Homomorphisms of complexes via homologies

Intan Muchtadi-Alamsyah

Abstract

By Rickard’s work, two rings are derived equivalent if there is a tilting complex, constructed from projective modules over the first ring such that the second ring is the endomorphism ring of this tilting complex. In this work I describe, under some conditions, the homomorphism space in the derived category between two complexes as iterated pullback of homomorphism spaces between the homologies of the complexes.

Introduction

In 1989, J.Rickard [6] and B.Keller [2] have given a necessary and sufficient criterion for the existence of derived equivalences between two rings. Rickard’s theorem says that for two rings Λ and Γ the derived categories D^{b}(Λ) and D^{b}(Γ) of Λ and Γ are equivalent as triangulated categories if and only if there exists an object T in D^{b}(Λ), named tilting complex, satisfying similar proprieties as those of a progenerator and such that Γ is isomorphic to the endomorphism ring of T in D^{b}(Λ).

To determine a priori the endomorphism ring Γ of a tilting complex over a given ring Λ is difficult. Hence it seems to be important to get an overview of all possible Γ by a simpler way than to calculate the endomorphism ring directly.

One possibility is to try to describe the endomorphism ring of tilting complexes via the endomorphism ring of the homologies of the complexes. The aim of this paper is to describe the endomorphism ring of complexes, more generally, the homomorphism space between two complexes as pullback of homomorphism spaces between their homologies.

S.König and A.Zimmermann [4] have used this approach to describe the endomorphism ring of a 2-term tilting complex with torsion free homology over a Gorenstein order Λ over a complete local Dedekind domain R (see [1, §37] for the relevant definitions).

Klaus Lux, in a private communication with A.Zimmermann, asked whether there exists a generalization of the above result for n-term complexes. We present here the description of Hom_{D^{b}(Λ)}(T, S) for two n-term complexes T and S over an algebra which is not necessarily an order and the homologies need not be torsion free.

For n = 2, we generalize König and Zimmermann’s result. In the absence of their assumpsions, the mapping ψ^{i} : Hom_{D^{b}(Λ)}(T, S) → Hom_{Λ}(H^{i+1}T, H^{i+1}S) is not surjective in general; however, the image of ψ^{i} can be described explicitly. This will give us explicit pull-backs.

The case where Hom_{D^{b}(Λ)}(T, S) is R-torsion free is interesting. Even though we don’t assume that the right-most homologies are R-torsion free, we can replace them by their torsion free parts if the homologies are torsion modules except at the ends of the complexes. In this case Hom_{D^{b}(Λ)}(T, S) is the pull-back of Im(ψ^{0}) and Hom_{D^{b}(Λ)}(H^{n}T/tors, H^{n}S/tors). In particular for S = T a tilting complex and Λ a symmetric order, Hom_{D^{b}(Λ)}(T, T) is R-torsion free by [11, Theorem 1]. Recall that a symmetric order is an order which is a symmetric algebra [11].

The paper is organized as follows. In section 1 we give the definition of derived categories and we review the Rickard’s Morita Theory for derived equivalence. In section 2, we describe Hom_{D^{b}(Λ)}(T, S) as iterated pull-back of homomorphism spaces between their homologies. The image of ψ^{i} and more explicit pull-backs will be explained in section 3. We will explain the torsion free Hom-spaces case in section 4. At the very end we give examples where the main result applies.

Acknowledgement I wish to thank Alexander Zimmermann for many discussions and for correcting this paper and also for having introduced me to derived categories.

1 Review of derived categories

1.1 Category of complexes

Let Λ be an associative ring with 1 and denote by Λ − mod the category of finitely generated left Λ-modules. A complex X in Λ − mod is a sequence of finitely generated Λ-modules (X^{k})_{k∈ℤ} and morphisms (α^{k})_{k∈ℤ} in (Hom_{Λ}(X^{k}, X^{k+1}))_{k∈ℤ}

⋯ → X^{k−1} ⟶^{α^{k−1}} X^{k} ⟶^{α^{k}} X^{k+1} → ⋯

such that α^{k−1}α^{k} = 0 for all k ∈ ℤ.

For two complexes X and Y, a morphism of complexes φ : X → Y is a sequence (φ^{k})_{k∈ℤ} of morphisms φ^{k} : X^{k} → Y^{k} such that

φ^{k}α_{Y}^{k} = α_{X}^{k}φ^{k+1}

for all k ∈ ℤ. With these definitions we form the category of complexes of finitely generated Λ-modules C(Λ).

The shift functor [i] is an endofunctor from C(Λ) to itself where the image of a complex X is another complex X[i] such that

X[i]^{k} = X^{k+i} and α_{X[i]}^{k} = (−1)^{i}α_{X}^{k+i}.

For a complex X in C(Λ) we define

H^{k}(X) = Ker(α_{X}^{k})/Im(α_{X}^{k−1})

and we called the k-th homology of X. We note that

H^{k}(X) ⊆ X^{k}/Im(α_{X}^{k−1}) = Coker(α_{X}^{k−1}) for all k ∈ ℤ.

The category Λ − mod embeds into C(Λ) by identifying a finitely generated Λ-module M with a complex with homology M concentrated in degree 1. We denote by C^{b}(Λ) the full subcategory of C(Λ) consisting of bounded complexes, i.e the complexes X where X^{k} = 0 for all k << 0 and all k >> 0.

1.2 Derived categories and derived equivalences

The objects of D^{b}(Λ), the derived category of Λ, are complexes

P = (⋯ → P^{k} ⟶^{α_{P}^{k}} P^{k+1} → ⋯)

of finitely generated projective left Λ-modules P^{k}, k ∈ ℤ such that

P^{k} = 0 for all k >> 0 and H^{k}(P) = 0 for all k << 0.

If P and Q are such complexes, morphisms φ and ψ from P to Q are said to be homotopic if there is some family of morphisms h^{k} : P^{k} → Q^{k−1}, k ∈ ℤ such that

h^{k}α_{Q}^{k−1} + α_{P}^{k} h^{k+1} = φ^{k} − ψ^{k} for all k ∈ ℤ.

This is an equivalence relation, and, by definition the equivalence classes form the morphisms from P to Q in D^{b}(Λ).

The category Λ − mod (resp. C^{b}(Λ)) embeds into D^{b}(Λ) by choosing for each module (resp. each bounded complex) a fixed projective resolution, which can then represent the module (resp. the complex) in D^{b}(Λ).

For two complexes X and Y in D^{b}(Λ) and k ∈ ℤ, we identify Hom_{D^{b}(Λ)}(X, Y [k]) with Ext_{Λ}^{k}(X, Y) ([3, Lemma 2]).

The derived category D^{b}(Λ) is not necessarily an exact category. However, it is triangulated ([5, Lemma 2.2.3]). To each X and Y in D^{b}(Λ) and each morphism φ : X → Y in D^{b}(Λ), we associate the mapping cone (C(φ), α_{C(φ)}) where for all k ∈ ℤ,

C(φ)^{k} := X^{k+1} ⊕ Y^{k}
α_{C(φ)}^{k} := ( \begin{matrix}−α_{X[1]}^{k} & φ^{k+1} \\ 0 & α_{Y}^{k} \end{matrix} )

and the standard triangle

X →^{φ} Y → C(φ) ⇝ X[1].

A triangle in D^{b}(Λ) is a sequence

X →^{φ} Y → Z ⇝ X[1]

isomorphic to a standard triangle, i.e. such that Z is isomorphic to C(φ) in D^{b}(Λ) [5, Proposition 2.5.3].

Denote by Λ−per the full subcategory of D^{b}(Λ) consisting of the perfect complexes, i.e complexes P of finitely generated projective modules P^{k} where P^{k} = 0 for all k << 0 and all k >> 0. Rickard [6] and Keller [2] have given a necessary and sufficient criterion for the existence of a derived equivalence between two rings Λ and Γ.

