Morphisms represented by monomorphisms

arXiv:math/0404243v1 [math.AC] 13 Apr 2004 Morphisms represented by monomorphisms Kiriko Kato Department of Applied Mathematics, Osaka Women’s University Sakai, Osaka 590-0035, JAPAN e-mail: [email protected] Abstract Every homomorphism of modules is projective-stably equivalent to an epimorphism but is not always to a monomorphism. We prove that a map is projectivestably equivalent to a monomorphism if and only if its kernel is torsionless, that is, a first syzygy. If it occurs although, there can be various monomorphisms that are projective-stably equivalent to a given map. But in this case there uniquely exists a ”perfect” monomorphism to which a given map is projective-stably equivalent. 2000 Mathematics Subjects Classification: 13D02, 13D25, 16D90 1 Introduction Let R be a semiperfect ring. A morphism f : A → B and f ′ : A′ → B ′ in mod R are said to be projective-stably equivalent if they are isomorphic in mod R; if there exist morphisms α : A → A′ and β : B → B ′ such that α and β are isomorphisms and β ◦ f = f ′ ◦ α in mod R. We say a morphism f is represented by monomorphisms (”rbm” for short) if there exists a monomorphism in mod R that is projective-stably equivalent to f . For any homomorphism f : A → B of R-modules, ( f ρB ) : A ⊕ PB → B is surjective with a projective cover ρB : PB → B. Thus every morphism is represented by epimorphisms. The choice of epimorphism is unique; if an epimorphism f ′ is projective-stably equivalent to f , then f ′ is isomorphic to (f ρB ) up to direct sum of projective modules. 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 would be two monomorphisms that are not isomorphic up to direct sum of projective modules and both of which are projectivestably equivalent to f . 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 projective-stably equivalent 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 1 whose R-dual is an epimorphism. This is our first focal point. We studied the situation where a map is rbm, especially the structure of monomorphisms into which a given map is modified. And we obtained the obstruction for a given map to be rbm. In the case that a map is rbm, the choice of a monomorphism is not unique, but then a perfect monomorphism projectivestably equivalent to the given map is uniquely determined up to 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 mod R and a subcategory of K(mod R). Due to this equivalence, we describe the obstruction of being rbm with a homology of a complex associated to the given map. 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 torsionlessness of the kernel is equivalent to rbm condition. This is our third point. Actually, for this we need Gorensteinness. Theorem 4.10 : 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 torsionless, equivalently, a first syzygy. 2 Preliminaries We shall fix the notations and give some review on the correspondence between stable module category and homotopy class category of complexes. We omit the proofs for results that are in [5]. Throughout the paper, R is a commutative semiperfect ring, equivalently a finite direct sum of local rings; that is, each finite module has a projective cover ( see [6] for semiperfect rings). 