ_{A})^{•} → C(ψ_{B})^{•} → C(f)^{•} → C(ψ_{A})^{•+1}

and its R-dual

C(ψ_{A})^{∗}_{•+1} → C(f)^{∗}_{•} → C(ψ_{B})^{∗}_{•} → C(ψ_{A})^{∗}_{•}

which induce an exact sequence of modules

0 → H_{−1}(C(f)_{•}^{∗}) → H_{−1}(C(ψ_{B})_{•}^{∗})

Note that H_{−1}(C(ψ_{B})^{∗})_{•}) = Ext_{R}^{1}(B, R). The assumption (Ext_{R}^{1}(B, R))^{∗} = 0 equivalently says Ext_{R}^{1}(B, R)_{p} = 0 for any associated prime ideal p of R. A submodule has the same property; H_{−1}(C(f)_{•}^{∗})_{p} = 0 for any associated prime ideal p of R therefore (H_{−1}(C(f)_{•}^{∗}))^{∗} = 0. (q.e.d.)

Proposition 4.4 Let f : A → B a morphism of mod R. Suppose (Ext_{R}^{1}(B, R))^{∗} = 0. Then f is rbm if and only if \underline{Ker}f is torsionless.

proof. We already get the ”only if ” part and have only to show the ”if” part. Adding a projective cover of B to f, we get an exact sequence

σ_{f} : 0 → \underline{Ker}\underline{f} \underrightarrow{\binom{n^{f}}{q}} A ⊕ P_{B} →^{(f ρ_{B})} B → 0.

Due to Theorem 3.6, we have a perfect exact sequence θ_{n^{f}}, because n^{f} is rbm:

θ_{n^{f}} : 0 → \underline{Ker}\underline{f} \underrightarrow{\binom{n^{f}}{ϵ}} A ⊕ F_{\underline{Ker}\underline{f}}^{1} \underrightarrow{(c^{n^{f}} π)} \underline{Cok} \underline{n^{f}} → 0
↓st ; ≅  ↓st ; ≅  ↓^{ω_{f}}
σ_{f} 0 → \underline{Ker}\underline{f} \underrightarrow{\binom{n^{f}}{q}} A ⊕ P_{B}  →^{(f ρ_{B})}  B  → 0.

From Theorem 3.6 3), we know \underline{Ker}\underline{ω_{f}} is projective. With the assumption (Ext_{R}^{1}(B, R))^{∗} = 0, we can apply Lemma 4.3 and get that ω_{f} is rbm. From the equation \underline{f} = \underline{ω_{f}} ∘ \underline{c^{n^{f}}}, f is rbm if c^{n^{f}} is rbm. Since

F_{\underline{Ker}\underline{f}}^{•} →^{n^{f}^{•}} F_{A}^{•} →^{c^{n^{f}} ^{•}} C(n_{f})^{•} → F_{\underline{Ker}\underline{f}}^{•+1}

is a triangle, C(c^{n^{f}})^{•} ≅ F_{\underline{Ker}\underline{f}}^{•+1}; H^{−1}(C(c^{n^{f}})^{•}) ≅ H^{0}(F_{\underline{Ker}\underline{f}}^{•}) ≅ Ext_{R}^{1}(Tr \underline{Ker}\underline{f}, R). Hence c^{n^{f}} is rbm if and only if \underline{Ker}\underline{f} is torsionless. (q.e.d.)

Lemma 4.5 Let the sequence of R-modules 0 → A →^{f} B →^{g} C → 0 be exact. Suppose (Ext_{R}^{1}(C, R))^{∗} = 0. If A and C are torsionless, then so is B. proof. From the assumption, A ≅ \underline{Ker}\underline{g} is torsionless. Due to Proposition 4.4, g is rbm; there exists an exact sequence

θ_{g} : 0 → B \underrightarrow{\binom{g}{ϵ}} C ⊕ Q → \underline{Cok}\underline{g} → 0

with a projective module Q and a map ϵ : B → Q. Since C is a submodule of some projective module, so is B. (q.e.d.) Proposition 4.6 The following are equivalent for a noetherian ring R.

1) Q(R) is Gorenstein.

2) Q(R) is Gorenstein of dimension zero.

3) (Ext_{R}^{1}(M, R))^{∗} = 0 for each M ∈ mod R.

4) Ψ_{M} is rbm for each M ∈ mod R. If R is a local ring with the maximal ideal m, the above conditions are also equivalent to the following.

5) Ψ_{R/m} is rbm. proof. As Q(R) is always of dimension zero, we get 1) ⇔ 2).

3) ⇔ 4) is already shown in Lemma 4.2.