Theorem 1.1 (Rickard [6] and Keller [2]) The derived categories D^{b}(Λ) and D^{b}(Γ) are equivalent as triangulated categories if and only if there is a complex T in Λ − per such that

1. Hom_{D^{b}(Λ)}(T, T [i]) = \{ \begin{matrix}0 & if i ≠ 0, \\ Γ & if i = 0. \end{matrix}

2. the smallest triangulated subcategory generated by direct summands of finite direct sums of copies of T inside Λ−per contains Λ as a complex concentrated in degree 1.

A complex T satisfying the condition 1 for i ≠ 0 and the condition 2 is called a tilting complex for Λ.

2 Description of Hom_{D^{b}(Λ)}(T, S) via homologies

Let R be a commutative ring and Λ an R-algebra. Let T and S be complexes

\begin{matrix}T = (0 ⟶^{α^{0}} P^{1} ⟶^{α^{1}} P^{2} ⟶^{α^{2}} ⋯ ⟶^{α^{n−1}} P^{n} → 0) and \\ S = (0 ⟶^{β^{0}} Q^{1} ⟶^{β^{1}} Q^{2} ⟶^{β^{2}} ⋯ ⟶^{β^{n−1}} Q^{n} → 0) \end{matrix}

of projective Λ-modules P^{i} and Q^{i}, with homologies concentrated in degree 1, ⋯, n. For all i ∈ {1, ⋯, n − 1}, we define T^{i} and S^{i} to be the following quotient-complexes of T and S respectively

T^{i} = (0 → Im(α^{i}) → P^{i+1} ⟶^{α^{i+1}} ⋯ ⟶^{α^{n−1}} P^{n} → 0)

S^{i} = (0 → Im(β^{i}) → Q^{i+1} ⟶^{β^{i+1}} ⋯ ⟶^{β^{n−1}} Q^{n} → 0).

We denote by T^{0} := T, S^{0} := S.

Theorem 2.1 If for all i ∈ {1, ⋯, n − 1},

Hom_{Λ}(Coker(α^{i}), Im(β^{i})) = Hom_{Λ}(Im(α^{i}), Coker(β^{i})) =
= Ext_{Λ}^{1}(Coker(α^{i}), H^{i}S) = Hom_{Λ}(Coker(α^{i}), Q^{i}) = 0,  (1)

then the diagram

Hom_{D^{b}(Λ)}(T^{i}, S^{i}) →  Hom_{D^{b}(Λ)}(T^{i+1}, S^{i+1})
↓ _{ψ^{i}}  ↓
Hom_{Λ}(H^{i+1}T, H^{i+1}S) → Hom_{D^{b}(Λ)}(Coker(α^{i+1}), H^{i+1}S[2])

is a pull-back diagram for each i ∈ {0, 1, ⋯, n − 2}.

Remarks

1. The condition (1) holds for i = 0 because α^{0} = β^{0} = 0 and Coker(α^{0}) = P^{1} is projective.

2. Recall that T^{0} = T and S^{0} = S, so that the theorem presents Hom_{D^{b}(Λ)}(T, S) as an iterated pull-back of homomorphism spaces between H^{i+1}T and H^{i+1}S for all i ∈ {0, 1, ⋯, n − 1}.

3. In general, the mapping ψ^{i} is not surjective. We will determine the image of ψ^{i} in section 2.

4. If R is a Dedekind domain with K = F rac(R) and Λ is an R-order, i.e. an R-algebra finitely generated projective as R-module such that K ⊗_{R} Λ is a semisimple K-algebra, then

Hom_{Λ}(Coker(α^{i}), Q^{i}) = 0 implies Hom_{Λ}(Coker(α^{i}), Im(β^{i})) = 0.

We abbreviate for any two complexes X and Y of Λ-modules (X, Y) := Hom_{D(Λ)}(X, Y), Ext^{i}(X, Y) = Ext_{Λ}^{i}(X, Y) and

φ ∘ (X, Y) := {φδ : δ ∈ (X, Y)}.

For all i ∈ {0, 1, ⋯, n − 1}, we denote by L^{i} := Coker(α^{i}) and by ι^{i} the inclusion H^{i+1}T ↪ L^{i}.

Lemma 2.2 For all i ∈ {0, 1, ⋯, n − 2} there exist triangles in D^{b}(Λ)

T^{i} → T^{i+1} → H^{i+1}T [−i + 1] ⇝ T^{i}[1] and
S^{i} → S^{i+1} → H^{i+1}S ⇝ S^{i}[1]

where T^{0} is T and S^{0} is S.

Proof We will show that

H^{i+1}T [−i] ⟶^{(ι̂)^{i}} T^{i} → T^{i+1} ⇝ H^{i+1}T [−i + 1]

is a triangle in D^{b}(Λ).

The map (ι̂)^{i} : H^{i+1}T [−i] → T^{i} is induced by the inclusion H^{i+1}T ↪^{ι^{i}} Coker(α^{i}). The mapping cone of (ι̂)^{i} is the following complex :

0 → Ker(α^{i+1}) → P^{i+1} ⟶^{α^{i+1}} ⋯ ⟶^{α^{n−1}} P^{n} → 0

which is isomorphic to T^{i+1}. Hence, the above sequence is a triangle. Similarly, H^{i+1}S[−i] → S^{i} → S^{i+1} ⇝ H^{i+1}S[−i + 1] is a triangle in D^{b}(Λ). The result follows from the axiom (TR3) in [5, Theorem 2.3.1]. □

From now on we assume that for all i ∈ {1, ⋯, n − 1},

Hom_{Λ}(Im(α^{i}), Coker(β^{i})) = Hom_{Λ}(L^{i}, Im(β^{i})) = 0
and Hom_{Λ}(L^{i}, Q^{i}) = 0.  (2)

The assumption Ext^{1}(L^{i}, H^{i}S) = 0 will be only needed for Lemma 2.9.

Let us fix an i ∈ {0, ⋯, n − 2} for the rest of this section and if i > 0 we denote by

⋯ → P̄^{0} → P̄^{1} → ⋯ → P̄^{i−1} → 0

the first terms of a projective resolution of Ker(α^{i})[−i + 2], and by

⋯ → P̃^{0} → P̃^{1} → ⋯ → P̃^{i−1} → P̃^{i} → 0

the first terms of a projective resolution of Ker(α^{i+1})[−i + 1].

We use similar notation for projective resolutions of Ker(β^{i})[−i+1] and Ker(β^{i+1})[−i] (with P replaced by Q).

Lemma 2.3 1. We have (T^{i}, H^{i+1}S[−i+1]) = 0 and (T^{i}, S^{i+1}[−1]) = 0.

2. There exists a short exact sequence

0 → (T^{i}, H^{i+1}S[−i]) → (T^{i}, S^{i}) → (T^{i}, S^{i+1}) → 0.

Proof We have a triangle

S^{i} → S^{i+1} → H^{i+1}S[−i + 1] ⇝ S^{i}[1]

to which we apply (T^{i}, −) and which then gives rise to a long exact sequence part of which looks as follows:

⋯ → (T^{i}, S^{i+1}[−1]) → (T^{i}, H^{i+1}S[−i]) →
→ (T^{i}, S^{i}) → (T^{i}, S^{i+1}) → (T^{i}, H^{i+1}S[−i + 1]) → ⋯  (3)

Proof of 1 :

(a) (T^{i}, H^{i+1}S[−i + 1]) = 0 For i = 0, the result follows from the fact that T^{0} is concentrated in degrees 1, ⋯, n, whereas the projective resolution of H^{1}S[1] is 0 in degrees 1, ⋯, n. For i > 0, a morphism in (T^{i}, H^{i+1}S[−i+1]) is given by a commutative diagram

⋯ → P̄^{i−1} → P^{i} → P^{i+1} → ⋯ → P^{n} → 0
↓ ↓ ↓
0 → H^{i+1}S → 0

the morphism P^{i} → H^{i+1}S factors through the cokernel of (P̄^{i−1} → P^{i}) which is Im(α^{i}). Hence we get

(T^{i}, H^{i+1}S[−i + 1]) = \frac{(Im(α^{i}), H^{i+1}S)}{j^{i} ∘ (P^{i+1}, H^{i+1}S)}

where j^{i} is the inclusion Im(α^{i}) ↪ P^{i+1} and

j^{i} ∘ (P^{i+1}, H^{i+1}S) = {j^{i}φ : φ ∈ (P^{i+1}, H^{i+1}S)} ⊆ (Im(α^{i}), H^{i+1}S).