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. By an R-module we mean ”a finitely generated R-module”. For an R-module M , ρM : PM → M denotes a projective cover of M . For an abelian category A, K(A) stands for the category of the homotopy equivalence class of complexes in A. A complex is denoted as d n−1 d n F F • : · · · → F n−1 F→ F n → F n+1 → · · · . A morphism in K(A) is a homotopy equivalence class of chain maps. A trivial complex is a split exact sequences of projective modules. Truncations of a 2 complex F • are defined as follows: d n−2 d n−1 τ≤n F • : · · · → F n−2 F→ F n−1 F→ F n → 0 → 0 → · · · , d n d n+1 F τ≥n F • : · · · → 0 → 0 → F n → F n+1 F→ F n+2 → · · · An R-dual F•∗ of a complex F • is the cocomplex such as Fn∗ = (Fn )∗ , dFn = ∗ (dF n−1 ) where ∗ means HomR ( , R). The projective stabilization mod R is defined as follows. ∗ • Each object of mod R is an object of mod R. • For A, B ∈ mod R, a set of morphisms from A to B is HomR (A, B) = HomR (A, B)/P(A, B) where P(A, B) := {f ∈ HomR (A, B) | f factors through some projective module}. Each element is denoted as f = f mod P(A, B). A morphism f : A → B in mod R is called a stable isomorphism if f is an isomorphism in mod R and we write st A∼ = B. For an R-module M , define a transpose Tr M of M to be Cok δ∗ where δ 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 ∈ HomR (M, N ), then f induces a map Tr N → Tr M , which represents a morphism Tr f ∈ HomR (Tr N, Tr M ). A kernel of projective cover of M is called the first syzygy module of M and denoted as Ω1R (M ). The first syzygy module of M is uniquely determined as an object of mod R. Inductively, we define Ω1R (M ) = n−1 Ω1R (ΩR (M )). If f ∈ HomR (M, N ), then f induces a map ΩnR (M ) → n ΩR (N ), which represents a morphism ΩnR (f ) ∈ HomR (ΩnR (M ), ΩnR (N )). Lemma 2.1 On the commutative diagram with exact rows in mod R 0 → 0 → A  α y A′ f → f′ → B   yβ B′ g → g′ → C  γ y C′ → 0 → 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. 3 2) β and γ are stable isomorphisms and γ is an epimorphism. 3) β and γ are stable isomorphisms. 1) Adding a projective cover ρA : PA → A to the given diagram, we get the following: 0 ↓ Q  0 0 → = v y A ⊕ PA  y(α w y ( f010 ) → B ⊕ PA  y(β ρA ) ′ A  → 0 ↓ Q  f′ C  ∼ y= g′ C′ →  y 0 f′ → f ′ ◦ρA ) ′ B  →  y (g 0) → 0 → 0 0 Since (β ◦ ρA ) 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 ρA ) 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 ′ : C ′ → C such that γ ◦ γ ′ = idC ′ . On the diagram 0 → A → k 0 → A → B′′ → C′ → 0 B → → 0,  ′ yβ f g  ′ yγ C Ext1R (γ ′ , A) is a stable isomorphism because : → Ext1R (C ′ , A) is an isomorphism. Connection of the above and the given diagrams yields another diagram: β′ 0 → A  α y A′ 0 → → f′ B′′ → → C′ k → 0 g′ C′ → 0  ′ yβ◦β B′ Ext1R (C, A) → β′ Since β ◦ is a stable isomorphism, we can apply 1). 3) Adding a projective cover ρB ′ : PB ′ → B ′ , we get 0 → 0 → A  α y A′ (f0) → B ⊕PB ′ f′ → (β y ( g0 ) 01 → ρB′ ) B′ g′ → 4 C ⊕PB ′  y(γ C′ → 0 g ′ ◦ρB ) → 0. Since β is a stable isomorphism and (γ g′ ◦ ρB ) is an epimorphic stable isomorphism, we can apply (2) to get the conclusion. (q.e.d.) Let L be a full subcategory of K(mod R) defined as L = {F • ∈ K(proj R) | Hi (F • ) = 0 (i < 0), Hj (F ∗ • ) = 0 (j ≥ 0)}. Lemma 2.2 ( [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 resolution of A. 2) For f ∈ HomR (A, B), there exists f • ∈ HomK(mod 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 . Theorem 2.3 ( [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 ∈ HomR (A, B), there exists a triangle n• f• c• C(f )•−1 → FA • → FB • → C(f )• . (2.1) In general, C • does not belong to L any more but it satisfies the following: Hi (C • ) = 0 (i < −1), Hj (C ∗ • ) = 0 (j > −1). Definition and Lemma 2.4 ([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 • ) are uniquely determined by f , up to isomorphisms. We call these the pseudokernel and the pseudo-cokernel of f . And we have Cok f = Tr Ker Trf . 