Isomorphisms of non noetherian down-up algebras
aa r X i v : . [ m a t h . R A ] N ov Isomorphisms of non noetherian down-upalgebras
Sergio Chouhy and Andrea Solotar ∗ Abstract
We solve the isomorphism problem for non noetherian down-up algebras A ( α, , γ ) by lifting isomorphisms between some of their non commutativequotients. The quotients we consider are either quantum polynomial algebrasin two variables for γ = 0 or quantum versions of the Weyl algebra A for nonzero γ . In particular we obtain that no other down-up algebra is isomorphic tothe monomial algebra A (0 , , . We prove in the second part of the article thatthis is the only monomial algebra within the family of down-up algebras. Ourmethod uses homological invariants that determine the shape of the possiblequivers and we apply the abelianization functor to complete the proof. down-up algebra, isomorphism, non noetherian, monomial. Let k be a fixed field of characteristic . Given parameters ( α, β, γ ) ∈ k , the as-sociated down-up algebra A ( α, β, γ ) , first defined in [2], is the quotient of the freeassociative algebra k h d, u i by the ideal generated by the relations d u − ( αdud + βud + γd ) ,du − ( αudu + βu d + γu ) . There are several well-known examples of down-up algebras such as A (2 , − , ,isomorphic to the enveloping algebra of the Heisenberg-Lie algebra of dimension , and, for γ = 0 , A (2 , − , γ ) , isomorphic to the enveloping algebra of sl (2 , k ) .A down-up algebra has a PBW basis given by { u i ( du ) j d k : i, j, k ≥ } . Note that A ( α, β, γ ) can be regarded as a Z -graded algebra where the degrees of u and d are respectively and − . The field k is the trivial module over A ( α, β, γ ) ,with d and u acting as .E. Kirkman, I. Musson and D. Passman proved in [7] that A ( α, β, γ ) is noethe-rian if and only it is a domain, if and only if β = 0 .The isomorphism problem for down-up algebras was posed in [2] where theauthors considered algebras A ( α, β, γ ) of four different types and proved, by study-ing one dimensional modules, that algebras of different types are not isomorphic.They considered the following types, ∗ This work has been supported by the projects UBACYT 20020130100533BA, UBACYT20020130200169BA, PIP-CONICET 11220150100483CO, and MATHAMSUD-REPHOMOL. The secondauthor is a research member of CONICET (Argentina). γ = 0 , α + β = 1 ,(b) γ = 0 , α + β = 1 , (c) γ = 0 , α + β = 1 ,(d) γ = 0 , α + β = 1 .As a consequence, we can restrict the question of whether two down-up alge-bras are isomorphic to each of the four types. In [3] the authors solved the isomor-phism problem for noetherian down-up algebras of types (a), (b) and (c) for everyalgebraically closed field k , and also for noetherian algebras of type (d) when inaddition char ( k ) = 0 . More precisely, they proved the following result. Theorem ([3]) . Let A = A ( α, β, γ ) and A ′ = A ( α ′ , β ′ , γ ′ ) be noetherian down-up alge-bras. Then A is isomorphic to A ′ if and only if1. γ = λγ ′ for some λ ∈ k × , and2. either α ′ = α , β ′ = β or α ′ = − αβ − , β ′ = β − . Their solution focuses mainly on the possible commutative quotients of down-up algebras. In contrast with this, there are very well studied non commutativealgebras that appear as quotients of non noetherian down-up algebras, for exam-ple, when α ∈ k × , the quantum plane k α [ x, y ] and the quantum Weyl algebra A α , k α [ x, y ] := k h x, y i / ( yx − αxy ) , A α := k h x, y i / ( yx − αxy − . In this article we describe isomorphisms amongst non noetherian down-up al-gebras by using these quotients. Our main result is the following.
Theorem 1.1.
Let k be an algebraically closed field and let α, α ′ , γ, γ ′ ∈ k . The algebras A ( α, , γ ) and A ( α ′ , , γ ′ ) are isomorphic if and only if1. γ = λγ ′ , for some λ ∈ k × , and2. α ′ = α . We obtain in particular that no other down-up algebra is isomorphic to A (0 , , .In Section 3 we prove Theorem 1.2.