Since (Im(α^{i}), H^{i+1}S) ⊆ (Im(α^{i}), Coker(β^{i})) = 0 we get (T^{i}, H^{i+1}S[−i + 1]) = 0.

(b) (T^{i}, S^{i+1}[−1]) = 0. Any such morphism is given by a commutative diagram as follows :

⋯ → P^{i+1} →^{α^{i+1}}  P^{i+2}  →^{α^{i+2}} ⋯ → P^{n}  →  0
↓  ↓  ↓  ↓
0  → Im(β^{i+1}) → ⋯ → Q^{n−1} →^{β^{n−1}} Q^{n} → 0

As the left-most square commutes, the morphism P^{i+2} → Im(β^{i+1}) factors through L^{i+1}. By assumption (2), (L^{i+1}, Im(β^{i+1})) = 0, hence we may assume that the homomorphism P^{i+2} → Im(β^{i+1}) is 0. Now since (L^{j}, Q^{j}) = 0 for all j ∈ {i + 2, ⋯, n − 1}, we can apply an analogous argument to the homomorphisms P^{j+1} → Q^{j} for all j ∈ {i + 2, ⋯, n − 1}. We therefore obtain (T^{i}, S^{i+1}[−1]) = 0.

Proof of 2 : The result follows immediately from 1 applied to the long exact sequence (3). □

We denote by ⋯ → P̂^{i−2} → P̂^{i−1} → P̃^{i} → 0 the first terms of a projective resolution of H^{i+1}T [−i + 1].

Lemma 2.4 We have

1. (H^{i+1}T [−i + 1], S^{i+1}) = 0,

2. (H^{i+1}T [−i], S^{i+1}) = 0,

3. (T^{i}, S^{i+1}) ≅ (T^{i+1}, S^{i+1}).

Proof We apply (−, S^{i+1}) to the triangle

T^{i} → T^{i+1} → H^{i+1}T [−i + 1] ⇝ T^{i}[1]

and get a long exact sequence part of which looks as follows:

⋯ → (H^{i+1}T [−i + 1], S^{i+1}) → (T^{i+1}, S^{i+1}) →
→ (T^{i}, S^{i+1}) → (H^{i+1}T [−i], S^{i+1}) → ⋯  (4)

1. (H^{i+1}T [−i + 1], S^{i+1}) = 0.

Given such a morphism, we get a commutative diagram

⋯ → P̂^{i−1} → P̃^{i} → 0
↓ ↓ ↓
⋯ → Q̃^{i−1} → Q̃^{i} → Q^{i+1} → ⋯ → Q^{n} → 0

Now, the morphism P̃^{i} → Q̃^{i} factors through the kernel of the morphism P̃^{i} → Q^{i+1}, hence through its projective cover Q̃^{i−1}. An analogous argument shows that for all j < i, the homomorphism P̂^{j} → Q̃^{j} factors through Q̃^{j−1}. Hence the chain map represented in the diagram above is homotopic to zero.

2. (H^{i+1}T [−i], S^{i+1}) = 0.

Again, such a morphism is given by a commutative diagram

⋯ → P̂^{i−1} → P̃^{i} → 0
↓ ↓ ↓
⋯ → Q̃^{i} → Q^{i+1} → Q^{i+2} → ⋯ → Q^{n} → 0

With the same argument as in 1, the homomorphism P̃^{i} → Q^{i+1} factors through Q̃^{i} and the homomorphism P̂^{j} → Q̃^{j+1} for all j < i factors through Q̃^{j}. Hence the chain map in the diagram above is homotopic to zero.

3. By applying 1 and 2 to the long exact sequence (4), we obtain (T^{i}, S^{i+1}) ≅ (T^{i+1}, S^{i+1}). □

Corollary 2.5 From Lemma 2.3 and Lemma 2.4 we get that

0 → (T^{i}, H^{i+1}S[−i]) → (T^{i}, S^{i}) → (T^{i+1}, S^{i+1}) → 0

is a short exact sequence.

Lemma 2.6 1. Ext^{1}(Im(α^{i+1}), H^{i+1}S) = (L^{i+1}, H^{i+1}S[2]).

2. The mapping H^{i+1}T ↪ L^{i} induces a short exact sequence

0 → ι^{i} ∘ (L^{i}, H^{i+1}S) → (H^{i+1}T, H^{i+1}S) →^{η^{i}} (L^{i+1}, H^{i+1}S[2]) → 0

Proof Apply (−, H^{i+1}S) to the exact sequence

0 → H^{i+1}T → L^{i} → Im(α^{i+1}) → 0

to get a long exact sequence part of which is :

0 → (Im(α^{i+1}), H^{i+1}S) → (L^{i}, H^{i+1}S) →  (5)

→ (H^{i+1}T, H^{i+1}S) →^{η^{i}} Ext^{1}(Im(α^{i+1}), H^{i+1}S) → Ext^{1}(L^{i}, H^{i+1}S) → ⋯

Proof of 1 : Ext^{1}(Im(α^{i+1}), H^{i+1}S) = (L^{i+1}, H^{i+1}S[2]) since Im(α^{i+1}) is the first syzygy of L^{i}. Proof of 2 :

(a) Ext^{1}(L^{i}, H^{i+1}S) = 0. For i = 0, Ext^{1}(L^{0}, H^{1}S) = 0 since L^{0} = P^{1} is projective. For i > 0, the hypothesis (Im(α^{i}), Coker(β^{i})) = 0 implies (Im(α^{i}), H^{i+1}S) = 0 and therefore Ext^{1}(L^{i}, H^{i+1}S) = 0.

(b) The kernel of (H^{i+1}T, H^{i+1}S) →^{η^{i}} (L^{i+1}, H^{i+1}S[2]) is the image of (L^{i}, H^{i+1}S) → (H^{i+1}T, H^{i+1}S) and this one is ι^{i} ∘ (L^{i}, H^{i+1}S).

From 1, (a), (b) and the long exact sequence (5), we obtain the exact se-

quence 0 → ι^{i} ∘ (L^{i}, H^{i+1}S) → (H^{i+1}T, H^{i+1}S) →^{η^{i}} (L^{i+1}, H^{i+1}S[2]) → 0. □

Lemma 2.7 1. We have (a) (T^{i+1}, H^{i+1}S[−i]) = Ext^{1}(L^{i+1}, H^{i+1}S) and

(b) (T^{i+1}, H^{i+1}S[−i + 1]) = (L^{i+1}, H^{i+1}S[2]).

2. There exists a short exact sequence 0 → Ext^{1}(L^{i+1}, H^{i+1}S) → (T^{i}, H^{i+1}S[−i]) → ι^{i} ∘ (L^{i}, H^{i+1}S) → 0 Proof We apply (−, H^{i+1}S[−i]) to the triangle

T^{i} → T^{i+1} → H^{i+1}T [−i + 1] ⇝ T^{i}[1]

to get a long exact sequence

⋯ → (H^{i+1}T [−i + 1], H^{i+1}S[−i]) → (T^{i+1}, H^{i+1}S[−i]) → (T^{i}, H^{i+1}S[−i]) →

→ (H^{i+1}T [−i], H^{i+1}S[−i]) → (T^{i+1}[−1], H^{i+1}S[−i]) → ⋯ (6)

Proof of 1 :

(a) (T^{i+1}, H^{i+1}S[−i]) = Ext^{1}(L^{i+1}, H^{i+1}S) A morphism T^{i+1} → H^{i+1}S[−i] gives rise to a commutative diagram

⋯ → P̃^{i−1} → P̃^{i} →^{α̃^{i}} P^{i+1}  →^{α^{i+1}} P^{i+2} → ⋯ → P^{n} → 0
↓ ↓ ↓
0 → H^{i+1}S →  0

and the morphism P^{i+1} → H^{i+1}S factors through Coker(α̃^{i}) which is Im(α^{i+1}). Hence, denoting the embedding Im(α^{i+1}) ↪ P^{i+2} by j^{i+1}, we have

(T^{i+1}, H^{i+1}S[−i]) = \frac{(Im(α^{i+1}), H^{i+1}S)}{j^{i+1} ∘ (P^{i+2}, H^{i+1}S)} = Ext^{1}(L^{i+1}, H^{i+1}S).