5 Notations. The pseudo-kernel or the pseudo-cokernel is determined as a stable equivalence class of modules, not as an isomorphic class of mod R. st st Although, for the simplicity, a module M with M ∼ = Kerf or M ∼ = Cokf is denoted by Kerf or Cokf respectively. Chain maps n• and c• in (2.1) induce cf : Cok f → A and nf : B → Ker f . These maps have the properties f ◦ nf = 0 and cf ◦ f = 0. The following lemma shows why Ker f and Cokf are called the pseudo-kernel and the pseudo-cokernel respectively. Lemma 2.5 ([5] Lemma 3.3, Lemma 3.5) Let f : A → B be a homomorphism of R-modules. 1) If x ∈ HomR (X, A) satisfies f ◦ x = 0, there exists hx ∈ HomR (X, Ker f ) such that x = nf ◦ hx . 2) If y ∈ HomR (B, Y ) satisfies y ◦ f = 0, there exists ey ∈ HomR (Cokf , Y ) such that y = ey ◦ cf . 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 p) Lemma 2.6 For a given f ∈ HomR (A, B), suppose both A ⊕ P → B and (f p′ ) A ⊕ P ′ → B are epimorphisms with projective modules P and P ′ . Then there are stable isomorphisms λ : A ⊕ P → A ⊕ P ′ and κ : Ker (f p) → Ker (f p) that make the following diagram commutative: 0 → Ker  (f p) → A⊕ P 0 → Ker (f p′ ) → A ⊕ P′ κ y  yλ (f p) → (f p′ ) → B k → 0 B → 0 proof. Set π : B → Cok f as a canonical map. Both π ◦ p and π ◦ p′ are projective covers of Cok f , hence there exists l : P → P ′ such that π ◦ p = π ◦ p′ ◦ l. Then Im (p′ ◦ l − p) ⊂ Im f , so we get h : P → Im f as h coincides with p′ ◦ l − p. Via A → Im f which is surjective, h can be lifted to a  map j  : P → A. This shows the equation f ◦ j + p′ ◦ l = p. The map 1 j λ= : A ⊕ P → A ⊕ P ′ yields the desired diagram . Obviously λ 0 l 6 is a stable isomorphism, which implies that κ is a stable isomorphism from Lemma 2.1. (q.e.d.) Lemma 2.7 ([5] Lemma 3.6) 1) There is an exact sequence 0 → Ker f → Ker f → Ω1R (Cok f ) → 0. 2) There is an exact sequence 0 → L → Cok f → Cok f → 0 such that Ω1R (L) is the surjective image of Ker f . Lemma 2.8 The following holds for f ∈ HomR (A, B). 1) Kerf is projective if and only if f • can be taken as f i are isomorphisms for i ≤ −1. 2) If Kerf is projective, then Ω1R (f ) is a stable isomorphism. proof. 1) The ”if” part is obvious. First notice that Kerf is projective if and only if we can choose C(f )• such that C(f )i = 0 (i ≤ −2) as an element of K(mod R). There is an exact sequence f ′• 0 → C(f )•−1 → E • → F • → 0 where E • ∼ = FB • in K(mod R) and via these isomorphisms, f • = FA • , F • ∼ • ′ is isomorphic to f . Easily we see that C(f )i = 0 if and only if f ′ i is an isomorphism in mod R. 2) The triangle n• f• c• C(f )•−1 → FA • → FB • → C(f )• induces an exact sequence of complexes • 0 → FA • → F˜B → C(f )• → 0 which again induces • 0 → τ≤−1 FA • → τ≤−1 F˜B → τ≤−1 C(f )• → 0 7 • where F˜B is a direct sum of FB • and a trivial complex. Hence the exact sequence of homology groups is 1 0 → Ω1R (A) (ΩR⋆(f )) 1 → ΩR (B) ⊕ P → Kerf → 0 1
with projective modules P. From the assumption, ΩR⋆(f ) is a split monomorphism, in particular, Ω1R (f ) is a stable isomorphism. (q.e.d.) Corollary 2.9 Suppose R is local. If a morphism f ∈ EndR (A) satisfies that Kerf 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 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. (q.e.d.) 3 Representation by monomorphisms and perfect exact sequences Definition 3.1 A morphism f : A → B in mod R is said to be represented by monomorphisms (rbm for short) if some monomorphism f ′ : A′ → B ′ in mod R is projective-stably equivalent to f , that is, there exist stable isomorphisms α : A → A′ and β : B → B ′ such that β ◦ f = f ′ ◦ 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 ϕN : N  → N the natural map. Then any map N ⊕ P → ϕN ⋆ N ∗∗ ⊕ Q of the form with projective modules P and Q, is 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. 