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 JAMP  Vol.5 No.1 , January 2017
Homological Properties in Relation to Nakayama Algebras
Abstract: The goal of this paper is to investigate whether the Ext-groups of all pairs (M, N) of modules over Nakayama algebras of type (n,n,n) satisfy the condition ExtΛn(M,N)=0 for n >> 0 ? ExtΛn(M,N)=0 for n >> 0. We achieve that by discussing the Ext-groups of Nakayama algebra with projectives of lengths 3n+1 and 3n+2 using combinations of modules of different lengths.

1. Introduction

In this paper, we describe homological properties of Nakayama algebras. The algebra Ʌ is a Nakayama algebra if every projective indecomposable and every injective indecomposable Ʌ-module is uniserial. In other words, these modules have a unique composition series, (see Schröer [1] ). Nakayama algebras are finite dimensional and representation-finite algebras that have a nice representation theory in the sense that the finite-dimensional indecomposable modules are easy to describe.

The main contribution of this paper is to investigate whether the Ext-groups of all pairs ( M , N ) of modules over Nakayama algebras of type ( n , n , n ) satisfy the condition E x t Λ n ( M , N ) = 0 for n 0 E x t Λ n ( N , M ) = 0 for n 0 .

2. Preliminary Notes

This section will briefly discuss the properties of Nakayama algebra and some related propositions. We consider also proofs of some of the propositions. All the information presented here can be found in deeper details in the book from Rotman [2] .

Definition: Let Λ be an artin algebra. A Λ -module A is called a uniserial module if the set of submodules is totally ordered by inclusion.

Proposition 1: The following are equivalent for Λ -module A

1. A is uniserial.

2. There is only one composition series for A .

3. The radical filtration of A is a composition series for A .

4. The socle filtration of A is a composition series for A .

5. l ( A ) = r l ( A ) , where l ( A ) is the length of A and r l ( A ) is the radical length of A .

Proof: See Rotman [2] for the details.

Proposition 2: The following are equivalent for an artin algebra Λ .

1. Λ is a sum of uniserial modules .

2. Λ / a is a sum of uniserial modules for all ideals a of Λ .

3. Λ / r 2 is a sum uniserial modules.

Proof: 1 2 and 2 3 are trivial. If Λ is the sum of uniserial modules, then Λ / a and Λ / r 2 which are factors of Λ are also a sum of uniserial modules.

3 1

Let P be an indecomposable projective Λ -module. We show that P / r n P is uniserial by induction on n when n 2 . When n = 2 , there is nothing to prove. Suppose n > 2 . Let the radical filtration of P be; P r P r 2 P r n 1 P r n P = 0 such that r i P / r i + 1 P is simple for i = 0 , 1 , , n 1 .

When n = 3 , we have r 3 P r 2 P r P P . Hence by induction hypothesis, P / r n 1 P is uniserial. Considering the exact sequence 0 r P P P / r P 0 , which also implies that P / r P is uniserial, hence P / r n 1 P is also uniserial.

If r n 1 P = 0 , then P / r n P is clearly uniserial, so we have to assume that r n 1 P 0 . From proposition 1, it follows that r i P / r i + 1 P is simple for i = 0 , 1 , , n 2 . To show that P / r n P is uniserial, then it is sufficient by proposition 1 to prove that r n 1 P / r n P is also simple.

Let Q r n 2 P be a projective cover. Since r n 2 P / r n 1 P is simple, Q must be indecomposable and so Q / r 2 Q is uniserial. But we have an epimorphism r Q / r 2 Q r n 1 P / r n P which shows that r n 1 P / r n P is simple.

Proposition3: Proposition 2 (1).

Let φ be a D Tr-orbit of ind Λ . Suppose there is a projective module P in φ . Then we have the following

1. φ of non-zero objects in { P , ( D T r ) 1 P , , ( D T r ) i P , } i N .

2. φ is finite if and only if ( D T r ) n P = ( T r D ) n is injective for some n in N . Moreover, if ( T r D ) n P is injective, then φ = { P , ( D T r ) 1 P , , ( D T r ) n P } .

