- Email: [email protected]
Morphisms represented by monomorphisms
Journal of Pure and Applied Algebra 208 (2007) 261–283 www.elsevier.com/locate/jpaa
Morphisms represented by monomorphisms Kiriko Kato Department of Mathematics and Information Sciences, Osaka Prefecture University Daisen, Sakai, Osaka 590-0035, Japan Received 28 July 2005; received in revised form 28 August 2005 Available online 20 January 2006 Communicated by I. Reiten
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, 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 projectivestably equivalent. c 2005 Elsevier B.V. All rights reserved.
MSC: 13D02; 13D25; 16D90
1. Introduction Let R be a commutative noetherian ring. Linear maps f : A → B and f 0 : A0 → B 0 of finite R-modules are said to be projective-stably equivalent (pse for short) if the following diagram is commutative 0 A⊕ P ∼ y=
→
A0 ⊕ P
f s t u
f 0 s0 t 0 u0
→
0 B⊕ Q ∼ y=
B0 ⊕ Q
with some projective modules P, Q, P 0 , Q 0 and R-linear maps s, t, u, s 0 , t 0 , u 0 . 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 ⊕ PB → B is surjective with a projective cover ρ B : PB → B. Thus every morphism is represented by epimorphisms. The choice of epimorphism is unique; if an f0
( f ρB )
epimorphism f 0 is pse to f , then two sequences 0 → Ker f 0 → A0 → B 0 → 0 and 0 → Ker( f ρ B ) → A⊕ PB → B → 0 become isomorphic after splitting off common projective summands. On the other hand, every morphism is not always represented by monomorphisms. Even if a morphism f is rbm, the choice of monomorphism is not unique; there may be two monomorphisms f 0 and f 00 both pse to f and that E-mail address: [email protected]. c 2005 Elsevier B.V. All rights reserved. 0022-4049/$ - see front matter doi:10.1016/j.jpaa.2005.12.009
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K. Kato / Journal of Pure and Applied Algebra 208 (2007) 261–283 f0
f 00
0 → A0 → B 0 → Cok f 0 → 0 and 0 → A00 → B 00 → Cok f 00 → 0 are not isomorphic by splitting off common projective summands. The purpose of the paper is finding a condition of a given map to be rbm. The problem was 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 first focal point. In the case where 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 a direct sum of projective modules (Theorem 3.9). Our next 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 the projective stabilization mod R of a category mod R of finite R-modules 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 with the given map. Looking at Theorem 3.9, 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 the 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 4.12: 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 ]]/(X Y ) with any field k, g : R/(X 2 ) → R/(X, Y ) a natural projection. Since Ker g ∼ = (X, Y )/(X 2 ) with depth zero is not a first syzygy, g is not rbm due to Theorem 4.12. 2 Example 2. Set R = k[[X, Y ]]/(X Y ) with any field k, f : R 2 / YX R → R 2 / YX 2 R with f ab mod YX R = 2 Xa X ∼ Y b mod Y 2 R. The map f is not a monomorphism; Ker f = R/(X ) ⊕ R/(Y ) is a first syzygy. By 2 Theorem 4.12, f is rbm. In fact, let f 0 : R 2 / YX R → R 2 / YX 2 R ⊕ R 2 be defined as f 0 ab mod YX R = 2 Xa X Ya mod . Obviously f 0 is a monomorphism that is pse to f . On the other hand, f 00 : Yb Y 2 R, X b 2 2 2 Xa X Y a R 2 / YX R → R 2 / YX 2 R ⊕ R 2 with f 00 ab mod YX R = is also a Y b mod Y 2 R, X 2 b monomorphism and pse to f . We have two exact sequences 2 X X X Y f0 2 2 2 2 2 θf : 0 → R R →R R⊕R →R R⊕R R → 0, Y Y2 Y X and σ :0→R
2
X Y
f 00
R →R
2
X2 Y2
2
R⊕R →R
2
X Y2
R⊕R
2
Y X2
R→0,
that are not isomorphic. We see θ f is perfect but σ is not. 2. Stable module category and homotopy category We shall fix the notations and give a review of the correspondence between stable module category and homotopy class category of complexes. We omit many of the proofs for results that are in [5]. 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 : PM → M denotes a projective cover of M. The category of finitely generated R-modules is denoted by mod R, and the category of finite projective R-modules is denoted by proj R. For an R-module M, M ∗ denotes an R-dual Hom R (M, R). For a morphism f : M → N in mod R, f ∗ denotes Hom R ( f, R) : N ∗ → M ∗ .
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263
For an additive subcategory A of an abelian category, the category of complexes in A is denoted as C(A). A complex F • in A is d F n−1
dF n
F • : · · · → F n−1 → F n → F n+1 → · · · . Truncations of a complex F • are defined as follows: d F n−2
d F n−1
τ≤n F • : · · · → F n−2 → F n−1 → F n → 0 → 0 → · · · , dF n
d F n+1
τ≥n F • : · · · → 0 → 0 → F n → F n+1 → F n+2 → · · · . In the case A = mod R, an R-dual F•∗ of a complex F • ∈ C(A) is the cocomplex such as Fn∗ = (F n )∗ , ∗ dnF = (d F n−1 )∗ . A trivial complex is a direct sum of split exact sequences of projective modules. The category of the homotopy equivalence class of complexes in A is denoted by K(A). A morphism in K(A) is a homotopy equivalence class of chain maps. Note that an object in K(proj R) is zero if and only if it is represented by a trivial complex. Definition 2.1 (Auslander–Bridger [1]). The projective stabilization mod R is defined as follows. • Each object of mod R is an object of mod R. • For objects A, B of mod R, a set of morphisms from A to B is 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 f = f mod P(A, B). Definition 2.2. A morphism f : A → A0 in mod R is called a stable isomorphism if f is an isomorphism in mod R. st
In this case we say A and A0 are stably isomorphic and write A ∼ = A0 . Lemma 2.3. Two R-modules A and A0 are stably isomorphic if and only if there exist projective modules P and P 0 such that A ⊕ P 0 ∼ = A0 ⊕ P in mod R. st
∼ A0 . Then there exist morphisms α : A → A0 and α 0 : A0 → A such Proof. The “if” part is obvious. Suppose A = 0 0 that α ◦ α = id A0 and α ◦ α = id A . In other words, there exist morphisms s : A → PA and s 0 : A0 → PA0 such that α ◦ α 0 + ρ A0 ◦ s 0 = id A0 and α 0 ◦ α + ρ A ◦ s = id A . Morphisms α and α0 induce a map α P : PA → PA0 and α ρ α 0P : PA0 → PA as ρ A0 ◦ α P = α ◦ ρ A and ρ A ◦ α 0P = α 0 ◦ ρ A0 . It is easy to see s −αA00 : A ⊕ PA0 → A0 ⊕ PA and P 0 α ρA 0 0 : A ⊕ PA → A ⊕ PA are isomorphisms. s0 −α P
Definition 2.4. In any category C, morphisms f : A → B and f 0 : A0 → B 0 are said to be isomorphic and denoted as f ∼ = f 0 if there exist isomorphisms α : A → A0 and β : B → B 0 such that β ◦ f = f 0 ◦ α. Definition 2.5. Morphisms f : A → B and f 0 : A0 → B 0 in mod R are said to be projective-stably equivalent (pse st for short) and denoted as f ∼ = f 0 if f ∼ = f 0 in mod R. Lemma 2.6. Morphisms f : A → B and f 0 : A0 → B 0 in mod R are pse if and only if there exist projective modules P, P 0 , Q, Q 0 , morphisms t, t 0 , χ , χ 0 , isomorphisms α˜ and β˜ in mod R such that the following diagram commutes in mod R. 0 A⊕ P yα˜
⊕P
f0 0 t0 χ
0 B⊕ Q ˜ yβ
→
A0
0 χ0
f t
→
B0 ⊕ Q
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Proof. There exist stable isomorphisms α : A → A0 and β : B → B 0 such that β ◦ f = f 0 ◦ α, which implies the existence of morphisms t : A → PB 0 such that −β ◦ f + f 0 ◦ α = ρ B 0 ◦ t. Let α 0 : A0 → A and β 0 : B 0 → B be morphisms such that α 0 = α −1 and β 0 = β −1 . Then from the equations α ◦ α 0 = id A0 and β 0 ◦ β = id B , we get s 0 : A0 → PA0 and u : B → PB with the property α ◦ α 0 + ρ A0 ◦ s 0 = id A0 and β 0 ◦ β + ρ B ◦ u = id B . Morphisms α, β 0 and f 0 induce morphismsα P : PA → PA0 , β P0 : PB 0 → PB and f P : PA → PB respectively. From the proof 0 β ρB0 of Lemma 2.3, we know β˜ = : B ⊕ PB 0 → B 0 ⊕ PB and α˜0 = α0 ρ A : A0 ⊕ PA → A ⊕ PA0 are 0 u
−β P
s
−α P
isomorphisms. Now the diagram
A ⊕ PA0 −1 yα˜0
˜ β◦
A0
⊕ PA
gives the conclusion.
f t
f t
0 f0P
B ⊕ PB 0 ˜ yβ
→
0 0 fP
◦α˜0
→
B 0 ⊕ PB
Definition 2.7 (Auslander–Bridger [1]). δ
(1) For an R-module M, define a transpose Tr M of M to be Cok δ ∗ where P → Q → M → 0 is a projective presentation of M. The transpose of M is uniquely determined as an object of mod R. If f ∈ Hom R (M, N ), then f induces a map Tr N → Tr M, which represents a morphism Tr f ∈ Hom R (Tr N , Tr M). (2) A kernel of projective cover of an R-module M is called the first syzygy module of M and denoted as Ω R1 (M). The first syzygy module of M is uniquely determined as an object of mod R. Inductively, we define Ω Rn (M) = Ω R1 (Ω Rn−1 (M)). If f ∈ Hom R (M, N ), then f induces a map Ω Rn (M) → Ω Rn (N ), which represents a morphism Ω Rn ( f ) ∈ Hom R (Ω Rn (M), Ω Rn (N )). Let L be a full subcategory of K(mod R) defined as L = {F • ∈ K(proj R) | Hi (F • ) = 0 (i < 0), H j (F ∗ • ) = 0 ( j ≥ 0)}. Lemma 2.8 ([5, Lemma 2.1]). For a morphism f • ∈ L, f • = 0 in K(mod R) if and only if H0 (τ≤0 f • ) = 0 in mod R. Proof. Let f • : F • → G • be a chain map with F • , G • ∈ L such that H0 (τ≤0 f • ) = 0. Then there exists g ∈ Hom R (H0 (τ≤0 F • ), G 0 ) that satisfies H0 (τ≤0 f • ) = ρ ◦ g where ρ : G 0 → H0 (τ≤0 G • ) is the natural projective cover. We get chain maps ρ • ∈ Hom R (G 0 , τ≤0 G • ) and g • ∈ Hom R (τ≤0 F • , G 0 ) such as H0 (ρ • ) = ρ and H0 (g • ) = g. From the assumption, τ≤0 f • is homotopic to ρ • ◦ g • , which implies f i = h i+1 ◦ d F i + dG i−1 ◦ h i ∗
with some h i+1 : F i+1 → G i for i ≤ −1. Similarly, since H0 (τ≤0 f • ) = 0, we have f j = h j+1 ◦ d F j + dG j−1 ◦ h j with some h j : F j → G j−1 for j ≥ 2. Therefore as an element of L, we may assume f i = 0 (i 6= 0, 1). Moreover, ∗ ∗ ∗ ∗ we may assume f i = 0 (i 6= 1); since d F −1 ◦ f 0 = 0, ( f 0 )∗ passes through Ker d F −1 = Im d F 0 hence we get ∗ ∗ s 1 : F 1 → G 0 with f 0 = s 1 ◦ d F 0 . Finally, to see f • = 0, observe d F 0 ◦ f 1 = 0, then we get u 2 : F 2 → G 1 with ∗ ∗ ∗ ∗ ∗ f 1 = u 2 ◦ d F 1 . Since d F 1 ◦ u 2 ◦ dG 1 = f 1 ◦ dG 1 = 0, there exists a map u 3 : F 3 → G 2 such that dG 1 ◦ u 2 + u 3 ◦ d F 2 = 0. Thus we obtain a homotopy map u : F • → G •−1 which shows that f • is homotopic to zero.
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Suppose a chain map f • : F • → G • is homotopic to zero with a homotopy map h • : F • → G •−1 . We construct chain maps δ • : τ≤0 F • → F 1 and η• : F 1 → τ≤0 G • as follows: ···
→
···
→
···
→
F−1 y 0 y G −1
d F −1
→
→ dG −1
→
F0 0 yd F F1 1 yh G0
It easy to see H0 (τ≤0 f • ) = H0 (δ • ◦ η• ) = H0 (δ • ) ◦ H0 (η• ), the righthand-side passes through F 1 .
Lemma 2.9 ([5, Proposition 2.3, Proposition 2.4]). (1) For A ∈ mod R, there exists FA • ∈ L that satisfies st
H0 (τ≤0 FA • ) ∼ = A. Such an FA • is uniquely determined by A up to isomorphisms. We fix the notation FA • and call this a standard complex of A. (2) For f ∈ Hom R (A, B), there exists f • ∈ HomK(proj R) (FA • , FB • ) that satisfies H0 (τ≤0 f • ) ∼ = f. Such an f • is uniquely determined by f up to isomorphisms, so we use the notation f • to describe a chain map with this property for given f . Proof. (1) First take a projective resolution PA • of A: · · · → PA−2 → PA−1 → PA0 → A → 0, ∗
and then a projective resolution PTr A • of Tr A = Cok d FA −1 as ∗
∗
−2 −3 0 ← Tr A ← PA−1 ← PA0 ← PTr A ← PTr A ← · · · .
Define a complex FA • as i PA (i ≤ −1) i FA = − 1−i ∗ PTr A (i ≥ 0), i d (i ≤ −1) d FA i = PA − 2−i ∗ d PTr A (i ≥ 0). We easily see FA • ∈ L and st
H0 (τ≤0 FA • ) ∼ = A. Suppose both FA • and FA0 • have this property. Adding some trivial complex P • of projective modules P• : · · · → 0 → P0 = P1 → 0 → · · · • if necessary, we may assume that H0 (τ≤0 FA • ) ∼ = H0 (τ≤0 F 0 A ) ∼ = A in mod R. Then there are chain maps • • • • 0 0 • • ϕ : τ≤0 FA → τ≤0 FA and γ : τ≤0 FA → τ≤0 FA such that ϕ • ◦ γ • = 1τ≤0 F 0 • and γ • ◦ ϕ • = 1τ≤0 FA • . As ∗
A
∗
∗ and τ 0 0∗ ∗ τ≥−1 FA• ≥−1 F A• are acyclic, H−1 (τ≥−1 ϕ• ) induces a chain map τ≥−1 F A • → τ≥−1 F A •. With this map for • • 0 • • the positive part, ϕ can be extended to a chain map f : FA → FA such that τ≤0 f • = ϕ • . Similarly we get a chain
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map g • : FA0 • → FA • such that τ≤0 g • = γ • . It is easy to see H0 (τ≤0 ( f • ◦ g • )) = 1 A and H0 (τ≤0 (g • ◦ f • )) = 1 A . From Lemma 2.8, we have f • ◦ g • = 1 F 0 • and g • ◦ f • = 1 FA • . A (2) An R-linear map f : A → B of R-modules is lifted to a chain map between projective resolutions f 0• : τ≤0 FA • → τ≤0 FB • which again induces an R-linear map H−1 (τ≥−1 f 0•∗ ) : Tr B → Tr A lifted to a chain map f 00• : τ≥−1 FA • → τ≥−1 FB • . We may choose f 00 as f 00i = f 0i (i = −1, 0), and obtain f • : FA • → FB • with f i = f 0i (i ≤ 0) and f j = f 00 j ( j ≥ −1). Uniqueness follows from Lemma 2.8. The above lemma is summed up in the following way. Theorem 2.10 ([5, Theorem 2.6]). The mapping A 7→ FA • gives a functor from mod R to K(mod R), and this gives a category equivalence between mod R and L. For f ∈ Hom R (A, B), there exists a triangle nf•
cf •
f•
C( f )•−1 → FA • → FB • → C( f )• .
