Han's conjecture for bounded extensions
Claude Cibils, Marcelo Lanzilotta, Eduardo N. Marcos, Andrea Solotar
aa r X i v : . [ m a t h . K T ] J a n Han’s conjecture for bounded extensions
Claude Cibils, Marcelo Lanzilotta, Eduardo N. Marcos,and Andrea Solotar ∗ Abstract
Let B ⊂ A be a bounded extension of finite dimensional algebras. We usethe Jacobi-Zariski long nearly exact sequence to show that B satisfies Han’sconjecture if and only if A does, regardless if the extension splits or not.We provide conditions ensuring that an extension by arrows and relations isbounded. Examples of non split bounded extensions are given and we obtaina structure result for the extensions of an algebra given by a quiver andadmissible relations. Keywords:
Hochschild, homology, relative, Han, quiver Introduction
The purpose of this paper is twofold. Firstly to provide results which may be usedto associate to a given algebra A for which we want to study if Han’s conjectureis true, an easier algebra B for which the property of satisfying Han’s conjectureis equivalent to the same property for A . In [9, 8] a theory is developed for splitextensions. Here we obtain results that allow us to deal with non split extensions.Even if there are algebras to which we cannot apply our methods, our guidingidea is that, starting from a given finite dimensional algebra, we can reduce thestudy of Han’s conjecture about it to the same problem for an algebra where thecomputation of the Hochschild homology is more tractable. The second aim ofthe paper is to provide examples of this procedure. For this purpose, we develop atheory of extensions by arrows and relations and we give conditions ensuring thatthe extension is bounded. Finally in the Appendix we show that all extensionsof an algebra given by a quiver and admissible relations are obtained through ourconstruction.We describe the structure of the paper in more detail in the next paragraphs.Let k be a field and let B ⊂ A be an extension of k -algebras. G. Hochschild de-fined Hochschild (resp. relative) homology with coefficients in an A -bimodule X in[12] (resp. [13]). These are vector spaces denoted H ∗ ( A, X ) (resp. H ∗ ( A | B, X ) ),for ∗ ≥ . ∗ This work has been supported by the projects UBACYT 20020170100613BA, PIP-CONICET 11220150100483CO, USP-COFECUB. The third mentioned author was supportedby the thematic project of FAPESP 2014/09310-5, a research grant from CNPq 302003/2018-5and acknowledges support from the ”Brazilian-French Network in Mathematics”. The fourthmentioned author is a research member of CONICET (Argentina), Senior Associate at ICTPand visiting Professor at Guangdong Technion-Israel Institute of Technology. k -algebra is left (resp. right) smooth if its left (resp. right) global dimension isfinite. Note that for noetherian rings the left and right global dimension coincide, see[2]. In this case we call smooth a left or right smooth k -algebra. Han’s conjecture(see [11]) states that for a finite dimensional algebra A , if H ∗ ( A, A ) = 0 in largeenough degrees, then A is smooth. In [8] we have included a brief account of themain advances towards proving this conjecture, see [7, 3, 5, 6, 4, 18, 17, 9].An extension B ⊂ A of k -algebras is called bounded if the B -bimodule A/B is B -tensor nilpotent, its projective dimension is finite and it is left or right B -projective. A main result of this paper is that for a bounded extension B ⊂ A , the k -algebra B satisfies Han’s conjecture if and only if A does. Notice that we do notassume the extension to be split.The proof of the above result relies on the normalised relative bar resolutionand the Jacobi-Zariski long nearly exact sequence obtained in [10]. We give a briefaccount of these results in Section 2.It is worth noting that in Sections 3 and 4 we consider extensions of algebraswhich may be infinite dimensional. We first show that if an extension B ⊂ A isbounded then the Hochschild homologies of A and B are isomorphic in large enoughdegrees. Then we prove in Section 4 that for a bounded extension, smoothness ispreserved. Note that for the latter, the normalised relative bar resolution is alsoa crucial tool. We end Section 4 by proving the main quoted result about Han’sconjecture for bounded extensions of finite dimensional algebras.In Section 5 we construct explicit extensions of an algebra B = kQ/I where Q is a quiver and I is an admissible ideal of the path algebra kQ . A structureresult is proved in the Appendix, which shows that this construction provides every B -finitely generated algebra extension of B . In detail, we first add new arrows to Q ,to obtain a tensor algebra extension T of B . Then we add new relations through anideal J ⊂ T verifying B ∩ J = 0 to obtain an extension B ⊂ A = T /J . In Section5 we focus on extensions B ⊂ A of this sort to give conditions ensuring that the B -bimodule A/B is tensor nilpotent and/or left (or right) projective over B . Weuse relative paths, they are alternate concatenations of new arrows and non-zeropaths of B . If every relative cycle contains a J -interrupter (see Definition 5.10),then A/B is B -tensor nilpotent. If the projection of J to the positive part of T isgenerated as a B -bimodule by relative paths ending in the same new arrow, then A/B is left projective as a B -module.Finally in Section 6 we consider examples of extensions by arrows and relations.The above criteria enables to prove that they are bounded, and we show that theyare not split. Normalised relative bar resolution and the Jacobi-Zariskisequence
In this section we recall some results from [10] that will be useful for the sequel.We first recall the normalised relative bar resolution.
