on the hochschild cohomology of truncated cycle algebras

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On the hochschild cohomology of truncated cyclealgebrasM.J. Bardzell a , Ana Claudia Locateli b & Eduardo N. Marcos ba Salisbury State University , Mariland, USA E-mail:b IME-USP , Caixa Postal 66281 (Ag. Cidade de São Paulo), São Paulo, SP, CEP 05315-970,BrazilPublished online: 27 Jun 2007.

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COMMUNICATIONS IN ALGEBRA, 28(3), 1615-1639 (2000)

ON THE HOCHSCHILD COHOMOLOGY OF TRUNCATED CYCLE

ALGEBRAS

hl. 3 . Bardzell

Salisbury State University, Alariland, USA,

email: [email protected]

Ana Claudia Locateli and Eduardo N. Marcos

IhIE-USP,

Caixa Postal 66281 (Ag. Cidade de S&o P a d o ) , CEP 05315-970 - SBo Paulo, SP,

Brazil

emails: [email protected], [email protected],

ABSTRACT: The purpose of this paper is to study the tlochschild col~omology ring I I e ( A )

of algebras of the form 11 = ~ z , / J " , where Z, is an oriented cycle with e vertices and J is the

ideal generated by the arrows, N 2 2. IVe provide a new description of the Yoneda product in

H O ( . l ) and prove that this is a finitely generated infinite dimensional ring. In addition we show

that algebras of the form A = k Z , / J N are not derived equivalent unless they are isomorphic.

Let k be a field and let ii be a finite dimensional k algebra. The nth Hochscliild

cohomology group of A is Hn(A) = E ~ t ; ~ ( i i , A). Here A' = h o p 81, A since we will

be considering right he - modules. The Hochschild cohomology ring is then defined

1615

Copyright Q 2888 by Marcei Dekker, inc.

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1616 BARDZELL, LOCATELI, AND MARCOS

to be H0(.i) = LI Hn(A) where multiplication is given by the Yoneda product (see n>O

[hl]). For most finite dimensional algebras little is known about the cohomology

groups and even less is known about the cohomology ring. The interested reader

should refer to [Ha] for an overview of Hochschild cohomology. In this paper we

study a class of algebras where we can compute all of the cohomology groups and

describe the cohomology ring using generators and relations. These algebras are

truncated quotients of path algebras where the underlying graph is a single oriented

cycle. Under these hypotheses, the Hochschild coho~nology ring is finitely generated

regardless of the field characteristic. These algebras, which are also studied in

[EH] using a different approach, are basic self-i~ijective Nakayanla algchras. Our

methods involve a new description of the Yoneda product based on the rnilii~nal :Ie-

resolution of the algebra A. We also provide examples of non-truncated algebras,

i.e. Nakayama but not self-injective, where the cohomology rings are not finitely

generated.

Let 2, denote the quiver with e vertices { v l , ..., u,) and e arrows { a l , ... a,) such

that the origin vertex o(a,) of the arrow ai equals the terminus vertex t (a i - l ) of a ,

(mod e).

Throughout this paper yi will denote a path of length J beginning at the vertex

ul. If the quiver is 2, then this path is unique. If A and B are sets of paths, then

( A / / B ) = { ( p , q) E -4 x B : o ( p ) = o(q) and t ( p ) = t ( q ) ) . If A is the set of all paths

of length j then we write ( A N B ) = ( j / / B ) . If p and q are paths, then the notation

(p, q) means that o(p) = o(q) and t ( p ) = t ( q ) . Throughout this paper J will always

denote the two-sided ideal generated by the arrows. Finally, if A is a set of linearly

independent vectors, then LA will denote the subspace spanned by A.

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HOCHSCHILD COHOMOLOGY 1617

In section 2 we discuss the cohomology groups for A = k Z e / J N . To do this

we introduce a group action on the terms P,' of a coboundary complex. This

complex is the dual complex of the minimal resolution of the algebra '1 over its

enveloping algebra l i e , i.e. it is the complex obtained by applying the functor

Hcnr~,~. (, .i). Here the group is the cyclic group of order e. In section 3 a new

product on P,' = LI P,' is described. This turns out to be a well-defined product n>O

on H 9 ( A ) . The fourth section explains why this product coi~icides with the Yoneda

product. In section 5 we provide the ring structure for H * ( ~ z , / J " ) . This includes

the cohomology ring of Brauer Tree Algebras as a special case. We show that

~ ' ( k Z , l J") is always finitely generated and provide a presentation of generators

and relations. In section 6 we apply our results to show that algebras of the form

k Z , / J N are not derived equivalent unless they are isomorphic. 111 the last section

we consider two examples of the form k Z , / I where I # J N . In other words, we

study two basic Nakayama algebras that are not self-injective. In both examples the

cohomology ring is infinite dimensional but the Yoneda product is zero in non-zero

degree. Hence the cohomology rings are not finitely generated.

Let A be a finite dimensional mononlial algebra. Then Hz(A) is the i t h colio-

mology group of the coboundary complex

where P; = LI , o ( p ) h t ( p ) . Here A P ( 0 ) = ( v l , ..., v,) , ilP(1) = { a l , ..., a , ) , and PE*P(I )

A P ( 2 ) is the minimal set of all paths that generate I. The AP( j ) sets for j > 2 are

sets of "longer paths" constructed inductively using the A P ( 2 ) paths. For a more

detailed description of these paths see [B] and [GHZ]. The 4; maps are described

in [Bh,I].

Now let us focus on the special case where A = ~ z , / J ~ where N > 2. In this

section we compute the kernels and images of these boundary maps t o provide a

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k-basis for the cohomology groups. For each 1 > 0 write P: = ,LI k( j // i lP(1)) (see J > O

[LO] ). Next let Ce =< g : ge = 1 > be the cyclic group of order e generated by g.