Proof: By proposition 1, D T r P = 0 if and only if P is projective. Since P is projective module in φ , ( D T r ) i P = 0 for all i > 0 . Hence the claim in 2 (1).

We claim that if ( D T r ) i P ( T D r ) ( i + j ) P 0 with j > 0 we have ( D T r ) i ( D T r ) ( i + j ) P ( D T r ) ( D T r ) i P which implies that P ( D T r ) j P = ( T r D ) j P which is not possible since j > 0 . φ can therefore be finite if ( D T r ) ( n + 1 ) P = 0 for some n 0 . Since ( D T r ) n P = ( T r D ) n P , then P is injective in φ . We know therefore that if ( D T r ) n P is injective, then φ = { P , ( D T r ) 1 P , , ( D T r ) n P } .

3. Results

The main goal of this work is to investigate whether the Ext-groups of all pairs (M, N) of modules over Nakayama algebras of type (n, n, n) satisfy the condition: E x t Λ n ( M , N ) = 0 for n 0 E x t Λ n ( N , M ) = 0 for n 0 , where n is a positive integer. We discuss the Ext-groups of Nakayama algebras with projectives of lengths 3 n + 1 and 3 n + 2 using combinations of modules of different lengths. Reader may refer to Auslander, M et al. [3] for ideas illustrated in this section.

We begin with the Ext-groups of Nakayama algebra with projectives of length 3 n + 1 using the combinations ( 3 i ,3 j ) , and ( 3 i ,3 j + 2 ) where 3 i and 3 j are modules of lengths 3 i and 3 j respectively, 3 j + 2 is also module of length 3 j + 2 .

Let Γ be a path with the relations

α γ β γ β α , β α γ α γ β and γ β α β α γ

where the length of each relation is 3 n + 1 . Let

Λ = k Γ / α γ β γ β α , β α γ α γ β , γ β α β α γ .

The projectives of the above path algebra are as follows:

P 1 = ( S 1 S 2 S 3 S 1 ) , P 2 = ( S 2 S 3 S 1 S 2 ) , P 3 = ( S 3 S 1 S 2 S 3 ) .

The above projectives P 1 , P 2 and P 3 each has length 3 n + 1 . The minimal projective resolution of the module M = ( S 1 , S 2 , , S 3 ) t of length 3 i , i = 1 , 2 , , n is given as;

Q 4 d 5 Q 3 d 4 Q 3 d 3 Q 2 d 2 Q 1 d 1 Q 0 d 0 M 0

where Q 6 i = Q 6 i + 1 = P 1 , Q 6 i + 2 = Q 6 i + 3 = P 2 and Q 6 i + 4 = Q 6 i + 5 = P 3 for i 0 .

The combinations have been reduced to four class because in the above minimal projective resolution of the module of length 3 i , we see that it is in the same group as the module of length 3 i + 1 . The modules of lengths 3 i and 3 i + 1 therefore have the same properties with respect to the conditions; E x t Λ n ( M , N ) = 0 for n 0 E x t Λ n = 0 for n 0 .

From the above resolution, we have;

Λ e 1 d 6 Λ e 3 d 5 Λ e 3 d 4 Λ e 2 d 3 λ e 2 d 2 Λ e 1 d 1 Λ e 1 d 0 ( S 1 , S 2 , , S 3 ) 0.

Let A = ( S 1 , S 2 , , S 3 ) t , the pd A = since the resolution is periodic. The period is 6. The truncation of the resolution is given as;

P Λ e 1 d 6 Λ e 3 d 5 Λ e 3 d 4 Λ e 2 d 3 λ e 2 d 2 Λ e 1 d 1 Λ e 1 d 0 0.

The map d 6 i + 1 is the multiplication by ( γ β α ) i , d 6 i + 2 is the multiplication by α ( γ β α ) n i , d 6 i + 3 is multiplication by ( α γ β ) i , d 6 i + 4 is the multiplication by β ( α γ β ) n i , d 6 i + 5 is the multiplication by ( β α γ ) i and d 6 i is the multiplication by γ ( β α γ ) n i .