The algebra A ( α, β, γ ) is monomial if and only if ( α, β, γ ) = (0 , , . So, the only monomial algebra in the family of down-up algebras is the evi-dent one. Our starting point is the fact that noetherian down-up algebras cannotbe monomial since they are a domain of global dimension [7]. The situationcan be related to -dimensional Sklyanin algebras. In both cases, an algebra A isnoetherian if and only if it is a domain. For Sklyanin algebras, these conditions areequivalent to A being monomial [9]. This is not the case for down-up algebras.Our proof uses homological invariants that determine the possible shapes ofthe quiver. We think that these methods may be useful for other families of alge-bras. The purpose of this section is to prove Theorem 1.1. Let k be an algebraically closedfield. Note that the condition γ = λγ ′ for λ ∈ k × is equivalent to the condition of γ and γ ′ being both zero or both non zero. We already know from [2] that if γ = 0 ,then A ( α, , γ ) is isomorphic to A ( α, , . This is done by rescaling d by γd . Also,observe that A ( α, , is not isomorphic to A ( α ′ , , for all α, α ′ ∈ k , since theybelong to different types. Gathering all this information, we deduce that Theorem1.1 is equivalent to the following two propositions:2 roposition 2.1. Let α, α ′ ∈ k . The algebras A ( α, , and A ( α ′ , , are isomorphic ifand only if α = α ′ . Proposition 2.2.
Let α, α ′ ∈ k . The algebras A ( α, , and A ( α ′ , , are isomorphic ifand only if α = α ′ . We will thus prove both of them in order to obtain our result.
Lemma 2.3.
Let α ∈ k × , γ ∈ k and let A = A ( α, , γ ) be a down-up algebra. Denote ω := du − αud − γ . The algebra A/ h ω i is isomorphic to k α [ x, y ] if γ = 0 and it isisomorphic to A α if γ = 0 . Moreover, in case γ = 0 or γ = 1 , the isomorphism maps theclass of d to y and the class of u to x .Proof. The algebra A ( α, , γ ) is the quotient of the free algebra generated by thevariables d and u subject to the relations d u − αdud − γd = 0 and du − αudu − γu =0 . Denote by Ω the element du − αud − γ in the free algebra. The projection of Ω onto A ( α, , γ ) is ω . The defining relations of A are d Ω = 0 and Ω u = 0 . Therefore, thealgebra A/ h ω i is isomorphic to the algebra freely generated by letters d, u subjectto the relation Ω = 0 . If γ = 0 , then this is exactly the definition of k α [ x, y ] . If γ = 0 ,then ω = γ − (( γd ) u − αu ( γd ) − , and so A/ h ω i is the quantum Weyl algebragenerated by x and y , with y = γd and x = u . (cid:3) In [8] the authors describe all isomorphisms and automorphisms for quantumWeyl algebras A α for α ∈ k × not a root of unity. In [10] this result is generalized tothe family of quantum generalized Weyl algebras , including the quantum plane andthe quantum Weyl algebra for all values of α ∈ k × . We recall some of their resultsin the cases relevant to us. Theorem 2.4 ([8],[10]) . Let α, α ′ ∈ k \ { , } .i) The algebras k α [ x, y ] and k α ′ [ x, y ] are isomorphic if and only if α ′ ∈ { α, α − } . More-over, if ϕ : k α [ x, y ] → k α − [ x, y ] is