THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS

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1 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS JOANNE LEADER AND NICOLE SNASHALL Abstract. We generalise Koszul and D-Koszul algebras by introducing a class of graded algebras called (D, A)-stacked algebras. We give a characterisation of (D, A)-stacked algebras and show that their Ext algebra is finitely generated as an algebra in degrees 0, 1, and 3. In the monomial case, we give an explicit description of the Ext algebra by quiver and relations, and show that the ideal of relations has a quadratic Gröbner basis; this enables us to give a regrading of the Ext algebra under which the regraded Ext algebra is a Koszul algebra. Introduction In this paper we introduce a class of graded algebras called (D, A)-stacked algebras which provide a natural generalisation of both Koszul algebras and D-Koszul algebras ([]) as well as the (D, A)-stacked monomial algebras of Green and Snashall ([11], [1]). There are many generalisations of Koszul algebras, and, in introducing (D, A)-stacked algebras, we retain the essential property that each projective in a minimal graded projective resolution of 0 is generated in a single degree, and show that the Ext algebra is finitely generated. Moreover, it is well-known that the Ext algebra of a Koszul algebra is again a Koszul algebra, and we show that this extends to (D, A)-stacked monomial algebras, in that there is a regrading of the Ext algebra so that the regraded Ext algebra is a Koszul algebra. Let K be a field and let = KQ/I be a finite-dimensional algebra where I is an admissible ideal of KQ. If the ideal I is generated by length homogeneous elements then there is an induced length grading on so that = 0 1. We suppose throughout this paper that = KQ/I is a graded finite-dimensional algebra with the length grading. Let r denote the graded Jacobson radical of, so r = 1 and 0 = /r is a finite product of copies of K. In this setting, graded -modules have minimal graded projective resolutions. So, viewing 0 as a graded right -module, there is a minimal graded projective resolution (P n, d n ) of 0. Recall that this resolution is minimal if Im d n P n 1 r for all n 1. The Ext algebra of, E(), is defined to be E() = n0 Extn ( 0, 0 ) where multiplication is given by the Yoneda product. The structure of the Ext algebra of a graded finitely-generated algebra is not in general well understood, and it is often difficult to check whether the Ext algebra is itself finitely 010 Mathematics Subject Classification. 16G0, 16S37, 16E30. Key words and phrases. Koszul, D-Koszul, Ext algebra, finite generation, quadratic Gröbner basis. Some of these results formed part of the first author s PhD thesis at the University of Leicester, which was supported by EPSRC. 1

2 LEADER AND SNASHALL generated as an algebra. It is well-known that the Ext algebra of a Koszul algebra is finitely generated in degrees 0 and 1 ([1, 9]), and this provides the starting point for the class of algebras we study in this paper. Recall that a graded algebra = 0 1 is said to be Koszul if 0 has a linear resolution, that is, if the nth projective module P n in a minimal graded projective resolution (P n, d n ) of 0 is generated in degree n. Moreover, if KQ/I is Koszul then I is a quadratic ideal, that is, I is generated by linear combinations of paths of length. The paper is structured as follows. In Section 1, we define the class of (D, A)-stacked algebras and discuss their relationship to Koszul algebras, the D-Koszul algebras introduced by Berger ([]), the (D, A)-stacked monomial algebras of Green and Snashall ([11]) and the δ-koszul algebras of Green and Marcos ([7]). Our first result is Theorem 1.4, where we show that the Ext algebra of a (D, A)-stacked algebra is finitely generated as an algebra in degrees 0, 1, and 3. We then study the properties of the Ext algebra of a (D, A)-stacked algebra and, in Theorem.5, provide a characterisation of (D, A)-stacked algebras. We give a regrading of the Ext algebra in Section 3 and recall some necessary Gröbner basis theory in Section 4. Section 5 focuses on the monomial case, where we give an explicit description of the Ext algebra E() by quiver and relations for a (D, A)-stacked monomial algebra, so that E() = K /I. In particular, we show in Proposition 5.6 that I has a quadratic reduced Gröbner basis in the case where A > 1 and D A. We then prove in Theorem 5.8 that the regraded Ext algebra of a (D, A)-stacked monomial algebra is a Koszul algebra, providing D A when A > 1. This generalises the previous results for Koszul and D-Koszul algebras ([8, Theorem 7.1]), and provides further justification for these algebras being considered a natural generalisation of Koszul algebras. The final section returns to the general case, where we give a non-monomial example of a (D, A)-stacked algebra where the regraded Ext algebra is also Koszul. We conclude the paper with some open questions. Throughout this paper, = KQ/I is a graded finite-dimensional algebra where I is length homogeneous. An arrow α starts at s(α) and ends at t(α); arrows in a path are read from left to right. A path p = α 1 α α n, where α 1, α,..., α n are arrows, is of length n with s(p) = s(α 1 ) and t(p) = t(α n ). If the ideal I is generated by paths in KQ then KQ/I is a monomial algebra. An element x in KQ is uniform if there exist vertices e i, e j Q 0 such that x = e i x = xe j. 1. (D, A)-Stacked Algebras In this section we introduce (D, A)-stacked algebras where D, A 1. The parameters D and A relate to the degree in which the projective modules P and P 3 are generated, in a minimal projective resolution (P n, d n ) of /r. We then prove that the Ext algebra of a (D, A)-stacked algebra is finitely generated. We denote the global dimension of by gldim. Definition 1.1. Let = KQ/I be a finite-dimensional algebra. Then is a (D, A)- stacked algebra if there is some D, A 1 such that, for all 0 n gldim, the

3 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS 3 projective module P n in a minimal projective resolution of /r is generated in degree δ(n), where 0 if n = 0 1 if n = 1 δ(n) = n D if n even, n (n 1) D + A if n odd, n 3. Remark. It is clear from the definition that if = KQ/I is a (D, A)-stacked algebra then the projective module P is generated in degree D. Thus the ideal I is length homogeneous and is generated by linear combinations of paths of length D, where D. Hence is necessarily a graded algebra with the length grading, and /r = 0. The (D, A)-stacked algebras with A = 1 are precisely the algebras where, for 0 n gldim, the projective module P n is generated in degree { n D if n even, n 0 (n 1) D + 1 if n odd, n 1. It follows that the (D, 1)-stacked algebras are precisely the finite-dimensional D-Koszul algebras of Berger ([]). Berger introduced D-Koszul algebras in order to include some cubic Artin-Schelter regular algebras and anti-symmetrizer algebras. It was shown by Green, Marcos, Martínez-Villa and Zhang in [8, Theorem 4.1] that the Ext algebra of a D-Koszul algebra is finitely generated in degrees 0, 1 and. Koszul algebras are D- Koszul algebras with D = and so the finite-dimensional Koszul algebras are precisely the (D, A)-stacked algebras where D =, A = 1. In this case, it is well-known that their Ext algebra is generated in degrees 0 and 1. We note also that the (D, A)-stacked algebras share the property of Koszul and D-Koszul algebras that each projective module in a minimal projective resolution of 0 is generated in a single degree. The (D, A)-stacked algebras clearly contain the (D, A)-stacked monomial algebras of Green and Snashall [11, Definition 3.1] (and see [1]). For monomial algebras of infinite global dimension, the (D, A)-stacked monomial algebras are precisely the monomial algebras for which every projective module P n in a minimal graded projective resolution (P n, d n ) of 0 is generated in a single degree and for which the Ext algebra E() is finitely generated. It was shown in [1, Theorem 3.6] that the Ext algebra of a (D, A)-stacked monomial algebra is finitely generated in degrees 0, 1, and 3. In [7], Green and Marcos defined a graded algebra = 0 1 to be δ-resolution determined if there is a function δ : N N such that, for all 0 n gldim, the nth projective P n in a minimal graded projective resolution (P n, d n ) of 0 over is generated in degree δ(n). Additionally, a δ-resolution determined algebra is said to be δ-koszul if E() is finitely generated as an algebra. It is clear that a (D, A)-stacked algebra is δ-resolution determined, with δ as in Definition 1.1. We now give an example of a (D, A)-stacked algebra with D = 4 and A = which is not monomial.