(b) (T^{i+1}[−1], H^{i+1}S[−i]) = (T^{i+1}, H^{i+1}S[−i + 1]) = (L^{i+1}, H^{i+1}S[2]). A mapping T^{i+1}[−1] → H^{i+1}S[−i] is given by a commutative diagram

⋯ → P̃^{i−1} →^{α̃^{i−1}}  P̃^{i}  →^{α̃^{i}} P^{i+1} → P^{i+2} → ⋯ → P^{n} → 0
↓  ↓ γ  ↓
0  → H^{i+1} → 0

Again, γ factors through Coker(α̃^{i−1}) = Im(α̃^{i}). Hence, denoting the embedding Im(α̃^{i}) ↪ P^{i+1} by ρ^{i}, we have :

(T^{i+1}, H^{i+1}S[−i+1]) = \frac{(Im(α̃^{i}), H^{i+1}S)}{ρ^{i} ∘ (P^{i+1}, H^{i+1}S)} = Ext^{1}(Im(α^{i+1}), H^{i+1}S).

From Lemma 2.6, we get Ext^{1}(Im(α^{i+1}), H^{i+1}S) = (L^{i+1}, H^{i+1}S[2]).

Proof of 2 : It is clear that

(H^{i+1}T [−i + 1], H^{i+1}S[−i]) = Ext^{−1}(H^{i+1}T, H^{i+1}S) = 0.  (7)

The image of (T^{i}, H^{i+1}S[−i]) → (H^{i+1}T, H^{i+1}S) is the kernel of

(H^{i+1}T, H^{i+1}S) →^{η^{i}} (L^{i}, H^{i+1}S[2]) which is ι^{i} ∘ (L^{i}, H^{i+1}S) by Lemma 2.6.

The result follows from 1, (7), and the exactness of (6). □

Lemma 2.8 The morphism T^{i} → T^{i+1} gives rise to an exact sequence

0 → (T^{i+1}, S^{i}) → (T^{i}, S^{i}) →^{ψ^{i}} (H^{i+1}T, H^{i+1}S).

Proof We apply (−, S^{i}) to the triangle

T^{i} → T^{i+1} → H^{i+1}T [−i + 1] ⇝ T^{i}[1]

to get a long exact sequence part of which is

⋯ → (H^{i+1}T [−i + 1], S^{i}) → (T^{i+1}, S^{i}) → (T^{i}, S^{i}) →
→ (H^{i+1}T [−i], S^{i}) → (T^{i+1}[−1], S^{i}) → ⋯

1. (H^{i+1}T [−i + 1], S^{i}) = 0.

For i = 0, the result follows from the fact that S^{0} is concentrated in degrees 1, ⋯, n whereas the projective resolution of H^{1}T [1] is 0 in degrees 1, ⋯, n. For i > 0, any morphism in (H^{i+1}T [−i + 1], S^{i}) is given by the commutative diagram as follows :

⋯ → P̃^{i−1} → P̃^{i} → 0
↓ ↓ ↓
⋯ → Q̄^{i−1} → Q^{i} → Q^{i+1} → ⋯ → Q^{n} → 0

With the same argument as in the first part of the proof of Lemma 2.4 applied to P̃^{i} → Q^{i} and P̃^{j} → Q̄^{j}, for all j < i, we get

(H^{i+1}T [−i + 1], S^{i}) = 0.

2. (H^{i+1}T [−i], S^{i}) = (H^{i+1}T, H^{i+1}S). We have a long exact sequence

⋯ → (H^{i+1}T [−i], S^{i+1}[−1]) → (H^{i+1}T [−i], H^{i+1}S[−i]) →

→ (H^{i+1}T [−i], S^{i}) → (H^{i+1}T [−i], S^{i+1}) → ⋯

coming from applying (H^{i+1}T [−i], −) to the triangle

S^{i} → S^{i+1} → H^{i+1}S[−i + 1] ⇝ S^{i}[1].

By Lemma 2.4, (H^{i+1}T [−i], S^{i+1}[−1]) = 0 and (H^{i+1}T [−i], S^{i+1}) = 0. Hence, (H^{i+1}T [−i], S^{i}) = (H^{i+1}T, H^{i+1}S).

We obtain an exact sequence

0 → (T^{i+1}, S^{i}) → (T^{i}, S^{i}) →^{ψ^{i}} (H^{i+1}T, H^{i+1}S).

□

Remark We will give an example in section 5 (Example 5.2) which shows that (T^{i}, S^{i}) → (H^{i+1}T, H^{i+1}S) is not surjective in general, i.e. the mapping (H^{i+1}T [−i], S^{i}) → (T^{i+1}[−1], S^{i}) is not always 0.

Lemma 2.9 If Ext^{1}(L^{i+1}, H^{i+1}S) = 0, then the morphism S^{i} → S^{i+1} gives rise to an exact sequence

0 → (T^{i+1}, S^{i}) → (T^{i+1}, S^{i+1}) → (L^{i+1}, H^{i+1}S[2]).

Proof Applying (T^{i+1}, −) to the triangle

S^{i} → S^{i+1} → H^{i+1}S[−i + 1] ⇝ S^{i}[1]

we get a long exact sequence

⋯ → (T^{i+1}, H^{i+1}S[−i]) → (T^{i+1}, S^{i}) → (T^{i+1}, S^{i+1}) →
→ (T^{i+1}, H^{i+1}S[−i + 1]) → (T^{i+1}, S^{i}[1]) → ⋯

From Lemma 2.7 and our hypothesis we get

(T^{i+1}, H^{i+1}S[−i]) = Ext^{1}(L^{i+1}, H^{i+1}S) = 0.

Now (T^{i+1}, H^{i+1}S[−i + 1]) = (L^{i+1}, H^{i+1}S[2]) by Lemma 2.7 and the result follows. □

PROOF OF THEOREM 2.1 :

As a consequence of Corollary 2.5, Lemma 2.6, Lemma 2.7, Lemma 2.8 and Lemma 2.9 we get the following diagram

0  0  0
↓  ↓  ↓
Ext^{1}(L^{i+1}, H^{i+1}S)  (T^{i+1}, S^{i})  (T^{i+1}, S^{i})
↓  ↓  ↓
0 → (T^{i}, H^{i+1}S[−i]) →  (T^{i}, S^{i})  → (T^{i+1}, S^{i+1}) → 0
↓  ↓ _{ψ^{i}}  ↓
0 → ι^{i} ∘ (L^{i}, H^{i+1}S) → (H^{i+1}T, H^{i+1}S) → (L^{i+1}, H^{i+1}S[2]) → 0

↓ 0

Using the identification (T^{i}, S^{i+1}) ≅ (T^{i+1}, S^{i+1}) shown in Lemma 2.3 and the hypothesis (Im(α^{i}), Coker(β^{i})) = 0, it follows that the bottom right square of the diagram is commutative. Similarly, the bottom left square is commutative.

If Ext^{1}(L^{i+1}, H^{i+1}S) = 0, then (T^{i}, H^{i+1}S[−i]) →^{≅} ι^{i} ∘ (L^{i}, H^{i+1}S) and

the snake lemma gives the isomorphism (T^{i+1}, S^{i}) →^{≅} (T^{i+1}, S^{i}) in the above diagram.