Proposition 2.1 [10, Proposition 2.3] Let B ⊂ A be an extension of k -algebrasand let σ be a k -linear section of the canonical projection π : A → A/B . Thefollowing is a B -relative resolution of A : · · · d → A ⊗ B ( A/B ) ⊗ B m ⊗ B A d → · · · d → A ⊗ B A/B ⊗ B A d → A ⊗ B A d → A → here the last d is the product of A and d ( a ⊗ α ⊗ · · · ⊗ α n − ⊗ a n ) = a σ ( α ) ⊗ α ⊗ · · · ⊗ α n − ⊗ a n + n − X i =1 ( − i a ⊗ α ⊗ · · · ⊗ π ( σ ( α i ) σ ( α i +1 )) ⊗ · · · ⊗ α n − ⊗ a n +( − n − a ⊗ α ⊗ · · · ⊗ σ ( α n − ) a n . Moreover the differential d does not depend on the choice of the section σ . The normalised relative bar resolution provides a chain complex for computingthe relative Hochschild homology as follows.
Theorem 2.2 [10, Corollary 2.4] Let B ⊂ A be an extension of k -algebras and let X be an A -bimodule. Let σ be a k -linear section of π : A → A/B . The homologyof the following chain complex is the relative Hochschild homology H ∗ ( A | B, X ) · · · b → X ⊗ B e ( A/B ) ⊗ B m b → · · · b → X ⊗ B e A/B b → X B → where b ( x ⊗ α ⊗ · · · ⊗ α n ) = xσ ( α ) ⊗ α ⊗ · · · ⊗ α n + n − X i =1 ( − i x ⊗ α ⊗ · · · ⊗ π ( σ ( α i ) σ ( α i +1 )) ⊗ · · · ⊗ α n +( − n σ ( α n ) x ⊗ α ⊗ · · · ⊗ α n − . The coboundary b does not depend on the choice of the section σ . Definition 2.3
Let B be a k -algebra. A B -bimodule M is tensor nilpotent if thereexists n such that M ⊗ B n = 0 . As usual, we write “ ∗ >> ” for “large enough ∗ ”, that is “all ∗ larger thansome positive integer”. Corollary 2.4
Let B ⊂ A be an extension of k -algebras with A/B tensor nilpotent.Let X be an A -bimodule. We have H ∗ ( A | B, X ) = 0 for ∗ >> . Proof.
The complex of Theorem 2.2 is zero in large enough degrees. ⋄ Next we recall the definition of a long sequence of vector spaces which is exacttwice in three, as introduced in [10].
Definition 2.5 A long nearly exact sequence is a complex of vector spaces . . . δ → U m I → V m K → W m δ → U m − I → V m − → . . . δ → U n I → V n K → W n ending at a fixed n ≥ which is exact at U ∗ and W ∗ .Its gap is the homology at V ∗ , namely the vector space ( Ker K/ Im I ) ∗ . . Kaygun has first obtained the Jacobi-Zariski long exact sequence in the non-commutative setting for a k -algebra extension B ⊂ A where A/B is flat as a B -bimodule, see [15, 16]. Note that for commutative algebras the Jacobi-Zariskilong exact sequence has been established for Andr´e-Quillen homology in 1974, seefor instance [1, p. 61] or [14]. The reason for giving the name “Jacobi-Zariski” tothis sequence is explained in [1, p. 102].We state now the Jacobi-Zariski long nearly exact sequence obtained in [10]. Theorem 2.6 [10, Theorem 4.4 and Theorem 5.1] Let B ⊂ A be an extension of k -algebras. Let X be an A -bimodule. There is a long nearly exact sequence · · · δ → H m ( B, X ) I → H m ( A, X ) K → H m ( A | B, X ) δ → H m − ( B, X ) I → · · · δ → H ( B, X ) I → H ( A, X ) K → H ( A | B, X ) which is exact at H ∗ ( B, X ) and H ∗ ( A | B, X ) . Moreover, there is a spectral sequence converging to its gap ( Ker K/ Im I ) ∗ for ∗ ≥ . Its terms at page are E p,q = Tor B e p + q ( X, ( A/B ) ⊗ B p ) for p, q > and elsewhere. Definition 2.7
An extension B ⊂ A of k -algebras is left (resp. right) bounded if- A/B is tensor nilpotent,-
A/B is of finite projective dimension as a B e -module,- A/B is a left (resp. right) projective B -module.The extension is called bounded if it is left or right bounded. Theorem 2.8 [10, Theorem 6.5]Let B ⊂ A be a bounded extension of k -algebras and let X be an A -bimodule.There is a Jacobi-Zariski long exact sequence ending at some n : . . . δ → H m ( B, X ) I → H m ( A, X ) K → H m ( A | B, X ) δ → H m − ( B, X ) I → . . . δ → H n ( B, X ) I → H n ( A, X ) K → H n ( A | B, X ) . Hochschild homology of bounded extensions
In this section we prove that for any bounded extension B ⊂ A the Hochschildhomology remains unchanged in large enough degrees. Theorem 3.1