4 4 LEADER AND SNASHALL Example 1.. Let Q be the quiver given by α 1 α α α 7 7 α α 3 α 5 5 and let = KQ/I, where I = (α 1 α α 7 α 8 )α 3 α 4, α 3 α 4 α 5 α 6, α 5 α 6 (α 1 α α 7 α 8 ). The ideal I is homogeneous with generators all of length 4. Thus is a length graded algebra. We show that is a (D, A)-stacked algebra with D = 4 and A =, by explicitly describing a minimal graded projective resolution of 0. Let S i be the simple module corresponding to the idempotent e i, for i = 1,..., 7. It is easy to see that S, S 4, S 6 and S 7 have finite projective dimension, whilst S 1, S 3 and S 5 have infinite projective dimension. Define projective -modules P 0 = 7 i=1 e i, P 1 = 8 i=1 t(α i) and P n = e 1 e 3 e 5 for n. Let d 0 : P 0 /r be the canonical surjection, and define -homomorphisms d n : P n P n 1, for n 1, as follows: d 1 : t(α i ) s(α i )α i t(α i ) e 1 e 4 α 4 α 5 α 6 d : e 3 e 6 α 6 (α 1 α α 7 α 8 ) e 5 e α α 3 α 4 e 7 α 8 α 3 α 4 e 1 e 5 α 5 α 6 for n 3, d n : e 3 e 1 (α 1 α α 7 α 8 ) e 5 e 3 α 3 α 4 where each path p lies in the s(p)-component of P n 1. It is straightforward to verify that (P n, d n ) is a minimal graded projective resolution of 0 as a right -module. It can be seen that the entries in a matrix representation of d 0 (resp. d 1, d ) are of length 0 (respectively, 1, 3) so P 0 (respectively, P 1, P ) is generated in degree 0 (respectively, 1, 4). Moreover, for n 3, the entries in a matrix representation of d n are of length, so P n is generated in degree more than that of P n 1. Thus, for n 0, the projective module P n is generated in degree 0 if n = 0 1 if n = 1 n if n. Hence is a (D, A)-stacked algebra with D = 4, A =. In order to prove that the Ext algebra of a (D, A)-stacked algebra is finitely generated, we require the following result from [8] which we state here for completeness. α 4

5 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS 5 Proposition 1.3. [8, Proposition 3.6] Let = KQ/I be a finite-dimensional graded algebra with the length grading and let P P 1 P be a minimal graded projective resolution of 0 as a right -module. Suppose that P i is finitely generated with generators in degree d i, for i = α, β, α + β. Assume that d α+β = d α + d β. Then the Yoneda map is surjective. Thus Ext α ( 0, 0 ) Ext β ( 0, 0 ) Ext α+β ( 0, 0 ) Ext α+β ( 0, 0 ) = Ext α ( 0, 0 ) Ext β ( 0, 0 ) = Ext β ( 0, 0 ) Ext α ( 0, 0 ). Theorem 1.4. Let = KQ/I be a (D, A)-stacked algebra with D and A 1. Then E() is generated in degrees 0, 1, and 3. Proof. The algebra = KQ/I is a (D, A)-stacked algebra, so I is generated by homogeneous elements of length D. Thus is a graded algebra with the length grading. Each projective P n is generated in degree δ(n) where δ is as in Definition 1.1. Let n 4 and recall that δ() = D. For n even, we have δ(n ) = (n )D/ and δ(n) = nd/ so δ() + δ(n ) = δ(n); similarly, for n odd, we have δ(n ) = (n 3)D/ + A and δ(n) = (n 1)D/ + A, so again we have δ() + δ(n ) = δ(n). Thus, from Proposition 1.3, Ext n ( 0, 0 ) = Ext ( 0, 0 ) Ext n ( 0, 0 ) for all n 4. Thus, E() is generated in degrees (at most) 0, 1, and 3. The following corollary is now immediate. Corollary 1.5. Let = KQ/I be a (D, A)-stacked algebra with D and A 1. Then E() is a δ-koszul algebra.. Properties of the Ext algebra In this section we investigate some of the general properties of the Ext algebra of a (D, A)-stacked algebra and give a full characterisation of (D, A)-stacked algebras in Theorem.5. We use the notation and results of [8] which we briefly recall here. Let = 0 1 be a graded algebra, and let M = i M i be a graded -module. The nth shift of M, denoted M[n], is the graded -module X = i X i, where X i = M i n. Hence, if M is generated in degree n, then M[ n] is generated in degree 0. Let Gr() be the category of graded right -modules together with the set of degree 0 homomorphisms and let F denote the forgetful functor, F : Gr() Mod-. In addition to the grading of the Ext algebra by homological degree, namely E() = n0 Extn ( 0, 0 ), we also have the shift-grading. Specifically, let M = i M i and N = i N i be graded -modules and suppose we have a graded projective resolution (Q n, d n ) of M where each Q n is finitely generated. Following [8], we define Ext n (F (M), F (N)) i = Ext n Gr() (M, N[i]),

6 6 LEADER AND SNASHALL which is the homology of the complex obtained by applying Hom Gr() (, N[i]) to (Q n, d n ), and is called the shift-grading. In our setting, we have M = N = 0 and a minimal graded projective resolution (P n, d n ) of 0 as a -module. We can then apply [8, Theorem.1] to see that Ext n (F ( 0 ), F ( 0 )) i = HomGr() (Ω n ( 0 ), 0 [i]) = Hom Gr() (Ω n ( 0 )[ i], 0 ), where Ω n ( 0 ) denotes the nth syzygy of 0 with respect to the resolution (P n, d n ). We will use this to investigate when products in the Ext algebra are zero. We assume that is a graded algebra and (P n, d n ) is a minimal graded projective resolution of 0. Proposition.1. Let be a (D, A)-stacked algebra with D >. Then Ext 1 ( 0, 0 ) Ext 1 ( 0, 0 ) = 0. Proof. The projective module P 1 is generated in degree 1 and Ext 1 ( 0, 0 ) = Hom (P 1, 0 ), so every element of Ext 1 ( 0, 0 ) can be viewed as a short exact sequence of graded modules of the form 0 0 [ 1] E 0 0. Let (1) 0 0 [ 1] E 0 0 and () 0 0 [ 1] Ê 0 0 be two short exact sequences in Ext 1 ( 0, 0 ). We can shift the sequence () by 1 to get (3) 0 0 [ ] Ê[ 1] 0[ 1] 0. We then splice the sequences (1) and (3) together to obtain 0 0 [ ] Ê[ 1] E 0 0. Thus the image of Ext 1 ( 0, 0 ) Ext 1 ( 0, 0 ) is contained in Ext ( 0, 0 ). However, we know that P is generated in degree D, so Ext ( 0, 0 ) = 0. Therefore, Ext 1 ( 0, 0 ) Ext 1 ( 0, 0 ) = 0, when D >. Proposition.. Let be a (D, A)-stacked algebra with D >. (i) If D A + 1 then Ext n ( 0, 0 ) Ext 1 ( 0, 0 ) = 0 = Ext 1 ( 0, 0 ) Ext n ( 0, 0 ), for all n odd, n 1. (ii) If A > 1 then Ext n ( 0, 0 ) Ext 1 ( 0, 0 ) = 0 = Ext 1 ( 0, 0 ) Ext n ( 0, 0 ), for all n even, n. Proof. The case n = 1 follows from Proposition.1. Thus we may assume n. The projective module P n is generated in degree δ(n). So, since Ext n ( 0, 0 ) = Hom (P n, 0 ), each extension can be viewed as an exact sequence of graded modules of the form 0 0 [ δ(n)] E n E Using the shift-grading, we can shift this sequence by 1 to obtain 0 0 [ δ(n) 1] E n [ 1] E 1 [ 1] 0 [ 1] 0.