(2.1)
In general, C( f )• does not belong to L but it satisfies the following: Hi (C( f )• ) = 0 (i < −1),
H j (C( f )∗ • ) = 0 ( j > −1).
Definition and lemma 2.11 ([5, Definition and Lemma 3.1]). As objects of mod R, Ker f := H−1 (τ≤−1 C( f )• ) and Cok f := H0 (τ≤0 C( f )• ) are uniquely determined by f , up to isomorphisms. We call these the pseudo-kernel and the pseudo-cokernel of f . And we have Cok f = Tr Ker Tr f . Notations. The pseudo-kernel or the pseudo-cokernel is determined as a stable equivalence class of modules, not as st st an isomorphic class of mod R. Although, for simplicity, a module M with M ∼ = Ker f or M ∼ = Cok f is denoted by Ker f or Cok f respectively. Chain maps n f • and c f • in (2.1) induce n f : Ker f → A and c f : B → Cok f . These maps have the properties f ◦ n f = 0 and c f ◦ f = 0. The following lemma explains why Ker f and Cok f are called the pseudo-kernel and the pseudo-cokernel respectively. Lemma 2.12 ([5, Lemma 3.3, Lemma 3.5]). Let f : A → B be a homomorphism of R-modules. (1) If x ∈ Hom R (X, A) satisfies f ◦ x = 0, there exists h x ∈ Hom R (X, Ker f ) such that x = n f ◦ h x . (2) If y ∈ Hom R (B, Y ) satisfies y ◦ f = 0, there exists e y ∈ Hom R (Cok f , Y ) such that y = e y ◦ c f . From (2.1), we have an exact sequence ( f ρ)
0 → Ker f → A ⊕ P → B → 0 with some projective module P. This characterizes the pseudo-kernel. ( f p0 ) ( f p) Lemma 2.13. For a given f ∈ Hom R (A, B), suppose both A ⊕ P → B and A ⊕ P 0 → B are epimorphisms with projective modules P and P 0 . Then there are stable isomorphisms λ : A ⊕ P → A ⊕ P 0 and κ : Ker ( f p) → 0 Ker f p that make the following diagram commutative: 0 →
0
→
Ker(f p) → κ y Ker f p 0 →
A⊕ P yλ A ⊕ P0
( f p)
→
B
→
0
→
0
k ( f p0 ) →
B
K. Kato / Journal of Pure and Applied Algebra 208 (2007) 261–283
267
Proof. With a canonical map π : B → Cok f , π ◦ p and π ◦ p 0 are both projective covers of Cok f , hence there exists l : P → P 0 such that π ◦ p = π ◦ p 0 ◦ l. Then Im p − p 0 ◦ l ⊂ Im f , so we get h : P → Im f as h coincides with p − p 0 ◦ l. Via A → Im f which issurjective, h can be lifted to a map j : P → A. This shows the equation f ◦ j + p 0 ◦ l = p. The map λ = 10 lj : A ⊕ P → A ⊕ P 0 yields the desired diagram. Obviously λ is a stable isomorphism, which implies that κ is a stable isomorphism from the following Lemma 2.14. Lemma 2.14. On the commutative diagram with exact rows in mod R f
0
→
A → α y
0
→
A0
f0
→
g
B → β y B0
0
C → γ y
g0
→ C0
0,
→
if β and γ are stable isomorphisms, so is α. Proof. We show that α is a stable isomorphism in the following case: (1) γ is an isomorphism and β is a stable isomorphism. (2) β and γ are stable isomorphisms and γ is an epimorphism. (3) β and γ are stable isomorphisms. (1) Adding a projective cover ρ A0 : PA0 → A0 to the given diagram, we get the following: 0 ↓ Q v y 0
→
0 ↓ Q w y
= f0 ( 01 )
(g 0)
A ⊕ PA0 → (α ρ 0 ) y A f0
0 A y 0
0 →
B ⊕ PA0 → (β f 0 ◦ρ ) y A0 g0
B0 y 0
→
→
C → ∼ y=
0
C0
0
→
Since (β f 0 ◦ ρ A0 ) is an epimorphism and a stable isomorphism at the same time, it is a split epimorphism and Q is projective. In other words, w is a split monomorphism. Hence v is also a split monomorphism and (α ρ A0 ) a split epimorphism with a projective kernel. In particular, α is a stable isomorphism. (2) Since γ is a split epimorphism with a projective kernel, there exists γ 0 : C 0 → C such that γ ◦ γ 0 = idC 0 . On the diagram 0
→
A
→
k 0
→
A
f
→
00
B → C0 → 0 0 yβ yγ g
B
→
C
→
0
0,
β 0 is a stable isomorphism because Cok γ 0 ∼ = Cok β 0 is projective. Connection of the above and the given diagrams yields another diagram: 0 →
A → α y
0
A0
→
f0
→
00
B → C 0 0 k yβ◦β B0
g0
→ C0
→
0
→
0
Since β ◦ β 0 is a stable isomorphism, we can apply (1).
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K. Kato / Journal of Pure and Applied Algebra 208 (2007) 261–283
(3) Adding a projective cover ρ B 0 : PB 0 → B 0 , we get 0
→
0
→
( 0f ) A → α y
( g0 01 )
B ⊕ PB 0 → (β ρ 0 ) y B
f0
A0
g0
B0
→
C ⊕ PB 0 → 0 y(γ g ◦ρ B ) C0
→
→
0
0.
Since β is a stable isomorphism and (γ g 0 ◦ ρ B ) is an epimorphic stable isomorphism, we can apply (2) to get the conclusion. Lemma 2.13 says that the sequences ( f p)
0 → Ker ( f p) → A ⊕ P → B → 0 and ( f p0 ) 0 → Ker f p 0 → A ⊕ P 0 → B → 0 are, isomorphic adding some split exact sequence of projective modules if necessary. This comes from the following Lemma 2.16. Definition 2.15. Two morphisms f : A → B and f 0 : A0 → B 0 are said to be isomorphic up to a direct sum of projective modules if there exist projective modules O, P, Q, O 0 , P 0 and Q 0 such that
f 0 0
A⊕O⊕P
0 0 0
0 1 0
B⊕P⊕Q
→
and f0 0 0
A0 ⊕ O 0 ⊕ P 0
0 0 0
0 1 0
!
B0 ⊕ P 0 ⊕ Q0
→
are isomorphic. The above definition equivalently says that the two exact sequences are isomorphic:
f 0 0
0 → Ker f ⊕ O → A ⊕ O ⊕ P
0 0 0
0 1 0
B ⊕ P ⊕ Q → Cok f ⊕ Q → 0,
→ f0 0 0
0 → Ker f 0 ⊕ O 0 → A0 ⊕ O 0 ⊕ P 0
0 0 0
0 1 0
!
→
B 0 ⊕ P 0 ⊕ Q 0 → Cok f ⊕ Q 0 → 0.
Each sequence is the direct sum of f
0 → Ker f → A → B → Cok f → 0 with a trivial complex 1 0
0 1 0 0
(0 1)
0→O → O⊕P → P⊕Q → Q→0 and so on.
st
Obviously f ∼ = f 0 if f and f 0 are isomorphic up to a direct sum of projective modules but the converse is not true. Lemma 2.16. The following are equivalent for two morphisms f : A → B and f 0 : A0 → B 0 in mod R. (1) f and f 0 are isomorphic up to a direct sum of projective modules.
K. Kato / Journal of Pure and Applied Algebra 208 (2007) 261–283
269
(2) There exist stable isomorphisms α, β and γ such that the next diagram commutes: f
B → β y
Cok f γ y
→
0
B0
Cok f 0
→
0
A → α y A0
f0
→
→
(3) There exist stable isomorphisms α, β, γ and δ such that the next diagram commutes: f
0
→
Ker f yδ
→
A → α y
B → β y
Cok f γ y
→ 0
0
→
Ker f 0
→
A0
B0
→
Cok f 0
→ 0
Cok f γ y Cok f 0
→
0
→
0,
f0
→
Proof. (1) ⇒ (2) is obvious. (2) ⇒ (3). On the diagram 0
→
0
→
Im f ε y Im f 0
B → β y B0 →
→ →
ε is a stable isomorphism by Lemma 2.14. From 0
→
0
→
Ker f yδ Ker f 0
Im f ε y Im f 0
A → α y A0 →
→ →
→
0
→
0,
δ is a stable isomorphism also by Lemma 2.14. (3) ⇒ (1). It is enough to show the following: A diagram with exact rows and stable isomorphisms α, β and γ f
0
→
A → α y
0
→
A0
f0
→
g
B → β y
C → γ y
g0
B0
→ C0
→
0
0
˜ γ˜ , and projective modules O, P, Q, O 0 , P 0 , Q 0 induces a diagram with exact rows, isomorphisms α, ˜ β,
0 →
A⊕ O yα˜
f 0 0
0 1 0
A0 ⊕ O 0
0 1 0
B⊕O ⊕P ˜ yβ
→ f0 0 0
0 →
g 0 0 0 0 1
g0 0 0 0 0 1
→
C⊕ P γ˜ y
→
0
C0 ⊕ P0
→
0.
!
B0 ⊕ O0 ⊕ P 0
→
→
First we get a diagram with exact rows:
0
→
0
→
f 0 0
0 1 0
A ⊕ PA0 → αρ y( A0 ) A0
f0
→
g 0 0 0 0 1
B ⊕ PA0 ⊕ PB 0 → β ρ ◦f0 ρ y( B 0 P B 0 ) B0
g0
→
C ⊕ PB 0 → 0 γ g0 ◦ρ y( B0 ) C0
→
0
where f 0 P : PA0 → PB 0 is a map induced by f 0 . All the vertical maps are split epimorphisms with projective kernels. Hence the upper row is isomorphic to the direct sum of the lower row and a split exact sequence of projective modules.
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Lemma 2.17 ([5, Lemma 3.6]). (1) There is an exact sequence 0 → Ker f → Ker f → Ω R1 (Cok f ) → 0. (2) There is an exact sequence 0 → L → Cok f → Cok f → 0 such that Ω R1 (L) is the surjective image of Ker f . Proof. (1) It follows from the diagram:
0
→
0
→
0
→
0 y Ker f y
0 y A 1 y 0
→
Ker f y 1 Ω R (Cok f) y 0
→ →
0 y Im f y
→ ( f ρB )
A ⊕ PB 0 y 1 PB y 0
B y Cok f y 0
→
→
→
0
→
0
→
0
(2) The transpose Tr f : Tr B → Tr A of a given map f : A → B makes an exact sequence 0 → Ker Tr f → Tr B ⊕ PTr A
(Tr f ρ Tr A )
→
Tr A → 0.
Dualizing this sequence with R, we get a long exact sequence 0 → (Tr A)∗ → (Tr B)∗ ⊕ (PTr A )∗ → (Ker Tr f )∗
→A
f Tr ρ Tr A
−→
B ⊕ Q → Cok f → 0
with some projective module Q. Put fˇ : A 0
→
0 →
Im fˇ → y Im f →
B⊕ Q → (1 0) y B →
f Tr ρ Tr A
→
Cok f y Cok f
B ⊕ Q. On the commutative diagram →
0
→
0
all the vertical maps are surjective. If we put L 0 the kernel of epimorphism Cok f → Cok f , then Ω R1 (L 0 ) is the kernel of the natural map A/Ker fˇ ∼ = Im fˇ → Im f ∼ = A/Ker f. Therefore Ω R1 (L 0 ) ∼ = Ker f /Ker fˇ.
Lemma 2.18. The following holds for f ∈ Hom R (A, B). (1) Ker f is projective if and only if there exists f 0• such that f • ∼ = f 0• in K(proj R) and f 0i = id for i ≤ −1. (2) If Ker f is projective, then Ω R1 ( f ) is a stable isomorphism. Proof. The chain map f • : FA • → FB • induces an exact sequence cf •
n f •+1
0 → FB • → C( f )• → FA •+1 → 0
K. Kato / Journal of Pure and Applied Algebra 208 (2007) 261–283
271
in C(proj R). There is a commutative diagram with exact rows: 0
0
→
FB • s • y f
→ cn f •
FA •
→
k f•
FA •
C(nf )• t • yf
cf•
C( f )•
→
n f •+1
→
FA •+1
→ 0
k n n f •+1
C( f )•
→
→ 0
FB •
→
We have t f • ◦ s f • = id FB • in C(proj R) and s f • ◦ t f • = idC(n f )• in K(proj R). We often write C yl( f )• for C(n f )• . (1) The “if” part is obvious. Suppose Ker f is projective. Then · · · → C( f )−2 → C( f )−1 → Ker f → 0 is a trivial complex, namely a direct sum of split exact sequences of projective modules. Hence there exists a complex C 0• such that C 0i = 0 (i ≤ −2) and C( f )• ∼ = C 0• in K(proj R). Let γ • : C( f )• → C 0• be a chain map that is an 0• isomorphism in K(proj R). Putting c : C(n f )• → C 0• as c0• = γ • ◦ n n f •+1 , we have a commutative diagram with exact rows in C(proj R): 0
0
→ C( f)•−1 → •−1 yγ →
C 0 •−1
C(nn f )• • yα
→ C(c0 )•−1
nnn
•+1 f
→
C(n f )•
→
0
→
0
k f 0•
→
C(n f )•
The following diagram commutes in K(proj R): C(nn f ) tn • y f FA •
nnn
•+1 f
→
k cn f •
→
k FA •
C(n f )•
f•
→
C(nf )• t • yf FB •
In K(proj R), α • , tn f • , and t f • are isomorphisms hence f • ∼ = f 0• . Clearly f 0i = id (i ≤ −1) and we get the conclusion. (2) Apply H−1 τ≤−1 to cn f •
n n f •+1
0 → FA • → C(n f )• → C( f )• → 0, we get 0 → Ω R1 (A) → Cok dC(n f ) −2 → Ker f → 0. If Ker f is projective, H−1 (τ≤−1 cn f • ) is a stable isomorphism. On the equation H−1 (τ≤−1 f • ) = H−1 (τ≤−1 t f • ) ◦ H−1 (τ≤−1 cn f • ), H−1 (τ≤−1 cn f • ) and H−1 (τ≤−1 t f • ) are stable isomorphisms, hence so is H−1 (τ≤−1 f • ) = Ω R1 ( f ). Corollary 2.19. Suppose R is local. If a morphism f ∈ End R (A) satisfies that Ker f is projective, then f is a stable isomorphism. Proof. We may assume C( f )i = 0 (i ≤ −2). Therefore the complex C( f )∗• has no cohomology except for H−1 (C( f )∗• ). The triangle C( f )∗• → FA ∗• → FA ∗• → C( f )∗•−1
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induces an exact sequence 0 → H−1 (C( f )∗• ) → H−1 (FA ∗• ) → H−1 (FA ∗• ) → 0. Since R is local, a surjective endomorphism on a finite module is always an automorphism. Thus we get H−1 (C( f )∗• ) = 0. It follows that C( f )∗• is an exact sequence of projective modules, equivalently, f is a stable isomorphism. 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 0 : A0 → B 0 in mod R is pse to f , that is, there exist stable isomorphisms α : A → A0 and β : B → B 0 such that β ◦ f = f 0 ◦ α. Each morphism is not always rbm. ∗∗ Example 3.2. Let R be a ring of dimension n ≥ 3, N an R-module with pd N = n, and ϕ N : N → N the natural ϕN ? ∗∗ map. Then any map N ⊕ P → N ⊕ Q of the form ? ? with projective modules P and Q, will never be monomorphic. If otherwise, N ⊕ P is a submodule of a projective module; this is a contradiction because N has a maximal projective dimension.