Let B ⊂ A be an extension of k -algebras, where the B -bimodule A/B is of finite projective dimension and tensor nilpotent. There is a linear injection H ∗ ( B, B ) ֒ → H ∗ ( A, A ) for ∗ >> . If moreover the extension is bounded then H ∗ ( B, B ) ≃ H ∗ ( A, A ) for ∗ >> . roof. Firstly we assert that if
A/B is of finite projective dimension as B -bimodule, then H ∗ ( B, B ) and H ∗ ( B, A ) are isomorphic for ∗ >> . Indeed, theexact sequence of B -bimodules → B → A → A/B → . (3.1) induces a long exact sequence in Hochschild homology · · · → H m ( B, B ) → H m ( B, A ) → H m ( B, A/B ) → H m − ( B, B ) → · · ·→ H ( B, B ) → H ( B, A ) → H ( B, A/B ) → . By definition for any B -bimodule MH ∗ ( B, M ) =
Tor B e ∗ ( B, M ) . Since M = A/B is of finite projective dimension as a B e -module, we infer that H ∗ ( B, A/B ) = 0 for ∗ >> . The assertion follows.Next we consider the Jacobi-Zariski long nearly exact sequence of Theorem 2.6for an A -bimodule X . It is exact at H ∗ ( B, X ) , that is Ker I = Im δ . Since A/B is B -tensor nilpotent, by Corollary 2.4 we have that H ∗ ( A | B, X ) = 0 for ∗ >> , thusfor large enough degrees we get δ = 0 . Hence the maps I : H ∗ ( B, X ) → H ∗ ( A, X ) are injective for ∗ >> . For X = A , we infer from the previous assertion that H ∗ ( B, B ) ֒ → H ∗ ( A, A ) for ∗ >> . Note that the gap of the Jacobi-Zariski long nearly exact sequence is H ∗ ( A, A ) /H ∗ ( B, B ) for ∗ >> . If the extension is bounded then by Theorem 2.8 the Jacobi-Zariski sequence isexact in degrees larger than some n . Hence the gap is zero in large enough degreesand H ∗ ( A, A ) ≃ H ∗ ( B, B ) for ∗ >> . ⋄ Bounded extensions, smoothness and Han’s conjecture
Smoothness
We consider extensions B ⊂ A with the aim of studying the relation betweensmoothness of B and A . For completeness we recall first the following. Lemma 4.1
Let B ⊂ A be an extension of k -algebras, with A/B of finite projectivedimension as a left B -module. Any left projective A -module is of finite projectivedimension as a left B -module. Proof.
The exact sequence (3.1) shows that A has finite projective dimension as aleft B -module. Then the result holds for any left projective A -module by standardarguments. ⋄ We recall that a k -algebra is left (resp. right) smooth if its left (resp. right)global dimension is finite. For noetherian rings the left and right global dimensionscoincide, see [2]. heorem 4.2 Let B ⊂ A be an extension of k -algebras where A is left smooth. If A/B is of finite projective dimension as a B -bimodule and projective as a B -rightmodule, then B is left smooth. Proof.
Let Y be a left B -module. The exact sequence of B -bimodules (3.1) isright split, hence the sequence of left B -modules → B ⊗ B Y → A ⊗ B Y → ( A/B ) ⊗ B Y → is exact. Of course B ⊗ B Y is canonically isomorphic to Y . In order to prove that Y is of finite projective dimension, we will prove that A ⊗ B Y and ( A/B ) ⊗ B Y are of finite projective dimension as left B -modules.Let P ∗ → A ⊗ B Y → be a finite projective resolution of the A -module A ⊗ B Y .We view it as an exact sequence of left B -modules by restricting the left actions to B . By hypothesis A/B is of finite projective dimension as a B -bimodule. Notethat a projective B -bimodule is projective as a left B -module, hence A/B is also offinite projective dimension as a left B -module. Then each P i is of finite projectivedimension as a left B -module by Lemma 4.1. We infer that A ⊗ B Y is of finiteprojective dimension as a left B -module.For ( A/B ) ⊗ B Y , let Q ∗ → ( A/B ) → be a finite projective resolution ofthe B -bimodule A/B . Each Q i is projective as a B -bimodule, then it is projectiveas a right B -module. Moreover A/B is projective as a right B -module. Hence Q ∗ ⊗ B Y → ( A/B ) ⊗ B Y → is exact, and it is a finite projective resolution ofthe left B -module ( A/B ) ⊗ B Y . ⋄ Theorem 4.3
Let B ⊂ A be an extension of k -algebras where B is left smooth.Suppose that the B -bimodule A/B is tensor nilpotent and that
A/B is projectiveas a B -right module. Then A is left smooth. Proof.