7 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS 7 We can then splice this with an extension 0 0 [ 1] E 0 0 from Ext 1 ( 0, 0 ) to obtain 0 0 [ δ(n) 1] E n [ 1] E 1 [ 1] E 0 0. Thus the image of Ext n ( 0, 0 ) Ext 1 ( 0, 0 ) lies in Ext n+1 ( 0, 0 ) δ(n)+1. However, the projective P n+1 is generated in degree δ(n+1), so Ext n+1 ( 0, 0 ) = Ext n+1 ( 0, 0 ) δ(n+1). If n = r + 1 is odd, then δ(n + 1) = δ(r + ) = (r + 1)D and δ(n) + 1 = δ(r + 1) + 1 = rd + A + 1, and since D A + 1, we have Ext n+1 ( 0, 0 ) δ(n)+1 = 0. Hence Ext n ( 0, 0 ) Ext 1 ( 0, 0 ) = 0 in this case. On the other hand, if n = r is even with r 1, then δ(n + 1) = δ(r + 1) = rd + A and δ(n) + 1 = δ(r) + 1 = rd + 1. Since A > 1, we have Ext n+1 ( 0, 0 ) δ(n)+1 = 0, and hence Ext n ( 0, 0 ) Ext 1 ( 0, 0 ) = 0. The case for Ext 1 ( 0, 0 ) Ext n ( 0, 0 ) = 0 is similar. Proposition.3. Let be a (D, A)-stacked algebra with D >, D A. Then for all m, n 1. Ext m+1 ( 0, 0 ) Ext n+1 ( 0, 0 ) = 0 Proof. Let m 1, n 1. The projective modules P m+1 and P n+1 are generated in degrees δ(m + 1) and δ(n + 1), respectively. So an extension in Ext m+1 ( 0, 0 ) can be viewed as an exact sequence of graded modules (4) 0 0 [ δ(m + 1)] E m+1 E and an extension in Ext n+1 ( 0, 0 ) can be viewed as an exact sequence of graded modules (5) 0 0 [ δ(n + 1)] E n+1 E Shifting the sequence (5) by δ(m + 1) and then splicing with (4), gives 0 0 [ δ(m + 1) δ(n + 1)] E n+1[ δ(m + 1)] E 1[ δ(m + 1)] E m+1 E which is in Ext m+n+ ( 0, 0 ) δ(m+1)+δ(n+1). However, P (m+n+1) is generated in degree δ((m + n + 1)) = (m + n + 1)D, whereas δ(m + 1) + δ(n + 1) = (m + n)d + A. Since D A, we have Ext (m+n+1) ( 0, 0 ) δ(m+1)+δ(n+1) = 0, so Ext m+1 ( 0, 0 ) Ext n+1 ( 0, 0 ) = 0. We summarise Propositions.1,. and.3 in the following result. Theorem.4. Let be a (D, A)-stacked algebra with D >. Then (i) Ext 1 ( 0, 0 ) Ext 1 ( 0, 0 ) = 0; (ii) if D A + 1, then Ext n ( 0, 0 ) Ext 1 ( 0, 0 ) = 0 = Ext 1 ( 0, 0 ) Ext n ( 0, 0 ), for all n odd, n 1; (iii) if A > 1, then Ext n ( 0, 0 ) Ext 1 ( 0, 0 ) = 0 = Ext 1 ( 0, 0 ) Ext n ( 0, 0 ), for all n even, n ; (iv) if D A, then Ext m+1 ( 0, 0 ) Ext n+1 ( 0, 0 ) = 0, for all n, m 1.

8 8 LEADER AND SNASHALL We now use Theorems 1.4 and.4 to give the following characterisation of (D, A)-stacked algebras. This uses the characterisation of D-Koszul algebras from [8, Theorem 4.1]. Theorem.5. Let = KQ/I where I is generated by homogeneous elements of length D. Then = 0 1 is length graded. Suppose, in the minimal projective resolution (P n, d n ) of 0 that P 3 is generated in a single degree, D + A, for A 1. Then is a (D, A)-stacked algebra if and only if E() is generated in degrees 0, 1, and 3 and the following conditions hold: (i) if D > then Ext 1 ( 0, 0 ) Ext 1 ( 0, 0 ) = 0; (ii) if D >, D A + 1, then Ext n ( 0, 0 ) Ext 1 ( 0, 0 ) = 0 = Ext 1 ( 0, 0 ) Ext n ( 0, 0 ), for all n odd, n 1; (iii) if D >, A > 1, then Ext n ( 0, 0 ) Ext 1 ( 0, 0 ) = 0 = Ext 1 ( 0, 0 ) Ext n ( 0, 0 ), for all n even, n ; (iv) if D >, D A, then Ext m+1 ( 0, 0 ) Ext n+1 ( 0, 0 ) = 0, for all m, n 1. Proof. Suppose = KQ/I is a (D, A)-stacked algebra. Then from Theorem 1.4 we know that E() is generated in degrees 0, 1, and 3. From Theorem.4 we know that conditions (i), (ii), (iii) and (iv) hold. To show the other direction, we consider 3 cases. Case 1: D =, A = 1. Assume I is generated by homogeneous elements of length, P 3 is generated in degree 3 and E() is generated in degrees 0, 1, and 3. We know that P 0 is generated in degree 0 and P 1 is generated in degree 1, and, since I is quadratic, we also have that P is generated in degree. By Proposition 1.3 with α = 1, β = 1 we have Ext 1 ( 0, 0 ) Ext 1 ( 0, 0 ) = Ext ( 0, 0 ), and, with α = 1, β = we have Ext 1 ( 0, 0 ) Ext ( 0, 0 ) = Ext 3 ( 0, 0 ). Therefore E() is generated in degrees 0 and 1. Hence is Koszul and therefore a (, 1)-stacked algebra. Case : D >, A = 1. Assume I is generated by homogeneous elements of length D, P 3 is generated in degree D + 1 and E() is generated in degrees 0, 1, and 3. We know that P 0 is generated in degree 0, P 1 is generated in degree 1 and P is generated in degree D. By Proposition 1.3 with α = 1, β = we have Ext 1 ( 0, 0 ) Ext ( 0, 0 ) = Ext 3 ( 0, 0 ). Therefore E() is generated in degrees 0, 1 and and by [8, Theorem 4.1] is D-Koszul. Hence is a (D, 1)-stacked algebra. Case 3: D >, A > 1. Suppose that P 3 is generated in degree D + A and E() is generated in degrees 0, 1, and 3, with conditions (i), (ii), (iii) and (iv) holding. We know that Ext 0 ( 0, 0 ) = Ext 0 ( 0, 0 ) 0 and Ext 1 ( 0, 0 ) = Ext 1 ( 0, 0 ) 1 since P 0 and P 1 are generated in degrees 0 and 1 respectively. By hypothesis, we have Ext ( 0, 0 ) = Ext ( 0, 0 ) D and Ext 3 ( 0, 0 ) = Ext 3 ( 0, 0 ) D+A. We continue by induction to show that Ext n ( 0, 0 ) = Ext n ( 0, 0 ) δ(n) for n 4, where δ is as in Definition 1.1. We assume, for m < n, that Ext m ( 0, 0 ) = Ext m ( 0, 0 ) δ(m)