8 It was Aulander and Bridger who first defined and studied ”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 ′ : A → B ⊕ P with a projective module P such that f = s ◦ f ′ via some split epimorphism s : B ⊕ P → B. 2) There exists a monomorphism f ′ : A → B ⊕ P with a projective module P such that f = s ◦ f ′ via some split epimorphism s : B ⊕ P → B, and f ′ ∗ is an epimorphism. 3) HomR (B, I) → HomR (A, I) is surjective if I is an injective module. Auslander and Bridger’s original definition of ”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 ′ : A → B ⊕ P with a projective module P such that f = s ◦ f ′ via some split epimorphism s : B ⊕ P → B. proof. The ” if” part is clear. We shall show ”only if” part. Suppose there exists a monomorphism f ′ : A′ → B ′ , stable isomorphisms α : A → A′ and β : B → B ′ such that β ◦ f = f ′ ◦ α. We first take projective covers ρA : PA → A and ρB : PB → B such that the induced map fP : PA → PB by f is a monomorphism. Since α is a stable isomorphism, there exists a morphism α′ : A′ → A such that α ◦ α′ = idA′ and α′ ◦ α = idA . From the last equation there exists a morphism
sA : A → PA such that ′ α ◦ α + ρA ◦ sA = idA , equivalently (α′ ρA ) ◦ sαA = idA . In particular,
(α′ ρA ) : A′ ⊕ PA → A is a split epimorphism and sαA : A → A′ ⊕ PA is a split monomorphism. Similarly we get morphisms β ′ : B ′ → B, sB :
B → PB ′ β
′ ′ and sB ′ : B → PB ′ such that (β ρB ) ◦ sB = idB and (β ρB ′ ) ◦ sβ ′ = idB ′ . B = f ′ ◦ α induces β ′ ◦ (β ◦ f ) ◦ α′ = Given equation β ◦ f ′ ′ ′ ′ ′ ′ β ◦ (f ◦ α) ◦ α , that is, f ◦ α = β ◦ f . Hence there exists a homomorphism t : A′ → PB such that f ◦ α′ − β ′ ◦ f ′ = ρB ◦ t. Now we get a commutative diagram A′ ⊕  PA  ′ y(α A  f′ t 0 fP −→ ρA ) f → 9  B′ ⊕  PB  ′ y(β B. ρB )   β′ 0 0  : B ′ ⊕ PB → B ⊕ PB ′ ⊕ PB Since the composite of maps  sB ′ 0 idPB   0 β ρB ′ and : B ⊕ PB ′ ⊕ PB → B ′ ⊕ PB is equal to idB ′ ⊕PB , 0 0 idPB   ′  f′ 0 f 0 ′ ′ ( β ρB ) ◦ = ( β ρB ) ◦ idB ′ ⊕PB ◦ = ( β ′ ρB ) ◦ t fP t f P     ′  β′ 0 0 β ρB ′ f 0   ◦ sB ′ 0 ◦ , and the following diagram 0 0 idPB t fP 0 idPB commutes: ′′ f A′ ⊕  PA −→ B ⊕ PB ′ ⊕ PB  ′ ρ y(α ρA ) . y f A where →  β′ ′′ f =  sB ′ 0   ′ 0 f 0 ◦ t idPB and ρ = ( β′ B ρB ) ◦  β 0 ρB ′ 0 0 fP  0 idPB  , . It is easy to see that f ′′ is a monomorphism and ρ is a split epimorphism.   α ′′′ ′′ Finally putting f = f ◦ which is a monomorphism, we have sA
f = f ◦ idA = f ◦ α′ ρA ◦  α sA  = ρ ◦ f ′′ ◦  α sA  = ρ ◦ f ′′′ . (q.e.d.) 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 to be perfectly exact if its R-dual 0 → HomR (C, R) → HomR (B, R) → HomR (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. 10 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 ∼ = Cokf . 6) FC • ∼ = C(f )• in K(proj R). proof. In [5] Lemma 2.7, we see the equivalence between 1) and 2). The implication 3) ⇒ 2) is obvious. For the rest of the proof, consider the following diagram: 0 → FA • k → 0 → • F A → 0 →  • yα C(g)•−1 f• → • C yl(f  )  ′• yβ • F B  ′′ • yβ • FfB → g• → • C(f )  ′• yγ • F C  ′′ • yγ → C yl(g)• → 0 → 0 → 0 The top-row and the bottom-row are exact. Chain maps β ′ • , β ′′ • and γ ′′ • are isomorphisms up to homotopy. 2) ⇒ 3),4),5) and 6). If the middle row is also exact, then C(γ ′ )• ∼ = C(β ′ )• , which are trivial complexes, hence γ ′ • is an isomorphism. Now β ′′ • ◦ β ′ • and γ ′′ • ◦ γ ′ • are isomorphisms, C(α)• = 0 follows from the exact sequence • • 0 → C(α)• → C(β ′′ ◦ β ′ ) → C(γ ′′ ◦ γ ′ ) → 0. 