Tilted symmetric orders are symmetric orders

Alexander Zimmermann

LAMFA, Université de Picardie, 33 rue St Leu, F-80039 Amiens Cedex 1, France

Abstract We prove in this note that a ring which is derived equivalent to a symmetric order is again a symmetric order. Group rings of finite groups over an integral domain of characteristic 0 are symmetric orders.

Introduction and statement of the result

Let R be a noetherian integral domain with field of fractions K. We call an R-algebra Λ with K ⊗_{R} Λ being semi-simple artinian and Λ being finitely generated projective as R-module a (classical) R-order. An R-algebra is called symmetric if Hom_{R}(Λ, R) ≃ Λ as Λ ⊗_{R} Λ^{op}- modules. The bounded derived category of a ring A is denoted by D^{b}(A). For more ample remarks to the definition see e.g. [3].

Theorem 1 Assume that R is Dedekind domain and Λ is a symmetric R-algebra, projective as R-module. If D^{b}(Λ) ≃ D^{b}(Γ) as triangulated categories for an R-algebra Γ, then Γ is a symmetric R-algebra as well. Especially, if Λ is an R-order, then Γ is an R-order as well.

In case Γ is supposed to be an order as well, Rickard proved this statement in [4]. His proof does not work if one does not know that Γ is R-projective. We get this fact as a consequence of the symmetry, which we prove first. The existence of a derived equivalence induced by a derived functor follows by a theorem of Keller (cf [1] or [3, Theorem 8.3.2]). König discovered that there are rings which are not orders but derived equivalent to a hereditary order [2]. Theorem 1 answers a question of K. W. Roggenkamp posed at the occasion of the ”Darstellungstag Bayreuth 1994”.

Acknowledgement : The author wishes to thank Bernhard Keller for much help in numerous discussions, especially for giving the advice to try to prove a statement like Lemma 1 first.

The proof

By a result of Keller [1] there are a complex X in D^{b}(Λ ⊗_{R} Γ^{op}) such that X ⊗_{Γ} Γ =: T and a complex X′ in D^{b}(Γ ⊗_{R} Λ^{op}) such that Γ ⊗_{Γ} X′ ≃ Hom_{Λ}(T, Λ) and furthermore

X ⊗_{Γ}^{𝕃} − : D^{b}(Γ) ⟶ D^{b}(Λ)

and

− ⊗_{Γ}^{𝕃} X′ : D^{b}(Γ^{op}) ⟶ D^{b}(Λ^{op})

are equivalences of triangulated categories. The identification of Γ as endomorphism ring of T and as endomorphism ring of Hom_{Λ}(T, Λ) is done coherently ; a fact which defines X′ in a coherent way once X is chosen. One should observe that since Λ is R-projective of finite type, one can choose a projective resolution of X and of X′ as bimodules, as described in [3, Proposition 6.3.1 and Lemma 6.3.12], representing the same object in the derived category. Any projective Λ ⊗_{R} Γ^{op} module is projective as Γ^{op}-module since Λ is projective as Rmodule. The analogous statement is valid for X′. Hence, the left derived tensor product can be replaced by the ordinary tensor product. Then, one gets

(X ⊗_{Γ} −) ⊗_{Γ} X′ ≃ X ⊗_{Γ} (− ⊗_{Γ} X′)

as functor D^{b}(Γ ⊗_{R} Γ^{op}) ⟶ D^{b}(Λ ⊗_{R} Λ^{op}). Hence, the functor

G := X ⊗_{Γ}^{𝕃} − ⊗_{Γ}^{𝕃} X′ : D^{b}(Γ ⊗_{R} Γ^{op}) ⟶ D^{b}(Λ ⊗_{R} Λ^{op})

is an equivalence.

Lemma 1 Let R be a commutative noetherian ring and let Λ be an R-algebra which is finitely generated projective as R-module. Let G be a derived equivalence as described above. Then,

G(Hom_{R}(Γ, R)) ≃ Hom_{R}(Λ, R).