9 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS 9 and P m is generated in degree δ(m). We have Ext n ( 0, 0 ) = Ext 1 ( 0, 0 ) Ext n 1 ( 0, 0 ) + Ext ( 0, 0 ) Ext n ( 0, 0 ) + + Ext m ( 0, 0 ) Ext n m ( 0, 0 ) + + Ext n ( 0, 0 ) Ext ( 0, 0 ) + Ext n 1 ( 0, 0 ) Ext 1 ( 0, 0 ). We consider the cases n even and n odd separately. First we suppose that n is even, n 4 and consider Ext m ( 0, 0 ) Ext n m ( 0, 0 ). If m is even then n m is even, P m is generated in degree m D and P n m is generated in degree (n m) D. An element of Ext m ( 0, 0 ) can be viewed as an exact sequence (6) 0 0 [ m D] E m E and an element of Ext n m ( 0, 0 ) can be viewed as an exact sequence (7) 0 0 [ (n m) D] E n m E Shifting (6) by (n m) D and then splicing the resulting sequence with (7) gives 0 0 [ n D] E m[ (n m) D] E 1 [ (n m) D] E n m E Thus the image of Ext m ( 0, 0 ) Ext n m ( 0, 0 ) is contained in Ext n ( 0, 0 ) n D. Now suppose m is odd with m 3 and n m 3; then n m is also odd, P m is generated in degree (m 1) D + A and P n m is generated in degree (n m 1) D + A. An element of Ext m ( 0, 0 ) can be viewed as an exact sequence (8) 0 0 [ (m 1) D A] E m E and an element of Ext n m ( 0, 0 ) as an exact sequence (9) 0 0 [ (n m 1) D A] E n m E Shifting (8) by (n m 1) D A and splicing the resulting sequence with (9) gives 0 0 [ (n ) D A] E m [ (n m 1) D A] E 1 [ (n m 1) D A] E n m E Thus the image of Ext m ( 0, 0 ) Ext n m ( 0, 0 ) is contained in Ext n ( 0, 0 ) (n ) D+A. If m = 1 or m = n 1, then a similar argument shows that the images of Ext 1 ( 0, 0 ) Ext n 1 ( 0, 0 ) Ext 1 ( 0, 0 ) are both contained in Ext n ( 0, 0 ) (n ) D+A+1. We now use conditions (ii) and (iv) and show that some products are necessarily zero. ( 0, 0 ) and Ext n 1 Suppose D = A. Then D A+1, so by (ii), we have Ext 1 ( 0, 0 ) Ext n 1 ( 0, 0 ) = 0 = Ext n 1 ( 0, 0 ) Ext 1 ( 0, 0 ). The image of Ext m ( 0, 0 ) Ext n m ( 0, 0 ) is contained in Ext n ( 0, 0 ) n D for m even and in Ext n ( 0, 0 ) (n ) D+A for m odd, m 3. Since D = A, we have that the image of Ext m ( 0, 0 ) Ext n m ( 0, 0 ) is contained in Ext n ( 0, 0 ) n D for all m 1. Hence Ext n ( 0, 0 ) = Ext n ( 0, 0 ) n D.

10 10 LEADER AND SNASHALL Suppose D A, D = A + 1. By (iv), Ext m ( 0, 0 ) Ext n m ( 0, 0 ) = 0 for m odd, m 3. The images of Ext 1 ( 0, 0 ) Ext n 1 ( 0, 0 ) and Ext 1 ( 0, 0 ) Ext n 1 ( 0, 0 ) are contained in Ext n ( 0, 0 ) (n ) D+A+1 = Ext1 ( 0, 0 ) n D, since D = A + 1. For m even, the image of Ext m ( 0, 0 ) Ext n m ( 0, 0 ) is contained in Ext n ( 0, 0 ) n D. Hence Ext n ( 0, 0 ) = Ext n ( 0, 0 ) n D. Suppose D A, D A+1. By (ii) and (iv), we have Ext m ( 0, 0 ) Ext n m = 0, for m odd, m 1. The image of Ext m ( 0, 0 ) Ext n m in Ext n ( 0, 0 ) n D, for m even. So Ext n ( 0, 0 ) = Ext n ( 0, 0 ) n D. ( 0, 0 ) ( 0, 0 ) is contained Hence for n even, we have that Ext n ( 0, 0 ) = Ext n ( 0, 0 ) n D. Thus P n is generated in degree n D = δ(n). Now suppose that n is odd with n 5. Then n 1 is even so, by (iii), we have that Ext 1 ( 0, 0 ) Ext n 1 ( 0, 0 ) = 0 = Ext n 1 ( 0, 0 ) Ext 1 ( 0, 0 ). We now consider Ext m ( 0, 0 ) Ext n m ( 0, 0 ) for m, n m. If m is odd, then n m is even, P m is generated in degree (m 1) D + A and P n m is generated in degree (n m) D. An element of Ext m ( 0, 0 ) has the form (10) 0 0 [ (m 1) D A] E m E and an element of Ext n m ( 0, 0 ) has the form (11) 0 0 [ (n m) D] E n m E Shifting (10) by (n m) D A and splicing with (11) we obtain 0 0 [ (n 1) D A] E m [ (n m) D A] E 1 [ (n m) D A] E n m E Thus the image of Ext m ( 0, 0 ) Ext n m ( 0, 0 ) is contained in Ext n ( 0, 0 ) (n 1) D+A. If m is even then n m is odd and a similar argument gives that Ext m ( 0, 0 ) ( 0, 0 ) is also contained in Ext n ( 0, 0 ) (n 1) D+A. Thus Extn ( 0, 0 ) = Ext n ( 0, 0 ) (n 1) D+A and hence P n is generated in degree (n 1) D + A = δ(n). Thus, for all n 0, P n is generated in degree δ(n), where δ is as in Definition 1.1, and hence is a (D, A)-stacked algebra. Ext n m 3. Regrading of the Ext Algebra Under Koszul duality, the Ext algebra of a Koszul algebra is again a Koszul algebra. For D-Koszul algebras, a regrading of the Ext algebra was introduced in [8, Section 7], called the hat-grading, in which the regraded Ext algebra of a D-Koszul algebra is a Koszul algebra. Specifically, given a D-Koszul algebra, Green, Marcos, Martínez-Villa and Zhang define the hat-grading of the Ext algebra Ê() = n0ê() n by Ê() 0 = Ext 0 ( 0, 0 ) Ê() 1 = Ext 1 ( 0, 0 ) Ext ( 0, 0 ) Ê() n = Ext n 1 ( 0, 0 ) Ext n ( 0, 0 ) for n,