Then each nonzero k( j / / . IP( l ) ) can naturally be considered as a free kce-module

of rank 1 and the maps 4; are Ce- module maps. The Ce action is defined as

follows. Given ( y , p ) E k ( j / / A P ( l ) ) , define g . ( y , p ) to he the ordered pair (y ' ,p ') ,

where y' and p' are the clockwise rotates of y and p by one arrow. In other words,

if p = aiai+l ... am (mod e ) , then p' = ai+l...a,,+l (mod e). Note that each ( y , p )

generates the module P;. N-1

Given i 2 0 we know by [Lo] that &+, : P;i + PC,,, where P& = j I l o k ( j / / N i )

N - 1 and PTi+, = ,LI k ( j / / N i + l ) , can be written in the form $;i+l = (Do, D l , ..., D N - I ) .

j=o For each j, with 0 5 j 5 iV - 2, Di : k ( j / / N i ) t k ( j + l / / N i + 1) is defined

by D, ( y , p) = ( a y , ap) - ( y p , pp) where a: and p are the unique arrows such that

t ( a ) = o(y ) = o(p) and o(p ) = t ( 7 ) = t (p) . hloreover DN-1 = 0. If we choose

yFi) and ( y i f ' , yFi+') to be generators of k ( j / / N i ) and k ( j + 1/ /Ni + I ) ,

respectively, as Ce-modules, then we can describe the map D j as the Ce- module

map given by Dj(-(6, # I ) = ( g - I - l)(?i+', yFi+l). This means that I m D j is the

augmentation ideal. That is, Im Dj 2 A(kCe) , which has co-dimension 1. It fol- e

lows that ker D j is one-dimensional and we can take { C (7; , yr')) as a k-basis for 1=1

ker D, .

Now, write N = me + t , 0 5 t < e. Let us desclibe the map $;, for i > 6. By N - 1 N - 1

[Lo] we know that we can write $I, : LI k ( j / / N ( i - 1) + 1) t LI k ( j / / N i ) as J=O ]=I

where Do : k(O//N(i - 1) + 1 ) ---+ k ( N - l / / N i ) is given by

is clearly a kce-map and we can write D ~ ( Z J O , y:(l-l ' fl ) = 7 1 1 ~ ( $ - 1 , f 2 ) + 1=1

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t (Yjhj-l,Yfia). We should point out here that gcd(N, e) = 1. Otherwise

1=1

(0//N(i - 1) + 1) = 0 and Do = 0.

Next let us compute the kernel and image of Do. Let el = m ($-l, Ty2)) + 2 ($-1,7r i) . Then cl and gcl are both in Im Do and 1=1

Since gcd(N,e) = 1,we also have gcd(t,e) = 1. This means gt generates C, and,

consequently, l-gt generates A(kCe). Once we have that k(N-l//Nz) is isomorphic

to kc , , we can conclude that Im Do contains a copy of A(kCe).

Let E be the augmentation homomorphism. Then &(el) = N and, if char k { N ,

then Im Do properly contains A(kCe). In this case A(kCe) must be k(N - l / /Ni) .

(Remember that A(kC,) has co-dimension 1). If char k N then In1 Do = A(kCe)

since cl E A(kCe). So if char k { N then Do is injective. If char k I N and

N i = N - 1 (mod e), then ker Do is one dimensional.

Let us now summarize the formulas we obtain from the discussion above and

[Lo]. Let N = me + t where 0 5 t 5 e - 1. Then

m + e if t = l m i f t = 0 , 1

if t = 0 and dim H 1 (11) = m + 1 otherwise

m + 1 otherwise

Given i 2 1, let Ci(N) = # { j ( 0 5 j < N - 2, j 5 t i (mod e)]. If char

k { N or N i $ N - 1 (mod e), then dim H2%(A) = dim H2'+'(.\) = Ci(N). If char

k I N and N i = N - 1 (mod e) , then dim H"(A) = dim H2i-1(A) = Ci(N) + 1. In

the last section we will provide alternate fornlulas for these dimensions. Formulas

for the characteristic zero case have also been provided recently by Pu Zhang in

[ZI.

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In this section we assume A: > 2.

Consider P: = LI P: = LI k ( r / / ; IP( i ) ) . We define a multiplication on P: as 120 120

follows. If ( y , p ) E ( r / / A P ( i ) ) and ( a , q) E (r/ / .JP(l)) with i , 1 2 0, let

( 7 , ~ ) v ( ~ 1 4 = otherwise

Now that the product is defined on the basis elements, we extend V linearly to a

product on P:. Since the AP(i) sets form a multiplicative basis of the Ext-algebra

E(A) when A is a monomial algebra ( see [GZ ] ), it is easy to see that this product e

is associative. It is also clear that C (e,, e,) is the unity element for this product. 2=1

So we now have a ring structure on P:. We should point out here that this is a

product for any monomial algebra ( see [Lo21 ).

Now we need to show that this product is well-defined on Ha(.i) = LI H i ( A ) . l>0

First we prove that K = LI ker4; is a subring of P,'. Then we will show that ? > I

I m = LI Im4: is a two-sided ideal of K. This will provide a ring structure for 221

H*(A). For the rest of this section we will be considering A = k Z , / J N .

L e m m a 3.1. The prdiluct e,f two e ? e ~ e % t s z odd Jey-ee is z e x .

Proof. See [Lo21 1

Proposition 3.2. K = ,LI ker 4; is a subring of P:. 22 1

Proof. Let x = C a l ( y l , p l ) E ker4: and y = Cbs(as ,qs ) E ker#j* for some i,j 2 1. I S

If i and j are even, then x E PCl and y E P C l . So x V y = 0 E k e r 4 L j p l .