This fact and the exactness of the horizontal sequences give us, for all i ∈ {0, 1, ⋯, n − 2}, the pull-back diagram

Hom_{D^{b}(Λ)}(T^{i}, S^{i}) → Hom_{D^{b}(Λ)}(T^{i+1}, S^{i+1})
↓  ↓
Hom_{Λ}(H^{i+1}T, H^{i+1}S) → Hom_{D^{b}(Λ)}(L^{i+1}, H^{i+1}S[2])

Theorem 2.1 is now proved. □

Remark In fact, the snake lemma shows that (T^{i+1}, S^{i}) → (T^{i+1}, S^{i}) is surjective. If R is a noetherian ring, a surjective endomorphism of a noetherian module is an isomorphism. Hence Ext^{1}(L^{i+1}, H^{i+1}S) = 0. But we have already used Ext^{1}(L^{i+1}, H^{i+1}S) = 0 to establish the right vertical exact sequence.

3 More explicit pull-backs

In the previous section, we got the pull-back diagrams

Hom_{D^{b}(Λ)}(T^{i}, S^{i}) → Hom_{D^{b}(Λ)}(T^{i+1}, S^{i+1})
↓ _{ψ^{i}}  ↓
Hom_{Λ}(H^{i+1}T, H^{i+1}S) → Hom_{D^{b}(Λ)}(L^{i+1}, H^{i+1}S[2])

for all i ∈ {0, 1, ⋯, n − 2}. In this section, we will determine the image of ψ^{i}, for all i ∈ {0, 1, ⋯, n−2}.

We denote for all i ∈ {0, 1, ⋯, n − 2},

A^{i} := Im(ψ^{i}) = Ker[(H^{i+1}T, H^{i+1}S) → (T^{i+1}, S^{i}[1])].

For all i ∈ {0, 1, ⋯, n − 2}, we denote by ι_{T}^{i} the inclusion H^{i+1}T ↪ Coker(α^{i}), by σ_{T}^{i} the composition L^{i} → Im(α^{i+1}) → P^{i+2} and by ι_{S}^{i} and σ_{S}^{i} the corresponding mappings for S. We remark that σ_{T}^{0} = α^{1}, σ_{S}^{0} = β^{1}, and ι_{T}^{i} = ι^{i}. Lemma 3.1 For all i ∈ {0, 1, ⋯, n − 2}, we have : A^{i} is the set of all φ^{i} ∈ (H^{i+1}T, H^{i+1}S) such that there exists a sequence (γ^{i+1}, γ^{i+2}, ⋯, γ^{n}) ∈ (Coker(α^{i}), Coker(β^{i}))×(P^{i+2}, Q^{i+2})×⋯×(P^{n}, Q^{n}) satisfying

1. φ^{i}ι_{S}^{i} = ι_{T}^{i} γ^{i+1},

2. γ^{i+1}σ_{S}^{i} = σ_{T}^{i} γ^{i+2}, and

3. for all j ∈ {i + 2, ⋯, n − 1}, γ^{j}β^{j} = α^{j}γ^{j+1}. Proof We denote by ϕ^{i} the projective cover mapping P̃^{i} → Ker(α^{i+1}), by ϵ^{i} the mapping Ker(α^{i+1}) → H^{i+1}T, and by α̃^{i} the mapping P̃^{i} → P^{i+1} (we remark that ϵ^{0} = id).

We have the following diagram :

⋯ → P̃^{i−1} →  P̃^{i}  →^{α̃^{i}} P^{i+1} →^{α^{i+1}} P^{i+2} → ⋯ → P^{n−1} → P^{n} → 0 T^{i+1}
↓  ↓ _{ϕ^{i}ϵ^{i}}  ↓  ↓
0 → H^{i+1}T →  0  H^{i+1}T
↓  ↓ _{φ^{i}}  ↓  ↓
0 → H^{i+1}S →  0  H^{i+1}S
↓  ↓ _{ι_{S}^{i}}  ↓  ↓
0 → Coker(β^{i}) → Q^{i+2} →^{β^{i+2}} Q^{i+3} → ⋯ → Q^{n} → 0  S^{i+1}

If φ^{i} ∈ A^{i} then φ^{i} ∈ (H^{i+1}, H^{i+1}S) and

1. there exists a mapping γ̄^{i+1} : P^{i+1} → Coker(β^{i}) such that ϕ^{i}ϵ^{i}φ^{i}ι_{S}^{i} = α̃^{i}γ̄^{i+1}, and

2. there exists a sequence (γ^{j})_{j∈{i+2,···,n−1}} γ^{j} : P^{j} → Q^{j} where (a) γ̄^{i+1}σ_{S}^{i} = α^{i+1}γ^{i+2} and

(b) for all j ∈ {i + 2, ⋯, n − 1} we have γ^{j}β^{j} = α^{j}γ^{j+1}.

Since (Im(α^{i}), Coker(β^{i})) = 0, there exists γ^{i+1} ∈ (Coker(α^{i}), Coker(β^{i})) such that

γ̄^{i+1} = π^{i}γ^{i+1}, φ^{i}ι_{S}^{i} = ι_{T}^{i} γ^{i+1} and γ^{i+1}σ_{S}^{i} = σ_{T}^{i} γ^{i+2},

where π^{i} is the mapping P^{i+1} → L^{i}. This finishes the proof of the lemma. □

Hence an element of A^{i} is a mapping H^{i+1}T → H^{i+1}S which induces a homomorphism between the complex

H^{i+1}T ↪ Coker(α^{i}) → P^{i+1} → ⋯ → P^{n}

and the complex

H^{i+1}S ↪ Coker(β^{i}) → Q^{i+1} → ⋯ → Q^{n}.

Moreover, for S = T, A^{i} has a multiplicative stucture as a subring of (H^{i+1}T, H^{i+1}S) for all i ∈ {0, ⋯, n − 2}.

We denote for all i ∈ {0, ⋯, n − 2},

Ω^{i} := Ker((L^{i+1}, H^{i+1}S[2]) → (T^{i+1}, S^{i}[1])).

We remark that if R is a Dedekind domain and Λ is an R-order, Ω^{i} is an R-torsion Λ-module and K ⊗_{R} Ω^{i} = 0 where K = F rac(R).

Lemma 3.2 For all i ∈ {0, 1, ⋯, n−2}, there exists a short exact sequence

0 → ι^{i} ∘ (L^{i}, H^{i+1}S) → A^{i} →^{ρ^{i}} Ω^{i} → 0.

Proof From Lemma 2.6 and the definitions of A^{i} and Ω^{i} we get the following commutative diagram.

0  0
↓  ↓
A^{i}  −−→^{ρ^{i}}  Ω^{i}
↓ _{β^{i}}  ↓ _{μ^{i}}
0 → ι^{i}.(L^{i}, H^{i+1}S) →^{ν^{i}} (H^{i+1}T, H^{i+1}S) →^{η^{i}} (L^{i+1}, H^{i+1}S[2]) → 0
↓ _{χ^{i}}  ↓ _{γ^{i}}
(T^{i+1}, S^{i}[1])  =  (T^{i+1}, S^{i}[1])

The restriction of η^{i} to A^{i} is the map ρ^{i} : A^{i} → Ω^{i} such that ρ^{i}μ^{i} = β^{i}η^{i}. Hence

Ker(ρ^{i}) ⊆ ι^{i} ∘ (L^{i}, H^{i+1}S).

On the other side, ι^{i} ∘ (L^{i}, H^{i+1}S) ⊆ A^{i}. In fact, ν^{i}χ^{i} = ν^{i}η^{i}γ^{i} = 0, hence

there exists ϕ^{i} : ι^{i} ∘ (L^{i}, H^{i+1}S) → A^{i} such that ϕ^{i}β^{i} = ν^{i}.

Since ν^{i} is injective, so is ϕ^{i}. Therefore Ker(ρ^{i}) = ι^{i} ∘ (L^{i}, H^{i+1}S) and this finishes the proof of the lemma. □

By Theorem 2.1 and Lemma 3.2 we get more explicit pull-backs.