Applying H o m Λ ( , M ) where M is the module

( S 1 S 2 S 3 )

of length 3 j , we have;

0 d 0 × H o m Λ ( Λ e 1 , M ) d 1 × H o m Λ ( Λ e 1 , M ) d 2 × H o m Λ ( Λ e 2 , M ) d 3 × H o m Λ ( Λ e 2 , M ) d 4 × H o m Λ ( Λ e 3 , M ) d 5 × H o m Λ ( Λ e 1 , M ) ,

where H o m Λ ( Λ e 1 , M ) e 1 M , H o m Λ ( Λ e 2 , M ) e 2 M and H o m Λ ( Λ e 3 , M ) e 3 M .

We have the following Ext-groups;

E x t Λ 1 ( N , M ) = ker ( e 1 M α ( γ β α ) n i e 2 M ) / ( e 1 M ( γ β α ) i e 1 M ) ,

where N and M are modules of length 3 i and 3 j respectively. We first compute the kernel for

d 2 × : ( e 1 M α ( γ β α ) n i e 2 M ) .

We have

d 2 × ( ( a 1 , a 2 , , a s , a s + 1 , ) t , 0 , 0 ) = ( 0 , ( 0 , 0 , 0 , , 0 , a 1 , a 2 , , a s ) t , 0 )

where a 1 is in the coordinate number n i + 1 , therefore a s is in the coordinate number n i + s . Hence s = j + i n and consequently s + 1 = j + i + 1 n ,

ker ( e 1 M α ( γ β α ) n i e 2 M ) = { ( ( 0 , 0 , , 0 , a s + 1 , , a j ) ) t , 0 , 0 ) | a r k , r = s + 1 , , j } .

Next we compute the image of

d 1 × : ( e 1 M ( γ β α ) i e 1 M ) .

We have that

d 1 × ( ( b 1 , b 2 , , b j ) t , 0 , 0 ) = ( ( 0 , 0 , , 0 , b 1 , , b j 1 ) t , 0 , 0 )

where b 1 is in the coordinate number i + 1 . This shows that

( e 1 M ( γ β α ) i e 1 M ) = { ( ( 0 , 0 , , 0 , b 1 , , b j 1 ) t , 0 , 0 ) | b t k , t = 1 , , j 1 } .

Hence dim k E x t Λ 1 ( ( S 1 , S 2 , , S 3 ) t , M ) = ( i + 1 ) ( j + i + 1 n ) = n j , for j n . We compute

E x t Λ 2 ( ( S 1 , S 2 , , S 3 ) t , M ) ) = ker ( e 2 M ( α γ β ) i e 2 M ) / ( e 1 M α ( γ β α ) n i e 2 M ) .

We compute the kernel for

d 3 × : ( e 2 M ( α γ β ) i e 2 M ) ,

we have

d 3 × ( 0 , ( a 1 , a 2 , , a j ) t , 0 ) = ( 0 , ( 0 , 0 , , 0 , a 1 , , a j i ) t , 0 )

where a 1 is the coordinate number i + 1 . This shows that

ker ( e 2 M ( α γ β ) i e 2 M ) = { ( 0 , ( 0 , 0 , , 0 , a 1 , , a j ) t , 0 ) | a r k , r = 1 , , j } ,

where a 1 is in the coordinate number j i + 1 . Next we compute the image of

d 2 × : ( e 1 M α ( γ β α ) n i e 1 M ) .

We see that

d 2 × ( ( b 1 , b 2 , , b s , , b j ) t , 0 ) = ( 0 , ( 0 , 0 , , 0 , b 1 , b 2 , , b s ) t , 0 )

where b 1 is in the coordinate number n i + 1 and therefore b s is in the coordinate number n i + s . This implies n i + s = j and hence s = j + i n ,

( e 1 M α ( γ β α ) n i e 2 M ) = { ( 0 , ( 0 , 0 , , 0 , b 1 , , b s ) t , 0 ) | b t k , t = 1 , , s } .