Now suppose i is even and j is odd. If char k 1 N, then 7 , E rad('1) for all 1. So

yla, ~ r a d ( A ) for every 1 , s and it follows that x V y E ker4:+,-,. If char k I .N Dow

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and (O/jAP(i - 1)) # 0, we can write x = xo + xl where xo E k(O//AP(i - 1)) N-1

and xl E LI k(j//..1P(z - 1)) . Since X I E ker4: we must have xo E kerg: also. ~ = l e

So we can suppose that x E k(O//.-IP(i - 1)) and write x = a x (vt,p,) where a is 1=1

a non-zero element of k . If we write y = yo + yl where yo E k(O//.-IP(1 - 1) ) and N-1

y1 E IJ k(j//AP(l - I ) ) , then we have x V y = x V yo + x V yl. Since x V yl E 1=1 N-1 e

LI k(j//.llP(i + j - 2)) we have x V y1 E ker 4:+,-, . Writing yo = C b(ut, q,), b E k , 3=1 i=l

P

then x V yo = Ca.b(vt ,ptqt) E ker Do. So we have x V y E ker q6:+,-, . The case t=l

where i is odd and 1 is even is analogous.

Finally, we need to consider the case where i and 1 are both odd. In this case N-l N-1

ker q6f = LI ker ~j and ker 4; = LI ker Df, so we can assume that x E ker D l J = O 1=0

and y E ker DL for some u. w with 0 5 u, w < N - 1. Thus we can write x =

a(-,;', yi2-1)), y = f: b($, -,ji-l)) where y i denotes the path starting a t vt and t=l t=l length j for 0 5 j < N - 1 and yl3) denotes the path starting a t vt belonging

to AP(s) . Then x V y = cab(-,:+"', yi2+1-" ) . I f u + w > N , t h e n r V y = O ~ t=l

ker@:+,-,. Otherwise, x V y E ker DL?;~. It follows that I( is a subring of P,*. I

Proposition 3.3. Im = ,LI Imq6: is a two-sided ideal of K. 221

Proof. Let 0 # x = E a t (7, ,pt) t im 4; P2' and 0 f y = Zb, (a,, q,) E ker 4; t S

PL, for some i , l 2 1. We will show that x V y and y V x are both in Im. First

suppose that i is odd. We can assume that x E In1 DL for some u, 0 5 u 5 N - 2.

Then the length of y, is greater than or equal to 1 for each t . If 1 is even, we have

X V y = 0 E Im since the product of two elements of odd degree is zero. If 1 is odd we

canassumey E kerDLforsomew,O < w < N - 1 . I f w = N-1, thenxVy = O E In1

since the length of o, is N - 1 for each s and the length of y, is greater than or equal

to 1 for each t . So suppose O 5 w < N - 1. Then we can write y = bC(y,W,

where b g k.We know that there exists z = 5 ~,(-y;,~!-l)) E P;_, such that n=l

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Therefore, x V y E Im C_ I m .

Now suppose that i is even. Then x E k ( N - l / / A P ( i ) ) and there is

z E k(O//AP(i - 1 ) ) such that z = $T(z) . ?Ve can assume that z = ( v j , y j ' - l ) ) and

If 1 is even we have y E k (O/ /AP( l -1) ) since x f k ( N - l / / A P ( i ) ) (as in Proposition

1). If char k t N then y = 0 and x V y = 0 E I m . If char k I N then we can write

e

But cn = t ( m + 1) + ( e - t )m = me + t = AT and, since char k I N , we have n=l

2 cn = 0. Now, i + 1 - I is odd so it follows that z V y E Im$:+,-,. Therefore n=l

If 1 is odd, we let y E ker D: for some 0 5 j 5 N - 1. If j > 1, we have x V y = 0 e

since x E k ( N - l / / .aP( i ) ) . So suppose y E ker Dk and write y = C (v , , &') . s=l

Then we have x V y = f: cn(y;- lr Ttti-i') and since n = l

m + l if j < n < j + t - 1 c;. = {

m otherwise

(z+l-2) we know that z V y = $~+,-,(v,, y, ) . S o z v y E I m .

Using a analogous argument one can show that y V x E I m . We conclude that

Im is a two-sided ideal of K. a

We now have a product on H 9 ( A ) . Note that H i ( A ) V H 1 ( ; i ) C_ Hi+' (A) . In the

next section we will show that V is actually the Yoneda product.

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In this section we will show that the product V just defined coilicictes with the

Yoneda product. Let

be a .Ie-projective resolution and A 1 ~ I b I o d ( , l ~ ) . Apply - $1) = ( - J l )

t o the resolution above t o obtain the complex

ker$, I Then Hn(.I,A1) = ,.,+ . Since P,, = LI Ao(p) 8 t(p)A we can rewrite this p € A P ( n )

complex as follows:

where CL[(mp,P)pEAP(n)] = pmp : LI Ao(p) @ t(p)h --+ A 1 is defined by P E A P ( ~ )

pnr,[(Xpo(p) @ t ( p ) ~ b ) ~ ] = C X ~ ~ ~ X ~ . Using the notation from [Lo] we have P

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1 4 1 , + 1 [ ( l n p r ~ ) p E ~ ~ ( r l ) l = (lqnzq2 - m q l q , q ) g E ~ ~ ( r , + ~ ) and

N

@ ; ~ [ ( m p , ~ ) p 6 A ~ ( n ) ] = ( C '-lqmq,qN-Jl Q ) ~ E A P ( ~ + I ) for 2 2 0. 1=1

Now, given n 2 0 and [f] = E HtL( . i , .I) co~isitler the map

of functors [f] V - : H O ( h , -) + Hn(.i, -) where, for each A l Eniod(Ae), the

It is not hard to see that ([f] V is well-defined and [f] V - is a map between

functors. One can also show that the definition of [f] v - does not depend on the

representation of f .

The universal property of derived functors asserts that there is a unique map of

cohomology functors f V - : Hm(A,-) t Hmt"(h, -) extending from the degree

zero component. Likewise, if we denote the Yoneda product on H*(A, -) by * , then [f] * - : Hm(A, -) ---i H m + n ( ~ I , -) is the unique map of cohomology functors

extending the degree zero component.

We want t o prove that in the case M = A, V coincides with the Yoneda product.