Let Y be a left B -module. First we will prove that the left A -module A ⊗ B Y is of finite projective dimension. Let P ∗ → Y be a finite projective resolutionof Y as a B -module. Observe that since the exact sequence (3.1) is right split, A isprojective as a right B -module. Hence A ⊗ B P ∗ → A ⊗ B Y → is exact. Moreover A ⊗ B P i is projective as a left A -module. Indeed this holds if P i is a free left B -module, thus it holds for projectives by standard arguments.Let now X be an A -module. We assert that there is an exact sequence of A -modules of the form → A ⊗ B Y n → A ⊗ B Y n − → · · · → A ⊗ B Y → X → where the Y i ’s are left B -modules. Note that the assertion shows that X is of finiteprojective dimension as an A -module.To prove the assertion, let n be such that ( A/B ) ⊗ B n = 0 . The normalised barresolution of Theorem 2.1 is zero in degrees ≥ n . We observe that the contractinghomotopy of this resolution is given by morphisms of right A -modules. Then thefollowing sequence is exact: → A ⊗ B ( A/B ) ⊗ B n − ⊗ B A ⊗ A X d → · · · d → A ⊗ B ( A/B ) ⊗ B A ⊗ A X d → A ⊗ B A ⊗ A X d → A ⊗ A X → , s well as the isomorphic sequence → A ⊗ B ( A/B ) ⊗ B n − ⊗ B X d → · · · d → A ⊗ B ( A/B ) ⊗ B X d → A ⊗ B X d → X → . The modules A ⊗ B (cid:0) ( A/B ) ⊗ B i ⊗ B X (cid:1) are of the form A ⊗ B Y . By the aboveresult they are of finite projective dimension as left A -modules. Hence X is of finiteprojective dimension as well. ⋄ Corollary 4.4
Let B ⊂ A be a right bounded extension of k -algebras. The algebra B is left smooth if and only if A is left smooth. Remark 4.5
Analogously, if an extension is left bounded then right smoothness ispreserved.
Han’s conjecture for bounded extensions
Next we use the results that we have obtained to prove the following. Recall thatif a k -algebra is finite dimensional then left smooth is equivalent to right smooth.In this case the algebra is called smooth. Theorem 4.6
Let B ⊂ A be a bounded extension of finite dimensional k -algebras.The algebra B satisfies Han’s conjecture if and only if A satisfies Han’s conjecture. Proof.
Assume that B satisfies Han’s conjecture. Suppose H ∗ ( A, A ) = 0 for ∗ >> . Since the extension verifies that the B -bimodule A/B is of finite pro-jective dimension and is tensor nilpotent, the first part of Theorem 3.1 gives that H ∗ ( B, B ) = 0 for ∗ >> , and then B is smooth. Moreover the B -bimodule A/B is of finite projective dimension and also that
A/B is a B -left (or right) projectivemodule. Hence Theorem 4.3 provides that A is smooth.Conversely, assume that A satisfies Han’s conjecture and H ∗ ( B, B ) = 0 for ∗ >> . The extension is bounded, hence the second part of Theorem 3.1 givesthat H ∗ ( A, A ) = 0 for ∗ >> and therefore the algebra A is smooth. The B -bimodule A/B is of finite projective dimension and
A/B is projective as a left orright projective module. Theorem 4.2 ensures shows that B is smooth. ⋄ Extensions by arrows and relations
In this section we consider extensions of algebras given by quivers and relations, inorder to provide conditions which ensure that the extensions are bounded.Let Q be a finite quiver, that is a finite set of vertices Q , a finite set of arrows Q and two maps s, t : Q → Q called source and target respectively. A pathof length n > is a sequence of concatenated arrows b n . . . b , that is verifying t ( b i ) = s ( b i +1 ) for i = 1 , . . . n − . The paths of length zero are the vertices. Let Q n be the set of paths of length n and consider the path algebra kQ = L i ≥ kQ i .Let h Q i i be the ideal generated by Q i . An ideal I of kQ is admissible if for some n ≥ we have h Q n i ⊂ I ⊂ h Q i . Let B = kQ/I . It is well known that if k is algebraically closed, any finite dimensional k -algebra is Morita equivalent to analgebra of this type. efinition 5.1 Let Q be a quiver. A set F of new arrows for Q is a finite setwith two maps s, t : F → Q . The quiver Q F is given by ( Q F ) = Q and ( Q F ) = Q ∪ F with the evident maps s, t : ( Q F ) → Q . Let B = kQ/I as above, let F be a set of new arrows for Q , and let h I i kQ F bethe ideal of kQ F generated by I . Let T = kQ F / h I i kQ F . Since h I i kQ F ∩ kQ = I we have B ⊂ T . It is useful to observe that T is a tensor algebra over B of aprojective B -bimodule as follows. Let N = M a ∈ F Bt ( a ) ⊗ s ( a ) B. We have T = T B ( N ) = B ⊕ N ⊕ ( N ⊗ B N ) ⊕ · · · Definition 5.2 An extension by arrows and relations B ⊂ A is given by • B = kQ/I as above, • F a set of new arrows for Q , • J an ideal of T = kQ F / h I i kQ F verifying J ∩ B = 0 .The extended algebra is A = T /J . Remark 5.3