It was Auslander and Bridger who first defined and studied the “represented by monomorphisms” property. Theorem 3.3 (Auslander–Bridger). The following are equivalent for a morphism f : A → B in mod R. (1) There exists a monomorphism f 0 : A → B ⊕ P with a projective module P such that f = s ◦ f 0 via some split epimorphism s : B ⊕ P → B. (2) There exists a monomorphism f 0 : A → B ⊕ P with a projective module P such that f = s ◦ f 0 via some split epimorphism s : B ⊕ P → B, and f 0∗ is an epimorphism. (3) Hom R (B, I ) → Hom R (A, I ) is surjective if I is an injective module. Auslander and Bridger’s original definition of the “represented by monomorphisms” condition is (1) of Theorem 3.3. Seemingly this is different from our definition. But we show that two conditions are equivalent. Lemma 3.4. For a morphism f : A → B in mod R, f is rbm if and only if there exists a monomorphism f 0 : A → B ⊕ P with a projective module P such that f = s ◦ f 0 via some split epimorphism s : B ⊕ P → B. Proof. The “if” part is clear. We shall show the “only if” part. Suppose there exists a monomorphism f 0 : A0 → B 0 , stable isomorphisms α : A → A0 and β : B → B 0 such that β ◦ f = f 0 ◦ α. We first take projective covers ρ A : PA → A and ρ B : PB → B such that the induced map f P : PA → PB by f is a monomorphism. Since α is a stable isomorphism, there exists a morphism α 0 : A0 → A such that α ◦ α 0 = id A0 and α 0 ◦ α = id A . 0 From the ◦ s A = id A , equivalently last equation there exists a morphism s A : A → PA such that α ◦ α + ρ A α 0 0 0 (α ρ A ) ◦ s A = id A . In particular, (α ρ A ) : A ⊕ PA → A is a split epimorphism and sαA : A → A0 ⊕ PA is
a split monomorphism. Similarly weget morphisms β 0 : B 0 → B, s B : B → PB and s B 0 : B 0 → PB 0 such that 0 β 0 (β ρ B ) ◦ s B = id B and (β ρ B 0 ) ◦ sβ 0 = id B 0 . B
Given equation β ◦ f = f 0 ◦ α induces β 0 ◦ (β ◦ f ) ◦ α 0 = β 0 ◦ ( f 0 ◦ α) ◦ α 0 , that is, f ◦ α 0 = β 0 ◦ f 0 . Hence there exists a homomorphism t : A0 → PB such that f ◦ α 0 − β 0 ◦ f 0 = ρ B ◦ t. Now we get a commutative diagram
f0 t
0 fP
−→ A0 ⊕ PA α0 ρ y( A ) A
f
→
B0 ⊕ PB β0 ρ y( B ) B.
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β0 Since the composite of maps
sB0 0
0 0
id PB
: B 0 ⊕ PB → B⊕ PB 0 ⊕ PB and
0 0 is equal to id B 0 ⊕PB , β 0 ρ B ◦ ft f0P = β 0 ρ B ◦id B 0 ⊕PB ◦ ft 0 f 0 t f P , and the following diagram commutes: f
= β0
ρB0 0
0
: B⊕ PB 0 ⊕ PB → B 0 ⊕ PB β0 0 β ρ 0 0 ρ B ◦ 0 0B 0 id P ◦ s B 0 ◦ id PB
B
0
id PB
00
A0 ⊕ PA −→ α0 ρ y( A )
B ⊕ PB0 ⊕ PB ρ y
f
A
0 fP
β 0
B
→
where β0 00 f = sB0 0
0 0 f 0 ◦ t id PB
0 , fP
and β ρB ◦ 0
ρ = β0
ρB0 0
0 . id PB
It is easy to see that f 00 is a monomorphism and ρ is a split epimorphism. Finally putting f 000 = f 00 ◦ sαA which is a monomorphism, we have f = f ◦ id A = f ◦ α ρ A 0
α α 00 ◦ =ρ◦ f ◦ = ρ ◦ f 000 . sA sA
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 3.5. An exact sequence 0 → A → B → C → 0 of R-modules is called a perfect exact sequence or said 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. Proposition 3.6 ([5, Lemma 2.7]). The following are equivalent for an exact sequence f
g
θ : 0 → A → B → C → 0. (1) θ is perfectly exact. f•
g•
(2) 0 → FA • → FB • → FC • → 0 is exact. f•
g•
(3) FC •−1 → FA • → FB • → FC • is a distinguished triangle in K(mod R). (4) FA • ∼ = C(g)•−1 in K(proj R). If these conditions are satisfied, we have the following. st
(5) C ∼ = Cok f . ∼ C( f )• in K(proj R). (6) FC • = Proof. In [5, Lemma 2.7], we see the equivalence between (1) and (2). The implication (3) ⇒ (2), (4), (5), (6) is obvious.
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(2) ⇒ (3). With the same notations as in the proof of Lemma 2.18, we have a diagram in C(proj R) with exact rows: cf •
FB • s • y f 0
→
FA •
cn f •
C(nf )• t • yf
→
k 0 →
FA •
f•
k n n f •+1
→ g•
FB •
→
C( f )•
→
C(f )• γ • y
→
0
FC •
→
0
→
Since s f • and t f • are isomorphisms, γ • is also an isomorphism and g • ∼ = c f • in K(proj R). (4) ⇒ (3). On the diagram in K(proj R), FA • • yα C(g)•−1
f•
→
FB • ,
k cg •
→
g•
FB •
→
FC •
α i = id (i ≤ 0). Hence α • is an isomorphism if FA • ∼ = cg • . = C(g)•−1 . Therefore f • ∼
If R is local, all the conditions above are equivalent. We shall give the proof later at the end of this section. Lemma 3.7. Let the sequence f
g
θ :0→ A→B→C →0 be exact. If R is local, the conditions (1)–(4) in Proposition 3.6 are equivalent to the conditions (5) and (6). st
(5) C ∼ = Cok f . ∼ C( f )• in K(proj R). (6) FC • = ( f ρB )
For a morphism f : A → B, A ⊕ PB −→ B is an epimorphism with a projective cover ρ B : PB → B. Thus each morphism is represented by epimorphisms. And the choice of the representing epimorphism is unique up to a direct sum of projective modules, as we have seen in Lemma 2.13. 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. We see it in Example 3.10. However, the uniqueness theorem is obtained in this way. Due to Theorem 3.3, 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 a direct sum of projective modules. These are the statements in Theorem 3.9, before which, we need some preparations. f
g
Suppose the sequence of modules A → B → C satisfies g ◦ f = 0, which implies that there exists a morphism f s
(g ρC )
s : A → PC such that A → B ⊕ C → C is a complex in mod R. We have a diagram of triangles FA • • yα C(g)•−1
f•
→
FB • k
→
FB •
→ C(f )• γ • y g•
→
n f •+1
→ cg
FC •
•
→
FA•+1 •+1 yα
(3.2)
C(g)•
which induces a diagram of complexes with term-wise exact rows in C(proj R) FA • • yα
0
→
0
→ C(g)•−1
→ C(nf )• β • y
→ C(f )• γ • y
→
→
C(cg )•
FC •
→
0
→
0.
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Applying τ≤0 to the diagram above and taking homology, we get the following diagram with exact rows.
θf :
0
→
H −1 (C( f )• )
σg :
0
f
→ A 0 yα
→
→ Ker g
→
B ⊕FA 1 0 yβ B ⊕ PC
(c f π)
→
(g ρC )
→
Cok f 0 yγ
→ 0
C
→ 0
(3.3)
We observe some facts below. • • • •
In K(mod R), β • is an isomorphism and C yl( f )• ∼ = FB • ∼ = C(cg )• . = C(n f )• ∼ •+1 ∼ • C(α) = C(γ ) . β 0 is a stable isomorphism. α 0 is the composite of natural maps A → Im sf and Im sf → Ker (g ρC ).
Lemma 3.8. With the notations above, the following holds. f
g
(1) If the sequence 0 → A → B → C → 0 is exact, then α 0 is a stable isomorphism and we may take α • and γ • as τ≤−1 C(α)• = 0 and τ≤−2 C(γ )• = 0. (2) If H−1 (C( f )• ) = 0, then the upper row of (3.3) is the short exact sequence f
(c f π)
θ f : 0 → A → B ⊕ FA 1 → Cok f → 0 which is a perfect exact sequence. st
Proof. (1) The map A → Im f is an isomorphism and Im f ∼ = Im a diagram with exact rows 0 →
Ker g ∼ y=
→
0
f0 01
B ⊕ PB ∼ y=
(g g◦ρ B )
→
C → ∼ y=
f s
since s = 0. We take ρC = g ◦ ρ B and have
0
(g 0)
B ⊕ PB → C → 0 from Lemma 2.13. It follows that Im sf = Im 0f → Ker g = Im 0f 10 is a stable isomorphism. In general α • →
A ⊕ PB
→
is the composite of α 0• : FA • → FKer g • induced by α 0 and the natural map δ • : FKer g • → C(g)•−1 with the property that δ i = id (i ≤ 0); C(δ) j = 0 ( j ≤ −1). Since α 0 is a stable isomorphism, α 0• is an isomorphism in K(proj R) hence C(α 0 )• = 0. From the triangle C(α 0 )• → C(α)• → C(δ)• → C(α 0 )•+1 , C(α)• ∼ = C(δ)• and we get the conclusion. (2) We have a term-wise exact sequence 0 → FA • → C yl( f )• → C( f )• → 0,
(3.4)
and C yl( f )• = C(n f )• is isomorphic to FB • in K(proj R). If H−1 (C( f )• ) = 0, then C( f )• = FCok f • , and the sequence θ f induced by (3.4) is perfect from Proposition 3.6. Theorem 3.9. Let f : A → B be a morphism in mod R. The following three conditions are equivalent. • f is rbm. • H−1 (C( f )• ) = 0. • Ext1R (Tr f, R) is an monomorphism in mod R. If this is the case, we have the following:
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(1) We have a perfect exact sequence f
(c f π)
θ f : 0 → A → B ⊕ FA 1 → Cok f → 0. (2) For any exact sequence of the form f q
(g p)
σ : 0 → A → B ⊕ P0 → C → 0 with some projective module P 0 , there is a commutative diagram
θf :
0 →
f
A → α y
B ⊕FA β y
cf π
1
→
Cok f γ y
→
0
C
→
0
σ :
0
→
f q
A
(g p)
B ⊕ P0
→
→
where α and β are stable isomorphisms. (3) There is an exact sequence with some projective module Q and Q 0 (γ ρ)
0 → Q 0 → Cok f ⊕ Q → C → 0. In other words, Ker γ is projective. (4) If σ is also perfectly exact, then σ is isomorphic to θ f up to a direct sum of trivial complexes, which are split exact sequences of projective modules. Proof. An equivalence between the second and the third conditions is easy to see: a triangle n f •+1
f•
FA • → FB • → C( f )• → FA •+1 induces an exact sequence H−1 (FB • ) → H−1 (C( f )• ) → H0 (FA • ) → H0 (FB • ). Since H−1 (FB • ) = 0, H−1 (C( f )• ) = Ker(H0 (FA • ) → H0 (FB • )) = Ker Ext1R (Tr f, R). Now suppose that f is rbm; there is an exact sequence f q
(g p)
σ : 0 → A → B ⊕ P 0 → C → 0. The maps f˜ = qf and g˜ = (g p) produce the similar diagram as (3.2): FA • • yα˜ C(g) ˜ •−1
→
FB⊕P 0 • k
→
FB⊕P 0 •
•
→ C(f˜) γ˜ • y → FC •
→ →
FA•+1 •+1 yα˜ C(g) ˜ •
Apply Lemma 3.8(1) to this sequence, and we get τ≤−2 C(γ˜ )• = 0. From the long exact sequence of homology groups • • • H−2 (C(γ˜ )• ) → H−1 (C( f˜) ) → H−1 (FC • ), we get H−1 (C( f˜) ) = 0. Obviously, H−1 (C( f˜) ) ∼ = H−1 (C( f )• ) • • −1 −1 hence H (C( f ) ) = 0. Conversely, suppose that H (C( f ) ) = 0. Then Lemma 3.8(2) shows that θ f is perfectly exact. Now it remains to prove (2)–(4) in the case H−1 (C( f )• ) = 0. (2) Applying the argument of Lemma 3.8 to the sequence σ , we get a similar diagram as (3.3):
θ f˜ :
0 →
σg˜ :
0
→
A 0 yα˜ Ker (g p)
f q
→
B ⊕ P 0 ⊕ FA 1 ˜0 yβ
→
B ⊕ P 0 ⊕ PC
→ (g p ρC )
→
Cok qf 0 yγ˜ C
→
0
→
0.
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Notice that β˜ 0 is a stable isomorphism. From Lemma 3.8(1), α˜ 0 is also a stable isomorphism. 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: f
0 →
cf π
B ⊕FA 1 0 yβ
A → 0 yα
→
Cok f γ y
→
0
C
→
0
0 →
f q
A
(g p)
B ⊕ P0
→
→
•
(3) As we see above, γ˜ • : C( f˜) = FCok f˜ • → FC • has τ≤−2 C(γ˜ )• = 0, which implies Q 0 = Ker γ˜ is projective from Lemma 2.18; (γ˜ 0 ρC ) 0 → Q 0 → Cok f˜ ⊕ PC → C → 0. st
st
Since Cok f˜ ∼ = Cok f and γ˜ 0 ∼ = γ , the above sequence is the desired sequence 0 → Q 0 → Cok f ⊕ Q → C → 0 with some projective module Q. f•
g•
(4) Suppose σ is perfect. From Proposition 3.6, FC •−1 → FA • → FB • → FC • is a distinguished triangle, and FC • ∼ = C( f )• , hence the induced sequence σ is isomorphic to θ f . Example 3.10. Let k be a field and R = k[[X, Y, Z ]]/(X 2 − Y Z ). Let M be an R-module defined as M = f
R/(X Y, Y 2 , Y Z ). The minimal Cohen–Macaulay approximation (see [2] for definition) θ f : 0 → Yk → X k → k → 0 of k is perfectly exact since Ext1R (k, R) = 0. On the other hand, the minimal Cohen–Macaulay approximation g
of M, σ : 0 → Y M → X M → M → 0 is not perfect since Ext1R (M, R) 6= 0. The map g is decomposed as st f q YM ∼ = X k ⊕ R and g = 0 Y . We easily check the statement of Theorem 3.9; k ∼ = Yk ⊕ R, X M ∼ = Cok f and there is an exact sequence 0→R→k⊕R→M →0 which clearly does not split. Proof of Lemma 3.7. (5) ⇔ (6). Since f is injective, we get H−1 (C( f )• ) = 0 from Theorem 3.9. In other words, • C( f )• ∼ = Fcok f . In this situation, (5) and (6) are clearly equivalent. (γ ρ)
(6) ⇒ (4). Since the assumption implies H−1 (C( f )• ) = 0, we get an exact sequence 0 → Q 0 → C ⊕ Q → C → 0 with projective modules Q and Q 0 , applying Theorem 3.9(3). Since R is local, we can apply Corollary 2.19, which shows that γ is a stable isomorphism, equivalently γ • is an isomorphism, hence α • is also an isomorphism. An exact sequence θ : 0 → A → B → C → 0 is perfectly exact if Ext1R (C, R) = 0. But as we see in the next example, the vanishing of Ext1R (C, R) is not the sufficient condition for θ to be perfectly exact. Example 3.11. Let R be R = k[[X, Y ]]/(X Y ) and let f be a map defined via projective resolutions as follows: (X Y )
···
→
R2 −→ (X Y ) y
0
→
R
( YX )
−→
R → X y( Y )
k f y
R2
Yk
→
Then the sequence 0 → X → k ⊕ R2
( f ρY k )
→ Yk → 0
with a projective cover ρY k of Y k is a perfect exact sequence since Ext1R (k, R) ∼ = Ext1R (Y k , R). But Ext1R (Y k , R) 6= 0.