Hence dim k E x t Λ 2 ( ( S 1 , S 2 , , S 3 ) t , M ) = ( n i + 1 ) ( j i + 1 ) = n j for j n .

dim k E x t Λ 2 ( ( S 1 , , S 3 ) t , M ) = n j

We compute

E x t Λ 3 ( ( S 1 , , S 3 ) t , M ) = ker ( e 2 M β ( α γ β ) n i e 3 M ) / ( e 2 M ( α γ β ) t e 2 M ) .

First we compute the kernel for

d 4 × : ( e 2 M β ( α γ β ) n i e 3 M ) .

We have

d 4 × ( 0 , ( a 1 , a 2 , , a s , , a j ) t , 0 ) = ( 0 , 0 , ( 0 , 0 , , 0 , a 1 , a 2 , , a s ) t , ) ,

where a 1 is in the coordinate number n i + 1 and therefore a s is in the coordinate number n i + s . Hence s = j + i n and consequently s + 1 = j + i + 1 n . This shows that

ker ( e 2 M β ( α γ β ) n i e 3 M ) = { ( 0 , ( 0 , 0 , , 0 , a s + 1 , a s + 2 , , a j ) t , 0 ) | a r k , r = s + 1 , , j } .

We compute the image of

d 3 × : ( e 2 M ( α γ β ) i e 2 M )

to have

d 3 × ( 0 , ( b 1 , b 2 , , b j ) t , 0 ) = ( 0 , ( 0 , 0 , , 0 , b 1 , b 2 , , b j i ) t , 0 )

where b 1 is in the coordinate number i + 1 . This shows that

( e 2 M ( α γ β ) i e 2 M ) = { ( 0 , ( 0 , 0 , , b 1 , , b j i ) t , 0 ) | b t k , t = 1 , , j i } .

Hence

dim k E x t Λ 3 ( ( S 1 , , S 3 ) t , M ) = i + 1 ( j + i + 1 n ) = n j

for j n .

Next we have

E x t Λ 4 ( ( S 1 , , S 3 ) t , M ) = ker ( e 3 M ( β α γ ) i e 3 M ) / ( e 2 M β ( α γ β ) n i e 3 M )

Similarly, we have

ker ( e 3 M ( β α γ ) i e 3 M ) = { ( 0 , 0 , ( 0 , 0 , , 0 , a 1 , a 2 , , a j ) t , 0 ) | a r k , r = 1 , , j }

where a 1 is in the coordinate number j + 1 i , then

( e 2 M β ( α γ β ) n i e 3 M ) = { ( 0 , 0 , ( 0 , 0 , , 0 , b 1 , b 2 , b s ) t ) | b t k , t = 1 , , s } ,

s = j + i n where b 1 is in the coordinate number n i + 1 . Hence

dim k E x t Λ 4 ( ( S 1 , , S 3 ) t , M ) = n i + 1 ( j + 1 i ) = n j , j n .

Again we have

E x t Λ 5 ( ( S 1 , , S 3 ) t , M ) = ker ( e 3 M γ ( β α γ ) n i e 1 M ) / ( e 3 M ( β α γ ) i e 3 M ) .

We compute the kernel for

d 6 × : ( e 3 M e 1 M )

to have

d 6 × ( 0 , 0 , ( a 1 , a 2 , , a s , , a j ) t ) = ( ( 0 , 0 , , 0 , a 1 , a 2 , , a s ) t , 0 , 0 )

where a 1 is in the coordinate number n i + 2 and therefore a s is in the coordinate number n i + 1 + s , n i + s + 1 = j s = j + i n 1 and consequently s + 1 = j + i n . This show that

ker ( e 3 M e 1 M ) = { ( 0 , 0 , ( 0 , 0 , , a s + 1 , , a j ) t ) | a r k , r = s + 1 , , j } .