Let B = kQ/I be an algebra as above. In the Appendix we provethat any extension of B which is B -finitely generated (see Definition 7.1) is givenby arrows and relations. Example 5.4
Let B = kQ/I where Q is the full arrows quiver a bα dβ c and I = h βα i . Let F = { a } be the dashed new arrow from to , and let J = h abcd − αβ i . Then A = kQ F / h βα, abcd − αβ i . Example 5.5
Let Q be a quiver and I be an admissible ideal of kQ . The alge-bra Λ = kQ/I is an extension by arrows and relations of its maximal semisimplesubalgebra E = kQ , where F = Q and J = I .In this case the extension E ⊂ Λ is split since Λ = E ⊕ r where r is theJacobson radical of Λ . Moreover it is well known and easy to prove that the E -bimodule Λ /E = r is E -tensor nilpotent if and only if Q has no oriented cycles. Let B ⊂ A be an extension by arrows and relations. To give a sufficient conditionfor A/B to be B -tensor nilpotent, we introduce the following definitions. Definition 5.6
Let B ⊂ A be an extension by arrows and relations. A relative pathof F -length n ≥ is a concatenated sequence β n a n . . . β a β a β where the β i ’s are paths of Q not in I (that is non-zero paths of B ) and the a i ’sare in F . emark 5.7 - A relative path of F -length is a path β of Q which is non-zero in B .- Relative paths span T over k .- The definition of a relative path only takes into account kQ/I and F . Definition 5.8 A relative cycle of F -length n ≥ is a concatenated sequence β n a n . . . β a β a where the β i ’s are paths of Q not in I , the a i ’s are in F , and t ( β n ) = s ( a ) . Remark 5.9
By abuse of language, we also call a relative cycle of F -length n ≥ a concatenated sequence a β n a n . . . β a β a where the β i ’s are paths of Q not in I and the a i ’s are in F . In Example 5.4, the sequence a ( bcd ) a is a relative cycle of length .In what follows, for γ ∈ T , we denote γ its class in A = T /J.
Definition 5.10
Let a β n a n . . . β a β a be a relative cycle. A J -interrupter ofthe relative cycle is an arrow a i such that in case ≤ i ≤ n we have β i a i β i − ∈ B, while for i = 1 we have β a β n ∈ B. In Example 5.4 the arrow a is a J -interrupter of the relative cycle a ( bcd ) a .Indeed ( bcd ) a ( bcd ) = bcdαβ ∈ B. Lemma 5.11
Let B ⊂ A be an extension by arrows and relations.If there is no relative cycles then A/B is B -tensor nilpotent. Before providing the proof, we underline that an oriented cycle in Q F is notnecessarily a relative cycle, as in the following example. Example 5.12
Let B = kQ/I where Q is the full arrows quiver b ca d and I = h dcb i . Let F = { a } be the dashed new arrow from to . The quiver Q F has oriented cycles but there are no relative cycles. roof of Lemma 5.11. Let B = kQ/I , where Q is a quiver and I is an admissibleideal.We observe first that A/B is spanned over k by { γ } , that is by the classesmodulo J and B of relative paths γ of positive F -length.Hence the set of m -tensors over B Γ = γ m ⊗ B · · · ⊗ B γ spans ( A/B ) ⊗ B m over k , where γ m , . . . , γ are relative paths of positive F -length.Note that since kQ ⊂ B , if the relative paths are not concatenated then Γ = 0 .Recall that I is admissible, this implies that the paths β not in I are of boundedlength. Hence there is a finite number n of relative paths of the form βa .Let Γ = γ n +1 ⊗ B · · ·⊗ B γ be a k -generator of ( A/B ) ⊗ B n +1 , where γ n +1 , . . . , γ are concatenated relative paths of positive F -length.We consider the concatenated sequence γ n +1 . . . γ and we reduce it as follows.For each occurrence · · · a ′′ β ′′ ⊗ B β ′ a ′ · · · we make the product β ′′ β ′ in the sequence.Suppose that some path β ′′ β ′ belongs to I . Since · · · a ′′ β ′′ ⊗ B β ′ a ′ · · · = · · · a ′′ β ′′ β ′ ⊗ B a ′ · · · (5.1) we infer that the sequence containing β ′′ β ′ is in h I i kQ F . Then the correspondingtensorand is zero, and Γ = 0 .On the contrary, in case none of the β ′′ β ′ is zero in B , we infer a relative pathof F -length at least n + 1 . Then there is a repetition of some relative path of theform βa and we obtain a relative cycle aβ m a m . . . β a βa for some m ≤ n . Therefore, this case does not occur. ⋄ Theorem 5.13
Let B ⊂ A be an extension by arrows and relations, where B = kQ/I and A = kQ F h I i kQF /J .If each relative cycle has a J -interrupter then A/B is B -tensor nilpotent. Proof.