4) ⇒ 5) is obvious.

5) ⇒ 1). The condition 5) is equivalent to Ext_{R}^{1}(R/m, R)⊗Q(R) ≅ Ext_{Q(R)}^{1}(R/m⊗ Q(R), Q(R)) = 0, which means Q(R) is Gorenstein. (q.e.d.)

In the case Q(R) is Gorenstein, every morphism in mod R satisfies the hypotheses of Proposition 4.4 and Lemma 4.5. Thus with the condition Q(R) is Gorenstein, when discussing rbm property, we can deal with normal kernel as well as pseudo-kernel.

Proposition 4.7 Suppose Q(R) is Gorenstein. For a given morphism f, Ker f is torsionless if and only if \underline{Ker}\underline{f} is torsionless.

proof. From Lemma 2.7, there is an exact sequence 0 → Ker f → \underline{Ker} f → Ω_{R}^{1}(Cok f) → 0. So the ”if” part is obvious, and the ”only if” part comes from Lemma 4.5. (q.e.d.)

Theorem 4.8 Suppose Q(R) is Gorenstein. The following are equivalent for a morphism f : A → B in mod R.

1) f is rbm.

2) Ker f is torsionless.

3) \underline{Ker}\underline{f} is torsionless.

4) H^{−1}(C(f)^{•}) = 0.

5) Ω_{R}^{1}(\underline{Cok}\underline{f}) ≅^{st} \underline{Ker}\underline{f} .

6) There exists f′ such that f′ ≅^{st} f and Ker f′ is torsionless.

7) For any f′ with f′ ≅^{st} f, Ker f′ is torsionless.

proof. Implications 5) ⇒ 3), 7) ⇒ 2) and 7) ⇒ 6) are obvious. We already showed 1) ⇔ 4) in Theorem 3.6, 1) ⇔ 3) in Proposition ??, and 3)⇔ 2) in Corollary 4.7. Implications 3) ⇒ 7) and 6) ⇒ 3) are obtained from ”if” and ”only if ” part of Corollary 4.7 respectively.

4) ⇒ 5). It comes directly from Cok d_{C(f)}^{0} = \underline{Ker}\underline{f} and Cok d_{C(f)}^{−1} = \underline{Cok}\underline{f}. (q.e.d.)

Remark 4.9 Takashima gives an easy proof for Theorem 4.8 using the torsion theory [7].

Corollary 4.10 The following are equivalent for a noetherian ring R.

1) Q(R) is Gorenstein.

2) Every morphism with torsionless kernel is rbm.

proof.

1) ⇒ 2). It comes directly from Theorem 4.8.

2) ⇒ 1). For every M ∈ mod R, Ker ψ_{M} is torsionless. Because \underline{Ker}\underline{ψ_{M}} is projective and Ker ψ_{M} is a submodule of \underline{Ker}\underline{ψ_{M}} from Lemma 2.7 1). So if 2) holds, ψ_{M} is rbm for any M ∈ mod R, which implies 1) from Proposition 4.6. (q.e.d.)

Acknowledgement. I thank Kazuhiko Kurano who suggested that conditions for rbm should be given in terms of the kernel not only by the pseudo-kernel. I also thank Shiro Goto who told me that the assumption of Theorem 4.8 is weakened and that Corollary 4.10 holds.

References

[1] M.Auslander and M.Bridger, “ The stable module theory,” Memoirs of AMS. 94., 1969.

[2] M.Auslander and R.O.Buchweitz, The homological theory of maximal Cohen-Macaulay approximations, Soc. Math. de France, Mem 38(1989), 5-37.

[3] M.Amasaki, Homogeneous prime ideals and graded modules fitting into long Bourbaki sequences, Journal of Pure and Applied Algebra, 8(2001), 1-21.

[4] E.G.Evans and P.Griffith, “Syzygies,” LMS Lecture Note Series. 106, 1985.

[5] K.Kato, Stable module theory with kernels, Math. J. Okayama Univ., 43 (2001), 31-41.

[6] K.Kato, Morphisms represented by monomorphisms, archive, math.AC/0404243 (2004).

[7] K.Kato and appendix by S.Takashima, Morphisms represented by monomorphisms, preprint (2004).

[8] J.Miyachi, Derived categories with applications to representations of algebras, Seminar Note, 2000.

[9] Y.Yoshino, Maximal Buchsbaum modules of finite projective dimension, J. Algebra, 159 (1993) , 240-264.

Department of Applied Mathematics Osaka Women’s University Sakai, Osaka 590-0035, JAPAN kiriko@appmath.osaka-wu.ac.jp