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Notice that the dual of a perfect exact sequence is not always perfect as we see in the next example. Example 3.12. Let (R, m, k) be a three-dimensional Gorenstein local ring. Set L = Tr k, TL = Ω R3 (Tr) Ω R3 (Tr)L and ϕ L : L → TL to be a natural map. Since Hi (FL • ) = 0 (i ≤ 1) and FTL • is exact, H−1 (C(ϕ L )• ) = 0. Putting N = Cok ϕ L , we have a perfectly exact sequence ( ϕL ) θϕ L : 0 → L → TL ⊕ P → N → 0 with some projective module P. Dualizing θ ∗ , we get an exact sequence 0 → L ∗∗ → TL ∗∗ → (N )∗∗ → Ext1R (L ∗ , R) → 0. But Ext1R (L ∗ , R) ∼ = Ext3R (Tr L , R) = Ext3R (k, R) 6= 0. Remark 3.13. Let θ : 0 → A → B → C → 0 be a perfect exact sequence. Then θ ∗ is also perfectly exact if and only if the induced map H2 (FA • ) → H2 (FB • ) is a monomorphism. 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, Ker f = Cok dC( f ) −2 is the first syzygy of Cok f = Cok dC( f ) −1 . So it is natural to ask the converse: Is a given map f represented by monomorphisms if Ker f is a first syzygy? 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 the rbm condition. We begin with seeing equivalent conditions for a module to be a first syzygy. Definition 4.1. An R-module M is said to be torsionless1 if the natural map φ : M → M ∗∗ is a monomorphism. The next theorem is well known. We use the proof in [1] and [4]. Lemma 4.2. The following are equivalent for an R-module M. (1) M is torsionless. (2) Ext1R (Tr M, R) = 0 (3) M is a first syzygy; there exists a monomorphism from M to a projective module. Proof. Let φ : M → M ∗∗ be the natural map. The well known formula Ker φ ∼ = Ext1R (Tr M, R) shows the equivalence between (1) and (2). d −2
d −1
(2) ⇒ (3). Let · · · → P −2 → P −1 → P 0 → Tr M be a free resolution of Tr M. Then (2) says (P 0 )∗ (d −2 )∗ (P −1 )∗ →
(P −2 )∗
(d −1 )∗
→
is exact. By definition, M ∼ d −2∗ ). = Cok (d −1 )∗ which is isomorphic to Ω R1 (Cok f1
(3) ⇒ (1). We may assume that M is a submodule of a free module
Rl .
f2 Let f = .. : M → R l be a . fl
monomorphisms. If m ∈ M is not zero, f (m) is not zero, so there exists some i such that 0 6= f i (m) = φ(m)( f i ), which implies φ(m) 6= 0. Thus Ker φ = (0). To solve our problem, the special kind of maps is a key. For M ∈ mod R, consider a module J 2 M = Tr Ω R1 TrΩ R1 M. Since Tr J 2 M is a first syzygy, we have Ext1R (J 2 M, R) = 0, which means H−1 (FJ 2 M ∗ • ) = 0 and τ≥−2 FJ 2 M ∗ • is a projective resolution of Tr Ω R1 M = Cok (d FJ 2 M −2 )∗ = Cok (d FM −2 )∗ . The identity map on Tr Ω R1 M induces a chain map (FM )∗• → (FJ 2 M )∗• and its R-dual ψ M • : FJ 2 M • → FM • subsequently. 1 In [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.
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Lemma 4.3. The map ψ M : J 2 M → M is rbm if and only if an R-module M has (Ext1R (M, R))∗ = 0. −1 −2 Proof. From Theorem 3.9, ψ M is rbm if and only if H−1 (C(ψ M )• ) = 0. By definition, ψ M and ψ M are i • identity maps hence ψ M are identity maps for i ≤ −1. We may assume τ≤−2 C(ψ M ) = 0, which implies H−1 (C(ψ M )• ) = Ker dC(ψ M ) −1 . As τ≥−1 C(ψ M )• ∗ is a projective resolution of Cok (dψ M −1 )∗ ∼ = H−1 (C(ψ M )∗• ), • ψM we get H−1 (C(ψ M )• ) ∼ = (H−1 (C(ψ M )∗• ))∗ . A triangle FJ 2 M • → FM • → C(ψ M )• → FJ 2 M •+1 induces an R-dual triangle FJ 2 M ∗•+1 → C(ψ M )∗• → FM ∗• → FJ 2 M ∗• which produces an exact sequence of modules
0 → H−1 (C(ψ M )∗• ) → H−1 (FM ∗• ) → H−1 (FJ 2 M ∗• ) → 0. As we see in the discussion above, H−1 (FJ 2 M ∗• ) = 0. Hence H−1 (C(ψ M )∗• ) ∼ = H−1 (FM ∗• ) = Ext1R (M, R), and we • ∼ 1 −1 ∗ get H (C(ψ M ) ) = (Ext R (M, R)) . The above result is generalized as follows: Lemma 4.4. Let f : A → B be a morphism in mod R. Suppose (Ext1R (B, R))∗ = 0. If Ker f is projective, then f is rbm. Proof. From Lemma 2.18, we may assume τ≤−2 C( f )• = 0. Similarly as in the proof of Lemma 4.3, we have st
H−1 (C( f )• ) ∼ = (H−1 (C( f )∗• ))∗ . Since Ker f is projective, f induces a stable isomorphism J 2 A ∼ = 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 )∗ )• ) = Ext1R (B, R). The assumption (Ext1R (B, R))∗ = 0 equivalently says Ext1R (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. Proposition 4.5. Let f : A → B a morphism of mod R. Suppose (Ext1R (B, R))∗ = 0. Then f is rbm if and only if 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
nf q
( f ρB )
σ f : 0 → Ker f → A ⊕ PB → B → 0. Due to Theorem 3.9, we have a perfect exact sequence θn f , because n f is rbm:
θn f : 0
→
Ker f st y∼ =
0
→
Ker f
→
σf
nf
nf q
1 A ⊕ FKer f st y∼ =
f cn π
→
Cok n f ω f y
→
0
B
→
0.
→
A ⊕ PB
( f ρB )
→
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From Theorem 3.9(3), we know Ker ω f is projective. With the assumption (Ext1R (B, R))∗ = 0, we can apply f
f
Lemma 4.4 and get that ω f is rbm. From the equation f = ω f ◦ cn , f is rbm if cn is rbm. Since nf
•
cn
f•
FKer f • → FA • → C(n f )• → FKer f •+1 • f •+1 • ∼ 1 −1 nf n f is rbm if and only if ∼ 0 is a triangle, C(cn )• ∼ = FKer f ; H (C(c ) ) = H (FKer f ) = Ext R (Tr Ker f , R). Hence c Ker f is torsionless. f
g
Lemma 4.6. Let the sequence of R-modules 0 → A → B → C → 0 be exact. Suppose (Ext1R (C, R))∗ = 0. If A and C are torsionless, then so is B. Proof. From the assumption, A ∼ = Ker g is torsionless. Due to Proposition 4.5, g is rbm; there exists an exact sequence ( g ) θg : 0 → B → C ⊕ Q → Cok g → 0 with a projective module Q and a map : B → Q. Since C is a submodule of some projective module, so is B.
Proposition 4.7. The following are equivalent for a noetherian ring R. (1) (2) (3) (4)
Q(R) is Gorenstein. Q(R) is Gorenstein of dimension zero. (Ext1R (M, R))∗ = 0 for each M ∈ mod R. 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. (1) ⇔ (2). Notice that Q(R) is always of depth zero, since each non-zero-divisor is a unit. Hence Gorensteinness of Q(R) implies that dim R = 0. (3) ⇔ (2). Recall that Supp X ∗ = Supp X ∩ Ass R for X ∈ mod R. Let M be a finite R-module. The condition (Ext1R (M, R))∗ = 0 is thus equivalent to (Ext1R (M, R))p = 0 for p ∈ Ass R, which is also equivalent to Ext1R (M, R) ⊗ Q(R) = Ext1Q(R) (M ⊗ Q(R), Q(R)) = 0. This gives the implication from (2) to (3). For the converse direction, we have only to notice ExtnQ(R) (M ⊗ Q(R), Q(R)) = Ext1Q(R) (Ω Rn−1 (M) ⊗ Q(R), Q(R)) for n ≥ 1. (3) ⇔ (4) is already shown in Lemma 4.3. (4) ⇒ (5) is obvious. (5) ⇒ (1). The condition (5) is equivalent to Ext1R (R/m, R) ⊗ Q(R) ∼ = Ext1Q(R) (R/m ⊗ Q(R), Q(R)) = 0, which means Q(R) is Gorenstein from the following lemma. Lemma 4.8 ([3, Lemma 3.4.7]). Let (R, m, k) be a noetherian local ring of depth zero. If Ext1R (k, R) = 0, then R is Gorenstein of dimension zero. For the reader’s convenience, we give a proof. Proof. First we claim l(M ∗ ) = l(M)l(k ∗ ) for an artinian module M ∈ mod R. We show this by an induction on l(M). It is obvious if l(M) = 0. If l(M) > 0, there is an exact sequence 0 → M 0 → M → k → 0. Since Ext1R (k, R) = 0, the sequence 0 → k ∗ → M ∗ → M 0∗ → 0 is also exact. Thus l(M ∗ ) = l(M 0∗ ) + l(k ∗ ) equals to l(M 0 )l(k ∗ ) + l(k ∗ ) = l(M)l(k ∗ ) from the inductive hypothesis.
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Our second assertion is that R is artinian. If dim R > 0, then limn→∞ l(R/mn ) = ∞, which implies limn→∞ l((R/mn )∗ ) = ∞ because l(k ∗ ) 6= 0. On the other hand Hom R (R/mn , R) = (0 : R mn ) make a sequence of ideals (0 : R m) ⊂ · · · ⊂ (0 : R mn ) ⊂ (0 : R mn+1 ) · · · which should stabilize. Finally we show that Ext1R (M, R) = 0 for every M ∈ mod R. It goes by an induction on l(M). If M 6= 0, there is an exact sequence 0 → M 0 → M → k → 0. The inductive hypothesis together with Ext1R (k, R) = 0 brings the conclusion.
In the case where Q(R) is Gorenstein, every morphism in mod R satisfies the hypotheses of Proposition 4.5 and Lemma 4.6. Proposition 4.9. Suppose Q(R) is Gorenstein. A morphism f of R-modules is rbm if and only if Ker f is torsionless. f
g
Lemma 4.10. Suppose Q(R) is Gorenstein. Let the sequence of R-modules 0 → A → B → C → 0 be exact. If A and C are torsionless, then so is B. Thus with the condition Q(R) is Gorenstein, when discussing the rbm property, we can deal with a normal kernel as well as a pseudo-kernel. Proposition 4.11. Suppose Q(R) is Gorenstein. For a given morphism f , Ker f is torsionless if and only if Ker f is torsionless. Proof. From Lemma 2.17, there is an exact sequence 0 → Ker f → Ker f → Ω R1 (Cok f ) → 0. So the “if” part is obvious, and the “only if” part comes from Lemma 4.10. Theorem 4.12. Suppose Q(R) is Gorenstein. The following are equivalent for a morphism f : A → B in mod R. (1) (2) (3) (4) (5)
f is rbm. Ker f is torsionless. Ker f is torsionless. H−1 (C( f )• ) = 0. Ext1R (Tr f, R) is an monomorphism in mod R. st
(6) Ω R1 (Cok f ) ∼ = Ker f . st
(7) There exists f 0 such that f 0 ∼ = f and Ker f 0 is torsionless. st ∼ f , Ker f 0 is torsionless. (8) For any f 0 with f 0 = Proof. Implications (6) ⇒ (3), (8) ⇒ (2) and (8) ⇒ (7) are obvious. We already showed the equivalence of (1), (4) and (5) in Theorem 3.9, (1) ⇔ (3) in Proposition 4.9, and (3) ⇔ (2) in Proposition 4.11. Implications (3) ⇒ (8) and (7) ⇒ (3) are obtained from “if” and “only if” part of Proposition 4.11 respectively. (4) ⇒ (6). It comes directly from Cok dC( f ) 0 = Ker f and Cok dC( f ) −1 = Cok f . Corollary 4.13. 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 4.12. (2) ⇒ (1). For every M ∈ mod R, Ker ψ M is torsionless. Because Ker ψ M is projective and Ker ψ M is a submodule of Ker ψ M from Lemma 2.17(1). So if (2) holds, ψ M is rbm for any M ∈ mod R, which implies (1) from Proposition 4.7.
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In the case Q(R) is not Gorenstein, let us see an example where a morphism with torsionless kernel is not rbm. Example 4.14. In the case R = k[[X, Y, Z ]]/(X Y, X 2 ) with any field k, consider the map ψk : k → J 2 k. We know Ker ψk is projective. We shall show that ψk is not rbm; that is, H−1 (C(ψk )• ) does not vanish. As we have seen, C(ψk )i = 0 (i ≤ −2), H−1 (C(ψk )• ) = (Ext1R (k, R))∗ . From a free resolution of k
R
3
Y X 0 0 0 X
(X Y )
−→ R 2 → R → k → 0,
we get a free resolution of Ext1R (k, R)
R
4
X Y 0 0 0 0 X Y
−→
R 2 → Ext1R (k, R) → 0.
We easily see that (Ext1R (k, R))∗ does not vanish. Acknowledgements I am grateful to Kazuhiko Kurano who suggested that conditions for rbm should be given in terms of the kernel not only by the pseudo-kernel. I thank Shiro Goto who told me that the assumption of Theorem 4.12 is weakened and that Corollary 4.13 holds. I also thank Charles Weibel for useful advice on the introduction. Appendix A Appendix A by Shinsuke Takashima. In this section we give an alternative proof of Theorem 4.12 by using the classical torsion theory. Definition A.1. Let R be a commutative ring and M an R-module. We define R-submodules ε M and τ M of M as follows: ε M = ∩ f ∈Hom R (M,R) Ker f τ M = ∩ f ∈Hom R (M,E(R)) Ker f. It is easy to see M is torsionless if and only if ε M is zero. Lemma A.2. Let M be an R-module over a commutative ring R, and K a submodule of M. Then τ K = K ∩ τ M . Proof. (See also [7].) The inclusion τ K ⊂ K ∩ τ M is obvious. Let x ∈ K ∩ τ M and f : K → E(R) be an Rhomomorphism. Then we can extend f to some R-homomorphism g : M → E(R). Thus 0 = g(x) = f (x). Lemma A.3. For a commutative noetherian ring R, the following are equivalent. (1) For M ∈ mod R, ε M = τ M . (2) Every finite R-submodule X of E(R) is torsionless. Proof (See also [6]). (1) ⇒ (2) Obvious. (2) ⇒ (1). Let M be a finitely generated R module. τ M ⊂ ε M is easy. Let x ∈ M \ τ M . Then we can find R-homomorphism f : M → E(R) such that f (x) 6= 0. Set X = f (M). By assumption we have ε X = 0, hence we can find a morphism g : X → R such that g( f (x)) 6= 0. Thus x 6∈ ε M . Lemma A.4. Let R be a commutative noetherian ring whose total ring of fractions Q(R) is Gorenstein. Then for any finitely generated R-submodule X of E(R) are torsionless. Proof. By assumption Q(R) is an injective Q(R)-module. From the exactness of the functor Hom R (−, Q(R)) ∼ = Hom R (−, Hom Q(R) (Q(R), Q(R))) ∼ = Hom Q(R) (− ⊗ R Q(R), Q(R)), Q(R) is also injective as an R-module. Thus E(R) is a submodule of Q(R). On the other hand, each non-zero divisor r ∈ R defines an R-free module X r = {x ∈ Q(R) | xr ∈ R}. Obviously Q(R) = ∪r X r hence any finitely generated R-submodule X of E(R) is contained in some X r , which means X is a torsionless R-module.
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An alternative proof of Theorem 4.12. We shall show the equivalence between (1) and (2). Set K = Ker f . It is easy to see f is rbm if and only if K ∩ ε A = 0. We claim K ∩ Ker ε A = ε K . From Lemmas A.3 and A.4, ε A = τ A . And K ∩ ε A = K ∩ τ A = τ K from Lemma A.2. Again from Lemmas A.3 and A.4, τ K = ε K . On the other hand, K is torsionless if and only if ε K = 0. References [1] M. Auslander, M. Bridger, The stable module theory, Mem. Amer. Math. Soc. 94 (1969). [2] M. Auslander, R.O. Buchweitz, The homological theory of maximal Cohen–Macaulay approximations, Soc. Math. de France, Mem 38 (1989) 5–37. [3] W. Bruns, J. Herzog, Cohen–Macaulay rings, Cambridge Stud. Adv. Math. 39 (1998). [4] E.G. Evans, P. Griffith, Syzygies, in: London Math. Soc. Lecture Note Series, vol. 106, Cambridge U.P., 1985. [5] K. Kato, Stable module theory with kernels, Math. J. Okayama Univ. 43 (2001) 31–41. [6] M. Hoshino, On Lambek torsion theories, Osaka J. Math. 29 (1992) 447–453. [7] B. Stenstr¨om, Rings of Quotients, Springer-Verlag, 1975.