Proposition 3.3 For all i ∈ {0, 1, ⋯, n − 2},

Hom_{D^{b}(Λ)}(T^{i}, S^{i}) → Hom_{D^{b}(Λ)}(T^{i+1}, S^{i+1})
↓  ↓
A^{i}  →^{ρ^{i}}  Ω^{i}

are pull-back diagrams.

Remark We get morphisms

Σ^{i} : Hom_{D^{b}(Λ)}(T, S) → Hom_{Λ}(H^{i+1}T, H^{i+1}S)

for all i ∈ {0, 1, ⋯, n − 2}. As Hom_{D^{b}(Λ)}(T^{i}, S^{i}) → Hom_{D^{b}(Λ)}(T^{i+1}, S^{i+1}) is surjective for all i ∈ {0, 1, ⋯, n − 2}, we get Im(Σ^{i}) = A^{i}. Hence Lemma 2.8 describes the cokernel of Σ^{i} as a submodule of Hom_{D^{b}(Λ)}(T^{i+1}[−1], S^{i}).

4 Torsion free Hom-spaces

In this section we assume that the R-module Hom_{D^{b}(Λ)}(T, S) is R-torsion free and the homologies of T and S are R-torsion modules except in degrees 1 and n. We will see that Hom_{D^{b}(Λ)}(T, S) is the pull-back of A^{0} defined in section 2 and Hom_{Λ}(H^{n}T/tors, H^{n}S/tors).

For all i ∈ {1, ⋯, n − 2}, we define the R-torsion part of (T^{i}, S^{i})to be

t(T^{i}, S^{i}) = {φ ∈ (T^{i}, S^{i}) : there exists r ∈ R, r ≠ 0 r.φ = 0}.

Let ξ : (T^{1}, S) → (T^{1}, S^{1}) be the mapping defined in Lemma 2.9.

Lemma 4.1 We have t(T^{1}, S^{1}) ∩ ξ((T^{1}, S)) = 0. Proof Since (T, S) is R-torsionfree, so is (T^{1}, S), as it embeds into (T, S). In particular ξ((T^{1}, S)) is also R-torsionfree since ξ is injective, while t(T^{1}, S^{1}) is R-torsion. □

We denote by θ : (T^{1}, S^{1}) → Ω^{0} the mapping defined in Proposition 3.3, and θ′ := θ|_{t(T^{1},S^{1})}. Lemma 4.2 The mapping θ′ : t(T^{1}, S^{1}) → Im(θ′) is bijective. Proof The surjectivity is clear, and the injectivity results from Lemma 4.1. □

The mapping θ induces θ̂: \frac{(T^{1},S^{1})}{t(T^{1},S^{1})} → \frac{Ω^{0}}{Im(θ′)}, where

θ(f + t(T^{1}, S^{1})) = θ(f) + Im(θ′).

Lemma 4.3 The mapping θ̂ is surjective and Ker(θ̂) ≅ (T^{1}, S).

Since θ is surjective, so is θ̂. It is not difficult to show that Ker(θ̂) is \frac{(T^{1},S)+t(T^{1},S^{1})}{t(T^{1},S^{1})} and by Lemma 4.1 we can identify this with (T^{1}, S).

As a consequence, we get a pull-back diagram

0  0
↓  ↓
(T^{1}, S) = \frac{(T^{1},S)+t(T^{1},T^{S})}{t(T^{1},S^{1})}
↓  ↓
0 → t(T^{1}, S^{1}) → (T^{1}, S^{1}) →  \frac{(T^{1},S^{1})}{t(T^{1},S^{1})}  → 0
||  ↓  ↓ θ̂
0 → Im(θ′) →  Ω^{0}  →  \frac{Ω^{0}}{Im(θ′)}  → 0
↓  ↓
0  0

(8)

From Lemma 3.3 we have the following diagram :

(T, S) → (T^{1}, S^{1})
↓  ↓
A^{0} → Ω^{0}  (9)

We compose the diagrams (8) and (9) to obtain the following diagram where the two small rectangles are pull-backs and where the kernels of the vertical mappings are isomorphic :

(T, S) → (T^{1}, S^{1}) → \frac{(T^{1},S^{1})}{t(T^{1},S^{1})}
↓  ↓  ↓
A^{0} →  Ω^{0}  → \frac{Ω^{0}}{Im(θ′)}

Corollary 4.4 The composition of diagram (8) and diagram (9) is also a pull-back :

(T, S) → \frac{(T^{1},S^{1})}{t(T^{1},S^{1})}
↓  ↓
A^{0} → \frac{Ω^{0}}{Im(θ′)}

Lemma 4.5 We have (T^{1}, S^{1})/t(T^{1}, S^{1}) ≅ (H^{n}T, H^{n}S)/t(H^{n}T, H^{n}S).

Proof Fix an i ∈ {1, ⋯, n − 2}. Let Φ : (T^{i}, S^{i}) → (T^{i+1}, S^{i+1}) be the mapping defined in Lemma 2.5 and let Φ′ := Φ|_{t(T^{i},S^{i})}.

We have the following diagram :

0  0
↓  ↓
t(T^{i}, S^{i}) →^{Φ′} Im(Φ′)
↓ _{ζ}  ↓ _{ν}
0 → ι^{i} ∘ (L^{i}, H^{i+1}S) →^{μ} (T^{i}, S^{i}) →^{Φ} (T^{i+1}, S^{i+1}) → 0
↓ _{λ}  ↓ _{τ}
(T^{i}, S^{i})/t(T^{i}, S^{i}) −−→^{δ} (T^{i+1}, S^{i+1})/Im(Φ′)
↓  ↓
0  0

Since ζΦτ = Φ′ντ = 0, there exists

δ : (T^{i}, S^{i})/t(T^{i}, S^{i}) → (T^{i+1}, S^{i+1})/Im(Φ′)

such that Φτ = λδ. Since Φ and τ are surjective, so is δ.

The kernel of δ is [μ(ι^{i}∘(L^{i}, H^{i+1}S))+t(T^{i}, S^{i})]/t(T^{i}, S^{i}). Since (H^{i+1}T, H^{i+1}S) is R-torsion, so is μ(ι^{i} ∘ (L^{i}, H^{i+1}S)), hence Ker(δ) = 0. This implies

(T^{i}, S^{i})/t(T^{i}, S^{i}) ≅ (T^{i+1}, S^{i+1})/Im(Φ′).

Since Im(Φ′) ⊆ t(T^{i+1}, S^{i+1}), we get a surjective mapping

(T^{i}, S^{i})/t(T^{i}, S^{i}) → (T^{i+1}, S^{i+1})/t(T^{i+1}, S^{i+1})

whose kernel is t(T^{i}, S^{i})/Im(Φ′).

This kernel is formed by R-torsion elements while (T^{i}, S^{i})/t(T^{i}, S^{i}) is R-torsionfree. Hence t(T^{i}, S^{i}) = Im(Φ′), and we get

(T^{i}, S^{i})/t(T^{i}, S^{i}) ≅ (T^{i+1}, S^{i+1})/t(T^{i+1}, S^{i+1})

for all i ∈ {1, ⋯, n − 2}.

As a consequence, (T^{1}, S^{1})/t(T^{1}, S^{1}) ≅ (T^{n−1}, S^{n−1})/t(T^{n−1}, S^{n−1}), and since T^{n−1} = Coker(α^{n}) = H^{n}T and S^{n−1} = Coker(β^{n}) = H^{n}S, we obtain (T^{1}, S^{1})/t(T^{1}, S^{1}) ≅ (H^{n}T, H^{n}S)/t(H^{n}T, H^{n}S). □

Lemma 4.6 (H^{n}T, H^{n}S)/tors ≅ (H^{n}T/tors, H^{n}S/tors).