an isomorphism and α = − , then there exist λ, µ ∈ k × such that ϕ ( x ) = λy and ϕ ( y ) = µx. ii) The algebras A α and A α ′ are isomorphic if and only if α ′ ∈ { α, α − } . If α = − ,then every isomorphism η : A α → A α − is of the form η ( x ) = λy and η ( y ) = − λ − α − x, for some λ ∈ k × .Proof of Proposition 2.1. Let α and α ′ be elements of k and suppose there exists anisomorphism of k -algebras ϕ : A ( α, , → A ( α ′ , , . Denote A := A ( α, , , A ′ := A ( α ′ , , and let d ′ and u ′ be the usual generators of A ′ .Suppose α, α ′ ∈ k \ { , } and α = α ′ . Let ω := du − αud and ω ′ := d ′ u ′ − α ′ u ′ d ′ .By Lemma 2.3, we can identify A/ h ω i with k α [ x, y ] , where the canonical projection π : A → k α [ x, y ] sends d to y and u to x , and similarly for A ′ / h ω ′ i and k α ′ [ x, y ] ; herewe denote by π ′ the canonical projection. Define ψ = π ′ ◦ ϕ . The equalities dω = 0 and ωu = 0 hold in A , so ψ ( d ) ψ ( ω ) = 0 = ψ ( ω ) ψ ( u ) . The algebra k α ′ [ x, y ] is a non commutative domain generated by ψ ( d ) and ψ ( u ) .Thus, ψ ( d ) and ψ ( u ) are not zero and from the above equations we deduce ψ ( ω ) = 0 . This implies that there exists an algebra map ψ : k α [ x, y ] → k α ′ [ x, y ] such that ψ = ψ ◦ π . In the other direction we obtain that ψ := π ◦ ϕ − factorsas ψ = ψ ◦ π ′ . Since ψ ◦ ψ ◦ π ′ = π ′ and ψ ◦ ψ ◦ π = π , we deduce ψ is an3somorphism. The situation is illustrated by the following commutative diagram, A π (cid:15) (cid:15) ϕ / / A ′ π ′ (cid:15) (cid:15) k α [ x, y ] ψ / / k α ′ [ x, y ] ψ o o By Theorem 2.4 and our assumption that α = α ′ , we obtain α ′ = α − . Theorem2.4 also says that there exist λ, µ ∈ k × and z , z ∈ h ω ′ i such that ϕ ( u ) = λd ′ + z and ϕ ( d ) = µu ′ + z . Note that A ′ is graded considering the generators d ′ and u ′ in degree . Since deg( ω ′ ) = 2 , it follows that z and z are either or sums ofhomogeneous elements of degree at least with respect to this grading. On theother hand d u − αdud is , and so ϕ ( d ) ϕ ( u ) − αϕ ( d ) ϕ ( u ) ϕ ( d ) . In particular the degree component of the right hand side of this equality, that is ( u ′ ) d ′ − αu ′ d ′ u ′ , must be . But the set { ( u ′ ) i ( d ′ u ′ ) j ( d ′ ) l : i, j, l ∈ N } is a k -basisof A ′ , and this is a contradiction.In case α = 0 , an argument similar to the above one shows that there is anepimorphism ψ : A → k α ′ [ x, y ] . As a consequence the elements ψ ( d ) and ψ ( u ) generate k α ′ [ x, y ] . If α ′ = 0 , then the algebra k α ′ [ x, y ] is a domain and it is notcommutative, thus it cannot be generated by one element. From the equality d u we obtain that ψ ( d u ) = ψ ( d ) ψ ( u ) , implying ψ ( d ) = 0 or ψ ( u ) = 0 . Thisis a contradiction and so α ′ = 0 .If α = 1 , then A belongs to type (a) and so does A ′ . This implies α ′ = 1 ,concluding the proof of the proposition. (cid:3) Now we turn our attention to Proposition 2.2. Let A = A ( α, , for α ∈ k .Recall that ω := du − αud − . Using Lemma 2.2 in [11], the set { u i ω j d l : i, j, l ≥ } is a k -basis of A . Lemma 2.5.