If there is are no relative cycles then Lemma 5.11 shows that
A/B is B -tensor nilpotent.Next we assume that each relative cycle contains a J -interrupter. As before,let Γ = γ n +1 ⊗ B · · · ⊗ B γ be a k -generator of ( A/B ) ⊗ B n +1 , where γ n +1 , . . . , γ are concatenated relative paths of positive F -length. By reducing the concatenatedsequence γ n +1 . . . γ as before we can assume that it is a relative path. Indeed,otherwise there is a product β ′′ β ′ arising from some · · · a ′′ β ′′ ⊗ B β ′ a ′ · · · which isin I , and by (5.1) we have Γ = 0 .Since n is the number of relative paths of the form βa , there is a repetition ofsome relative path βa , that is we have a relative cycle aβ m a m . . . β a βa. (5.2)10 et c be a J -interrupter of it, with β ′ cβ ∈ B . If in the expression of Γ we havethat β ′ and β belong to different tensorands, for instance · · · β ′ c ⊗ B β · · · , usingthe equality · · · β ′ c ⊗ B β · · · = · · · β ′ cβ ⊗ B · · · we obtain an expression of Γ which contains, in one of its tensorands, the relativepath β ′ cβ . Modulo J , we replace β ′ cβ by a linear combination of paths of Q which are non-zero in B . Hence Γ is a linear combination of n -tensors, each of thesummands having one less new arrow. If a summand has a tensorand which is in B ( i.e. it has no more new arrows), then this summand is zero. For the others, werestart the process. This way we end with Γ = 0 . ⋄ Next we will give conditions ensuring that
A/B is projective as a left B -module.Of course reversing the arguments in the following proof gives conditions ensuringthat A/B is right B -projective. Lemma 5.14
Let C be a k -algebra and let X and Y be respectively a left and aright C -module. Let Y ′ be a right C -subbimodule of Y .The C -bimodules X ⊗ YY ′ and X ⊗ YX ⊗ Y ′ are isomorphic. Proof.
Consider the exact sequence → Y ′ → Y → YY ′ → and apply the exactfunctor X ⊗ − ⋄ Lemma 5.15
Let B ⊂ A be an extension by arrows and relations, where A is finitedimensional over k . There exists n > such that any relative path of F -length m ≥ n is J -equivalent to a linear combination of paths of F -length smaller than n . Proof.
The classes of relative paths k -span A , and we extract a finite basis fromit. Then n − is the maximal F -length of the relative paths in the basis. ⋄ Definition 5.16
For a finite dimensional extension by arrows and relations B ⊂ A ,the length index of the extension is the smallest n as above. Recall that for an extension by arrows and relations B ⊂ A , the algebra T isthe tensor algebra T B ( N ) = B ⊕ M i> N ⊗ B i where N = ⊕ a ∈ F Bt ( a ) ⊗ s ( a ) B . Let T > = ⊕ i> N ⊗ B i be its positive ideal, andlet T ]0 ,n [ be the B -bimodule ⊕
Let B ⊂ A be an extension by arrows and relations of length index n . Suppose that the projection J ]0 ,n [ of J on T ]0 ,n [ is generated as a B -bimoduleby relations of the form r = a P λ γ γ where a ∈ F , the γ ’s are relative pathsconcatenated with a and λ γ ∈ k .Then A/B is left B -projective. Before proving the theorem, we observe that the hypothesis is verified in Example5.4. Indeed, the ideal J is generated by abcd − αβ . Note that abcda is a relativepath of F -length two, and αβa is a relative path of F -length one. In A , we have hat abcda = αβa . The relative paths of F -length ≥ are zero in A . Thereforethe length index of the extension is . Moreover J ]0 , is generated by abcd as a B -bimodule. Proof.