11 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS 11 and show in [8, Theorem 7.1], that Ê() is a Koszul algebra with this regrading. Recall that a (D, A)-stacked algebra which is neither Koszul nor D-Koszul necessarily has D > and A > 1. In this section, we define a grading on a (D, A)-stacked algebra with D >, A > 1, D A, D A + 1 which we also call the hat-degree grading. This will be used in Section 5, to show (with these conditions on D, A) that the regraded Ext algebra of a (D, A)-stacked monomial algebra is a Koszul algebra. Definition 3.1. Let = KQ/I be a (D, A)-stacked algebra, with D >, A > 1, D A and D A + 1. We define the hat-degree grading on the Ext algebra of by Let Ê() = n0ê() n. Ê() 0 = Ext 0 ( 0, 0 ) Ê() 1 = Ext 1 ( 0, 0 ) Ext ( 0, 0 ) Ext 3 ( 0, 0 ) Ê() n = Ext n ( 0, 0 ) Ext n+1 ( 0, 0 ) for n. First we show that the hat-degree does indeed give a well defined grading. Theorem 3.. Let be a (D, A)-stacked algebra with D >, A > 1, D A, D A + 1. Then the Ext algebra Ê() is a graded algebra with the hat-degree grading of Definition 3.1. Proof. We need to show Ê() m Ê() n = Ê() m+n for all m, n 0. This is clearly true if m = 0 or n = 0. In the case m = n = 1 and using Theorem.4, we have Ê() 1 Ê() 1 = Ext ( 0, 0 ) Ext ( 0, 0 ) Ext ( 0, 0 ) Ext 3 ( 0, 0 ) Ext 3 ( 0, 0 ) Ext ( 0, 0 ). Now, by Proposition 1.3, Ext ( 0, 0 ) Ext ( 0, 0 ) = Ext 4 ( 0, 0 ) and Ext ( 0, 0 ) Ext 3 ( 0, 0 ) = Ext 5 ( 0, 0 ) = Ext 3 ( 0, 0 ) Ext ( 0, 0 ). Thus Ê() 1 Ê() 1 = Ê(). Now let m = 1, n. Using Proposition 1.3, we have Ê() 1 Ê() n = Ext ( 0, 0 ) Ext n ( 0, 0 ) Ext ( 0, 0 ) Ext n+1 ( 0, 0 )+ Ext 3 ( 0, 0 ) Ext n ( 0, 0 ). Theorem.4 then gives Ê() 1 Ê() n = Ê() n+1. Similar arguments show that Ên() Ê 1 () = Ên+1() for n, and that Ê() m Ê() n = Ê() m+n for m, n. We now discuss the case where D = A and show that there is no regrading of the Ext algebra if gldim 6. Proposition 3.3. Let be a (D, A)-stacked algebra, with D = A, A > 1 and gldim 6. Then there is no regrading such that the Ext algebra of is Koszul. Proof. For E() to be Koszul we would need a hat-degree such that Ê() is generated in degrees 0 and 1. From Theorem 1.4, we know that E() is generated in degrees 0, 1,, 3. Now, Ext 1 ( 0, 0 ) Ext ( 0, 0 ) = 0 from Theorem.4(iii), so E() cannot be generated in degrees 0, 1 and. So we may assume that Ê0() = Ext 0 ( 0, 0 ) and

12 1 LEADER AND SNASHALL Ê 1 () = Ext 1 ( 0, 0 ) Ext ( 0, 0 ) Ext 3 ( 0, 0 ). From Proposition 1.3 together with D = A, we have Ext 6 ( 0, 0 ) = Ext 3 ( 0, 0 ) Ext 3 ( 0, 0 ) = Ext ( 0, 0 ) Ext ( 0, 0 ) Ext ( 0, 0 ). Since gldim 6, we have that Ext 6 ( 0, 0 ) 0. So we may choose 0 z Ext 6 ( 0, 0 ) with z = x 1 x x 3 = y i y i, for x j Ext ( 0, 0 ) and y i, y i Ext3 ( 0, 0 ). But x j, y i, y i Ê 1 () so y i y i (Ê1()) whereas x 1 x x 3 (Ê1()) 3. Thus any definition of a hatdegree would require z Ê() Ê3() which contradicts the definition of a grading. The hypothesis gldim 6 in the previous result is necessary, but to see this we require some Gröbner basis theory. We return to this in Example A review of Gröbner basis theory We give a brief introduction to Gröbner bases following [3], [4] and [6]. Let Γ be a finite quiver, and let B be the basis of the path algebra KΓ which consists of all paths in KΓ. We remark that B is a multiplicative basis of KΓ, that is, if p, q B then either pq B or pq = 0. An admissible order on B is a well-order > on B that satisfies the following properties: (1) if p, q, r B and p > q then pr > qr if both are not zero and rp > rq if both are not zero; () if p, q, r B and p = qr then p q and p r. From [4], the left length lexicographic order is an admissible order and is defined as follows. Arbitrarily order the vertices and arrows such that every vertex is less than every arrow. For paths of length greater than 1, if p = α 1 α α n and q = β 1 β β m where the α i and β i are arrows and p, q B, then p > q if n > m or, if n = m, then there is some 1 i n with α j = β j for j < i and α i > β i. Let KΓ be a path algebra and let B be the basis of all paths, with admissible order >. Let x be an element of KΓ, so x is a linear combination of paths p 1,..., p n in B (all with non-zero coefficients). Then Tip(x) is the largest p i occurring in x with respect to the ordering >. We let CTip(x) denote the coefficient of Tip(x). The paths p 1,..., p n are called the support of x, denoted Supp(x). If I is an ideal in KΓ then Tip(I) is the set of paths that occur as tips of non-zero elements of I. We let Nontip(I) be the set of finite paths in KΓ that are not in Tip(I). A Gröbner basis for I is a non-empty subset G of I such that the tip of each non-zero element of I is divisible by the tip of some element in G. Now let a be a non-zero element of KΓ. A simple (algebra) reduction for a is determined by a 4-tuple (λ, u, f, v) where λ K\{0}, f KΓ\{0} and u, v B, satisfying: (1) u Tip(f)v Supp(a); () u Tip(f)v / Supp(a λufv). We say that a reduces over f to a λufv and write a f a λufv. More generally, a reduces to a over a set X = {f 1,..., f n }, denoted a X a, if there is a finite sequence of

13 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS 13 reductions so that a reduces to a 1 over f 1, a i reduces to a i+1 over f i+1 for i = 1,..., n, and a n 1 reduces to a over f n. For a, b B, we say that a divides b if there exist u, v B such that b = uav. Let h 1, h KΓ and suppose there are elements p, q B such that: (1) Tip(h 1 )p = q Tip(h ); () Tip(h 1 ) does not divide q and Tip(h ) does not divide p. Then the overlap difference of h 1 and h by p, q is defined to be o(h 1, h, p, q) = (1/ CTip(h 1 ))h 1 p (1/ CTip(h ))qh. It is clear that the overlap difference of two elements in B is always zero. We are now ready to state the main results we will need in Sections 5 and 6. We keep the above notation. Theorem 4.1. [3, Theorem 13] Let KΓ be a path algebra and let H = {h j : j J } be a set of non-zero uniform elements in KΓ which generates the ideal I. Assume that the following conditions hold, for all i, j J : (i) CTip(h j ) = 1; (ii) h i does not reduce over h j if i j; (iii) every overlap difference for two (not necessarily distinct) members of H always reduces to zero over H. Then H is a reduced Gröbner basis of I. Theorem 4.. [6, Theorem 3] Let KΓ be a path algebra, let I be a quadratic ideal of KΓ and set A = KΓ/I. Then: (i) The reduced Gröbner basis of I consists of homogeneous elements. (ii) If the reduced Gröbner basis of I consists of quadratic elements then A is a Koszul algebra. 5. Koszulity of the Ext algebra of a (D, A)-stacked monomial algebra In this section we show that there is a regrading so that the Ext algebra of a (D, A)- stacked monomial algebra is a Koszul algebra, providing we have D A when A > 1. Let be a (D, A)-stacked monomial algebra. In the case where D = then is a Koszul algebra, in which case it is known that the Ext algebra is again Koszul, so no regrading of the Ext algebra is required. Moreover, the structure of E() by quiver and relations was given in [10]. If D > and A = 1 then is a D-Koszul algebra, and it was shown in [8, Theorem 7.1] that there is a regrading of the Ext algebra under which the regraded Ext algebra is a Koszul algebra. Thus we may assume that is a (D, A)-stacked monomial algebra with D >, A > 1 and D A; we use the regrading of Section 3. We note that if gldim 4, then [11, Proposition 3.3(3)] shows that A D. Thus D A + 1 (for otherwise D = da so that da = A + 1, which cannot occur since A > 1). So the conditions of Definition 3.1 are satisfied and we can consider the hat-degree grading on E(). On the other hand, if