The set { u i ω j d l : i, l ≥ and j ≥ } is a k -linear basis of the two sided ideal h ω i , and, for each n ∈ N , the set { u i ω j d l : i, l ≥ and j ≥ n } is a k -linear basis of h ω i n .Proof. Every element of the form u i ω j d l with j ≥ belongs to h ω i , so it onlyremains to prove that h ω i is contained in the k -vector space with basis the set { u i ω j d l : i, l ≥ and j ≥ } . Given z ∈ h ω i , write z = P i,j,l λ i,j,l u i ω j d l with i, j, l ≥ and λ i,j,l ∈ k . By Lemma 2.3 we can identify A/ h ω i with A α , and thecanonical projection π : A → A α sends u to x and d to y . The set { x i y l : i, l ≥ } isa basis of A α . From the equalities P i,l λ i, ,l x i y l = π ( z ) = 0 , we deduce λ i, ,l = 0 for all i, l ≥ .Taking into account the description we now have of h ω i , we see that the ele-ments of h ω i are linear combinations of monomials of type u i ω j d l u i ′ ω j ′ d l ′ , with j, j ′ ≥ . Similarly, the elements of h ω i n are linear combinations of n -fold productsof the same type. Therefore, to prove the second claim, it is sufficient to show thatfor every r, s ≥ there exist λ i ∈ k such that ωd r u s ω = P i ≥ λ i ω i . Indeed, thereexist λ i,j,l ∈ k such that d r u s = X i,j,l ≥ λ i,j,l u i ω j d l . So ωd r u s ω = X i,j,l ≥ λ i,j,l ωu i ω j d l ω = X j ≥ λ ,j, ω j +2 . The last equality follows from dω = 0 and ωu = 0 . (cid:3) orollary 2.6. The set { [ u i ωd l ] : i, l ≥ } , where [ p ] denotes the class of an element p in h ω i / h ω i , is a k -linear basis of the A -bimodule h ω i / h ω i . Moreover, in case α = 1 , thefollowing equalities hold [ u i ωd l u ] = α l − α − u i ωd l − ] , [ du i ωd l ] = α i − α − u i − ωd l ] , where the terms on the right are considered to be zero for l = 0 or i = 0 .Proof. The first claim is a direct consequence of Lemma 2.5. To prove the firstformula, we fix i ≥ and proceed by induction on l , the case l = 0 being trivialfrom the equalities ωu = 0 = dω . On the other hand, since ω = ω ( du − αud −
1) = ωdu − ω , we obtain that ωdu = ω + ω . Similarly duω = ω + ω . Therefore, [ u i ωdu ] = [ u i ω ] . Now, for l ≥ u i ωd l u ] = [ u i ωd l − ( αdud + d )] = α [ u i ωd l − ud ] + [ u i ωd l − ]= α α l − − α − u i ωd l − d ] + [ u i ωd l − ]= α l − α − u i ωd l − ] . The second formula can be proved analogously. (cid:3)
Proof of Proposition 2.2.
Let α, α ′ ∈ k . Denote A := A ( α, , , A ′ := A ( α ′ , , . Let d ′ , u ′ be the generators of A ′ . Suppose there exists an isomorphism of k -algebras ϕ : A → A ′ . Recall that ω = du − αud − and ω ′ = d ′ u ′ − α ′ u ′ d ′ − .If α = 1 , then A belongs to type (a), and so does A ′ , hence α ′ = 1 . Now suppose α = 0 and α ′ = 0 . By Lemma 2.3 the algebra A ′ / h ω ′ i is isomorphic to A α ′ and, if π ′ denotes the canonical projection, then π ′ ( u ′ ) = x and π ′ ( d ′ ) = y . Let ψ = π ′ ◦ ϕ .Since dω = ωu = 0 , we have ψ ( d ) ψ ( ω ) = ψ ( ω ) ψ ( u ) = 0 . Note that ψ ( d ) and ψ ( u ) generate A α ′ , and therefore they cannot belong to k ; in particular they cannot bezero. We deduce ψ ( ω ) = ψ ( d ) ψ ( u ) − . The algebra A α ′ has a filtration whoseassociated graded algebra Gr ( A α ′ ) is k α ′ [ x, y ] . The equality ψ ( d ) ψ ( u ) = 1 impliesthat Gr ( A α ′ ) is not a domain, which is a contradiction since α ′ = 0 .Suppose α, α ′ ∈ k \ { , } and α = α ′ . By the same arguments as in the proofof Proposition 2.1, the map ψ := π ′ ◦ ϕ : A → A α ′ induces an isomorphism of k -algebras ψ : A α → A α ′ . Theorem 2.4 implies α ′ = α or α ′ = α − . Since we areassuming α = α ′ , we deduce α ′ = α − . Again, by Theorem 2.4 we obtain thatthere exist λ ∈ k × and z , z ∈ h ω i such that ϕ − ( d ′ ) = − λ − αu + z ,ϕ − ( u ′ ) = λd + z . By rescaling the variables