Morphisms represented by monomorphisms Kiriko Kato Department of Mathematics and Information Sciences, Osaka Prefecture University Daisen, Sakai, Osaka 590-0035, Japan Received 28 July 2005; received in revised form 28 August 2005 Available online 20 January 2006 Communicated by I. Reiten
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, 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 projectivestably equivalent. c 2005 Elsevier B.V. All rights reserved.
MSC: 13D02; 13D25; 16D90
1. Introduction Let R be a commutative noetherian ring. Linear maps f : A → B and f 0 : A0 → B 0 of finite R-modules are said to be projective-stably equivalent (pse for short) if the following diagram is commutative 0 A⊕ P ∼ y=
→
A0 ⊕ P
f s t u
f 0 s0 t 0 u0
→
0 B⊕ Q ∼ y=
B0 ⊕ Q
with some projective modules P, Q, P 0 , Q 0 and R-linear maps s, t, u, s 0 , t 0 , u 0 . 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 ⊕ PB → B is surjective with a projective cover ρ B : PB → B. Thus every morphism is represented by epimorphisms. The choice of epimorphism is unique; if an f0
( f ρB )
epimorphism f 0 is pse to f , then two sequences 0 → Ker f 0 → A0 → B 0 → 0 and 0 → Ker( f ρ B ) → A⊕ PB → B → 0 become isomorphic after splitting off common projective summands. On the other hand, every morphism is not always represented by monomorphisms. Even if a morphism f is rbm, the choice of monomorphism is not unique; there may be two monomorphisms f 0 and f 00 both pse to f and that E-mail address: [email protected]. c 2005 Elsevier B.V. All rights reserved. 0022-4049/$ - see front matter doi:10.1016/j.jpaa.2005.12.009
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K. Kato / Journal of Pure and Applied Algebra 208 (2007) 261–283 f0
f 00
0 → A0 → B 0 → Cok f 0 → 0 and 0 → A00 → B 00 → Cok f 00 → 0 are not isomorphic by splitting off common projective summands. The purpose of the paper is finding a condition of a given map to be rbm. The problem was 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 first focal point. In the case where 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 a direct sum of projective modules (Theorem 3.9). Our next 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 the projective stabilization mod R of a category mod R of finite R-modules 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 with the given map. Looking at Theorem 3.9, 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 the 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 4.12: 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 ]]/(X Y ) with any field k, g : R/(X 2 ) → R/(X, Y ) a natural projection. Since Ker g ∼ = (X, Y )/(X 2 ) with depth zero is not a first syzygy, g is not rbm due to Theorem 4.12. 2 Example 2. Set R = k[[X, Y ]]/(X Y ) with any field k, f : R 2 / YX R → R 2 / YX 2 R with f ab mod YX R = 2 Xa X ∼ Y b mod Y 2 R. The map f is not a monomorphism; Ker f = R/(X ) ⊕ R/(Y ) is a first syzygy. By 2 Theorem 4.12, f is rbm. In fact, let f 0 : R 2 / YX R → R 2 / YX 2 R ⊕ R 2 be defined as f 0 ab mod YX R = 2 Xa X Ya mod . Obviously f 0 is a monomorphism that is pse to f . On the other hand, f 00 : Yb Y 2 R, X b 2 2 2 Xa X Y a R 2 / YX R → R 2 / YX 2 R ⊕ R 2 with f 00 ab mod YX R = is also a Y b mod Y 2 R, X 2 b monomorphism and pse to f . We have two exact sequences 2 X X X Y f0 2 2 2 2 2 θf : 0 → R R →R R⊕R →R R⊕R R → 0, Y Y2 Y X and σ :0→R
2
X Y
f 00
R →R
2
X2 Y2
2
R⊕R →R
2
X Y2
R⊕R
2
Y X2
R→0,
that are not isomorphic. We see θ f is perfect but σ is not. 2. Stable module category and homotopy category We shall fix the notations and give a review of the correspondence between stable module category and homotopy class category of complexes. We omit many of the proofs for results that are in [5]. 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 : PM → M denotes a projective cover of M. The category of finitely generated R-modules is denoted by mod R, and the category of finite projective R-modules is denoted by proj R. For an R-module M, M ∗ denotes an R-dual Hom R (M, R). For a morphism f : M → N in mod R, f ∗ denotes Hom R ( f, R) : N ∗ → M ∗ .
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263
For an additive subcategory A of an abelian category, the category of complexes in A is denoted as C(A). A complex F • in A is d F n−1
dF n
F • : · · · → F n−1 → F n → F n+1 → · · · . Truncations of a complex F • are defined as follows: d F n−2
d F n−1
τ≤n F • : · · · → F n−2 → F n−1 → F n → 0 → 0 → · · · , dF n
d F n+1
τ≥n F • : · · · → 0 → 0 → F n → F n+1 → F n+2 → · · · . In the case A = mod R, an R-dual F•∗ of a complex F • ∈ C(A) is the cocomplex such as Fn∗ = (F n )∗ , ∗ dnF = (d F n−1 )∗ . A trivial complex is a direct sum of split exact sequences of projective modules. The category of the homotopy equivalence class of complexes in A is denoted by K(A). A morphism in K(A) is a homotopy equivalence class of chain maps. Note that an object in K(proj R) is zero if and only if it is represented by a trivial complex. Definition 2.1 (Auslander–Bridger [1]). The projective stabilization mod R is defined as follows. • Each object of mod R is an object of mod R. • For objects A, B of mod R, a set of morphisms from A to B is 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 f = f mod P(A, B). Definition 2.2. A morphism f : A → A0 in mod R is called a stable isomorphism if f is an isomorphism in mod R. st
In this case we say A and A0 are stably isomorphic and write A ∼ = A0 . Lemma 2.3. Two R-modules A and A0 are stably isomorphic if and only if there exist projective modules P and P 0 such that A ⊕ P 0 ∼ = A0 ⊕ P in mod R. st
∼ A0 . Then there exist morphisms α : A → A0 and α 0 : A0 → A such Proof. The “if” part is obvious. Suppose A = 0 0 that α ◦ α = id A0 and α ◦ α = id A . In other words, there exist morphisms s : A → PA and s 0 : A0 → PA0 such that α ◦ α 0 + ρ A0 ◦ s 0 = id A0 and α 0 ◦ α + ρ A ◦ s = id A . Morphisms α and α0 induce a map α P : PA → PA0 and α ρ α 0P : PA0 → PA as ρ A0 ◦ α P = α ◦ ρ A and ρ A ◦ α 0P = α 0 ◦ ρ A0 . It is easy to see s −αA00 : A ⊕ PA0 → A0 ⊕ PA and P 0 α ρA 0 0 : A ⊕ PA → A ⊕ PA are isomorphisms. s0 −α P
Definition 2.4. In any category C, morphisms f : A → B and f 0 : A0 → B 0 are said to be isomorphic and denoted as f ∼ = f 0 if there exist isomorphisms α : A → A0 and β : B → B 0 such that β ◦ f = f 0 ◦ α. Definition 2.5. Morphisms f : A → B and f 0 : A0 → B 0 in mod R are said to be projective-stably equivalent (pse st for short) and denoted as f ∼ = f 0 if f ∼ = f 0 in mod R. Lemma 2.6. Morphisms f : A → B and f 0 : A0 → B 0 in mod R are pse if and only if there exist projective modules P, P 0 , Q, Q 0 , morphisms t, t 0 , χ , χ 0 , isomorphisms α˜ and β˜ in mod R such that the following diagram commutes in mod R. 0 A⊕ P yα˜
⊕P
f0 0 t0 χ
0 B⊕ Q ˜ yβ
→
A0
0 χ0
f t
→
B0 ⊕ Q
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Proof. There exist stable isomorphisms α : A → A0 and β : B → B 0 such that β ◦ f = f 0 ◦ α, which implies the existence of morphisms t : A → PB 0 such that −β ◦ f + f 0 ◦ α = ρ B 0 ◦ t. Let α 0 : A0 → A and β 0 : B 0 → B be morphisms such that α 0 = α −1 and β 0 = β −1 . Then from the equations α ◦ α 0 = id A0 and β 0 ◦ β = id B , we get s 0 : A0 → PA0 and u : B → PB with the property α ◦ α 0 + ρ A0 ◦ s 0 = id A0 and β 0 ◦ β + ρ B ◦ u = id B . Morphisms α, β 0 and f 0 induce morphismsα P : PA → PA0 , β P0 : PB 0 → PB and f P : PA → PB respectively. From the proof 0 β ρB0 of Lemma 2.3, we know β˜ = : B ⊕ PB 0 → B 0 ⊕ PB and α˜0 = α0 ρ A : A0 ⊕ PA → A ⊕ PA0 are 0 u
−β P
s
−α P
isomorphisms. Now the diagram
A ⊕ PA0 −1 yα˜0
˜ β◦
A0
⊕ PA
gives the conclusion.
f t
f t
0 f0P
B ⊕ PB 0 ˜ yβ
→
0 0 fP
◦α˜0
→
B 0 ⊕ PB
Definition 2.7 (Auslander–Bridger [1]). δ
(1) For an R-module M, define a transpose Tr M of M to be Cok δ ∗ where P → Q → M → 0 is a projective presentation of M. The transpose of M is uniquely determined as an object of mod R. If f ∈ Hom R (M, N ), then f induces a map Tr N → Tr M, which represents a morphism Tr f ∈ Hom R (Tr N , Tr M). (2) A kernel of projective cover of an R-module M is called the first syzygy module of M and denoted as Ω R1 (M). The first syzygy module of M is uniquely determined as an object of mod R. Inductively, we define Ω Rn (M) = Ω R1 (Ω Rn−1 (M)). If f ∈ Hom R (M, N ), then f induces a map Ω Rn (M) → Ω Rn (N ), which represents a morphism Ω Rn ( f ) ∈ Hom R (Ω Rn (M), Ω Rn (N )). Let L be a full subcategory of K(mod R) defined as L = {F • ∈ K(proj R) | Hi (F • ) = 0 (i < 0), H j (F ∗ • ) = 0 ( j ≥ 0)}. Lemma 2.8 ([5, Lemma 2.1]). For a morphism f • ∈ L, f • = 0 in K(mod R) if and only if H0 (τ≤0 f • ) = 0 in mod R. Proof. Let f • : F • → G • be a chain map with F • , G • ∈ L such that H0 (τ≤0 f • ) = 0. Then there exists g ∈ Hom R (H0 (τ≤0 F • ), G 0 ) that satisfies H0 (τ≤0 f • ) = ρ ◦ g where ρ : G 0 → H0 (τ≤0 G • ) is the natural projective cover. We get chain maps ρ • ∈ Hom R (G 0 , τ≤0 G • ) and g • ∈ Hom R (τ≤0 F • , G 0 ) such as H0 (ρ • ) = ρ and H0 (g • ) = g. From the assumption, τ≤0 f • is homotopic to ρ • ◦ g • , which implies f i = h i+1 ◦ d F i + dG i−1 ◦ h i ∗
with some h i+1 : F i+1 → G i for i ≤ −1. Similarly, since H0 (τ≤0 f • ) = 0, we have f j = h j+1 ◦ d F j + dG j−1 ◦ h j with some h j : F j → G j−1 for j ≥ 2. Therefore as an element of L, we may assume f i = 0 (i 6= 0, 1). Moreover, ∗ ∗ ∗ ∗ we may assume f i = 0 (i 6= 1); since d F −1 ◦ f 0 = 0, ( f 0 )∗ passes through Ker d F −1 = Im d F 0 hence we get ∗ ∗ s 1 : F 1 → G 0 with f 0 = s 1 ◦ d F 0 . Finally, to see f • = 0, observe d F 0 ◦ f 1 = 0, then we get u 2 : F 2 → G 1 with ∗ ∗ ∗ ∗ ∗ f 1 = u 2 ◦ d F 1 . Since d F 1 ◦ u 2 ◦ dG 1 = f 1 ◦ dG 1 = 0, there exists a map u 3 : F 3 → G 2 such that dG 1 ◦ u 2 + u 3 ◦ d F 2 = 0. Thus we obtain a homotopy map u : F • → G •−1 which shows that f • is homotopic to zero.
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Suppose a chain map f • : F • → G • is homotopic to zero with a homotopy map h • : F • → G •−1 . We construct chain maps δ • : τ≤0 F • → F 1 and η• : F 1 → τ≤0 G • as follows: ···
→
···
→
···
→
F−1 y 0 y G −1
d F −1
→
→ dG −1
→
F0 0 yd F F1 1 yh G0
It easy to see H0 (τ≤0 f • ) = H0 (δ • ◦ η• ) = H0 (δ • ) ◦ H0 (η• ), the righthand-side passes through F 1 .
Lemma 2.9 ([5, Proposition 2.3, Proposition 2.4]). (1) For A ∈ mod R, there exists FA • ∈ L that satisfies st
H0 (τ≤0 FA • ) ∼ = A. Such an FA • is uniquely determined by A up to isomorphisms. We fix the notation FA • and call this a standard complex of A. (2) For f ∈ Hom R (A, B), there exists f • ∈ HomK(proj R) (FA • , FB • ) that satisfies H0 (τ≤0 f • ) ∼ = f. Such an f • is uniquely determined by f up to isomorphisms, so we use the notation f • to describe a chain map with this property for given f . Proof. (1) First take a projective resolution PA • of A: · · · → PA−2 → PA−1 → PA0 → A → 0, ∗
and then a projective resolution PTr A • of Tr A = Cok d FA −1 as ∗
∗
−2 −3 0 ← Tr A ← PA−1 ← PA0 ← PTr A ← PTr A ← · · · .
Define a complex FA • as i PA (i ≤ −1) i FA = − 1−i ∗ PTr A (i ≥ 0), i d (i ≤ −1) d FA i = PA − 2−i ∗ d PTr A (i ≥ 0). We easily see FA • ∈ L and st
H0 (τ≤0 FA • ) ∼ = A. Suppose both FA • and FA0 • have this property. Adding some trivial complex P • of projective modules P• : · · · → 0 → P0 = P1 → 0 → · · · • if necessary, we may assume that H0 (τ≤0 FA • ) ∼ = H0 (τ≤0 F 0 A ) ∼ = A in mod R. Then there are chain maps • • • • 0 0 • • ϕ : τ≤0 FA → τ≤0 FA and γ : τ≤0 FA → τ≤0 FA such that ϕ • ◦ γ • = 1τ≤0 F 0 • and γ • ◦ ϕ • = 1τ≤0 FA • . As ∗
A
∗
∗ and τ 0 0∗ ∗ τ≥−1 FA• ≥−1 F A• are acyclic, H−1 (τ≥−1 ϕ• ) induces a chain map τ≥−1 F A • → τ≥−1 F A •. With this map for • • 0 • • the positive part, ϕ can be extended to a chain map f : FA → FA such that τ≤0 f • = ϕ • . Similarly we get a chain
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map g • : FA0 • → FA • such that τ≤0 g • = γ • . It is easy to see H0 (τ≤0 ( f • ◦ g • )) = 1 A and H0 (τ≤0 (g • ◦ f • )) = 1 A . From Lemma 2.8, we have f • ◦ g • = 1 F 0 • and g • ◦ f • = 1 FA • . A (2) An R-linear map f : A → B of R-modules is lifted to a chain map between projective resolutions f 0• : τ≤0 FA • → τ≤0 FB • which again induces an R-linear map H−1 (τ≥−1 f 0•∗ ) : Tr B → Tr A lifted to a chain map f 00• : τ≥−1 FA • → τ≥−1 FB • . We may choose f 00 as f 00i = f 0i (i = −1, 0), and obtain f • : FA • → FB • with f i = f 0i (i ≤ 0) and f j = f 00 j ( j ≥ −1). Uniqueness follows from Lemma 2.8. The above lemma is summed up in the following way. Theorem 2.10 ([5, Theorem 2.6]). The mapping A 7→ FA • gives a functor from mod R to K(mod R), and this gives a category equivalence between mod R and L. For f ∈ Hom R (A, B), there exists a triangle nf•
cf •
f•
C( f )•−1 → FA • → FB • → C( f )• .