14 14 LEADER AND SNASHALL gldim < 4 then E() = 3 i=0 Exti 0( 0, 0 ) and the definitions of Ê() 0 and Ê() 1 in Definition 3.1 trivially give a grading in this case. The Ext algebra of a monomial algebra was described by Green and Zacharia in [14, Theorem B], using the concept of overlaps from [5]. We start by recalling their work before turning our attention to the case of (D, A)-stacked monomial algebras. (The reader may also see [1].) Let = KQ/I be a monomial algebra. Fix a set R which is a minimal generating set of paths for I. Recall that a path p is a prefix of a path q if there is some path p such that q = pp. Definition 5.1. (1) A path q overlaps a path p with overlap pu if there are paths u and v such that pu = vq and 1 l(u) < l(q). We may illustrate the definition with the following diagram. v p q Note that we allow l(v) = 0 here. () A path q properly overlaps a path p with overlap pu if q overlaps p and l(v) 1. (3) A path p has no overlaps with a path q if p does not properly overlap q and q does not properly overlap p. We use these overlaps to define sets R n recursively. Let R 0 = the set of vertices of KQ, R 1 = the set of arrows of KQ, R = the minimal generating set of paths for I. For n 3, we say R R maximally overlaps R n 1 R n 1 with overlap R n = R n 1 u if (1) R n 1 = R n p for some path p; () R overlaps p with overlap pu; (3) no proper prefix of pu is an overlap of an element of R with p. We may also say that R n is a maximal overlap of R R with R n 1 R n 1. We let R n denote the set of all overlaps R n formed in this way. The construction of the paths in R n may be illustrated with the following diagram of R n. R n R n 1 Recall from [5] that if R n 1 p = Rn q, for Rn 1, Rn Rn and paths p, q, then R n 1 = Rn and p = q. These sets R n were used in [5] to construct a minimal graded projective resolution (P n, d n ) of 0 as a right -module as follows. Define P n = R n R n t(rn ) for all n 0. Let d 0 : P 0 0 be the canonical surjection and, for n 1, define -homomorphisms d n : P n P n 1 by t(r n ) t(r n 1 )p where R n = R n 1 p and t(r n 1 )p is in the component of P n 1 corresponding to t(r n 1 ). p u R u

15 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS 15 We identify the set R n with a basis f n for Ext n ( 0, 0 ) in the following way. List the elements of R n as R1 n,..., Rn r for some r. Let n 0 and define fi n to be the - homomorphism P n 0 given by { fi n : t(rj n t(r n ) i ) + r if i = j 0 otherwise. We use right modules throughout this paper, together with the standard convention that we compose module homomorphisms from right to left. Thus the composition f g means we apply g first then f. Recall that we write paths in a quiver from left to right. So, if fi n corresponds to the path Ri n R n and if Ri n = eri ne, where e = s(ri n) and e = t(ri n) are in R 0, then fi n = fe 0 f n i f e 0 where fe 0 (respectively, fe 0 ) denotes the element of f 0 that corresponds to e (respectively, e ). With this notation, we have from [14] that fj mf i n 0 in E() if and only if Ri nrm j = R n+m k R n+m for some k and fj mf i n = f n+m k. Now let = KQ/I be a (D, A)-stacked monomial algebra with D >, A > 1 and D A. The set f 0 is a basis of Ê() 0 and the set f 1 f f 3 is a basis of Ê() 1. From Theorem 1.4, we have that E() is generated in degrees 0, 1, and 3, and, from Theorem.4, we know Ext 1 ( 0, 0 ) Ext 1 ( 0, 0 ) = 0 and Ext 1 ( 0, 0 ) Ext ( 0, 0 ) = 0 = Ext ( 0, 0 ) Ext 1 ( 0, 0 ) = 0. Thus, the set f 0 f 1 f f 3 is a minimal generating set for E() as a K-algebra. Let be the quiver with vertex set 0 = f 0 and arrow set 1 = f 1 f f 3. Following [14, Definition.1], let I be the ideal of K generated by elements of the form f n 1 f n 1 f nr i r f nr i r f m 1 where the path R nr i r R n 1 f ms i s is not in R n 1+ +n r and by elements of the form = R ms i s R m 1 and where, as paths in KQ, we have R nr i r R n 1 n n r = m m s. The Ext algebra of our (D, A)-stacked monomial algebra can now be described by quiver and relations. Theorem 5.. [14, Theorem B] Let = KQ/I be a (D, A)-stacked monomial algebra with D >, A > 1 and D A. Keeping the above notation, we have Ê() = K /I. We now describe a minimal generating set H for I. The aim is to show in Proposition 5.6, that it is also a quadratic reduced Gröbner basis of I, and hence use Theorem 4. to show that K /I is a Koszul algebra. Note that since is a (D, A)-stacked monomial algebra, each element of R n has length δ(n) where δ is as in Definition 1.1. We start with the following proposition. Proposition 5.3. Let = KQ/I be a (D, A)-stacked monomial algebra with D >, A > 1 and D A. (i) If R i R3 j R5 for some i, j, then R i R3 j = R3 k R l for some R l R, R 3 k R3. (ii) If R 3 i R j R5 for some i, j, then R 3 i R j = R k R3 l for some R k R, R 3 l R 3. Proof. We prove the first statement and leave the second to the reader. Without loss of generality, we may assume that R 1 R3 1 R5. Then R 1 R3 1 = R5 k R5 for some k. The element R 3 1 is a maximal overlap of R R with R 3 R so that we may illustrate R 3 1

16 16 LEADER AND SNASHALL as follows: v R R3 From the construction of Rk 5 R5 via overlaps, there is some Ri 3 R3 which is a prefix of Rk 5, so we may write R5 k = R3 i p for some path p. Now, R5 k has length D + A and R3 i has length D + A, so p has length D. However, Rk 5 = R 1 R3 1 = R 1 vr as paths in KQ and R has length D, so we have that p = R and thus R5 k = R3 i R as required. Note that we may illustrate the element Rk 5 as follows: u R 3 i v R u R 1 R 3 For ease of notation, we now use a, b, c to denote the elements of f 1, f, f 3 respectively. It follows from Proposition 5.3, for b f and c f 3, that if bc 0 in Ext 5 ( 0, 0 ) then there is some b f and c f 3 such that bc = c b in Ext 5 ( 0, 0 ). Similarly, if cb 0 in Ext 5 ( 0, 0 ) then there is some b f and c f 3 such that cb = b c in Ext 5 ( 0, 0 ). Definition 5.4. Let = KQ/I be a (D, A)-stacked monomial algebra with D >, A > 1 and D A. Let K be as above. Let H be the subset of K containing all elements of the form aa, ab, ba, ac, ca, cc together with bb if bb = 0 in Ext 4 ( 0, 0 ) bc if bc = 0 in Ext 5 ( 0, 0 ) cb if cb = 0 in Ext 5 ( 0, 0 ) bc c b if bc 0 and where bc = c b in Ext 5 ( 0, 0 ) for all a, a f 1, b, b f and c, c f 3. Proposition 5.5. Let = KQ/I be a (D, A)-stacked monomial algebra with D >, A > 1 and D A. Keeping the above notation, the set H is a minimal generating set for I. Proof. We have D >, A > 1 and D A and, from the discussion at the start of this section, we may also assume that D A + 1. So, from Theorem.4, aa, ab, ba, ac, ca, cc are in I for all a, a f 1, b, b f and c, c f 3. Thus the ideal H is contained in I. The next step is to give two expressions for an arbitrary element η of f n where n. We may write η as a finite product of monomials in the generators f 1 f f 3. Moreover, η is a non-zero element in Ext n ( 0, 0 ) so η H. For ease of notation, let b denote an arbitrary element of f so that b r denotes a product of r elements in f. Since aa, ab, ba, ac, ca, cc H, we have that η = b t 1 c ε 1 b t c ε b tr c εr for some t j 0 and ε j {0, 1}. Suppose ε 1 = 1 and t 1. Then cb 0 in Ext 5 ( 0, 0 ) so, by Proposition 5.3, we