d, u , we may assume λ = 1 . The equality ( d ′ ) u ′ − α − d ′ u ′ d ′ − d ′ = 0 implies ϕ − ( d ′ ) ϕ − ( u ′ ) − α − ϕ − ( d ′ ) ϕ − ( u ′ ) ϕ − ( d ′ ) − ϕ − ( d ′ )= α u d − αudu + αu + α u z − αuz d − αz ud + udz − αuz u + z du − z + z = − αuω + α ( αu z − uz d ) + ( udz − αuz u ) + z ω + z ∈ h ω i , z denotes the sum of all terms in which at least two factors z or z areinvolved. Note that z ω and z belong to h ω i . Taking classes modulo h ω i , α [ uω ] + α ([ uz d ] − α [ u z ]) = [ udz ] − α [ uz u ] . Write now z = P i,l ≥ ,j ≥ λ i,j,l u i ω j d l and z = P i,l ≥ ,j ≥ µ i,j,l u i ω j d l . Using theformulas of Corollary 2.6 we obtain α [ uω ]+ X i,l ≥ α ( λ i, ,l [ u i +1 ωd l +1 ] − αµ i, ,l [ u i +2 ωd l ]) == X i ≥ ,l ≥ λ i, ,l α i − α − u i ωd l ] − X i ≥ ,l ≥ αµ i, ,l α l − α − u i +1 ωd l − ] . By Corollary 2.6, the set { [ u i ωd l ] : i, l ≥ } is a k -linear basis of h ω i / h ω i . For m ≥ , define Λ m := λ m +1 , ,m − αµ m, ,m +1 . Looking at the coefficient correspond-ing to the term [ u m +1 ωd m ] in the last equation for each m ≥ , we deduce α = Λ ,α Λ m − = α m +1 − α − m , for m ≥ . The fact that Λ = α = 0 implies, by an inductive argument, that Λ m = 0 for all m ∈ N . As a consequence, either λ m +1 , ,m = 0 for infinitely many values of m ∈ N ,or µ m, ,m +1 = 0 for infinitely many values of m ∈ N . This is a contradiction thatcomes from the assumption α = α ′ . (cid:3) An algebra is monomial if it is isomorphic to an algebra of the form k Q/I , where Q is a quiver with a finite number of vertices and I is a two-sided ideal in k Q generated by paths of length at least . The algebra A (0 , , is monomial and noother down-up algebra is isomorphic to it. However, other monomial down-upalgebras may exist. In this section we prove Theorem 1.2. Before doing it we provea series of preparatory lemmas.We will make use of the abelianization functor defined on k -algebras as A A ab := A/J A , where J A is the two sided ideal in A generated by the set { xy − yx : x, y ∈ A } . Thecanonical projection π A : A → A ab is a natural transformation from the identity tothe abelianization functor.In order to state the next lemma we need some previous definitions. Given aquiver Q with a finite number of vertices and e, e ′ ∈ Q , define eQ e ′ := { α ∈ Q : t ( α ) = e, s ( α ) = e ′ } , where t and s are the usual target and source maps.Also, denote by B e the k -algebra k [ X α : α ∈ eQ e ] . That is, B e is the polynomialalgebra in variables indexed by the elements of the set eQ e . In case eQ e = ∅ weset B e = k . If I is a two-sided ideal in k Q generated by paths of length at least ,define I e to be the ideal in B e generated by the set [ n ≥ { X α n · · · X α : α n · · · α ∈ I, α i ∈ eQ e } . emma 3.1. Let Q be a quiver with a finite number of vertices and I a two-sided ideal in k Q generated by paths of length at least . There is an isomorphism of k -algebras (cid:18) k QI (cid:19) ab ∼ = M e ∈ Q B e I e . Proof.
The classes e in k Q ab of the vertices e in Q are a complete set of centralorthogonal idempotents and e ( k Q ab ) e is isomorphic to ( e k Qe ) ab , and thus isomor-phic to B e . As a consequence, there is an isomorphism θ : k Q ab → L e ∈ Q B e suchthat θ ( π ( α )) = X α for all α ∈ ekQe and e ∈ Q . Let g be the map f ab ◦ θ − . Thecommutativity of the diagram k Q π / / f (cid:15) (cid:15) k Q ab f ab (cid:15) (cid:15) θ / / L e ∈ Q B eg x x rrrrrrrrrr k Q/I / / ( k Q/I ) ab implies that g is surjective and that its kernel is ( θ ◦ π )( I + J k Q ) = ( θ ◦ π )( I ) = ⊕ e ∈ Q I e , and the lemma follows. (cid:3) The following lemma is a well known result for finite dimensional algebrasreplacing
Tor by Ext . Here we give the proof for completeness.