Proof We denote by L := H^{n}T, M := H^{n}S, and by tL and tM the torsion parts of L and M respectively. We have :

1. 0 → (L, tM) → (L, M) → (L, M/tM) → 0 is an exact sequence. We apply (L, −) to the exact sequence 0 → tM → M → M/tM → 0 to get

0 → (L, tM) → (L, M) → (L, M/tM) → Ext^{1}(L, tM) → ⋯ (10)

Since (Im(α^{n}), tM) ⊆ (Im(α^{n}), M) = 0, we have Ext^{1}(L, tM) = 0 and the statement holds.

2. (L, M/tM) ≅ (L/tL, M/tM).

We apply (−, M/tM) to the exact sequence 0 → tL → L → L/tL → 0 to get

0 → (L/tL, M/tM) → (L, M/tM) → (tL, M/tM) → ⋯

Since (tL, M/tM) = 0, we get (L, M/tM) = (L/tL, M/tM).

3. t(L, M) = (L, tM)

First we need that t(L, M) ⊆ (L, tM). In fact, if φ ∈ t(L, M), there exists an r ∈ R − {0}, such that rφ = 0. Hence there exists an r such that for all l ∈ L, rφ(l) = 0. This implies that for all l ∈ L, φ(l) ∈ tM.

On the other hand, if f ∈ (L, tM) and x ∈ L, f(x) ∈ tM i.e there exists an r ∈ R−{0} such that rf(x) = 0. Since L is finitely generated, there exists an s ∈ R − {0} such that sf = 0. Therefore, f ∈ t(L, M).

By 1, 2, and 3, the rows of the following diagram are exact :

0 → (L, tM) → (L, M) → (L/tL, M/tM) → 0
||  ||
0 → t(L, M) → (L, M) → (L, M)/t(L, M) → 0

and this proves the lemma. □

From Corollary 4.4, Lemma 4.5 and Lemma 4.6 we get the following pull-back

Hom_{D^{b}(Λ)}(T, S) → Hom_{Λ}(H^{n}T/t(H^{n}T), H^{n}S/t(H^{n}S)
↓  ↓
A^{0}  →  \frac{Ω^{0}}{Im(θ′)}

We have just proved the following theorem

Theorem 4.7 If for all i ∈ {1, ⋯, n − 1},

Ext^{1}(L^{i+1}, H^{i+1}T) = Hom_{Λ}(L^{i}, Q^{i}) = Hom_{Λ}(L^{i}, Im(β^{i})) =
= Hom_{Λ}(Im(α^{i}), Coker(β^{i})) = 0,

the R-module Hom_{D^{b}(Λ)}(T, S) is R-torsion free and H^{i}T and H^{i}S are Rtorsion for all i ∈ {2, ⋯, n−2}, then, we get the following pull-back diagram

Hom_{D^{b}(Λ)}(T, S) → Hom_{Λ}(H^{n}T/t(H^{n}T), H^{n}S/t(H^{n}S))
↓  ↓
A^{0}  →  \frac{Ω^{0}}{Im(θ′)}

where A^{0} is formed by those morphisms of Hom_{Λ}(H^{1}T, H^{1}S) that induce morphisms in Hom_{D(Λ)}(T, S) and Ω^{0} = A^{0}/(ι^{0} ∘ (P^{1}, H^{1}S)).

Remarks

1. If R is a Dedekind domain, Λ is a symmetric order, i.e. an R-order which is also a symmetric algebra, and S = T is a tilting complex, then Hom_{D^{b}(Λ)}(T, T) is an R-order by [11, Theorem 1], hence R-torsionfree.

2. If for all i ∈ {1, ⋯, n − 2}, the R-modules Hom_{D^{b}(Λ)}(T^{i}, S^{i}) are Rtorsionfree, we can get Hom_{D^{b}(Λ)}(T^{i}, S^{i}) as pull-back

(T^{i}, S^{i}) → (T^{i+1}, S^{i+1})/t(T^{i+1}, S^{i+1})
↓  ↓
A^{i}/tA^{i} →  Ω̄^{i}

where A^{i}/tA^{i} is the R-torsionfree part of A^{i}. But even though Λ is symmetric, Hom_{D^{b}(Λ)}(T^{i}, S^{i}) can be R-torsion, as is showed in Example 5.1 with n = 2, S = T and T^{1} = Coker(α).

3. For i = 0, Ext^{1}(L^{i+1}, H^{i+1}S) = 0 is automatic, because we have the inclusions Ext^{1}(L^{1}, H^{1}S) ↪ (T, H^{1}S) ↪ (T, S) by Lemma 2.7 and Lemma 2.8, and (T, S) is an R-torsionfree Λ-module.

5 Examples

We will give some examples demonstrating how to use Theorem 2.1. We use rings Λ which are orders and we take S = T. The presentations of Λ in Example 5.1 and Example 5.2 can be found in [8] or in [5, section 4.4].

We abbreviate (X, Y) := Hom_{D^{b}(Λ)}(X, Y) and Ext^{1}(X, Y) := Ext_{Λ}^{1}(X, Y) for two complexes X and Y.

Let K be F rac(R) and Λ be an R-order. Let χ_{1}, ⋯, χ_{m} be the irreducible characters of K ⊗_{R} Λ. Then we write L_{i} for a Λ-lattice L such that K ⊗_{R} Λ acts on K ⊗_{R} L as χ_{i}.

Example 5.1 Let R be the 5-adic integers ℤ_{5}, and π = Rad(R) =< 5 > . Let Λ be the R-order

\{(d_{0}, ( \begin{matrix}a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix} ) , ( \begin{matrix}a_{2} & b_{2} \\ c_{2} & d_{2} \end{matrix} ) , ( \begin{matrix}a_{3} & b_{3} \\ c_{3} & d_{3} \end{matrix} ) , a_{4}) , a_{i}, b_{i}, c_{i}, d_{i} ∈ R, 5|(d_{i} − a_{i+1}), 5|c_{i}\}

which we write (see [8] or [5, section 4.4]) as

[figure: R_{1}; ( \begin{matrix}R & R \\ π & R \end{matrix} )_{2}; ( \begin{matrix}R & R \\ π & R \end{matrix} )_{3}; ( \begin{matrix}R & R \\ π & R \end{matrix} )_{4}; R_{5},]

where the indices indicate the characters involved in the indecomposable projective modules. Then, by [8], Λ is Morita equivalent to B_{0}(ℤ_{5}S_{5}), the block principal of the group ring of the symmetric group of degree 5 over the 5-adic integers. Let

[figure: P_{0} =; R_{1}; ( \begin{matrix}R \\ π \end{matrix} )_{2} ,; P_{1} =; ( \begin{matrix}R \\ R \end{matrix} )_{2}; ( \begin{matrix}R \\ π \end{matrix} )_{3} ,]

[figure: P_{2} =; ( \begin{matrix}R \\ R \end{matrix} )_{3}; ( \begin{matrix}R \\ π \end{matrix} )_{4} ,; P_{3} =; ( \begin{matrix}R \\ R \end{matrix} )_{4}; R_{5}.]

By definition, the characters of P_{0} are χ_{1} and χ_{2}, those of P_{1} are χ_{2} and χ_{3}, those of P_{2} are χ_{3} and χ_{4}, and those of P_{3} are χ_{4} and χ_{5}.

Let T be the complex T_{1} ⊕ T_{2} ⊕ T_{3} :

T_{1} : 0 → P_{1} ⊕ P_{3} →^{α} P_{2} → 0
⊕  ⊕
T_{2} : 0 → P_{1} ⊕ P_{3} → 0
⊕  ⊕
T_{3} : 0 →  P_{0}  → 0

The complex T is a tilting complex according to the proof of [5, Lemma 5.1.2].

We define ᾱ = (α, 0, 0) and observe that End_{Λ}(Ker(ᾱ)) decomposes as a direct product of two rings :

[figure: R_{4}; ( \begin{matrix}R & R \\ π & R \end{matrix} )_{5}; ⊕; ( \begin{matrix}R & R & π \\ π & R & π \\ R & R & R \end{matrix} )_{2}; R_{3}; R_{1}]

If the mapping ψ defined in Lemma 2.8 was surjective,then since End_{Λ}(Coker(ᾱ)) is an R-torsion module, we would get End_{D^{b}(Λ)}(T) ≅ End_{Λ}(Ker(ᾱ)) which is decomposable. This is impossible. Hence, ψ cannot be surjective.