By the universal property of derived functors, it suffices to show that for each [f] E

Hrl(A), the maps [f] V - and [f] * - coincide in degree zero. So take f ker C

Horn,,. (P,,, A) and g E ker 4; E Horn*. (Po, A) where f E [f] and g E [g]. The

Yoneda product is given by the following commutative diagram:

@ 0,-1 P,, 4 Pn-1 ---+ . . . $2. PI 3 Po -+ A --+ 0

L f A

where the yi are the liftings of g. Then f * g = f 0 y, ( see [W] ). Define the maps

yi : Pn -+ Pn on the idempotents by yi : o(p) 8 t(p) ++ o(p) 8 g(t(p) 63 t(p)).

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5 . THE RING STRUCTURE O F H * ( A )

In this section we will provide the ring structure of He(] ! ) for A = k Z e / J N . First

we find the group generators for H i ( A ) and then we determine the ring structure by

providing generators and relations. For another presentation of these rings see [EH].

By Corollary 3.3 and the fact that tlie Yoneda product is graded co~linlutative, we

know that H*( , i ) is graded commutative. Throughout this section we will provide

k- bases of equivalence classes for cohomology groups. To keep the notation simple,

a representative of each equivalence class will be used to denote the class itself. We

begin with the following result.

P r o p o s i t i o n 5.1. If A = ~ z , / J ~ with N > 2, then H O ( A ) has k-basis B =

(?ae, vr); 0 5 a I [v] i f N $ 1 (mod e ) or B U { ($ ' - l , u l ) , 1 5 1 5 e ) if

N e 1 (mod e ) .

Proof. This is a direct consequence of the fact that H O ( l l ) = ker4; and

( N - 1 / / 1 ) = 0 if N f. 1 (mod e ) . I

P r o p o s i t i o n 5.2. If char k I N and N i z N - 1 (mod e) , then

{ 5 ( 7 : . 7 ~ ' ) : 0 5 j 5 N - 2 , j = N - l [=I ( m o d ~ ) } U { ~ ( ~ ~ - ~ , ~ ~ ~ ) ) i = 1

6s a basis of H 2 ' ( A ) for z 2 1. Otherwise, we have that Dow

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j N - 2 , j = N i ( m o d e ) I is a basis of H 2 2 ( ~ Z ) for i 2 1.

Proof. We have already seen that

If char k I N and N i N - 1 (mod e ) , then we have that

Otherwise, if char k ( N , then I m & = k ( N - l / / N i ) . In this way, if char k I N and e

N i r N - 1 (niod e), then it is a straightforward verification that (??- I , 7:') 4 1=1

Im &. Since H 2 ' ( A ) = ker &+, / Im &, the result follows.

Proposition 5.3. If char k I N and N i E N - 1 (mod e ) , then

{ ~ ( 7 ~ i ' , ? r ( ' - l ) + ' ) : 0 < j < N - 2, j + 1 r 0 (mod e ) ) U { e h , ?r('-l)+l 1=1 1=1

is a basis o f H2i-1(! i) for i 2 1. Otherwise we have that

) : 0 5 j 5 N - 2 , j E N ( i - 1 ) (mod e ) 1=1

Proof. We know

N(i-l)+l Im # L 1 = k{(y;+' , yl ) - (?3+1 N(l-l)+l

1 + 1 , ~ 1 + 1 ) : l < l < e - 1 ,

0 I j 5 N - 2, j + 1 = N ( i - 1) (mod e)}

If char k I N and N i I N - 1 (mod e ) , then ker #li = k (5 ( v l , -yy('-')il 1=1

N-1 ( ' l l k ( j / / ~ ( i - 1 + 1) . Otherwise ker& = ,lI k ( j / / N ( i - 1) + 1). It is now

j=1

easy t o verify that , for 1 < j < N - 1, f:(7!, yy('-l'+l ) 4 I m & - l . I 1=1

We are now ready to provide generators and relations for the Hochschild co-

homology algebras. In particular, we show that if A = k Z e / J N , then He( . i ) is

finitely generated.

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e e Theorem 5.4. Suppose that N 5 e and consider x = C (ul, vl), y = C (al, al),

1=1 1=1

and z, = 2 (-/:, 7r'~ ), where i j is the smallest integer greater than zero such that 1=1

Ni, = j (mod e). Then if char k .1. N or char k I N and gcd(N,e) # 1, then H0(A)

is generated by {x, y, z, : 0 5 j < N - 2) subject to the following relations:

U" 0, z?"] = 0 , and zq = z,, if 1 < u j 5 N - 2 and i a j = ai, for 1 < j 5 N - 2 .

Proof. Since N < e all of the cohomology groups are one-dimensional. So x and

y clearly generate HO(:\) and H1(:i), respectively. For i 2 1, Hz(:\) is ge~imited

by 2 (7: , r r ' ) where iVi j (mod e), 0 < j 5 N - 2. For each such j, let i j be 1=1

the sniallest integer greater than zero such that N i j r j (mod e) . Note that if i is

such that N i - 0 (mod e ) and i > i o , then we have i = Xio + i f with 0 5 i' < io.

So N i = NXio + Nil = 0 (mod e ) and since NioX - 0 (mod e ) we have Nit r 0

(mod e). However, io is the smallest integer satisfying this property. It follows

that i' = 0. Thus, i = X i o and e = Xio. Note also that if i is such that hr i j

(mod e ) 1 5 J 5 N - 2 and i > a , , then we can write i = il + i' where a' 2 1. So

N i = Nil + ~ i ' r j (mod e) and since Ni , = j (mod e ) , we have Ni t r 0 (mod e)

and i t is a multiple of io. That is, i = i j + aio.

Now, if H2'(A) # 0, then N i r j (mod e ) for some 0 5 j 5 N - 2. Since H Z i ( h ) e

' Ni ,+Nnio is generated by C ($, yp') = C ($, yl ) = z: V zj and HZi+'(A) # 0 is

1=1 1=1 P

generated by ($+I, yyZ+') = z: V z , V y , we have that {z, y, 2, : 0 5 j 5 N - 2 ) 1=1

generates as a k-algebra.