4) ⇒ 3). On the above diagram αi = id for i ≤ 0, so FA • ∼ = C(g)•−1 • • • ′′ ′ implies that α is an isomorphism. Therefore γ ◦γ is also an isomorphism. (q.e.d.) If R is local, all the conditions above are equivalent. We shall give the proof later at the end of this section. 11 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 ∼ = Cokf . 6) FC • ∼ = C(f )• in K(proj R). (f ρB ) For a morphism f : A → B, A ⊕ PB −→ B is an epimorphism with a projective cover ρB : PB → B. Thus each morphism is represented by epimorphisms. And the choice of the representing epimorphism is unique up to direct sum of projective modules, as we have seen in Lemma 2.6. Unlikely, we already know an example of a morphism that is not rbm. And moreover, even if a given map is represented by a monomorphism, there would be another representing monomorphism. We see it in Example 3.10. However, 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 stable isomorphisms. These are the statements in Theorem 3.9, before which, we need some preparations. f g For given exact sequence of modules A → B → C, we have a diagram of triangles f• • • • F → FB • → C(f → F A A )  •+1  •  • k yα yγ yα (3.2) C(g)•−1 FB • → g• → FC • → C(g)• which induces a diagram with exact rows • 0 → H−1 (C(f ) ) 0 → A  α y → Kerg (fǫ ) → B ⊕FA 1 β y → B ⊕ PC (cf π) → (g ρC ) → Cokf  γ y C We observe some facts below. Lemma 3.8 With the notations above, the following holds. 1) β is a stable isomorphism. 12 → 0 (3.3) → 0 • 2) C(α)•+1 ∼ = C(γ) . 3) α is the composite of natural maps A → Im f = Ker g and Ker g → Kerg. So if f is injective and g is surjective, then from Lemma 2.7, α is a stable isomorphism, τ≤−1 C(α)• = 0, and τ≤−2 C(γ)• = 0. 4) If H−1 (C(f )• ) = 0, then the upper row of (3.3) is the short exact sequence (fǫ ) (cf π) θf : 0 → A → B ⊕ FA 1 → Cokf → 0 which is a perfect exact sequence. Theorem 3.9 Let f : A → B be a morphism in mod R. Then f is rbm if and only if H−1 (C(f )• ) vanishes. If this is the case, we have the following: 1) We have a perfect exact sequence (fǫ ) (cf π) θf : 0 → A → B ⊕ FA 1 → Cokf → 0. 2) For any exact sequence of the form f q
(g p) σ : 0 → A → B ⊕ P′ → C → 0 with some projective module P ′ , there is a commutative diagram θf : σ: 0 0 → → A   yα̃ A ( fǫ ) → f q
→ (cf π) → B ⊕FA 1  yβ̃ (g p) B ⊕ P′ → Cokf  → 0 C → 0 γ̃ y where α̃ and β̃ are stable isomorphisms. 3) There is an exact sequence with some projective module Q and Q′ (γ̃ ⋆) 0 → Q′ → Cokf ⊕ Q → C → 0. 4) If σ is also perfectly exact, then σ is isomorphic to θf up to direct sum of trivial complexes. 13 proof. Suppose that f is rbm; there is an exact sequence f q
(g p) σ : 0 → A → B ⊕ P ′ → C → 0. The maps f˜ = f
q and g̃ = (g p) produce the same diagram as (3.2) because • ˜ we may consider f = f • and g̃• = g• . Apply Lemma 3.8 3) to this sequence, and we get τ≤−2 C(γ)• = 0 as for γ • : C(f )• → FC • . 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. Conversely, suppose that H−1 (C(f )• ) = 0. Then Lemma 3.8 4) shows that θf 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 ) : 0 → A f  → B ⊕ P ′ ⊕ FA 1 → 0 → Ker(g p) → B ⊕ P ′ ⊕ PC (g p ρC ) q ǫ  α y  β y Cok → f
q  γ y C → 0 → 0. The upper row is a direct sum of θf and a trivial complex, and the lower row is that of σ and a trivial complex. Hence we get a desired diagram. 0 → 0 → A   yα̃ A ( fǫ ) → f q