Proof. This is a consequence of adjointness formulas and the fact that T is perfect over Λ. Indeed,

Hom_{D^{b}(Γ⊗_{R}Γ^{op})}(Γ ⊗_{R} Γ^{op}, G^{−1}(Hom_{R}(Λ, R)[n]))
≃ Hom_{D^{b}(Λ⊗_{R}Λ^{op})}(T ⊗_{R} Hom_{Λ}(T, Λ), Hom_{R}(Λ, R)[n])
≃ Hom_{D^{b}(Λ)}(T, Hom_{Λ^{op}}(Hom_{Λ}(T, Λ), Hom_{R}(Λ, R))[n])
≃ Hom_{D^{b}(Λ)}(T [−n], Hom_{R}(Hom_{Λ}(T, Λ) ⊗_{Λ} Λ, R))
≃ Hom_{D^{b}(Λ)}(T [−n], Hom_{R}(Hom_{Λ}(T, Λ), R))
≃ Hom_{R}(RHom_{D^{b}(Λ)}(T, Λ) ⊗_{Λ} T [−n], R)
≃ Hom_{R}(Hom_{D^{b}(Λ)}(T, T [−n]), R)
≃ \{ \begin{matrix}Hom_{R}(Γ, R) & if n = 0 \\ 0 & if n ≠ 0 \end{matrix}

Here the first isomorphism is applying G, the second is an adjointness formula between Hom_{Λ^{op}}(Hom_{Λ}(T, Λ), −) and − ⊗_{Λ} Hom_{Λ}(T, Λ), the third is again a similar adjointness formula, the fourth is the fact that − ⊗_{Λ} Λ is equivalent to the identity functor, and then we use again the adjointness of Hom_{R}(Hom_{Λ}(T, Λ), −) and Hom_{Λ}(T, Λ) ⊗_{Λ} −. Finally we use the fact that T is a tilting complex. This proves that the homology of G^{−1}(Hom_{R}(Λ, R)) is concentrated in degree 0 and isomorphic to Hom_{R}(Γ, R). This finishes the proof of the lemma.

Proof of theorem 1. Since Λ is symmetric, Λ ≃ Hom_{R}(Λ, R) in D^{b}(Λ⊗_{R} Λ^{op}). Moreover,

Hom_{D^{b}(Γ⊗_{R}Γ^{op})}(Γ ⊗_{R} Γ^{op}, G^{−1}(Λ)[n])) = Hom_{D^{b}(Λ⊗_{R}Λ^{op})}(T ⊗_{R} Hom_{Λ}(T, Λ), Λ)[n])
= Hom_{D^{b}(Λ)}(T, Hom_{Λ}(Hom_{Λ}(T, Λ), Λ[n]))
= Hom_{D^{b}(Λ)}(T, T [n])
= \{ \begin{matrix}Γ & if n = 0 \\ 0 & else \end{matrix}

applying G^{−1} gives that Γ ≃ Hom_{R}(Γ, R) in D^{b}(Γ ⊗_{R} Γ^{op}).

If R is an integral domain, Hom_{R}(Γ, R) is R-torsion free, hence so is Γ. Tensoring the derived equivalence induced by T with the field of fractions K of R (see [3, Proposition 6.2.4]), one gets a Morita equivalence between A := K ⊗_{R} Λ and B := K ⊗_{R} Γ. Since Γ is R-torsion free, Γ is an R-order in B. This proves the theorem.

References

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

[2] S. König and A. Zimmermann; Tilting hereditary orders; Comm. in Alg. 24 (1996) 1897-1913.

[3] S. König and A. Zimmermann; Derived Equivalences for Group Rings; with contributions by Bernhard Keller, Markus Linckelmann, Jeremy Rickard and Raphaël Rouquier. Springer Lecture Notes in Mathematics 1685, Springer Verlag Berlin-Heidelberg 1998.

[4] J. Rickard; Derived equivalences as derived functors; J. London Math. Soc. 43 (1991) 37-48.