We know that the product of elements of odd degree is always zero. In particular,

y2 = 0. Since paths of length at least N are zero, we also have z[1"'" = 0. It is easy

t o see that if 0 5 j 5 N - 2, and if a is such that 1 < a j 5 N - 2 with iag = uij,

then we must have ZP = z,j.

Corollary 5.5. If N = e and char k j N , then Ho(A) is isomorphic to the poly-

nomial ring v. Proof. If N = e then io = 1 and Ni r 0 (mod e ) for all i > 1.

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Theorem 5.6. Suppose that N < e , char k I N , and gcd(N,e) = 1 and let I be

the smallest integer such that N ( I - 1) + 1 = 0 (mod e). Consider x, y, and zj e

N(I-1)f 1 for 0 < j 5 N - 2 as in Theorem 5.4. Consider also w = C (ul,yl 1x1

E

If2'-' (.I), and u = 5 ( 7 , ~ - ' , 7 ~ ' ) t HZ'(A). Then Ho(A) is generated by the set 1=1

{ x , y, z,, w , u : 0 < j 5 N - 2) . These generators satisfy the relations described in

Theorem 5.4 and u2 = 0 , u V y = u V z, = 0, 1 5 j 5 N - 2.

Proof. Suppose char k I N . 4 s we have seen in Theorem 5.4, x, y, and zJ, 0 5 j 5

N - 2 , generate the cohomology groups HZi(h) and HZi-l (A) for all i 2 0 such that,

Ni j (mod e). If gcd(N, e ) # 1, there is no integer i such that N i N - 1 (mod

e ) and so Ho(;i) is generated by these elements. If gcd(N, e ) = 1 then there exists

integers i such that N i r N - 1 (mod e) . Since char k I N we have that , for those i,

H2'(A) and HZ'-'(A) are nonzero. By Propositions 5.2 and 5.3, Hz'-'(A) is gener- e N(i-l)+l e

ated by C (vi,-Yl ) and H2i(.A) is generated by 2 (??-I, $ l ) . Let I be the 1=1 1=1

e N(I-l)+l smallest integer such that N I = N - 1 (mod e) . Consider ru = C (ul, y1 1=1

) E

Hz'-'(A) and u = k ( 7 ~ - 1 , ? y ' ) 6 H 2 ' ( I ) . AS we did in the proof of Theorem 1=1

5.1, we note that if Ni - N-1 (mod e ) then i = I+aio for some integer a 2 0. Then

addition; H2'(A) is generated by 2 (7y-1; #') = (-$-I ! ?:'+N~'~) = U V T ~ It 1=1 1=1

follows that {z, y, z J : w , u : 0 5 j < N - 2) generates He(,\) as an algebra. hlore-

over, since every path of length greater than or equal to N is zero, it is easy to see

that these elements must satisfy the relations u2 = 0 and u V y = u V zj = 0 for

1 5 j < N - 2 . 1

Theorem 5.7. Suppose N > e and N f 1 (mod e) . If char k t N or if char k I N

and gcd(N, e ) # 1, then Hg(A) is generated by {xo, x l , y , zr : 0 5 r 5 e - 1) where

each 0 5 r < e - 1, ZT = c(?:, 7Yir) with ir the smallest integer greater than 1=1

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zero such that A'z, T (mod e). Moreover, these generators satzsfy 4" = 0 zf

N - 0 (mod e); x;'" = 0 zf N f 0 (mod e); y2 = 0, zINIr1 = 0, and z: = zbr zf

1 < br < e - 1 and ibr = bi,.

Proof. By Proposition 5.1, HO(A) has k-basis C ( y a e , v l ) : 0 5 a < [y] .Then L1 1 we have that xo and x l generate HO(l l ) . For each i 2 1, H2"A) has k -basis

x (y{ ,Y;v l ) : 0 5 j 5 N - 2, j N i (mod e) . For each r, with 0 5 r < e - 1, {/Il 1 let z, be the smallest integer greater than zero such that Ni , = r (mod e). For each

such r consider z. = 2 ( 5 ; , r p i F ) t H2lv(.l). So if N i - r (mod e ) , i > 0, then 1=1

e e ,Ni,+Na,ro we have i = i, + a,io for some a, > 0. Then x ($,?Yi) = C (yr, , ( ) = 1=1 1=1

z, V 2:". If 0 5 j 5 N - 2 and j r r (mod e) then j = r + b,e for some e .

bj > 0. SO E (r:,rri) = 5 (7:+b1e, y ~ i v + " a F i O ) = z, v 2:' v x:'. ~t follows that 1=1 1=1

XI and z,, 0 < r 5 e - 1, generate H2'(A) for i 2 1. For each i 2 1, HZi-'(A)

e sider y = C (cul,cur) E H 1 (,I). Note that if 0 5 j 5 N - 2 and j 0 (mod e) ,

1=1 then j = bje = b;io for some b; 2 0 (remember that e = Xio). So wc can write

e ' b,1o+l ( I ) = ( 7 , U I ) = y V xi'. We conclude that y and xi generate 1=1 1=1 H 1 (A).

Then 5 ($+I, -yl\'('-')+l ) = (?;.+I, $'tp+Nai0+l ) = z, V z l V y. Since for each j 1=1 1=1

such that 0 < j 5 N - 2, j N( i - 1) - r (mod e) we have j = r + bJe for some

b > 0, 5 ( $ + 1 , y y - l ) + l '+b~e+l, 7 ~ i r + ~ a i o + l ) = v x? 3 -

1=1 = C ( r l

1=1 follows that X I , y , and z,, 0 5 r 5 e - 1, generates Hz-'(A) for every i > 1. We

conclude that {xo,xl, y , z, : 0 5 r 5 e - 1) generates He(A) as an algebra. Since

the paths of length a t least N are zero and the product of elements of odd degree

is zero, it is easy t o check that the generators satisfy the claimed relations. I

Theorem 5.8. If N > e, N $ 1 (mod e), char k ( N, and gcd(N,e) = 1, then

H 0 ( A ) is generated US an algebra by {xo, X I , y , z,, w : 0 5 r < e - 1) where xo, X I , y,

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z , are as i n Theorem 5.7 and w = 5 (u l , 7 ~ ( i " - 1 - ' ) + 1 ) with in-, the snmllest 1=1

integer greater than zero svch that I V L , ~ - ~ = N - 1 (mod e ) . Furthermore, these

generators satisfy the same relations described i n Theorem 5.7.