→ B ⊕FA 1  yβ̃ B⊕P (cf π) → Cokf  (g p) → γ̃ y C → 0 → 0. Notice that β̃ is a stable isomorphism. From Lemma 3.8 3), α̃ is also a stable isomorphism. 3) Consider the exact sequence of complex • γ• g ) → F • → 0, 0 → C(γ)•−1 → C(f C • g )• ∼ where C(f = C(f ) in K(projR). Applying the truncation τ≤0 , we get • g ) → τ F • → 0, 0 → (τ≤−1 C(γ))•−1 → τ≤0 C(f ≤0 C which induces an exact sequence of homology 0 → Q′ → Cokf ⊕Q → C → 0 with projective modules Q′ = C(γ)−1 and Q from Lemma 3.8 3). 14 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 θf . (q.e.d.) 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 = R/(XY, Y 2 , Y Z). The minimal Cohenf Macaulay approximation θf : 0 → Yk → Xk → k → 0 of k is perfectly exact since Ext1R (k, R) = 0. On the other hand, the minimal Cohen-Macaulay g approximation (see [2] for definition) of M , σ : 0 → YM → XM → M → 0 is ∼ not perfect since Ext1R (M,  R) 6=0. The map g is decomposed as YM = Yk ⊕R, f q XM ∼ . We easily check the statement of Theorem = Xk ⊕R and g = 0 Y st ∼ Cokf and there is an exact sequence 3.9; k = 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 • . In this situation, 5) and from Theorem 3.9. In other words, C(f )• ∼ = FCokf 6) are clearly equivalent. 6) ⇒ 4). Since the assumption implies H−1 (C(f )• ) = 0, we get an ex(γ ⋆) act sequence 0 → Q′ → C ⊕ Q → C → 0 with projective modules Q and Q′ , applying Theorem 3.9 3). Since R is local, we can apply Corollary 2.9, which shows that γ is a stable isomorphism, equivalently γ • is an isomorphism, hence α• is also an isomorphism. (q.e.d.) 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 ]]/(XY ) and let f be a map defined via projective resolutions as follows: ··· 0 → → R2 (X y R (X Y ) −→ Y) (X ) Y −→ R  → R2 →  X y( Y ) k  f y Yk Then the sequence 0 → X → k ⊕ R2 15 (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. 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 three-dimensional Gorenstein local ring. Set L = Tr k, TL = Ω3R (Tr)Ω3R (Tr)L and ϕL : L → TL to be a natural map. Since Hi (FL • ) = 0 (i ≤ 1) and FTL • is exact, H−1 (C(ϕL )• ) = 0. Putting N = CokϕL , we have a perfectly exact sequence (ϕ∗L ) θϕL : 0 → L → TL ⊕ P → N → 0 with some projective module P . It is easy to see that (N )∗ is free. 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, Kerf = Cok dC(f ) −2 is the first syzygy of Cokf = Cok dC(f ) −1 . So it is natural to ask the converse: Is a given map f represented by monomorphisms if Kerf is a first sygyzy? This section deals with the problem. As a conclusion, the answer is yes if the total ring of fractions Q(R) of R is Gorenstein. Notice that if Q(R) is Gorenstein, then Q(R) is Artinian as we see in Lemma 4.3. What is more, if Q(R) is Gorenstein, instead of a pseudo-kernel, we can use a (usual) kernel to describe 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 torsionless map φ : M → M ∗∗ is a monomorphism. 1 1 if the natural 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. 16 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). 2) ⇒ 3). We may assume that M is a submodule of a free module Rl . Let d−2 d−1 · · · P −2 → P −1 → P 0 → Tr M be a free resolution of Tr M . Then 2) says ∗ ∗ ∗ (d−1 ) ∗ (d−2 ) ∗ ∗ (P 0 ) → (P −1 ) → (P −2 ) is exact. By definition, M ∼ = Cok (d−1 ) ∗ which is isomorphic to Ω1R (Cok d−2 ). 3) ⇒ 1). We may assume that M is a submodule of a free module Rl .   f1  f2    Let  ..  : M → Rl be a monomorphisms. If m ∈ M is not zero, f (m)  .  