17 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS 17 may write cb = b c for some b f, c f 3. Thus η = b t 1 (b c h)b t 1 c ε b tr c εr where h = b c cb H. Thus η = b t1+1 c b t 1 c ε b tr c εr + h for some h H. Continued use of Proposition 5.3 and that cc H, shows that we may write η = b t c ε + h where t = t t r, ε {0, 1} and h H. Similarly, we may write η in the form c ε b t + h where ε {0, 1} and h H. To show the reverse inclusion that I H, we suppose first that f n 1 f nr i r is in I. Without loss of generality, we may assume that f n 1 f n i I with f n 1, f n i both non-zero in E(). Thus R n 1 R n 1, R n i R n but the path R n i R n 1 is not in R n 1+n. We consider the case where n 1, n ; the other cases are straightforward and are left to the reader. From the argument above, we may write f n 1 = b t 1 c ε 1 + h 1 and f n i = c ε b t + h, where ε 1, ε {0, 1} and h 1, h H. Then f n 1 f n i = b t 1 c ε 1 c ε b t + h for some h H. Note that the path corresponding to b t 1 c ε 1 is in R n 1 so that n 1 = t 1 + 3ε 1, the path corresponding to c ε b t is in R n so that n = t + 3ε, but the path corresponding to b t 1 c ε 1 c ε b t is not in R n 1+n. We now show that b t 1 c ε 1 c ε b t H. If ε 1 = ε = 1 then c H so b t 1 c ε 1 c ε b t H. If ε 1 = ε = 0 then b t 1 = b 1 b t1 corresponds to the path r t1 r 1 R t 1 and b t = b 1 b t corresponds to the path r t r 1 Rt where b i (respectively, b i ) is represented by the path r i (respectively, r i ) in R but r t r 1 r t 1 r 1 R (t 1+t ). The path r t r 1 is represented by the overlap sequence s 1 s t 1 u r t r t 1 r 1 and the path r t1 r 1 by the overlap sequence v r t1 s 1 r t1 1 s t1 1 r 1 where l(u) = D A, l(v) = A, and s i, s i R. We claim that r 1 r t 1 R 4. For otherwise, there is an overlap sequence representing r 1 r t 1 R 4 of the form s r 1 r t1 for some s R. But then s r 1 represents an element in R 3 and is thus of length D + A. Hence s = uv as paths in KQ, and so r t r 1 r t 1 r 1 R (t 1+t ), a contradiction. Thus, r 1 r t 1 R 4 and we have that b t1 b 1 = 0 in Ext4 ( 0, 0 ). So b t1 b 1 H and b t 1 b t = b 1 b t1 b 1 b t H. If ε 1 = 1, ε = 0, then, from the above discussion, we may write b t 1 c in the form cb t 1 + h where h H, and we need to show that cb t 1 b t H. Keeping the above notation, cb t 1 = cb 1 b t1 corresponds to the path r t1 r 1 R 3 R t1+3, b t = b 1 b t corresponds

18 18 LEADER AND SNASHALL to the path r t r 1 Rt but r t r 1 r t 1 r 1 R 3 R (t 1+t )+3, where c is represented by R 3 R 3. The path r t r 1 is represented by the overlap sequence s 1 s t 1 u r t r t 1 r 1 and the path r t1 r 1 R 3 by the overlap sequence v r t1 s 1 r t1 1 s t1 1 r 1 s t1 r s R 3 where l(u) = D A, l(v) = A, and r, s, s i, s i R. Using the same arguments as in the case ε 1 = ε = 0, we see that r 1 r t 1 R 4. Thus b t1 b 1 H and cb t 1 b t = cb 1 b t1 b 1 b t H. If ε 1 = 0, ε = 1 then, from the above discussion, we may write cb t in the form b t c + h where h H, and we need to show that b t 1 b t c H. Keeping the above notation, b t 1 = b 1 b t1 corresponds to the path r t1 r 1 R t 1, b t c = b 1 b t c corresponds to the path R 3 r t r 1 Rt+3 but R 3 r t r 1 r t 1 r 1 R (t 1+t )+3, where c is represented by R 3 R 3. The path R 3 r t r 1 is represented by the overlap sequence s R 3 r s 1 r t s t r 1 u and the path r t1 r 1 by the overlap sequence v r t1 s 1 r t1 1 s t1 1 r 1 where l(u) = l(v) = A, and r, s, s i, s i R. We claim that r 1 r t 1 R 4. For otherwise, there is an overlap sequence representing r 1 r t 1 R 4 of the form r 1 s w r t1 for some s R and with r t1 = vw as paths. But then r t1 overlaps s with overlap s w of length D A. By assumption, D A, so, as every element of R 3 is of length D + A, this is not a maximal overlap. Thus there is s R 3 and a maximal overlap of r t1 with s as follows: u s r t1

19 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS 19 Hence, the path R 3 r t r 1 r t 1 is in R (t +1)+3 and is represented by the maximal overlap sequence s R 3 r s 1 r t Now, we may continue inductively and show that R 3 r t r 1 r t 1 r 1 is in R (t +t 1 )+3, which is a contradiction. Hence r 1 r t 1 R 4. Thus b t1 b 1 H and b t 1 b t c = b 1 b t1 b 1 b t c H. This shows that, in all cases, f n 1 f n i H. Now suppose that f n 1 f nr i r f m 1 f ms i s is in I. Then, as paths in KQ, we have R nr i r R n 1 = R ms i s R m 1 and n n r = m m s. Set n = n n r. Clearly we may assume that neither f n 1 f nr i r nor f m 1 f ms i s is in I. So, we may write f n 1 f nr i r = b t 1 c ε 1 + h 1 and f m 1 f ms i s = b t c ε + h, where ε 1, ε {0, 1} and h 1, h H. However, as paths in KQ, the element of R n corresponding to b t 1 c ε 1 is equal to R nr i r R n 1, and the element of R n corresponding to b t c ε is equal to R ms i s R m 1. Since these paths are equal, it follows that b t 1 c ε 1 = b t c ε. Thus f n 1 f nr i r f m 1 f ms i s = h 1 h H. Thus all the generators of I lie in H, and we have shown that I = H. It is now clear that H is a minimal generating set for I. To show that H is a reduced Gröbner basis for I, we need an admissible order. Let B be the basis of K which consists of all paths. Label the elements of each set f 0, f 1, f and f 3 as follows: f 0 = {e 1, e,..., e m } f 1 = {a 1, a,..., a r } f = {b 1, b,..., b s } f 3 = {c 1, c,..., c t }. Order the vertices and arrows of by a 1 > a > > a r > b 1 > b > > b s > c 1 > c > > c t > e 1 > e > > e m. Let > be the left length lexicographic order on B as given in Section 4, so that > is an admissible order. Proposition 5.6. Let = KQ/I be a (D, A)-stacked monomial algebra with D >, A > 1 and D A. Keeping the above notation, the set H is a quadratic reduced Gröbner basis of I. Proof. For ease of notation within the proof, write H for H. From Proposition 5.5, the set H generates the ideal I, and it clearly consists of uniform quadratic elements of K. We apply Theorem 4.1 to show that H is a reduced Gröbner basis of I. By inspection, CTip(h) = 1 for all h H so (i) holds. To show (ii), let h, h be distinct elements of H. Since both h and h have length, if h reduces over h then Tip(h ) Supp(h). But, by inspection of H we see this is never the case. Thus h does not reduce over h, and (ii) holds. s t r 1 s r t1