Proof. The proof is very similar to the proof of the last theorem , except for the

case N i N - 1 (mod e ) . Since gcd(N,e) = 1, there exists integers i such that

N i - N - 1 (mod e). For these i , HZZ(.l) is generated by z~ and z l where

N i r N - 1 R (mod e ) with 0 5 R _< e -- 1, and H2'-'('1) is generated

as a vector space by ) : O ~ j < N - 2 , j + l ~ O (mode)

{ ~ ( u ~ . ~ ~ ( ' - ' ) ~ ~ ) ) . Since j + 1 0 (mod e) inlplies j z e - 1 (mod e), w have I = : 5 ( 7 f , 7 ; ( 4 + l ) = 2,-1 V z,0 V y for some a 2 0. We can generate the first set

I = 1 . - in the union above using re-:, zo, y, and x l . However, we cannot generate the

last generator above from the set {xo, X I , y , zr : 0 5 r 5 e - 1 ) . So consider iNPl

the smallest integer greater than zero such that NiN-l - N - 1 (mod e). Let

'U = f : ( w , 7 ~ ( ~ ~ - 1 - l ) + 1 ). ~h~~ H ~ N - I - ' (A) is generated by 2,-1, zo, y, XI , and 1=1

Now if i is such that N i r N - 1 (mod e ) then i can be written as i = iN-1 + biO

. w V z i . Thus we have the desired generating set for Ha-'(A). I

For our last result we give a ring presentation for the Hochschild cohomology ring

of a Brauer tree algebra. For a definition and examples of Brauer tree algebras, see

[ARS]. In [HI Holm shows that the even cohomology ring of a Brauer tree algebra is

finitely generated. He also gives a ring presentation. One can compute Hochschild

cohomology groups for Brauer tree algebras using truncated cycle algebras. Let

B ( e , m ) = kZe/Jme+' . In [Ri] Rickard shows the following:

Theorem 5.9. (R i ckard ) Let 11 be a Brauer tree algebra for a Brauer tree T with

e edges and multiplicity m. T h e n H 1 ( h ) H i ( ~ ( e , m)) for every i > 0.

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Using this result we obtain the cohomology ring for a Brauer tree algebra.

Theorem 5.10. I f N > 1 and N = 1 (mode) then the algebra H O ( A ) zs generated

by {xo ,x l ,? , , y , z , : 1 5 1 < e , O < r S e - 1 ) z f char k { N or by

{ ~ 0 , x ~ , ~ ~ , y , ~ ~ , u ~ : 1 < ~ ~ e ~ O ~ r ~ e - l } zf chark I N . Herexo,xl ,y ,z , , and

w are descrzbed zn Theorem 5.7 and, for each j wzth 1 5 j < e, ZJ = ((7r-1, v,) E

HO(.I) . Furthermore, these generators must satzsfy the relntzons zn Theorem 5 7

and xlV ZJ = y v Z, = z,V ZJ = 0 for 1 5 j 5 e and 1 5 r 5 e - 1. Also,

x , v xJ> = 0 for 1 5 j , j' 5 e.

Corollary 5.11. Let .I be a Brauer tree algebra for a Brauer tree T with e edges

and ntultiplzcity m. Then

i) dimk H O ( A ) = m + e

ii) diml, H2'(11) = dimk H2i-1 (A) = m if z $ 0 (mod e )

iii) If i = 0 (mod e), then

{ I ij c h a r k l m e + l dimk H2'(11) = dimk ~ ~ i - 1 ( A ) =

otherwise

111 this sectim we will show t!?at if niod k:ZC,/JN and mod kZeg / J N are derived

equivalent then e = e' and N = N'. To show this we first note that e is the

rank of the Grothendieck group of ~ z , / J ~ and this rank is invariant by derived

equivalence. So e and e' are equal. Now let A = k Z , / J N and A' = ~ z , / J N ' .

We will prove that if H O ( A ) is isomorphic to H * ( A ' ) as graded rings, then A is

isomorphic t o A'. Since the Hochschild cohomology ring is an invariant of derived

categories. by Rickard7s theorem [Ri2] the desired result will follow.

Assume H O ( A ) 2 H O ( A ' ) . Initially we will also suppose that either char k C NN' or the condition gcd(N, e) # 1 and g c d ( ~ ' , e ) # 1 holds true. We refer to

this condition as Property 1. Unless otherwise stated we shall assume Property 1

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throughout this section. After this case has been established we resolve the general

case. Note that if e = 1 then .I = k [ x ] / x N is commutative and is determined by its

derived category. So we may assume e > 1. Write N = me + t and N' = m'e + t' where 0 < t , t' 5 e - 1. We have already seen that

m + e if t = l 111 if t = O o r l

if t = o and dirnH1(2i) = { nz + 1 otherwise

rn + 1 otherwise

Recall that C,(N) = #{j : 0 5 j 2 N-2 and j - ti (mod e) ). Then dim Ha+' (A) =

dim Ha(A) = C,(N) for i 2 1. From this formula we obtain our first corollary.

Corollary 6.1. i) dim HO(A) # dim H1(A) if and only if t = 1.

ti) If t = 1 (i.e. N = me + 1) and He(A) r ~ ' ( i l ' ) then '1 2 '1'.

iii) If e = 2 and H 0 ( h ) EZ H'(A') then A 2 A'.