fl is not zero, so there exists some i such that 0 6= fi (m) = φ(m)(fi ), which implies φ(m) 6= 0. Thus Ker φ = (0). (q.e.d.) Lemma 4.3 If Q(R) is Cohen-Macaulay, then Q(R) is of dimension zero. proof. Suppose there exists a non-minimal prime ideal q. Then [ q 6∈ Ass Q(R), since Q(R) has no embedded prime. This implies q 6⊂ p, p∈Ass Q(R) hence q contains a non-zero-divisor which is a unit. (q.e.d.) Lemma 4.4 Let R be a Noetherian ring and f : A → B be a morphism in mod R. Suppopse Q(R) is Gorenstein. If Kerf is projective, then f is rbm. proof. The assumption says τ≤−2 C(f )• = 0. From Theorem 3.9, f is rbm if and only if H−1 (C(f )• ) = 0, which means that dC(f ) −1 is injective. ∗ ∗ So we have only to show Ker dC(f ) −1 = (Cok (dC(f ) −1 ) ) = 0. A triangle f• FA • → FB • → C(f )• → FA •+1 induces an exact sequence 0 → H−1 (C(f )∗• ) → H−1 (FB ∗• ) → H−1 (FA ∗• ) → 0. 17 ∗ Note that Cok (dC(f ) −1 ) Since Q(R) is Gorenstein So if p is any associated ∗ ∗ ∗ Cok (dC(f ) −1 ) ∗ ∼ = H−1 (C(f )• ) and H−1 (FB ∗• ) = Ext1R (B, R). of dimension zero, Ext1R (B, R) ⊗ Q(R) = 0. prime ideal of R, Ext1R (B, R)p = 0. Hence p = 0, which implies (Cok (dC(f ) −1 ) ) = 0. (q.e.d.) To solve our problem, the special kind of maps is a key. For M ∈ mod R, consider a module J 2 M = Tr Ω1R Tr Ω1R M . Since Tr J 2 M is a first sygyzy, 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 Ω1R M = Cok (dFJ 2 M −2 ) = Cok (dFM −2 ) . The identity map on Tr Ω1R M induces a chain map (FM )∗• → (FJ 2 M )∗• i and ψM • : FJ 2 M • → FM • subsequently. The maps ψM are identity • maps for i ≤ −1, in other words, τ≤−2 C(ψM ) = 0 and KerψM is projective. Thus we can apply the argument in Lemma 4.4 for f = ∗ ψM ; H−1 (C(ψM )• ) ∼ = (H−1 C(ψM )∗• ) . Since H−1 (FJ 2 M ∗ • ) = 0, we have H−1 (C(ψM )∗• ) ∼ = Ext1R (M, R). Therefore we have = H−1 ((FM )∗• ) ∼ ∗ • ∼ −1 1 H (C(ψM ) ) = (ExtR (M, R)) . Now we get a result as follows: Corollary 4.5 The map ψM : J 2 M → M is rbm if and only if an R-module ∗ M has Ext1R (M, R) = 0. For given morphism of R-modules f : A → B, adding a projective cover of B to f , we get an exact sequence (n⋆f ) A ⊕ PB 0 → Kerf → (f ρB ) → B → 0. Due to Theorem 3.9, we have a perfect exact sequence θnf , because nf is rbm: θ nf : 0 → Kerf (n⋄f ) 1 → A ⊕ FKerf 0 → Kerf (n⋆f ) →  st y∼ =  st y∼ = A ⊕ PB → (f ρB ) → Cok nf → 0 B → 0   ωf y ∗ Lemma 4.6 With notation as above, suppose (Ext1R (Cokf , R)) = 0. Then the following conditions are equivalent. 1) f is rbm. 2) Kerf is torsionless and ωf : Cok nf → B is rbm. 18 proof. On the diagram of triangles nf • FKerf • →   χ• y f FA • → C(nf )• f• FB • k C(f )•−1 → FA • →  ω • y f → FKerf •+1 →  χ•+1 y f C(f )• , • • we observe C(χf )•+1 ∼ = C(ωf ) . We have H−1 (C(nf ) ) = 0 because nf is rbm. There is an exact sequence H0 (FKerf • ) → H−1 (C(f )• ) → H−1 (C(ωf )• ) . 2) ⇒ 1). Since Kerf is torsionless, H0 (FKerf • ) = Ext1R (Tr Kerf , R) = 0. And H−1 (C(ωf )• ) = 0 because ωf is rbm. From the above exact sequence, we have H−1 (C(f )• ) = 0. st 1) ⇒ 2). From the assumption, H−1 (C(f )• ) = 0 which implies Kerf ∼ = Ω1R (Cokf ). We show that H−1 (C(ωf )• ) = H0 (C(χf )• ) vanishes. The equast tion Kerf ∼ = FJ•2 (Cokf ) . On the other hand, = Ω1R (Cokf ) implies FKerf •+1 ∼ ∼ FCokf • . Via these isomorphisms, χ•+1 is regarded as ψ • . C(f )• = Cokf f • ∼ 0 −1 Hence from the proof of Corollary 4.5, H (C(χf ) ) = H (C(ψCokf )• ) ∼ = ∗ (Ext1R (Cokf , R)) = 0. (q.e.d.) Proposition 4.7 Suppose Q(R) is Gorenstein. modules is rbm if and only if Kerf is torsionless. A morphism f of R- proof. In the previous section, we already have ”only if” part. Apply Theorem 3.9 to nf which is rbm, Theorem 3.9 3) says that Kerωf is projective. Therefore ωf is rbm from Lemma 4.4. Since Q(R) is Gorenstein, we can use Lemma 4.6, which completes the proof. (q.e.d.) Now we go to the next stage; we