20 0 LEADER AND SNASHALL For (iii), we consider the overlap difference for two (not necessarily distinct) members of H. By inspection of H, it is clear that in order to have an overlap difference of elements h, h H, then at least one of h, h is a monomial. Moreover, we note that the overlap difference of two monomial elements always reduces to zero. Thus we only need to consider the overlap difference of one monomial element and one non-monomial element. We have the following six cases. First, suppose h = bc c b. If h = ca, then o(h, h, a, b) = c b a H 0 since b a H. For h = c b we have o(h, h, b, b) = c b b. We show that b b H. Let r, r, r in R represent b, b, b respectively, and let R, R in R 3 represent c, c respectively. Then, bc corresponds to the path Rr R 5, c b corresponds to the path r R R 5 with Rr = r R as paths in KQ, and rr R 5. We can represent Rr and r R by the overlap sequence R u r s 1 with r 1, s 1 R and l(u) = A. If b b H so that rr R 4, then rr is represented by the maximal overlap sequence r s for some s R. Then it is immediate that rr is represented by the maximal overlap sequence r s u and so rr R 5 ; this is a contradiction. Hence b b H and o(h, h, b, b) = c b b H 0. For h = c c we have o(h, h, c, b) = c b c. If b c H then c b c H 0. Otherwise, if b c H then, from Proposition 5.3(ii), we have that b c ĉˆb H for some ˆb f, ĉ f 3. Thus c b c H c ĉˆb. But c ĉ H and thus c b c H 0. Hence, on both occasions, o(h, h, c, b) H 0. The remaining cases are where h = bc c b and h {ab, bb, cb}; these are similar to the above and are left to the reader. Thus (iii) holds and H is a quadratic reduced Gröbner basis of I. The next result now follows immediately from Theorem 4.(ii). Theorem 5.7. Let = KQ/I be a (D, A)-stacked monomial algebra with D >, A > 1 and D A. Let Ê() be the Ext algebra of with the hat-degree grading. Then Ê() is a Koszul algebra. Following the discussion at the beginning of this section and using [8, Theorem 7.1], we have the following theorem. u r 1 r R r R r s 1

21 THE EXT ALGEBRA AND A NEW GENERALISATION OF D-KOSZUL ALGEBRAS 1 Theorem 5.8. Let = KQ/I be a (D, A)-stacked monomial algebra with D, A 1. Assume that D A when A > 1. Then there is a regrading of the Ext algebra of so that the regraded Ext algebra is a Koszul algebra. We conclude this section by showing the necessity of the condition gldim 6 in Proposition 3.3. Example 5.9. Let = KQ/I where Q is the quiver 1 α 1 α 3 10 α and I = α 1 α α 3 α 4, α 3 α 4 α 5 α 6, α 5 α 6 α 7 α 8, α 7 α 8 α 9 α 10. Then is a monomial algebra, so using [11], we have that is a (4, )-stacked monomial algebra. Moreover, has global dimension 5. The set R is the minimal generating set for I above, and the set R 3 is Let R 3 = {α 1 α α 3 α 4 α 5 α 6, α 3 α 4 α 5 α 6 α 7 α 8, α 5 α 6 α 7 α 8 α 9 α 10 }. Ê() 0 = Ext 0 ( 0, 0 ) Ê() 1 = Ext 1 ( 0, 0 ) Ext ( 0, 0 ) Ext 3 ( 0, 0 ) Ê() = Ext 4 ( 0, 0 ) Ext 5 ( 0, 0 ). Then the Ext algebra of is Ê() = Ê() 0 Ê() 1 Ê(). Using Theorem.4 and that Ext 6 ( 0, 0 ) = 0, it follows that Ê1() Ê1() = Ê(). Thus the Ext algebra is a graded algebra with the hat-degree. It may be easily verified that Ê() = K /I where is the quiver b 1 b b 3 b a 1 a a 3 a 4 a 5 a 6 a 7 a 8 a 9 a 10 and I is generated by: c 1 c a i a i 1 for i =,..., 10 a 5 b 1, a 7 b, a 9 b 3, b a, b 3 a 4, b 4 a 6, a 7 c 1, a 9 c, c a, c 3 a 4 b 4 c 1 c 3 b 1 where the a i, b i, c i correspond to elements in Ext 1 ( 0, 0 ), Ext ( 0, 0 ), Ext 3 ( 0, 0 ) respectively. Let H be the set of generators for I given above. Order the vertices and arrows of by a 1 > > a 10 > b 1 > > b 4 > c 1 > c > c 3 > e 1 > > e 11. Let > be the left length lexicographic order on B as given in Section 4, so that > is an admissible order. There are no overlaps of b 4 c 1 c 3 b 1 with any element of H; so the only overlaps of elements of H are of monomials in H, and these immediately reduce to zero. It follows from Theorem 4.1 that H is a reduced Gröbner basis for I. From Theorem 4., we then have that Ê() is a Koszul algebra with this regrading. In the final section we consider the non-monomial case. c 3

22 LEADER AND SNASHALL 6. An example of a non-monomial (D, A)-stacked algebra We conclude this paper with an example of a (D, A)-stacked algebra which is not monomial but where the regraded Ext algebra is still Koszul. Our example has D A and is of infinite global dimension. Example 6.1. Let Q be the quiver given by 3 α 3 4 α α α 14 1 α 1 α 7 7 α 8 8 α9 9 α10 10 α 5 α α 6 α 13 α 1 1 α α α 15 α 16 and let = KQ/I where I is generated by the elements g 1 = α 1 α α 3 α 4 α 5 α 6 α 7 α 8 α 9 α 10 α 11 α 1, g = α 3 α 4 α 5 α 6 α 13 α 14, g 3 = α 5 α 6 α 13 α 14 α 15 α 16, g 4 = α 9 α 10 α 11 α 1 α 13 α 14, g 5 = α 11 α 1 α 13 α 14 α 15 α 16, g 6 = α 13 α 14 α 15 α 16 α 17 α 18, g 7 = α 15 α 16 α 17 α 18 α 13 α 14, g 8 = α 17 α 18 α 13 α 14 α 15 α 16. The first step is to show that is a (6, )-stacked algebra. We start by finding a minimal projective resolution of 0 using the approach of [13]. We define sets g n in KQ for each n 0 inductively as follows: g 0 = {gi 0 = e i for i = 1,..., 17}, the set of vertices of Q. g 1 = {g = α i for i = 1,..., 18}, the set of arrows of Q. g = {g1,..., g 8 }, the minimal generating set for I as described above. For n = r + 1 with r 1, g n = {g1 n = gn 1 1 α 13 α 14, g n = gn 1 α 15 α 16, g3 n = gn 1 3 α 17 α 18, g4 n = gn 1 4 α 15 α 16, g5 n = gn 1 5 α 17 α 18, g6 n = gn 1 6 α 13 α 14, g7 n = gn 1 7 α 15 α 16, g8 n = gn 1 8 α 17 α 18 }. For n = r with r, g n = {g1 n = gn 1 1 α 15 α 16 α 17 α 18, g n = gn 1 α 17 α 18 α 13 α 14, g3 n = gn 1 3 α 13 α 14 α 15 α 16, g4 n = gn 1 4 α 17 α 18 α 13 α 14, g5 n = gn 1 5 α 13 α 14 α 15 α 16, g6 n = gn 1 6 α 15 α 16 α 17 α 18, g7 n = gn 1 7 α 17 α 18 α 13 α 14, g8 n = gn 1 1 α 13 α 14 α 15 α 16 }. Note that we may write g1 = g1 1 α α 3 α 4 α 5 α 6 g7 1α 8α 9 α 10 α 11 α 1. It follows that each x g n may be written as x = i gn 1 i r i for some r i KQ, and all n 1. Let P n be the projective -module defined by P n = i t(gn i ). Let d0 : P 0 /r be the canonical surjection and, for n 1, define -homomorphisms d n : P n P n 1 by t(gi n ) t(gn 1 i )r i where t(gi n 1 )r i is in the component of P n 1 corresponding to t(gi n 1 ). It may be checked