Now we assume that e > 3. Then it is not difficult to show the following:

Lemma 6.2. i) If t = 0, then Ci(N) = m for every i 2 0 .

ii) If t = 1 , Ci(N) = m.

Lemma 6.3. Assume e > 3, and t # 1 . Let A be the following e-tuple of nnturnl

numbers: A = (C1 (N), ..., C,-1 (N)) . Then

i) If gcd(t, e) = 1 then the e-tuple A has t-1 entries equal to m+l and e-t+l entries

equal to m.

ii) If gcd(t, e) = a # 1 or t = 0, then the e-tuple A has t entries equal to m+l and

e-t entries equal to m.

Proof. i) Since gcd(t, e) = 1 = gcd(N, e) the set NO, N1, ..., N(e - 1) of residue {- - -7

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classes modulo e is a complete residue class. Using this fact and part (ii) of Lemma

6.2 the result follows.

ii) The result is clear if t = 0. So assume t # 0 and cu = gcd(t,e). Write -

e = ELI and t = L. Then we have N i z t i a(t i) (mod e) and, for each j with

0 2 j 5 E - 1, there is some integer ij such that ti, zi Ni r ja (mod e). We now -- -

consider the e-tuple B = (NO, N1, ..., N ( e - 1)) where I denotes the representative

of I (mod e) between 0 and e - 1. In the e-tuple B each j n appears 0 times because -- - -- -

( X O : N l , ..., Ar(e - 1)) = (3, ..., (E- l ) a t , to, ..., (E- l ) n t , ...). Since cu > 2 wc have

that the liu111her of times which all entry in the e-tuple B is less than or q u a 1 to

t - 2 is a t = t . Since this is precisely the number of times that we obtain an entry

equal to m + 1 in the e-tuple A, we have the desired result. I

We conclude the following

Corollary 6.4. i ) dim Hi(A) = dim Hj(,l) for every i,j if and only if t = 0.

ii) If t = 0 and dim Hi(A) = dim ~ ' ( ' 4 ' ) for every i then A r A'.

Corollary 6.5. If dim H1(A) = dim H ~ ( A ' ) for every i then m = mJ

Proof. From corollary 6.1 and corollary 6.4 we can assume t # 0 and t # 1. Then

m + 1 = dim H Y A ) . I

From now on we assume that A = k Z , / J N and A' = k ~ , / J N ' , where N = me+t,

N' = r n e + t l , t $ { 0 , 1 ) , a n d e > 3 . supPoset s t ' .

Corollary 6.6. If dim H i ( A ) = dim H ~ ( A ' ) for all i then A 2 A' or

i ) t' = t + 1 and

ii) gcd(t + 1, e) = 1 and gcd(t, e) # 1.

Proof. Since we know rn we can determine the number of times that m + 1 appears

in the e-tuple ,A. Let z be this number. Then t , t' E { x , x + 1) . .Assume that t # t ' .

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Then we can conclude (i). If t' is not coprime with e then by Lemma 6.3 we have

t' = a: which is not possible. I

Proposition 6.7. Assume that dim Hf(A) = dim ~ ' ( 1 1 ' ) for all i. Then '1 E '1'.

Proof. Assume that A 11'. Corollary 6.6 allows us to assume that

1) N' = N + 1

2) gcd(N' , e) = 1 and gcd(N, e) # 1.

Let cuE = e and a t = t where (Y = gcd(N, e) and 2 5 e , Z We have E

{ o , ~ , ..., (?- 1 ) ~ ) if and only if (t + 1)i E {g,T, ..., (t - 1)). Now let j be such that

( t + l ) j - 1 (mod e). Then for 0 5 k 5 e - 1 we have Ck, ( N ) = m + 1 if and only

if 0 5 k 5 t - 1 = t' - 2. If we consider the e-tuple

then the first t entries of this n-tuple are equal to m + 1 and the rest are equal t o

m.

The hypothesis implies that C = (Co(N), C,(N), Cz, (N), ..., C(,-l)j(N)). Since

( m + 1 ) j = 1 (mod e) we have that gcd(j, e) = 1. So the subgroup generated by F in Z/Ze is the group generated by 2 which is, in turn, the group generated by Z.

Therefore, for some r with 0 5 r 5 t - 1 we have t r j TQ t (mod e). To see this

consider the set (0, t j , 2tj, ..., (t - 2) t j ) . Then there exists some X < A' such that

Xtj = A' t j . So (A' - X)tj 0 (mod e) and {n, G, ..., (A' - X)tj} is the subgroup of

Z/Ze generated by 2. Therefore there is some .u 5 t - 2 such that u t j r f (mod e).

Then C,(N) = m and we get a contradiction. I

Now do not assume Property 1. As previously noted, if N r 0 or 1 (mod e), then

~ ' ( k 2 , I . l ~ ) E H'(LZ, / J~ ' ) if and only if N = N'. So assume N = me + t and

N' = m'e + t' where 2 5 t, < e - 1. Using previous arguments it is not difficult

t o show that m = m'. As a consequence of the ring structure of Hm(i1) and H'(A')

we have that char k I N and gcd(N, e) = 1 if char k I N 'and . g r d ( ~ ' , e) = 1. So we

can always assume that char k NN' or that gcd(N,ej # 1 and gcd(N',e) # 1.

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Finally we note that it is possible for H ' ( ~ Z , / J ~ ) H*(kZ, t /J" ' ) with e # e'

as the following example shows.

E x a m p l e 6.1. Given any integer i 2 2 , H e ( k Z r , + l / ~ ' ) E ~ * ( k Z d , + z / J ? as

graded rings. See [Ci].

In this section we provide some non-truncated examples. As a niotivatio~i, let us

first recall a result of Cibils. In [Ci] algebras of the fonri 11 = k - r / JL are considered.

Here r need not be a 2, cycle. Cibils shows that H e ( A ) is finitely generated if

arid only if J? = 2,. At this point one may ask whether He( i i ) is finitely generated

when r = 2, and I is any admissible ideal. It turns out that the answer is no as

the following example shows.