(2.1)
In general, C( f )• does not belong to L but it satisfies the following: Hi (C( f )• ) = 0 (i < −1),
H j (C( f )∗ • ) = 0 ( j > −1).
Definition and lemma 2.11 ([5, Definition and Lemma 3.1]). As objects of mod R, Ker f := H−1 (τ≤−1 C( f )• ) and Cok f := H0 (τ≤0 C( f )• ) are uniquely determined by f , up to isomorphisms. We call these the pseudo-kernel and the pseudo-cokernel of f . And we have Cok f = Tr Ker Tr f . Notations. The pseudo-kernel or the pseudo-cokernel is determined as a stable equivalence class of modules, not as st st an isomorphic class of mod R. Although, for simplicity, a module M with M ∼ = Ker f or M ∼ = Cok f is denoted by Ker f or Cok f respectively. Chain maps n f • and c f • in (2.1) induce n f : Ker f → A and c f : B → Cok f . These maps have the properties f ◦ n f = 0 and c f ◦ f = 0. The following lemma explains why Ker f and Cok f are called the pseudo-kernel and the pseudo-cokernel respectively. Lemma 2.12 ([5, Lemma 3.3, Lemma 3.5]). Let f : A → B be a homomorphism of R-modules. (1) If x ∈ Hom R (X, A) satisfies f ◦ x = 0, there exists h x ∈ Hom R (X, Ker f ) such that x = n f ◦ h x . (2) If y ∈ Hom R (B, Y ) satisfies y ◦ f = 0, there exists e y ∈ Hom R (Cok f , Y ) such that y = e y ◦ c f . From (2.1), we have an exact sequence ( f ρ)
0 → Ker f → A ⊕ P → B → 0 with some projective module P. This characterizes the pseudo-kernel. ( f p0 ) ( f p) Lemma 2.13. For a given f ∈ Hom R (A, B), suppose both A ⊕ P → B and A ⊕ P 0 → B are epimorphisms with projective modules P and P 0 . Then there are stable isomorphisms λ : A ⊕ P → A ⊕ P 0 and κ : Ker ( f p) → 0 Ker f p that make the following diagram commutative: 0 →
0
→
Ker(f p) → κ y Ker f p 0 →
A⊕ P yλ A ⊕ P0
( f p)
→
B
→
0
→
0
k ( f p0 ) →
B
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267
Proof. With a canonical map π : B → Cok f , π ◦ p and π ◦ p 0 are both projective covers of Cok f , hence there exists l : P → P 0 such that π ◦ p = π ◦ p 0 ◦ l. Then Im p − p 0 ◦ l ⊂ Im f , so we get h : P → Im f as h coincides with p − p 0 ◦ l. Via A → Im f which issurjective, h can be lifted to a map j : P → A. This shows the equation f ◦ j + p 0 ◦ l = p. The map λ = 10 lj : A ⊕ P → A ⊕ P 0 yields the desired diagram. Obviously λ is a stable isomorphism, which implies that κ is a stable isomorphism from the following Lemma 2.14. Lemma 2.14. On the commutative diagram with exact rows in mod R f
0
→
A → α y
0
→
A0
f0
→
g
B → β y B0
0
C → γ y
g0
→ C0
0,
→
if β and γ are stable isomorphisms, so is α. Proof. We show that α is a stable isomorphism in the following case: (1) γ is an isomorphism and β is a stable isomorphism. (2) β and γ are stable isomorphisms and γ is an epimorphism. (3) β and γ are stable isomorphisms. (1) Adding a projective cover ρ A0 : PA0 → A0 to the given diagram, we get the following: 0 ↓ Q v y 0
→
0 ↓ Q w y
= f0 ( 01 )
(g 0)
A ⊕ PA0 → (α ρ 0 ) y A f0
0 A y 0
0 →
B ⊕ PA0 → (β f 0 ◦ρ ) y A0 g0
B0 y 0
→
→
C → ∼ y=
0
C0
0
→
Since (β f 0 ◦ ρ A0 ) is an epimorphism and a stable isomorphism at the same time, it is a split epimorphism and Q is projective. In other words, w is a split monomorphism. Hence v is also a split monomorphism and (α ρ A0 ) a split epimorphism with a projective kernel. In particular, α is a stable isomorphism. (2) Since γ is a split epimorphism with a projective kernel, there exists γ 0 : C 0 → C such that γ ◦ γ 0 = idC 0 . On the diagram 0
→
A
→
k 0
→
A
f
→
00
B → C0 → 0 0 yβ yγ g
B
→
C
→
0
0,
β 0 is a stable isomorphism because Cok γ 0 ∼ = Cok β 0 is projective. Connection of the above and the given diagrams yields another diagram: 0 →
A → α y
0
A0
→
f0
→
00
B → C 0 0 k yβ◦β B0
g0
→ C0
→
0
→
0
Since β ◦ β 0 is a stable isomorphism, we can apply (1).
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(3) Adding a projective cover ρ B 0 : PB 0 → B 0 , we get 0
→
0
→
( 0f ) A → α y
( g0 01 )
B ⊕ PB 0 → (β ρ 0 ) y B
f0
A0
g0
B0
→
C ⊕ PB 0 → 0 y(γ g ◦ρ B ) C0
→
→
0
0.
Since β is a stable isomorphism and (γ g 0 ◦ ρ B ) is an epimorphic stable isomorphism, we can apply (2) to get the conclusion. Lemma 2.13 says that the sequences ( f p)
0 → Ker ( f p) → A ⊕ P → B → 0 and ( f p0 ) 0 → Ker f p 0 → A ⊕ P 0 → B → 0 are, isomorphic adding some split exact sequence of projective modules if necessary. This comes from the following Lemma 2.16. Definition 2.15. Two morphisms f : A → B and f 0 : A0 → B 0 are said to be isomorphic up to a direct sum of projective modules if there exist projective modules O, P, Q, O 0 , P 0 and Q 0 such that
f 0 0
A⊕O⊕P
0 0 0
0 1 0
B⊕P⊕Q
→
and f0 0 0
A0 ⊕ O 0 ⊕ P 0
0 0 0
0 1 0
!
B0 ⊕ P 0 ⊕ Q0
→
are isomorphic. The above definition equivalently says that the two exact sequences are isomorphic:
f 0 0
0 → Ker f ⊕ O → A ⊕ O ⊕ P
0 0 0
0 1 0
B ⊕ P ⊕ Q → Cok f ⊕ Q → 0,
→ f0 0 0
0 → Ker f 0 ⊕ O 0 → A0 ⊕ O 0 ⊕ P 0
0 0 0
0 1 0
!
→
B 0 ⊕ P 0 ⊕ Q 0 → Cok f ⊕ Q 0 → 0.
Each sequence is the direct sum of f
0 → Ker f → A → B → Cok f → 0 with a trivial complex 1 0
0 1 0 0
(0 1)
0→O → O⊕P → P⊕Q → Q→0 and so on.
st
Obviously f ∼ = f 0 if f and f 0 are isomorphic up to a direct sum of projective modules but the converse is not true. Lemma 2.16. The following are equivalent for two morphisms f : A → B and f 0 : A0 → B 0 in mod R. (1) f and f 0 are isomorphic up to a direct sum of projective modules.
K. Kato / Journal of Pure and Applied Algebra 208 (2007) 261–283
269
(2) There exist stable isomorphisms α, β and γ such that the next diagram commutes: f
B → β y
Cok f γ y
→
0
B0
Cok f 0
→
0
A → α y A0
f0
→
→
(3) There exist stable isomorphisms α, β, γ and δ such that the next diagram commutes: f
0
→
Ker f yδ
→
A → α y
B → β y
Cok f γ y
→ 0
0
→
Ker f 0
→
A0
B0
→
Cok f 0
→ 0
Cok f γ y Cok f 0
→
0
→
0,
f0
→
Proof. (1) ⇒ (2) is obvious. (2) ⇒ (3). On the diagram 0
→
0
→
Im f ε y Im f 0
B → β y B0 →
→ →
ε is a stable isomorphism by Lemma 2.14. From 0
→
0
→
Ker f yδ Ker f 0
Im f ε y Im f 0
A → α y A0 →
→ →
→
0
→
0,
δ is a stable isomorphism also by Lemma 2.14. (3) ⇒ (1). It is enough to show the following: A diagram with exact rows and stable isomorphisms α, β and γ f
0
→
A → α y
0
→
A0
f0
→
g
B → β y
C → γ y
g0
B0
→ C0
→
0
0
˜ γ˜ , and projective modules O, P, Q, O 0 , P 0 , Q 0 induces a diagram with exact rows, isomorphisms α, ˜ β,
0 →
A⊕ O yα˜
f 0 0
0 1 0
A0 ⊕ O 0
0 1 0
B⊕O ⊕P ˜ yβ
→ f0 0 0
0 →
g 0 0 0 0 1
g0 0 0 0 0 1
→
C⊕ P γ˜ y
→
0
C0 ⊕ P0
→
0.
!
B0 ⊕ O0 ⊕ P 0
→
→
First we get a diagram with exact rows:
0
→
0
→
f 0 0
0 1 0
A ⊕ PA0 → αρ y( A0 ) A0
f0
→
g 0 0 0 0 1
B ⊕ PA0 ⊕ PB 0 → β ρ ◦f0 ρ y( B 0 P B 0 ) B0
g0
→
C ⊕ PB 0 → 0 γ g0 ◦ρ y( B0 ) C0
→
0
where f 0 P : PA0 → PB 0 is a map induced by f 0 . All the vertical maps are split epimorphisms with projective kernels. Hence the upper row is isomorphic to the direct sum of the lower row and a split exact sequence of projective modules.
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Lemma 2.17 ([5, Lemma 3.6]). (1) There is an exact sequence 0 → Ker f → Ker f → Ω R1 (Cok f ) → 0. (2) There is an exact sequence 0 → L → Cok f → Cok f → 0 such that Ω R1 (L) is the surjective image of Ker f . Proof. (1) It follows from the diagram:
0
→
0
→
0
→
0 y Ker f y
0 y A 1 y 0
→
Ker f y 1 Ω R (Cok f) y 0
→ →
0 y Im f y
→ ( f ρB )
A ⊕ PB 0 y 1 PB y 0
B y Cok f y 0
→
→
→
0
→
0
→
0
(2) The transpose Tr f : Tr B → Tr A of a given map f : A → B makes an exact sequence 0 → Ker Tr f → Tr B ⊕ PTr A
(Tr f ρ Tr A )
→
Tr A → 0.
Dualizing this sequence with R, we get a long exact sequence 0 → (Tr A)∗ → (Tr B)∗ ⊕ (PTr A )∗ → (Ker Tr f )∗
→A
f Tr ρ Tr A
−→
B ⊕ Q → Cok f → 0
with some projective module Q. Put fˇ : A 0
→
0 →
Im fˇ → y Im f →
B⊕ Q → (1 0) y B →
f Tr ρ Tr A
→
Cok f y Cok f
B ⊕ Q. On the commutative diagram →
0
→
0
all the vertical maps are surjective. If we put L 0 the kernel of epimorphism Cok f → Cok f , then Ω R1 (L 0 ) is the kernel of the natural map A/Ker fˇ ∼ = Im fˇ → Im f ∼ = A/Ker f. Therefore Ω R1 (L 0 ) ∼ = Ker f /Ker fˇ.
Lemma 2.18. The following holds for f ∈ Hom R (A, B). (1) Ker f is projective if and only if there exists f 0• such that f • ∼ = f 0• in K(proj R) and f 0i = id for i ≤ −1. (2) If Ker f is projective, then Ω R1 ( f ) is a stable isomorphism. Proof. The chain map f • : FA • → FB • induces an exact sequence cf •
n f •+1
0 → FB • → C( f )• → FA •+1 → 0
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271
in C(proj R). There is a commutative diagram with exact rows: 0
0
→
FB • s • y f
→ cn f •
FA •
→
k f•
FA •
C(nf )• t • yf
cf•
C( f )•
→
n f •+1
→
FA •+1
→ 0
k n n f •+1
C( f )•
→
→ 0
FB •
→
We have t f • ◦ s f • = id FB • in C(proj R) and s f • ◦ t f • = idC(n f )• in K(proj R). We often write C yl( f )• for C(n f )• . (1) The “if” part is obvious. Suppose Ker f is projective. Then · · · → C( f )−2 → C( f )−1 → Ker f → 0 is a trivial complex, namely a direct sum of split exact sequences of projective modules. Hence there exists a complex C 0• such that C 0i = 0 (i ≤ −2) and C( f )• ∼ = C 0• in K(proj R). Let γ • : C( f )• → C 0• be a chain map that is an 0• isomorphism in K(proj R). Putting c : C(n f )• → C 0• as c0• = γ • ◦ n n f •+1 , we have a commutative diagram with exact rows in C(proj R): 0
0
→ C( f)•−1 → •−1 yγ →
C 0 •−1
C(nn f )• • yα
→ C(c0 )•−1
nnn
•+1 f
→
C(n f )•
→
0
→
0
k f 0•
→
C(n f )•
The following diagram commutes in K(proj R): C(nn f ) tn • y f FA •
nnn
•+1 f
→
k cn f •
→
k FA •
C(n f )•
f•
→
C(nf )• t • yf FB •
In K(proj R), α • , tn f • , and t f • are isomorphisms hence f • ∼ = f 0• . Clearly f 0i = id (i ≤ −1) and we get the conclusion. (2) Apply H−1 τ≤−1 to cn f •
n n f •+1
0 → FA • → C(n f )• → C( f )• → 0, we get 0 → Ω R1 (A) → Cok dC(n f ) −2 → Ker f → 0. If Ker f is projective, H−1 (τ≤−1 cn f • ) is a stable isomorphism. On the equation H−1 (τ≤−1 f • ) = H−1 (τ≤−1 t f • ) ◦ H−1 (τ≤−1 cn f • ), H−1 (τ≤−1 cn f • ) and H−1 (τ≤−1 t f • ) are stable isomorphisms, hence so is H−1 (τ≤−1 f • ) = Ω R1 ( f ). Corollary 2.19. Suppose R is local. If a morphism f ∈ End R (A) satisfies that Ker f is projective, then f is a stable isomorphism. Proof. We may assume C( f )i = 0 (i ≤ −2). Therefore the complex C( f )∗• has no cohomology except for H−1 (C( f )∗• ). The triangle C( f )∗• → FA ∗• → FA ∗• → C( f )∗•−1
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induces an exact sequence 0 → H−1 (C( f )∗• ) → H−1 (FA ∗• ) → H−1 (FA ∗• ) → 0. Since R is local, a surjective endomorphism on a finite module is always an automorphism. Thus we get H−1 (C( f )∗• ) = 0. It follows that C( f )∗• is an exact sequence of projective modules, equivalently, f is a stable isomorphism. 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 0 : A0 → B 0 in mod R is pse to f , that is, there exist stable isomorphisms α : A → A0 and β : B → B 0 such that β ◦ f = f 0 ◦ α. Each morphism is not always rbm. ∗∗ Example 3.2. Let R be a ring of dimension n ≥ 3, N an R-module with pd N = n, and ϕ N : N → N the natural ϕN ? ∗∗ map. Then any map N ⊕ P → N ⊕ Q of the form ? ? with projective modules P and Q, will never be monomorphic. If otherwise, N ⊕ P is a submodule of a projective module; this is a contradiction because N has a maximal projective dimension.