Lemma 3.2.
Let B = k Q/I be a monomial algebra. For each e ∈ Q denote by T e thesimple B -module corresponding to e . If e, e ′ ∈ Q , then eQ e ′ = dim k Tor B ( T e , T e ′ ) .Moreover, if Q has only one vertex e , then the dimension of Tor B ( T e , T e ) is sup { dim k Tor B ( T , T ) : T , T are one dimensional B -modules } . Proof.
Let e, e ′ ∈ Q . The first terms of the minimal projective bimodule resolutionof B are · · · → B ⊗ E k Q ⊗ E B → B ⊗ E B → B → , where E = k Q .Applying the functor T e ⊗ B ( − ) ⊗ B T e ′ to this resolution we obtain the followingcomplex · · · → T e ⊗ k k eQ e ′ ⊗ k T e ′ → T e ⊗ k T e ′ → , whose homology is isomorphic to Tor B • ( T e , T e ′ ) . The minimality of the resolu-tion implies that every arrow in the above complex is zero. As a consequence, Tor B ( T e , T e ′ ) ∼ = T e ⊗ k k eQ e ′ ⊗ k T e ′ ∼ = e k Q e ′ , from where we deduce that thedimension of Tor B ( T e , T e ′ ) is eQ e ′ . As for the second assertion, the same argu-ment shows that if T , T are one dimensional B -modules, then the homology ofthe complex · · · → T ⊗ k k Q ⊗ k T → T ⊗ k T → , is isomorphic to Tor B • ( T , T ) . It follows that dim k Tor B ( T , T ) ≤ dim k ( k Q ) = Q = dim k Tor B ( T e , T e ) , where e is the only vertex of Q . (cid:3) Lemma 3.3.
Let A = A ( α, , γ ) and let T , T be one dimensional A -modules.(i) If γ = 0 and α = 1 , then dim k Tor A ( T , T ) = 0 .(ii) If γ = 0 and α = 1 , then dim k Tor A ( T , T ) ≤ . iii) If γ = 0 , then dim k Tor A ( T , T ) ≤ and dim k Tor A ( k , k ) = 2 . Moreover, if α = 1 and T = k , then dim k Tor A ( T , T ) = 1 .Proof. Let T , T be one dimensional A -modules with bases { v } and { v } , respec-tively. Let δ , δ , µ , µ ∈ k be such that d · v i = δ i v i and u · v i = µ i v i for i = 1 , .From the equalities d u − αdud − γd = 0 = du − αudu − γu we deduce δ i ((1 − α ) δ i µ i − γ ) = 0 ,µ i ((1 − α ) δ i µ i − γ ) = 0 , (3.1)for i = 1 , . Consider the following resolution of A as A -bimodule [4] → A ⊗ k Ω ⊗ k A d → A ⊗ k R ⊗ k A d → A ⊗ k V ⊗ k A d → A ⊗ k A → , where V , R and Ω are the subspaces of the free algebra k h d, u i spanned, respec-tively, by the sets { d, u } , { d u, du } and { d u } . The differentials are d (1 ⊗ v ⊗
1) = v ⊗ − ⊗ v, for all v ∈ V,d (1 ⊗ d u ⊗
1) = 1 ⊗ d ⊗ du + d ⊗ d ⊗ u + d ⊗ u ⊗ − α (1 ⊗ d ⊗ ud + d ⊗ u ⊗ d + du ⊗ d ⊗ − β (1 ⊗ u ⊗ d + u ⊗ d ⊗ d + ud ⊗ d ⊗ − γ ⊗ d ⊗ ,d (1 ⊗ du ⊗
1) = 1 ⊗ d ⊗ u + d ⊗ u ⊗ u + du ⊗ u ⊗ − α (1 ⊗ u ⊗ du + u ⊗ d ⊗ u + ud ⊗ u ⊗ − β (1 ⊗ u ⊗ ud + u ⊗ u ⊗ d + u ⊗ d ⊗ − γ ⊗ u ⊗ , and d (1 ⊗ d u ⊗
1) = d ⊗ du ⊗ β ⊗ du ⊗ d − ⊗ d u ⊗ u − βu ⊗ d u ⊗ . (3.2)Applying the functor T ⊗ A ( − ) ⊗ A T to this resolution of A we obtain the fol-lowing complex of k -vector spaces whose homology is isomorphic to Tor A • ( T , T ) , / / k f / / k f / / k f / / k / / , where f = (cid:0) δ − δ µ − µ (cid:1) and f = (cid:18) (1 − α ) δ µ + δ ( µ − αµ ) − γ µ ( µ − αµ ) δ ( δ − αδ ) (1 − α ) δ µ + µ ( δ − αδ ) − γ (cid:19) . The claims of the lemma follow from these formulas and Equation (3.1). (cid:3)