E x a m p l e 7.1 Let r = Z5 and I =< alaza3, a2a3a4,adnSal > . Then the projec-

tives from the minimal projective resolution of i i = k Z 5 / I satisfy P2 = P5 = P8 =

. . . , P3 = P6 = P9 = ... , a n d P4 = P7 = Plo = .. . . Also, F'; = 0, P3f = ku2@kvq,

and P,' = Lal $ ka2as $ ka4. So the Hochschild cohomology of A = kZ , / I is

determined from the following complex:

5 * 4' $* $* 0 --+ U kv, -% U ka, 4 0 4 kvp $ ku4 -% kal $ kaaa3 @ ka4

z= 1 1=1

4. 4' Using the boundary maps from [BM] we see 2.2 -% ( a l , aza3,O) and v4 -%

0; @ * ( 0 , anas, ad). Similarly, vz ---+ ( a l , -aza3,0) and v4 4 (0, anas, -a4) . From this

complex we can now compute H1'(A) = ker$i+l / Ini4; for all n 2 0. It is easy

to check the non-zero cohomology groups are H O ( h ) and H n ( A ) for n 1 (mod

3). Moreover, these non-zero groups all have dimension 1. However, this means

that if i , j > 0, then Hi(.k) V H3(A) H i f j ( A ) = 0 since i + j f 1 (mod 3).

So multiplication in non-zero degrees is always zero. It follows that H e ( A ) is not

finitely generated.

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In example 7.1, the cohomology ring was not finitely generated because there

are "gaps" between the non-zero cohomology groups. However, even if all of the

groups are non-zero, the cohomology ring need not be finitely generated. We show

this with the following example.

Example 7.2 Let h = kZe/ < y t f l > . Recall that is the unique path of

length e + 1 beginning a t the vertex vl . Then the minimal Ae- resolution of A is

given by Q 4 - - - + P n - - I t . - % P 1 4 P 0 4 ! i , O

e e where Po = LI iiv @ vil, Pl = ,II iio(a) @ t(a).i, and P,, = Awl @ v z h for all n > 2.

r = l 2=1

Here T is the nlultiplication map, 41(o(a) @ t ( a ) ) = n ~i t (a ) - o(a) @ a for each

arrow n, 4,,(vl @ 212) = .UL CP a2a3 . . . nl + ala2 . . . ne c% u:! if 11 > 2 is ovel~, and

4 n ( v l @ ~ 2 ) = ala? . . . ae@va -vl@aza3 . . . al if n 2 2 isodd. So P,' = vliiv2 = kal

for each n 2 2. So the Hochschild cohomology of A is determined by the following

complex:

6;"-1 Q- . . . --+ kal --+ kal -% kal t .. .

Note that = q5Zn = 0 for each n 2 2. So H2(A) r Horna. ( P I , A) is one-

dimensional for i > 3.

Now let q €Horn,\. (Pi, A) for some i 2 2 with q(ul @ u2) = a. Define ?ji : Pi --+

PO by q,(vl@v2) = a1 @vz , 7i+l : Pa+l ---i PI by q2+1(v1 @vq) = v1 @a2 . . . a1 ,and - qi+j : Pi+j --+ Pj by qi+j(vl @ v 2 ) = ~1 @ a z . . . a l .

Consider the following diagrams:

P,+l #i+! Pa P. $+I . @-~-tr Pi+j-l

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So the left diagram commutes. For the right diagram, consider

71i+J-141+j(t11 @up) = qif jPl (vl @ a 2 . . . al -a1 ... a n @ v 2 ) = -a1 . . . a , @ a p . . . al.

Also, @,r11+j(r9 8 vz) = dJ(vl @ a' . . . a l ) = -a1 ... a , 8 a? . . . a l . Thus, the right

diagram also commutes. Note that v , + ~ is not an isomorpliisn~. Let [ : Pi -+ A

where E # 0. Then [ qi+, = A[ since dim Hom(Pi, A) = 1. So <(qi+, - X I ) = 0.

Now, qi+, E rad End(P,+j). So if X # 0 then qi+j - X is invertible. Then it must

be the case that X = 0. This means 6 v ~ + ~ = 0. Iterating this argument we have

that H'( . I )HJ(: i ) = 0 for i j j > 1. Notice that in this example not even the even

cohomology ring He"(A) = LI HZn(,I) is finitely generated. n>O

7.1. Concluding Remarks. In [Ha] Happel shows that if a finite dimensional

algebra I\ has an infinite dimensional Hochschild cohomology ring, then A has

infinite global dimension. He then states that the converse is not known. For the

algebras is this paper the converse does hold (see [Lo] for the truncated case). X

related question is the following. When is the Hochschild cohomology ring of a finite

dimensional monomial algebra finitely generated ? A particularly nice relation set

is J'. In this case A P ( n ) is precisely all paths of length n for each n 2 0. So

this case seems like a natural candidate to look for finitely generated cohomology

rings. However, Cibils shows (see [Ci]) that if r contains an oriented cycle, then

H ' ( k r / J 2 ) is finitely generated if and only if r = 2,. We have also seen that

having the underlying quiver of A be a cycle is not sufficient to guarantee that

H e ( A ) is finitely generated. As a consequence, it appears that it is fairly special

for the cohoinology ring of a finite dimensional monomial algebra of infinite global

dimension to be finitely generated. It would be interesting to classify the monomial

algebras for which the Hochschild cohomology rings are finitely generated.

The authors would like to thank Bernhard Keller for providing valu-

able suggestions for this work.

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The second author would like to thank CNPq from Brazil for a PhD

scholarship.

The third author would like to thank CNPq from Brazil for a research

grant and FAPESP from Brazil for a scholarship.

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[hi] S. hIacLane, Homology, Springer-Verlag,1963.

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1'01. 40 No. 12, 1272-1278, 1997.

Received: January 1999

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on the hochschild cohomology of truncated cycle algebras

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