We compute the image for

d 5 × : ( e 3 M ( β α γ ) i e 3 M )

to get

d 5 × ( 0 , 0 , ( b 1 , b 2 , , b j i ) t ) = ( 0 , 0 , ( 0 , 0 , , 0 , b 1 , b 2 , , b j i ) t )

where b 1 is in the coordinate number i + 1 . This shows that

( e 3 M ( β α γ ) i e 3 M ) = { ( 0 , 0 , ( 0 , 0 , , , b 1 , , b j i ) t ) | b t k , t = 1 , , j 1 } .

Hence dim k E x t Λ 5 ( ( S 1 , , S 3 ) t , M ) = i + 1 ( j + i n ) = n + 1 j . This shows that dim k E x t Λ t ( 3 i , 3 j ) = n j for t = 1 , 2 , 3 , 4 and n j + 1 for t = 5 , n j = 0 , n j + 1 = 0 which implies n = j and n + 1 = j .

It follows from the computation of the fifth Ext-group that the condition E x t Λ n ( M , N ) = 0 for n 0 E x t Λ n ( N , M ) = 0 for n 0 holds.

We calculate the Ext-groups for the combination ( 3 i ,3 j + 2 ) .

E x t Λ 1 ( ( S 1 , , S 3 ) t , M )

where M is the module of length 3 j + 2

E x t Λ 1 ( ( S 1 , , S 3 ) t , M ) = ker ( e 1 M α ( γ β α ) n i e 2 M ) / ( e 1 M ( γ β α ) i e 1 M ) .

We compute the kernel for

d 2 × : ( e 1 M α ( γ β α ) n i e 2 M ) ,

we have

d 2 × ( ( a 1 , a 2 , , a s , , a j + 1 ) t , 0 , 0 ) = ( 0 , ( 0 , 0 , , 0 , a 1 , a 2 , , a s ) t , 0 )

where a 1 is in the coordinate number n i + 1 and therefore a s is in the coordinate number n i + s = j + 1 . Hence s = j + i n and consequently s + 1 = j + i + 2 n . This shows that

ker ( e 1 M α ( γ β α ) n i e 2 M ) = { ( ( 0 , 0 , , 0 , a s + 1 , , a j + 1 ) t , 0 , 0 ) | a r k , r = s + 1 , , j + 1 } .

Computing the image for

d 1 × : ( e 1 M ( γ β α ) i e 1 M )

we have

d 1 × ( ( b 1 , b 2 , , b j + 1 ) t , 0 , 0 ) = ( ( 0 , 0 , , 0 , b 1 , , b j + 1 i ) t , 0 , 0 ) ,

where b 1 is in the coordinate number i + 1 . This shows that

( e 1 M ( γ β α ) i e 1 M ) = { ( ( 0 , 0 , , b 1 , , b j + 1 i ) t , 0 , 0 ) | b t k , t = 1 , , j + 1 i } .

We therefore have

dim k E x t Λ 1 ( ( S 1 , , S 3 ) t , M ) = i + 1 ( j + i + 2 n ) = n ( j + 1 )

for j + 1 n .

Next we have

E x t Λ 2 ( ( S 1 , , S 3 ) t , M ) = ker ( e 2 M ( α γ β ) i e 2 M ) / ( e 1 M v e 2 M ) .

We compute the kernel for

d 3 × : ( e 2 M ( α γ β ) i e 2 M ) ,

we have

d 3 × ( 0 , ( a 1 , a 2 , , a j + 1 ) t , 0 ) = ( 0 , ( 0 , 0 , , 0 , a 1 , , a j + 1 i ) t )

where a 1 is in the coordinate number i + 1 . This shows that

ker ( e 2 M ( α γ β ) i e 2 M ) = { ( 0 , ( 0 , 0 , , 0 , a 1 , , a j + 1 ) t , 0 ) | a r k , r = 1 , , j + 1 } ,

where a 1 is in the coordinate number j + 2 i . We compute the image for

d 2 × : ( e 1 M α ( γ β α ) n i e 2 M ) .