We now turn to the proof of Theorem 1.2. Suppose ( α, β, γ ) = (0 , , . Denoteas usual A ( α, β, γ ) by A . Let B = k Q/I be a monomial algebra and suppose thereexists an isomorphism of k -algebras ϕ : A → B . Since every down-up algebra hasglobal dimension [6], we deduce that I = 0 and so B is not a domain, thus weget β = 0 .Suppose γ = 0 . In this case α = 0 since we are assuming ( α, β, γ ) = (0 , , .Note that 8 ab = k [ d, u ] / h (1 − α ) d u, (1 − α ) du i , in particular it is connected and so is B ab . By Lemma 3.1, the quiver Q has onlyone vertex e . Moreover, by Lemmas 3.2 and 3.3 we deduce k ( T e , T e ) = Q , thus, Q has exactly two arrows a and b . By Lemma 2.3 the quantum plane k α [ x, y ] is a quotient of A and so there exists an epimorphism ϕ : B → k α [ x, y ] . Since α = 0 ,the quantum plane k α [ x, y ] is a domain. Given a non zero path p in I , ϕ ( p ) = 0 ,implying that either ϕ ( a ) or ϕ ( b ) is zero. As a consequence, the quantum plane isgenerated as algebra by one variable, which is a contradiction.Now suppose γ = 0 . If α = 1 , then Tor A ( T , T ) = 0 for every pair of onedimensional A -modules T , T , from Lemma 3.3. Since A ∼ = B , this is also true for B and one dimensional B -modules. By Lemma 3.2, the quiver Q has no arrows.This is impossible, and so α = 1 . The same lemmas imply in this case that there isat most one arrow between each pair of vertices in Q . In particular, there is at mostone element in eQ e for every vertex e .Define V = { e ∈ Q : eQ e = 1 } and for every e ∈ V , denote a e the uniqueelement in eQ e . Since A , and thus B , is of global dimension , Bardzell’s resolu-tion [1] of B is of finite length. So a en / ∈ I for all n ∈ N . This implies B e = k [ X ] and I e = 0 for all e ∈ V , and B e = k for all e / ∈ V . By Lemma 3.1, B ab ∼ = M e/ ∈ V k ⊕ M e ∈ V k [ X ] . In particular, its group of units is contained in the finite dimensional vector space k Q . On the other hand, the fact that the ideals h d, u i and h (1 − α ) du − γ ) i in k [ d, u ] are coprime implies A ab ∼ = k ⊕ k [ d, u ] h (1 − α ) du − γ i . The group of units of this algebra is contained in no finite dimensional space. Since B ab is isomorphic to A ab , this is a contradiction and we conclude the proof of The-orem 1.2. References [1] Michael J. Bardzell,
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1, 2[3] Paula A. A. B. Carvalho and Ian M. Musson,
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Automorphisms and isomorphisms of quantum generalizedWeyl algebras , J. Algebra (2015), 540–552. ↑ Centers of down-up algebras , J. Algebra (1999), 103–121. ↑ [email protected] A.S.:Departamento de Matemática, Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires,Ciudad Universitaria, Pabellón I, 1428 Buenos Aires, Argentina; andIMAS, UBA-CONICET, Consejo Nacional de Investigaciones Científicas y Técnicas, Argentina [email protected]@dm.uba.ar