It was Auslander and Bridger who first defined and studied the “represented by monomorphisms” property. Theorem 3.3 (Auslander–Bridger). The following are equivalent for a morphism f : A → B in mod R. (1) There exists a monomorphism f 0 : A → B ⊕ P with a projective module P such that f = s ◦ f 0 via some split epimorphism s : B ⊕ P → B. (2) There exists a monomorphism f 0 : A → B ⊕ P with a projective module P such that f = s ◦ f 0 via some split epimorphism s : B ⊕ P → B, and f 0∗ is an epimorphism. (3) Hom R (B, I ) → Hom R (A, I ) is surjective if I is an injective module. Auslander and Bridger’s original definition of the “represented by monomorphisms” condition is (1) of Theorem 3.3. Seemingly this is different from our definition. But we show that two conditions are equivalent. Lemma 3.4. For a morphism f : A → B in mod R, f is rbm if and only if there exists a monomorphism f 0 : A → B ⊕ P with a projective module P such that f = s ◦ f 0 via some split epimorphism s : B ⊕ P → B. Proof. The “if” part is clear. We shall show the “only if” part. Suppose there exists a monomorphism f 0 : A0 → B 0 , stable isomorphisms α : A → A0 and β : B → B 0 such that β ◦ f = f 0 ◦ α. We first take projective covers ρ A : PA → A and ρ B : PB → B such that the induced map f P : PA → PB by f is a monomorphism. Since α is a stable isomorphism, there exists a morphism α 0 : A0 → A such that α ◦ α 0 = id A0 and α 0 ◦ α = id A . 0 From the ◦ s A = id A , equivalently last equation there exists a morphism s A : A → PA such that α ◦ α + ρ A α 0 0 0 (α ρ A ) ◦ s A = id A . In particular, (α ρ A ) : A ⊕ PA → A is a split epimorphism and sαA : A → A0 ⊕ PA is
a split monomorphism. Similarly weget morphisms β 0 : B 0 → B, s B : B → PB and s B 0 : B 0 → PB 0 such that 0 β 0 (β ρ B ) ◦ s B = id B and (β ρ B 0 ) ◦ sβ 0 = id B 0 . B
Given equation β ◦ f = f 0 ◦ α induces β 0 ◦ (β ◦ f ) ◦ α 0 = β 0 ◦ ( f 0 ◦ α) ◦ α 0 , that is, f ◦ α 0 = β 0 ◦ f 0 . Hence there exists a homomorphism t : A0 → PB such that f ◦ α 0 − β 0 ◦ f 0 = ρ B ◦ t. Now we get a commutative diagram
f0 t
0 fP
−→ A0 ⊕ PA α0 ρ y( A ) A
f
→
B0 ⊕ PB β0 ρ y( B ) B.
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β0 Since the composite of maps
sB0 0
0 0
id PB
: B 0 ⊕ PB → B⊕ PB 0 ⊕ PB and
0 0 is equal to id B 0 ⊕PB , β 0 ρ B ◦ ft f0P = β 0 ρ B ◦id B 0 ⊕PB ◦ ft 0 f 0 t f P , and the following diagram commutes: f
= β0
ρB0 0
0
: B⊕ PB 0 ⊕ PB → B 0 ⊕ PB β0 0 β ρ 0 0 ρ B ◦ 0 0B 0 id P ◦ s B 0 ◦ id PB
B
0
id PB
00
A0 ⊕ PA −→ α0 ρ y( A )
B ⊕ PB0 ⊕ PB ρ y
f
A
0 fP
β 0
B
→
where β0 00 f = sB0 0
0 0 f 0 ◦ t id PB
0 , fP
and β ρB ◦ 0
ρ = β0
ρB0 0
0 . id PB
It is easy to see that f 00 is a monomorphism and ρ is a split epimorphism. Finally putting f 000 = f 00 ◦ sαA which is a monomorphism, we have f = f ◦ id A = f ◦ α ρ A 0
α α 00 ◦ =ρ◦ f ◦ = ρ ◦ f 000 . sA sA
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 3.5. An exact sequence 0 → A → B → C → 0 of R-modules is called a perfect exact sequence or said 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. Proposition 3.6 ([5, Lemma 2.7]). The following are equivalent for an exact sequence f
g
θ : 0 → A → B → C → 0. (1) θ is perfectly exact. f•
g•
(2) 0 → FA • → FB • → FC • → 0 is exact. f•
g•
(3) FC •−1 → FA • → FB • → FC • is a distinguished triangle in K(mod R). (4) FA • ∼ = C(g)•−1 in K(proj R). If these conditions are satisfied, we have the following. st
(5) C ∼ = Cok f . ∼ C( f )• in K(proj R). (6) FC • = Proof. In [5, Lemma 2.7], we see the equivalence between (1) and (2). The implication (3) ⇒ (2), (4), (5), (6) is obvious.
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(2) ⇒ (3). With the same notations as in the proof of Lemma 2.18, we have a diagram in C(proj R) with exact rows: cf •
FB • s • y f 0
→
FA •
cn f •
C(nf )• t • yf
→
k 0 →
FA •
f•
k n n f •+1
→ g•
FB •
→
C( f )•
→
C(f )• γ • y
→
0
FC •
→
0
→
Since s f • and t f • are isomorphisms, γ • is also an isomorphism and g • ∼ = c f • in K(proj R). (4) ⇒ (3). On the diagram in K(proj R), FA • • yα C(g)•−1
f•
→
FB • ,
k cg •
→
g•
FB •
→
FC •
α i = id (i ≤ 0). Hence α • is an isomorphism if FA • ∼ = cg • . = C(g)•−1 . Therefore f • ∼
If R is local, all the conditions above are equivalent. We shall give the proof later at the end of this section. Lemma 3.7. Let the sequence f
g
θ :0→ A→B→C →0 be exact. If R is local, the conditions (1)–(4) in Proposition 3.6 are equivalent to the conditions (5) and (6). st
(5) C ∼ = Cok f . ∼ C( f )• in K(proj R). (6) FC • = ( f ρB )
For a morphism f : A → B, A ⊕ PB −→ B is an epimorphism with a projective cover ρ B : PB → B. Thus each morphism is represented by epimorphisms. And the choice of the representing epimorphism is unique up to a direct sum of projective modules, as we have seen in Lemma 2.13. 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. We see it in Example 3.10. However, the uniqueness theorem is obtained in this way. Due to Theorem 3.3, 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 a direct sum of projective modules. These are the statements in Theorem 3.9, before which, we need some preparations. f
g
Suppose the sequence of modules A → B → C satisfies g ◦ f = 0, which implies that there exists a morphism f s
(g ρC )
s : A → PC such that A → B ⊕ C → C is a complex in mod R. We have a diagram of triangles FA • • yα C(g)•−1
f•
→
FB • k
→
FB •
→ C(f )• γ • y g•
→
n f •+1
→ cg
FC •
•
→
FA•+1 •+1 yα
(3.2)
C(g)•
which induces a diagram of complexes with term-wise exact rows in C(proj R) FA • • yα
0
→
0
→ C(g)•−1
→ C(nf )• β • y
→ C(f )• γ • y
→
→
C(cg )•
FC •
→
0
→
0.
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Applying τ≤0 to the diagram above and taking homology, we get the following diagram with exact rows.
θf :
0
→
H −1 (C( f )• )
σg :
0
f
→ A 0 yα
→
→ Ker g
→
B ⊕FA 1 0 yβ B ⊕ PC
(c f π)
→
(g ρC )
→
Cok f 0 yγ
→ 0
C
→ 0
(3.3)
We observe some facts below. • • • •
In K(mod R), β • is an isomorphism and C yl( f )• ∼ = FB • ∼ = C(cg )• . = C(n f )• ∼ •+1 ∼ • C(α) = C(γ ) . β 0 is a stable isomorphism. α 0 is the composite of natural maps A → Im sf and Im sf → Ker (g ρC ).
Lemma 3.8. With the notations above, the following holds. f
g
(1) If the sequence 0 → A → B → C → 0 is exact, then α 0 is a stable isomorphism and we may take α • and γ • as τ≤−1 C(α)• = 0 and τ≤−2 C(γ )• = 0. (2) If H−1 (C( f )• ) = 0, then the upper row of (3.3) is the short exact sequence f
(c f π)
θ f : 0 → A → B ⊕ FA 1 → Cok f → 0 which is a perfect exact sequence. st
Proof. (1) The map A → Im f is an isomorphism and Im f ∼ = Im a diagram with exact rows 0 →
Ker g ∼ y=
→
0
f0 01
B ⊕ PB ∼ y=
(g g◦ρ B )
→
C → ∼ y=
f s
since s = 0. We take ρC = g ◦ ρ B and have
0
(g 0)
B ⊕ PB → C → 0 from Lemma 2.13. It follows that Im sf = Im 0f → Ker g = Im 0f 10 is a stable isomorphism. In general α • →
A ⊕ PB
→
is the composite of α 0• : FA • → FKer g • induced by α 0 and the natural map δ • : FKer g • → C(g)•−1 with the property that δ i = id (i ≤ 0); C(δ) j = 0 ( j ≤ −1). Since α 0 is a stable isomorphism, α 0• is an isomorphism in K(proj R) hence C(α 0 )• = 0. From the triangle C(α 0 )• → C(α)• → C(δ)• → C(α 0 )•+1 , C(α)• ∼ = C(δ)• and we get the conclusion. (2) We have a term-wise exact sequence 0 → FA • → C yl( f )• → C( f )• → 0,
(3.4)
and C yl( f )• = C(n f )• is isomorphic to FB • in K(proj R). If H−1 (C( f )• ) = 0, then C( f )• = FCok f • , and the sequence θ f induced by (3.4) is perfect from Proposition 3.6. Theorem 3.9. Let f : A → B be a morphism in mod R. The following three conditions are equivalent. • f is rbm. • H−1 (C( f )• ) = 0. • Ext1R (Tr f, R) is an monomorphism in mod R. If this is the case, we have the following:
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(1) We have a perfect exact sequence f
(c f π)
θ f : 0 → A → B ⊕ FA 1 → Cok f → 0. (2) For any exact sequence of the form f q
(g p)
σ : 0 → A → B ⊕ P0 → C → 0 with some projective module P 0 , there is a commutative diagram
θf :
0 →
f
A → α y
B ⊕FA β y
cf π
1
→
Cok f γ y
→
0
C
→
0
σ :
0
→
f q
A
(g p)
B ⊕ P0
→
→
where α and β are stable isomorphisms. (3) There is an exact sequence with some projective module Q and Q 0 (γ ρ)
0 → Q 0 → Cok f ⊕ Q → C → 0. In other words, Ker γ is projective. (4) If σ is also perfectly exact, then σ is isomorphic to θ f up to a direct sum of trivial complexes, which are split exact sequences of projective modules. Proof. An equivalence between the second and the third conditions is easy to see: a triangle n f •+1
f•
FA • → FB • → C( f )• → FA •+1 induces an exact sequence H−1 (FB • ) → H−1 (C( f )• ) → H0 (FA • ) → H0 (FB • ). Since H−1 (FB • ) = 0, H−1 (C( f )• ) = Ker(H0 (FA • ) → H0 (FB • )) = Ker Ext1R (Tr f, R). Now suppose that f is rbm; there is an exact sequence f q
(g p)
σ : 0 → A → B ⊕ P 0 → C → 0. The maps f˜ = qf and g˜ = (g p) produce the similar diagram as (3.2): FA • • yα˜ C(g) ˜ •−1
→
FB⊕P 0 • k
→
FB⊕P 0 •
•
→ C(f˜) γ˜ • y → FC •
→ →
FA•+1 •+1 yα˜ C(g) ˜ •
Apply Lemma 3.8(1) to this sequence, and we get τ≤−2 C(γ˜ )• = 0. From the long exact sequence of homology groups • • • H−2 (C(γ˜ )• ) → H−1 (C( f˜) ) → H−1 (FC • ), we get H−1 (C( f˜) ) = 0. Obviously, H−1 (C( f˜) ) ∼ = H−1 (C( f )• ) • • −1 −1 hence H (C( f ) ) = 0. Conversely, suppose that H (C( f ) ) = 0. Then Lemma 3.8(2) shows that θ f is perfectly exact. Now it remains to prove (2)–(4) in the case H−1 (C( f )• ) = 0. (2) Applying the argument of Lemma 3.8 to the sequence σ , we get a similar diagram as (3.3):
θ f˜ :
0 →
σg˜ :
0
→
A 0 yα˜ Ker (g p)
f q
→
B ⊕ P 0 ⊕ FA 1 ˜0 yβ
→
B ⊕ P 0 ⊕ PC
→ (g p ρC )
→
Cok qf 0 yγ˜ C
→
0
→
0.
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Notice that β˜ 0 is a stable isomorphism. From Lemma 3.8(1), α˜ 0 is also a stable isomorphism. 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: f
0 →
cf π
B ⊕FA 1 0 yβ
A → 0 yα
→
Cok f γ y
→
0
C
→
0
0 →
f q
A
(g p)
B ⊕ P0
→
→
•
(3) As we see above, γ˜ • : C( f˜) = FCok f˜ • → FC • has τ≤−2 C(γ˜ )• = 0, which implies Q 0 = Ker γ˜ is projective from Lemma 2.18; (γ˜ 0 ρC ) 0 → Q 0 → Cok f˜ ⊕ PC → C → 0. st
st
Since Cok f˜ ∼ = Cok f and γ˜ 0 ∼ = γ , the above sequence is the desired sequence 0 → Q 0 → Cok f ⊕ Q → C → 0 with some projective module Q. f•
g•
(4) Suppose σ is perfect. From Proposition 3.6, FC •−1 → FA • → FB • → FC • is a distinguished triangle, and FC • ∼ = C( f )• , hence the induced sequence σ is isomorphic to θ f . Example 3.10. Let k be a field and R = k[[X, Y, Z ]]/(X 2 − Y Z ). Let M be an R-module defined as M = f
R/(X Y, Y 2 , Y Z ). The minimal Cohen–Macaulay approximation (see [2] for definition) θ f : 0 → Yk → X k → k → 0 of k is perfectly exact since Ext1R (k, R) = 0. On the other hand, the minimal Cohen–Macaulay approximation g
of M, σ : 0 → Y M → X M → M → 0 is not perfect since Ext1R (M, R) 6= 0. The map g is decomposed as st f q YM ∼ = X k ⊕ R and g = 0 Y . We easily check the statement of Theorem 3.9; k ∼ = Yk ⊕ R, X M ∼ = Cok f and there is an exact sequence 0→R→k⊕R→M →0 which clearly does not split. Proof of Lemma 3.7. (5) ⇔ (6). Since f is injective, we get H−1 (C( f )• ) = 0 from Theorem 3.9. In other words, • C( f )• ∼ = Fcok f . In this situation, (5) and (6) are clearly equivalent. (γ ρ)
(6) ⇒ (4). Since the assumption implies H−1 (C( f )• ) = 0, we get an exact sequence 0 → Q 0 → C ⊕ Q → C → 0 with projective modules Q and Q 0 , applying Theorem 3.9(3). Since R is local, we can apply Corollary 2.19, which shows that γ is a stable isomorphism, equivalently γ • is an isomorphism, hence α • is also an isomorphism. An exact sequence θ : 0 → A → B → C → 0 is perfectly exact if Ext1R (C, R) = 0. But as we see in the next example, the vanishing of Ext1R (C, R) is not the sufficient condition for θ to be perfectly exact. Example 3.11. Let R be R = k[[X, Y ]]/(X Y ) and let f be a map defined via projective resolutions as follows: (X Y )
···
→
R2 −→ (X Y ) y
0
→
R
( YX )
−→
R → X y( Y )
k f y
R2
Yk
→
Then the sequence 0 → X → k ⊕ R2
( f ρY k )
→ Yk → 0
with a projective cover ρY k of Y k is a perfect exact sequence since Ext1R (k, R) ∼ = Ext1R (Y k , R). But Ext1R (Y k , R) 6= 0.