In fact, End_{D^{b}(Λ)}(T) ≅ A^{0} is the ring :

[figure: R_{4}; ( \begin{matrix}R & R \\ π & R \end{matrix} )_{5}; ( \begin{matrix}R & R & π \\ π & R & π \\ R & R & R \end{matrix} )_{2}; R_{3}; R_{1}]

where the congruence R_{2} − R_{5} is a consequence of the congruence in P_{2}.

Example 5.2 Let R be the 7-adic integers ℤ_{7}, the 7-adic integers, π = Rad(R) =< 7 >, and let Λ be the R-order

[figure: R; ( \begin{matrix}R & R \\ π & R \end{matrix} ); ( \begin{matrix}R & R \\ π & R \end{matrix} ); ( \begin{matrix}R & R \\ π & R \end{matrix} ); ( \begin{matrix}R & R \\ π & R \end{matrix} ); ( \begin{matrix}R & R \\ π & R \end{matrix} ); R]

By [8] Λ is Morita equivalent to B_{0}(ℤ_{7}S_{7}), the block principal of the group ring of the symmetric group of degree 7 over the 7-adic integers.

Let T be the complex

0 → P_{0} ⊕ P_{0} →^{(α,0)}  P_{1}  → 0
⊕  ⊕  ⊕
0  →  P_{2} ⊕ P_{4}  →^{β} P_{3} → 0
⊕  ⊕  ⊕
0  → P_{2} ⊕ P_{4} ⊕ P_{5} → 0

where P_{1}, P_{2}, P_{3}, P_{4} and P_{5} are projective Λ-modules :

[figure: P_{0} =; R_{1}; ( \begin{matrix}R \\ π \end{matrix} )_{2} ,; P_{1} =; ( \begin{matrix}R \\ R \end{matrix} )_{2}; ( \begin{matrix}R \\ π \end{matrix} )_{3} ,]

[figure: P_{2} =; ( \begin{matrix}R \\ R \end{matrix} )_{3}; ( \begin{matrix}R \\ π \end{matrix} )_{4} ,; P_{3} =; ( \begin{matrix}R \\ R \end{matrix} )_{4}; ( \begin{matrix}R \\ π \end{matrix} )_{5} ,]

[figure: P_{4} =; ( \begin{matrix}R \\ R \end{matrix} )_{5}; ( \begin{matrix}R \\ π \end{matrix} )_{6} ,; P_{5} =; ( \begin{matrix}R \\ R \end{matrix} )_{6}; R_{7},]

and the indices indicate the characters involved in the modules.

We define by ᾱ := ((α, 0), 0, 0) and β̄ := (0, β, 0). Since (Im(ᾱ), Ker(ᾱ)) = (Im(β̄), H^{2}T) = 0, we get

Ext^{1}(Coker(ᾱ), Ker(ᾱ)) = Ext^{1}(Coker(β̄), H^{2}T) = 0.

Let T^{1} be the complex

0 → Im(ᾱ) → ( \begin{matrix}P_{1} \\ P_{2} ⊕ P_{4} \\ P_{2} ⊕ P_{4} ⊕ P_{5} \end{matrix} ) → ( \begin{matrix}0 \\ P_{3} \\ 0 \end{matrix} ) → 0.

By Theorem 2.1, End_{D^{b}(Λ)}(T^{1}) is a pull-back of End_{Λ}(H^{2}T) and End_{Λ}(Coker(β̄)) over Ext^{2}(Coker(β̄), H^{2}T).

The mapping (Coker(β̄), Coker(β̄)) →^{θ^{1}} Ext^{2}(Coker(β̄), H^{2}T) is induced

by Ker(β̄) → H^{2}T and this is multiplication by an r ∈ R.

Hence, θ^{1} is injective, and

End_{D^{b}(Λ)}(T^{1}) ≅ Im(ψ^{1})

where ψ^{1} : (T^{1}, T^{1}) → (H^{2}T, H^{2}T). In fact, A^{1} = Im(ψ^{1}) is the following ring :

[figure: ( \begin{matrix}R & R & R \\ π & R & R \\ π & π & R \end{matrix} )_{3}; R_{4}; ( \begin{matrix}R & R & R \\ π & R & R \\ π & π & R \end{matrix} )_{6}; R_{5}; R_{7}]

where the congruence of R_{3} − R_{6} comes from the one in P_{3}.

Now, End_{D^{b}(Λ)}(T) is a pull-back of End_{Λ}(Ker(β̄)) and End_{D^{b}(Λ)}(T^{1}) over Ext^{2}(Coker(ᾱ), Ker(ᾱ)).

The only non-zero component of Ext^{2}(Coker(ᾱ), Ker(ᾱ)) is R/π, which gives the congruence R_{1} − R_{3}. Hence, End_{D^{b}(Λ)}(T) is the following ring :

[figure: R_{2}; ( \begin{matrix}R & R \\ π & R \end{matrix} )_{1}; ( \begin{matrix}R & R & R \\ π & R & R \\ π & π & R \end{matrix} )_{3}; R_{4}; ( \begin{matrix}R & R & R \\ π & R & R \\ π & π & R \end{matrix} )_{6}; R_{5}; R_{7}]

Remarks

1. The complex T is a tilting complex. If ψ^{1} : (T^{1}, T^{1}) → (H^{2}T, H^{2}T) was surjective, we would get (T^{1}, T^{1}) ≅ (H^{2}T, H^{2}T) which is decomposable and this would impliy that (T, T) is decomposable, which is a contradiction. Hence, in this case, ψ^{1} cannot be surjective.

2. Nevertheless, the mapping ψ^{0} : (T, T) → (H^{1}T, H^{1}T) is surjective.

References

[1] C.W.Curtis and I.Reiner, Methods in Representation Theory Vol I, John Wiley, NY (1982).

[2] B.Keller, A remark on tilting theory and DG algebras, Manus. Math. 79 (1993), 247-253.

[3] B.Keller, Introduction to abelian and derived categories, in “Representations of Reductive Groups,” edited by R. W. Carter and M. Geck, Cambridge University Press (1998), 41-62.

[4] S.König and A.Zimmermann, Tilting self-injective algebras and Gorenstein orders, Oxford Quaterly Journal of Maths. 48 (1997) 351-361.

[5] S.König and A.Zimmermann, Derived equivalences for group rings, Springer Lecture Notes in Mathematics 1685 (1998).

[6] J.Rickard, Morita theory for derived categories, J. London Math. Soc. 39 (1989), 436-456.

[7] J.Rickard, Derived categories and stable equivalences, J. of Pure and Appl. Alg. 61 (1989), 303-317.

[8] K.W.Roggenkamp, Blocks of cyclic defect and Green orders, Comm. in Alg. 20 (1992), 1715-1734.

[9] J.L.Verdier, Des catégories dérivées des catégories abéliennes, Astérisque 239 (1997)

[10] C.A.Weibel, An introduction to homological algebra, Cambridge studies in advanced mathematics 38, Cambridge University Press (1994).

[11] A.Zimmermann, Tilted symmetric orders are symmetric orders, Archiv der Mathematik 73 (1999), 15-17.

Address : Faculté de Mathématiques et CNRS (LAMFA UMR 6140), Université de Picardie, 33 rue St Leu, 80039 Amiens Cedex, France Email address : intan.muchtadi@u-picardie.fr

Current address : Department of Mathematics, University of Leicester, University road, Leicester LE1 7RH, United Kingdom Email address : idma1@le.ac.uk

Permanent address : Departemen Matematika, Institut Teknologi Bandung (ITB), Jl. Ganesha no. 10, Bandung 40132, Indonesia Email address : ntan@dns.math.itb.ac.id, intan@alamsyah.net