Let J > be the projection of J on T > . Since B ∩ J = 0 , as B -bimoduleswe have that AB = T > J > . Moreover, since the length index is nT > J > = T ]0 ,n [ J ]0 ,n [ . The relative path aβ corresponds to t ( a ) ⊗ s ( a ) β = t ( a ) ⊗ β ∈ T . The B -bimodulein T generated by aβ is k -spanned by β ′′ t ( a ) ⊗ ββ ′ where β ′′ and β ′ are paths of Q which are non-zero in B .We observe that for i ≥ we have N ⊗ B i = M a ∈ F Bt ( a ) ⊗ s ( a ) B ⊗ B N ⊗ B i − = M a ∈ F Bt ( a ) ⊗ s ( a ) N ⊗ B i − . Therefore T ]0 ,n [ = M a ∈ F Bt ( a ) ⊗ s ( a ) T We will use the tools that we have developed in order to consider examples ofextensions by arrows and relations. We will study if the extension is bounded andif it is split or not. e recall that an extension B ⊂ A is split if there is an ideal M of A suchthat A = B ⊕ M. This is of course equivalent to the fact that the algebra inclusion B ֒ → A splits. Example 6.1 We recall the algebra extension provided in Example 5.4, where B = kQ/I with Q the full arrows quiver a bα dβ c and I = h βα i . Let F = { a } be the dashed new arrow from to , and let J = h abcd − αβ i . Then A = kQ F / h βα, abcd − αβ i . As mentioned, a is a J -interrupter of the relative cycle abcda , and the sameholds for longer relative cycles such as abcdabcda . So A/B is tensor nilpotent byTheorem 5.13. Let e be the idempotent corresponding to the vertex t ( a ) . The left B -module A/B is projective by Theorem 5.17. More precisely, as noticedbefore, the length index of the extension is . A basis of s ( a ) T < Y a = e BbcdB is { e , b, bc } . Hence by Theorem 5.17 we have that A/B ≃ Be ⊕ Be ⊕ Be asleft B -modules.The algebra B is smooth, then B e is smooth and the B -bimodules are of finiteprojective dimension, in particular A/B is of finite projective dimension as a B -bimodule. Hence this extension by arrows and relations is bounded.We claim it is not split. In fact we will prove that there is not even a right B -submodule of A such that A = B ⊕ X . Let X be a right B -submodule of A .Next we show that B ∩ X = 0 .Let B be a basis of paths for B with Q ⊂ B . The vector space A/B is k -spanned by relative paths of F -positive length, including a . Hence we can complete B to a basis A of A , with relative paths of positive F -length, including a . Let D be the k -vector space with basis A \ { a } . We claim that there exists x ∈ X suchthat in the basis A , the coefficient x a of a is not . Indeed, otherwise X ⊂ D andsince B ⊂ D we would have B ⊕ X ⊂ D . This cannot hold since D ( A . Then x = x a a + y with y ∈ D and x a = 0 .Since X is a right B -module, we have that xbcd = x a abcd + ybcd ∈ X . Weassert that ybcd = λbcd for λ ∈ k . Indeed, y is a k -linear combination of the basis A \ { a } of D . The only possible path of A \ { a } which concatenates with bcd onits left is e , and this proves the assertion.Then xbcd = x a abcd + λbcd = x a βα + λbcd . Note that βα and bcd are linearlyindependent in B . Hence we obtain = xbcd ∈ X ∩ B . Example 6.2 Let Q be the full arrows quiver dµ a b c nd let B = kQ . Let a be the dashed new arrow from to , let J = h da − cb i and let A be the corresponding extension by arrows and relations.We assert that the extension is bounded and not split. There are no relativecycles, so A/B is tensor nilpotent. Moreover the conditions are satisfied to ensurethat A/B is a right projective B -module by the dual version of Theorem 5.17.The algebra B is hereditary, hence B e is smooth and A/B is of finite projectivedimension as a B -bimodule. As in the previous example the extension B ⊂ A isnot split. Appendix Let B ⊂ A be an extension of k -algebras. Equivalently, in the sequel we call A a B -algebra. Definition 7.1 Let A be a B -algebra. A subset G of A is a B -generating set for A as a B -algebra if every element of A is a sum of products of the form b n g n . . . b g b g b (7.1) where g i ∈ G and b i ∈ B for all i . The extension is B -finitely generated if thereexists a finite B -generating set for A . Remark 7.2 - Assume B is central in A . Let G ′ be a set in one-to-one correspondence with G , and let B { G ′ } be the polynomial algebra in the non commuting variables G ′ . There is an algebra surjection B { G ′ } → A , given by B ⊂ A and g ′ g .