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Notice that the dual of a perfect exact sequence is not always perfect as we see in the next example. Example 3.12. Let (R, m, k) be a three-dimensional Gorenstein local ring. Set L = Tr k, TL = Ω R3 (Tr) Ω R3 (Tr)L and ϕ L : L → TL to be a natural map. Since Hi (FL • ) = 0 (i ≤ 1) and FTL • is exact, H−1 (C(ϕ L )• ) = 0. Putting N = Cok ϕ L , we have a perfectly exact sequence ( ϕL ) θϕ L : 0 → L → TL ⊕ P → N → 0 with some projective module P. Dualizing θ ∗ , we get an exact sequence 0 → L ∗∗ → TL ∗∗ → (N )∗∗ → Ext1R (L ∗ , R) → 0. But Ext1R (L ∗ , R) ∼ = Ext3R (Tr L , R) = Ext3R (k, R) 6= 0. Remark 3.13. Let θ : 0 → A → B → C → 0 be a perfect exact sequence. Then θ ∗ is also perfectly exact if and only if the induced map H2 (FA • ) → H2 (FB • ) is a monomorphism. 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, Ker f = Cok dC( f ) −2 is the first syzygy of Cok f = Cok dC( f ) −1 . So it is natural to ask the converse: Is a given map f represented by monomorphisms if Ker f is a first syzygy? 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 the rbm condition. We begin with seeing equivalent conditions for a module to be a first syzygy. Definition 4.1. An R-module M is said to be torsionless1 if the natural map φ : M → M ∗∗ is a monomorphism. The next theorem is well known. We use the proof in [1] and [4]. Lemma 4.2. The following are equivalent for an R-module M. (1) M is torsionless. (2) Ext1R (Tr M, R) = 0 (3) M is a first syzygy; there exists a monomorphism from M to a projective module. Proof. Let φ : M → M ∗∗ be the natural map. The well known formula Ker φ ∼ = Ext1R (Tr M, R) shows the equivalence between (1) and (2). d −2
d −1
(2) ⇒ (3). Let · · · → P −2 → P −1 → P 0 → Tr M be a free resolution of Tr M. Then (2) says (P 0 )∗ (d −2 )∗ (P −1 )∗ →
(P −2 )∗
(d −1 )∗
→
is exact. By definition, M ∼ d −2∗ ). = Cok (d −1 )∗ which is isomorphic to Ω R1 (Cok f1
(3) ⇒ (1). We may assume that M is a submodule of a free module
Rl .
f2 Let f = .. : M → R l be a . fl
monomorphisms. If m ∈ M is not zero, f (m) is not zero, so there exists some i such that 0 6= f i (m) = φ(m)( f i ), which implies φ(m) 6= 0. Thus Ker φ = (0). To solve our problem, the special kind of maps is a key. For M ∈ mod R, consider a module J 2 M = Tr Ω R1 TrΩ R1 M. Since Tr J 2 M is a first syzygy, we have Ext1R (J 2 M, R) = 0, which means H−1 (FJ 2 M ∗ • ) = 0 and τ≥−2 FJ 2 M ∗ • is a projective resolution of Tr Ω R1 M = Cok (d FJ 2 M −2 )∗ = Cok (d FM −2 )∗ . The identity map on Tr Ω R1 M induces a chain map (FM )∗• → (FJ 2 M )∗• and its R-dual ψ M • : FJ 2 M • → FM • subsequently. 1 In [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.
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Lemma 4.3. The map ψ M : J 2 M → M is rbm if and only if an R-module M has (Ext1R (M, R))∗ = 0. −1 −2 Proof. From Theorem 3.9, ψ M is rbm if and only if H−1 (C(ψ M )• ) = 0. By definition, ψ M and ψ M are i • identity maps hence ψ M are identity maps for i ≤ −1. We may assume τ≤−2 C(ψ M ) = 0, which implies H−1 (C(ψ M )• ) = Ker dC(ψ M ) −1 . As τ≥−1 C(ψ M )• ∗ is a projective resolution of Cok (dψ M −1 )∗ ∼ = H−1 (C(ψ M )∗• ), • ψM we get H−1 (C(ψ M )• ) ∼ = (H−1 (C(ψ M )∗• ))∗ . A triangle FJ 2 M • → FM • → C(ψ M )• → FJ 2 M •+1 induces an R-dual triangle FJ 2 M ∗•+1 → C(ψ M )∗• → FM ∗• → FJ 2 M ∗• which produces an exact sequence of modules
0 → H−1 (C(ψ M )∗• ) → H−1 (FM ∗• ) → H−1 (FJ 2 M ∗• ) → 0. As we see in the discussion above, H−1 (FJ 2 M ∗• ) = 0. Hence H−1 (C(ψ M )∗• ) ∼ = H−1 (FM ∗• ) = Ext1R (M, R), and we • ∼ 1 −1 ∗ get H (C(ψ M ) ) = (Ext R (M, R)) . The above result is generalized as follows: Lemma 4.4. Let f : A → B be a morphism in mod R. Suppose (Ext1R (B, R))∗ = 0. If Ker f is projective, then f is rbm. Proof. From Lemma 2.18, we may assume τ≤−2 C( f )• = 0. Similarly as in the proof of Lemma 4.3, we have st
H−1 (C( f )• ) ∼ = (H−1 (C( f )∗• ))∗ . Since Ker f is projective, f induces a stable isomorphism J 2 A ∼ = 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 )∗ )• ) = Ext1R (B, R). The assumption (Ext1R (B, R))∗ = 0 equivalently says Ext1R (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. Proposition 4.5. Let f : A → B a morphism of mod R. Suppose (Ext1R (B, R))∗ = 0. Then f is rbm if and only if 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
nf q
( f ρB )
σ f : 0 → Ker f → A ⊕ PB → B → 0. Due to Theorem 3.9, we have a perfect exact sequence θn f , because n f is rbm:
θn f : 0
→
Ker f st y∼ =
0
→
Ker f
→
σf
nf
nf q
1 A ⊕ FKer f st y∼ =
f cn π
→
Cok n f ω f y
→
0
B
→
0.
→
A ⊕ PB
( f ρB )
→
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From Theorem 3.9(3), we know Ker ω f is projective. With the assumption (Ext1R (B, R))∗ = 0, we can apply f
f
Lemma 4.4 and get that ω f is rbm. From the equation f = ω f ◦ cn , f is rbm if cn is rbm. Since nf
•
cn
f•
FKer f • → FA • → C(n f )• → FKer f •+1 • f •+1 • ∼ 1 −1 nf n f is rbm if and only if ∼ 0 is a triangle, C(cn )• ∼ = FKer f ; H (C(c ) ) = H (FKer f ) = Ext R (Tr Ker f , R). Hence c Ker f is torsionless. f
g
Lemma 4.6. Let the sequence of R-modules 0 → A → B → C → 0 be exact. Suppose (Ext1R (C, R))∗ = 0. If A and C are torsionless, then so is B. Proof. From the assumption, A ∼ = Ker g is torsionless. Due to Proposition 4.5, g is rbm; there exists an exact sequence ( g ) θg : 0 → B → C ⊕ Q → Cok g → 0 with a projective module Q and a map : B → Q. Since C is a submodule of some projective module, so is B.
Proposition 4.7. The following are equivalent for a noetherian ring R. (1) (2) (3) (4)
Q(R) is Gorenstein. Q(R) is Gorenstein of dimension zero. (Ext1R (M, R))∗ = 0 for each M ∈ mod R. 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. (1) ⇔ (2). Notice that Q(R) is always of depth zero, since each non-zero-divisor is a unit. Hence Gorensteinness of Q(R) implies that dim R = 0. (3) ⇔ (2). Recall that Supp X ∗ = Supp X ∩ Ass R for X ∈ mod R. Let M be a finite R-module. The condition (Ext1R (M, R))∗ = 0 is thus equivalent to (Ext1R (M, R))p = 0 for p ∈ Ass R, which is also equivalent to Ext1R (M, R) ⊗ Q(R) = Ext1Q(R) (M ⊗ Q(R), Q(R)) = 0. This gives the implication from (2) to (3). For the converse direction, we have only to notice ExtnQ(R) (M ⊗ Q(R), Q(R)) = Ext1Q(R) (Ω Rn−1 (M) ⊗ Q(R), Q(R)) for n ≥ 1. (3) ⇔ (4) is already shown in Lemma 4.3. (4) ⇒ (5) is obvious. (5) ⇒ (1). The condition (5) is equivalent to Ext1R (R/m, R) ⊗ Q(R) ∼ = Ext1Q(R) (R/m ⊗ Q(R), Q(R)) = 0, which means Q(R) is Gorenstein from the following lemma. Lemma 4.8 ([3, Lemma 3.4.7]). Let (R, m, k) be a noetherian local ring of depth zero. If Ext1R (k, R) = 0, then R is Gorenstein of dimension zero. For the reader’s convenience, we give a proof. Proof. First we claim l(M ∗ ) = l(M)l(k ∗ ) for an artinian module M ∈ mod R. We show this by an induction on l(M). It is obvious if l(M) = 0. If l(M) > 0, there is an exact sequence 0 → M 0 → M → k → 0. Since Ext1R (k, R) = 0, the sequence 0 → k ∗ → M ∗ → M 0∗ → 0 is also exact. Thus l(M ∗ ) = l(M 0∗ ) + l(k ∗ ) equals to l(M 0 )l(k ∗ ) + l(k ∗ ) = l(M)l(k ∗ ) from the inductive hypothesis.
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Our second assertion is that R is artinian. If dim R > 0, then limn→∞ l(R/mn ) = ∞, which implies limn→∞ l((R/mn )∗ ) = ∞ because l(k ∗ ) 6= 0. On the other hand Hom R (R/mn , R) = (0 : R mn ) make a sequence of ideals (0 : R m) ⊂ · · · ⊂ (0 : R mn ) ⊂ (0 : R mn+1 ) · · · which should stabilize. Finally we show that Ext1R (M, R) = 0 for every M ∈ mod R. It goes by an induction on l(M). If M 6= 0, there is an exact sequence 0 → M 0 → M → k → 0. The inductive hypothesis together with Ext1R (k, R) = 0 brings the conclusion.
In the case where Q(R) is Gorenstein, every morphism in mod R satisfies the hypotheses of Proposition 4.5 and Lemma 4.6. Proposition 4.9. Suppose Q(R) is Gorenstein. A morphism f of R-modules is rbm if and only if Ker f is torsionless. f
g
Lemma 4.10. Suppose Q(R) is Gorenstein. Let the sequence of R-modules 0 → A → B → C → 0 be exact. If A and C are torsionless, then so is B. Thus with the condition Q(R) is Gorenstein, when discussing the rbm property, we can deal with a normal kernel as well as a pseudo-kernel. Proposition 4.11. Suppose Q(R) is Gorenstein. For a given morphism f , Ker f is torsionless if and only if Ker f is torsionless. Proof. From Lemma 2.17, there is an exact sequence 0 → Ker f → Ker f → Ω R1 (Cok f ) → 0. So the “if” part is obvious, and the “only if” part comes from Lemma 4.10. Theorem 4.12. Suppose Q(R) is Gorenstein. The following are equivalent for a morphism f : A → B in mod R. (1) (2) (3) (4) (5)
f is rbm. Ker f is torsionless. Ker f is torsionless. H−1 (C( f )• ) = 0. Ext1R (Tr f, R) is an monomorphism in mod R. st
(6) Ω R1 (Cok f ) ∼ = Ker f . st
(7) There exists f 0 such that f 0 ∼ = f and Ker f 0 is torsionless. st ∼ f , Ker f 0 is torsionless. (8) For any f 0 with f 0 = Proof. Implications (6) ⇒ (3), (8) ⇒ (2) and (8) ⇒ (7) are obvious. We already showed the equivalence of (1), (4) and (5) in Theorem 3.9, (1) ⇔ (3) in Proposition 4.9, and (3) ⇔ (2) in Proposition 4.11. Implications (3) ⇒ (8) and (7) ⇒ (3) are obtained from “if” and “only if” part of Proposition 4.11 respectively. (4) ⇒ (6). It comes directly from Cok dC( f ) 0 = Ker f and Cok dC( f ) −1 = Cok f . Corollary 4.13. 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 4.12. (2) ⇒ (1). For every M ∈ mod R, Ker ψ M is torsionless. Because Ker ψ M is projective and Ker ψ M is a submodule of Ker ψ M from Lemma 2.17(1). So if (2) holds, ψ M is rbm for any M ∈ mod R, which implies (1) from Proposition 4.7.
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In the case Q(R) is not Gorenstein, let us see an example where a morphism with torsionless kernel is not rbm. Example 4.14. In the case R = k[[X, Y, Z ]]/(X Y, X 2 ) with any field k, consider the map ψk : k → J 2 k. We know Ker ψk is projective. We shall show that ψk is not rbm; that is, H−1 (C(ψk )• ) does not vanish. As we have seen, C(ψk )i = 0 (i ≤ −2), H−1 (C(ψk )• ) = (Ext1R (k, R))∗ . From a free resolution of k
R
3
Y X 0 0 0 X
(X Y )
−→ R 2 → R → k → 0,
we get a free resolution of Ext1R (k, R)
R
4
X Y 0 0 0 0 X Y
−→
R 2 → Ext1R (k, R) → 0.
We easily see that (Ext1R (k, R))∗ does not vanish. Acknowledgements I am grateful to Kazuhiko Kurano who suggested that conditions for rbm should be given in terms of the kernel not only by the pseudo-kernel. I thank Shiro Goto who told me that the assumption of Theorem 4.12 is weakened and that Corollary 4.13 holds. I also thank Charles Weibel for useful advice on the introduction. Appendix A Appendix A by Shinsuke Takashima. In this section we give an alternative proof of Theorem 4.12 by using the classical torsion theory. Definition A.1. Let R be a commutative ring and M an R-module. We define R-submodules ε M and τ M of M as follows: ε M = ∩ f ∈Hom R (M,R) Ker f τ M = ∩ f ∈Hom R (M,E(R)) Ker f. It is easy to see M is torsionless if and only if ε M is zero. Lemma A.2. Let M be an R-module over a commutative ring R, and K a submodule of M. Then τ K = K ∩ τ M . Proof. (See also [7].) The inclusion τ K ⊂ K ∩ τ M is obvious. Let x ∈ K ∩ τ M and f : K → E(R) be an Rhomomorphism. Then we can extend f to some R-homomorphism g : M → E(R). Thus 0 = g(x) = f (x). Lemma A.3. For a commutative noetherian ring R, the following are equivalent. (1) For M ∈ mod R, ε M = τ M . (2) Every finite R-submodule X of E(R) is torsionless. Proof (See also [6]). (1) ⇒ (2) Obvious. (2) ⇒ (1). Let M be a finitely generated R module. τ M ⊂ ε M is easy. Let x ∈ M \ τ M . Then we can find R-homomorphism f : M → E(R) such that f (x) 6= 0. Set X = f (M). By assumption we have ε X = 0, hence we can find a morphism g : X → R such that g( f (x)) 6= 0. Thus x 6∈ ε M . Lemma A.4. Let R be a commutative noetherian ring whose total ring of fractions Q(R) is Gorenstein. Then for any finitely generated R-submodule X of E(R) are torsionless. Proof. By assumption Q(R) is an injective Q(R)-module. From the exactness of the functor Hom R (−, Q(R)) ∼ = Hom R (−, Hom Q(R) (Q(R), Q(R))) ∼ = Hom Q(R) (− ⊗ R Q(R), Q(R)), Q(R) is also injective as an R-module. Thus E(R) is a submodule of Q(R). On the other hand, each non-zero divisor r ∈ R defines an R-free module X r = {x ∈ Q(R) | xr ∈ R}. Obviously Q(R) = ∪r X r hence any finitely generated R-submodule X of E(R) is contained in some X r , which means X is a torsionless R-module.
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An alternative proof of Theorem 4.12. We shall show the equivalence between (1) and (2). Set K = Ker f . It is easy to see f is rbm if and only if K ∩ ε A = 0. We claim K ∩ Ker ε A = ε K . From Lemmas A.3 and A.4, ε A = τ A . And K ∩ ε A = K ∩ τ A = τ K from Lemma A.2. Again from Lemmas A.3 and A.4, τ K = ε K . On the other hand, K is torsionless if and only if ε K = 0. References [1] M. Auslander, M. Bridger, The stable module theory, Mem. Amer. Math. Soc. 94 (1969). [2] M. Auslander, R.O. Buchweitz, The homological theory of maximal Cohen–Macaulay approximations, Soc. Math. de France, Mem 38 (1989) 5–37. [3] W. Bruns, J. Herzog, Cohen–Macaulay rings, Cambridge Stud. Adv. Math. 39 (1998). [4] E.G. Evans, P. Griffith, Syzygies, in: London Math. Soc. Lecture Note Series, vol. 106, Cambridge U.P., 1985. [5] K. Kato, Stable module theory with kernels, Math. J. Okayama Univ. 43 (2001) 31–41. [6] M. Hoshino, On Lambek torsion theories, Osaka J. Math. 29 (1992) 447–453. [7] B. Stenstr¨om, Rings of Quotients, Springer-Verlag, 1975.