are to state rbm condition in terms of normal kernel. Lemma 4.8 Suppose Q(R) is Gorenstein. Let the sequence of R-modules f g 0 → A → B → C → 0 be exact. If A and C are torsionless, then so is B. ∼ Kerg is torsionless. Due to Proposition proof. From the assumption, A = 4.7, g is rbm; there exists an exact sequence (gq) θg : 0 → B → C ⊕ Q → Cokg → 0 with a projective module Q and a map q : B → Q. Since C is a submodule of some projective module, so is B. (q.e.d.) 19 Corollary 4.9 Suppose Q(R) is Gorenstein. For a given morphism f , Ker f is torsionless if and only if Kerf is torsionless. proof. From Lemma 2.7, there is an exact sequence 0 → Ker f → Ker f → Ω1R (Cok f ) → 0. So the ”if” part is obvious, and the ”only if” part comes from Lemma 4.8. (q.e.d.) Theorem 4.10 Suppose Q(R) is Gorenstein. The following are equivalent for a morphism f : A → B in mod R. 1) f is rbm. 2) Ker f is torsionless. 3) Kerf is torsionless. 4) H−1 (C(f )• ) = 0. st 5) Ω1R (Cokf ) ∼ = Kerf . st 6) There exists f ′ such that f ′ ∼ = f and Ker f ′ is torsionless. st 7) For any f ′ with f ′ ∼ = f , Ker f ′ is torsionless. proof. Implications 5) ⇒ 3), 7) ⇒ 2) and 7) ⇒ 6) are obvious. We already showed 1) ⇔ 4) in Theorem 3.9, 1) ⇔ 3) in Proposition 4.7, and 3)⇔ 2) in Corollary 4.9. Implications 3) ⇒ 7) and 6) ⇒ 3) are obtained from ”if” and ”only if ” part of Corollary 4.9 respectively. 4) ⇒ 5). It is clear since Cok dC(f ) 0 = Kerf and Cok dC(f ) −1 = Cokf . (q.e.d.) The statement of Theorem 4.10 does not hold for ring R with Q(R) non-Gorenstein. Corollary 4.11 The following are equivalent for a Noetherian ring R. 1) Q(R) is Gorenstein. 2) Every morphism with torsionless kernel is rbm. ∗ 3) Ext1R (M, R) = 0 for each M ∈ mod R. 20 proof. 1) ⇒ 2). It comes directly from Therem 4.10. ∗ For M ∈ mod R, Ext1R (M, R) = 0 means Ext1Rp (Mp , Rp ) = 0 for every p ∈ Ass R. Hence 3) says that Rp is a zero-dimensional Gorenstein for each p ∈ Ass R, which is equivalent to 1) from Lemma 4.3. 2) ⇒ 3). With no assumption, Ker ψM is torsionless. Because KerψM is projective and Ker ψM is a submodule of KerψM from Lemma 2.7 1). So if ∗ 2) holds, ψM is rbm and Ext1R (M, R) = 0 from Corollary 4.5. (q.e.d.) Example 4.12 In the case R = k[[X, Y, Z]]/(XY, X 2 ) with any field k, consider the map ψk : k → J 2 k. We know Kerψk 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 R3 ( Y0 X0 X0 ) 2 (X Y ) −→ R → R → k → 0, we get a free resolution of Ext1R (k, R) R4 ( X0 Y 0 0 0 X Y ) −→ R2 → Ext1R (k, R) → 0. ∗ • We easily see that (Ext1R (k, R)) ∼ = H−1 (C(ψk ) ) does not vanish. Acknowledgement. I thank Kazuhiko Kurano who suggested that conditions for rbm should be given in terms of the kernel not only by the pseudokernel. I also thank Shiro Goto who told me that the assumption of Theorem 4.10 is weakened and that Corollary 4.11 holds. References [1] M.Auslander and M.Bridger, “ The stable module theory,” Memoirs of AMS. 94., 1969. [2] M.Auslander and R.O.Buchweitz, The homological theory of maximal Cohen-Macaulay approximations, Soc. Math. de France, Mem 38(1989), 5-37. [3] M.Amasaki, Homogeneous prime ideals and graded modules fitting into long Bourbaki sequences, Journal of Pure and Applied Algebra, 8(2001), 1-21. 21 [4] E.G.Evans and P.Griffith, “Syzygies,” LMS Lecture Note Series. 106, 1985.@ [5] K.Kato, Stable Module Theory with Kernels, Math. J. Okayama Univ., 43 (2001), 31-41. [6] J.Miyachi, Derived categories with applications to representations of algebras, Seminar Note, 2000. [7] Y.Yoshino, Maximal Buchsbaum modules of finite projective dimension, J. Algebra, 159 (1993) , 240-264. 22