- If G is a generating set for the B -bimodule A , then every element of A is asum of elements of the form b gb . In particular G is a B -generating set of the algebra A .- If A/B is finitely generated as a B -bimodule (for instance if A is finite di-mensional), then the extension is B -finitely generated. Definition 7.3 Let A be a B -algebra. A B -algebra generating set G is free if eachelement of A can be written as a sum of elements of the form (7.1) in a uniqueway. Let B be a k -algebra and N a B -bimodule. We recall that the tensor al-gebra T B ( N ) has the following universal property. A morphism of k -algebras ϕ : T B ( N ) → X is uniquely determined by a morphism of k -algebras ϕ B : B → X (which endows X with a B -bimodule structure), and a morphism of B -bimodules ϕ N : N → X . Definition 7.4 Let B be a k -algebra and let H be a set. The free B -bimodulewith basis H is BHB = M h ∈ H BhB where BhB denotes the free rank one B -bimodule B ⊗ B whose generator ⊗ isidentified to h . he following result is clear. Proposition 7.5 Let A be B -algebra, G a subset of A and G ′ a copy of G . Considerthe morphism of algebras ϕ : T B ( BG ′ B ) → A determined by ϕ B : B ֒ → A and bythe B -bimodule map ϕ BG ′ B : BG ′ B → A given by g ′ g for all g ∈ G .- G is a B -generating set of A if and only if ϕ is surjective.- G is a free B -generating set of A if and only if ϕ is an isomorphism. We turn now to the case where B = kQ/I , for Q a quiver and I an admissibleideal. Note that the vertices are a complete set of orthogonal idempotents of a B algebra A . The associated Peirce decomposition of A is A = M x,y ∈ Q yAx. Definition 7.6 Let A be a B -algebra where B is as above. A subset F of A is aPeirce subset if every element of F belongs to one of the summands of the abovePeirce decomposition. Theorem 7.7 Let B = kQ/I where Q is a quiver and I is an admissible ideal.Every finitely generated B -algebra A is an extension by arrows and relations. Proof. Let G be a finite B -generating set of the extension B ⊂ A . We firstobserve that G provides a Peirce B -generating set F = { ygx = 0 | g ∈ G and x, y ∈ Q } . Note that ⊗ X x,y ∈ Q y ⊗ x. Therefore B ⊗ B = M x,y ∈ Q By ⊗ xB. Recall that G ′ is a set in one-to-one correspondence with G . Hence for g ∈ G , wehave Bg ′ B = M x,y ∈ Q Byg ′ xB where Byg ′ xB is the B -direct summand of Bg ′ B which corresponds to By ⊗ xB in B ⊗ B .Let F ′ be a copy of F . If a ∈ F belongs to yAx , we set t ( a ′ ) = y and s ( a ′ ) = x .This way F ′ is a set of new arrows for Q . Let N = M a ′ ∈ F ′ Bt ( a ′ ) a ′ s ( a ′ ) B = M a ′ ∈ F ′ Bt ( a ′ ) ⊗ s ( a ′ ) B. Clearly N is a direct summand of the B -bimodule BG ′ B with complement M g ∈ Gygx =0 Byg ′ xB. et ϕ : T B ( BG ′ B ) → A be the surjective algebra morphism given by Proposition7.5. Note that if ygx = 0 then by construction of ϕ we have ϕ ( yg ′ x ) = yϕ ( g ′ ) x = ygx = 0 . Hence Byg ′ xB ⊂ Ker ϕ . Therefore ψ = ϕ | TB ( N ) has the same image than ϕ , then ψ is surjective.For J = Ker ψ we infer an isomorphism of algebras T B ( N ) /J ψ → A . Thecomposition B → T B ( N ) ψ → A is the inclusion B ⊂ A , hence J ∩ B = 0 .Recall that B = kQ/I , and that Q F ′ is the union of the quiver Q with the newarrows F ′ . 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Math. Soc. 138 (2010), 861–869. C.C.:Institut Montpelli´erain Alexander Grothendieck, CNRS, Univ. Montpellier, France. M.L.:Instituto de Matem´atica y Estad´ıstica “Rafael Laguardia”, Facultad de Ingenier´ıa, Universidad dela Rep´ublica, Uruguay. [email protected] E.N.M.:Departamento de Matem´atica, IME-USP, Universidade de S˜ao Paulo, Brazil. [email protected] A.S.:IMAS-CONICET y Departamento de Matem´atica, Facultad de Ciencias Exactas y Naturales,Universidad de Buenos Aires, Argentina. [email protected]@dm.uba.ar