We have

d 2 × ( ( b 1 , b 2 , , b s , , b j + 1 ) t , 0 , 0 ) = ( 0 , ( 0 , 0 , , b 1 , b 2 , , b s ) t , 0 )

where b 1 is in the coordinate number n i + 1 and therefore b s is in the coordinate number n i + s = j + 1 . Hence s = j + 1 + i n . This shows that

( e 1 M α ( γ β α ) n i e 2 M ) = { ( 0 , ( 0 , 0 , , 0 , b 1 , , b j + 1 ) t , 0 ) | b t k , t = 1 , , j + 1 } .

We therefore have dim k E x t Λ 2 ( ( S 1 , , S 3 ) t , M ) = n i + 1 ( j + 2 i ) = n ( j + 1 ) for j + 1 n . Next we have

E x t Λ 3 ( ( S 1 , , S 3 ) t , M ) = ker ( e 2 M β ( α γ β ) n i e 3 M ) / ( e 2 M ( α γ β ) i e 2 M ) .

Similarly, we have;

ker ( e 2 M β ( α γ β ) n i e 3 M ) = { ( 0 , ( 0 , 0 , , a 1 , , a j + 1 ) t , 0 ) | a r k , r = 1 , , j + 1 } ,

where a 1 is in the coordinate number j + i + 2 n . We also have

( e 2 M e 2 M ) = { ( 0 , ( 0 , 0 , , 0 , b 1 , , b j + 1 i ) t , 0 ) | b t k , t = 1 , , j + 1 i } ,

where b 1 is in the coordinate number i + 1 . We therefore have dim k E x t Λ 3 ( ( S 1 , , S 3 ) t , M ) = i + 1 ( j + i + 2 n ) = n ( j + 1 ) for j + 1 n . Next we have

E x t Λ 4 ( ( S 1 , , S 3 ) t , M ) = ker ( e 3 M ( β α γ ) i e 3 M ) / ( e 2 M β ( α γ β ) n i e 3 M ) .

We compute the kernel for

d 5 × : ( e 3 M ( β α γ ) i e 3 M ) .

We have

d 5 × ( 0 , 0 , ( a 1 , a 2 , , a j ) t ) = ( 0 , 0 , ( 0 , 0 , , 0 , a 1 , , a j i ) t )

where a 1 is in the coordinate number i + 1 . This shows that

ker ( e 3 M ( β α γ ) i e 3 M ) = { ( 0 , 0 , ( 0 , 0 , , a 1 , , a j ) t , ) | a r k , r = 1 , , j } ,

where a 1 is in the coordinate number j + 1 i . We compute the image for

d 4 × : ( e 2 M β ( α γ β ) n i e 3 M ) .

We have

d 4 × ( 0 , ( b 1 , b 2 , , b s , b j + 1 ) t , 0 , ) ( 0 , 0 ( ( 0 , 0 , , b 1 , b 2 , b s ) t )

where b 1 is in the coordinate number n i + 1 and therefore b s is in the coordinate number n i + s = j + 1 and therefore s = j + i n ,

( e 2 M β ( α γ β ) i e 3 M ) = { ( 0 , 0 , ( 0 , 0 , , b 1 , , b j i ) t ) | b t k , t = 1 , , j } .

We therefore have dim k E x t Λ 4 ( ( S 1 , , S 3 ) t , M ) = n i + 1 ( j + 1 i ) n j for j n . Finally, we have

E x t Λ 5 ( ( S 1 , , S 3 ) t , M ) = ker ( e 3 M γ ( β α γ ) n i e 1 M ) / ( e 3 M ( β α γ ) i e 3 M ) .

We compute the kernel for

d 6 × : ( e 3 M γ ( β α γ ) n i e 1 M ) ,

we have

d 6 × ( 0 , 0 , ( a 1 , a 2 , , a s , a j ) t ) = ( ( 0 , 0 , , 0 , a 1 , , a s ) t , 0 , 0 )

where a 1 is in the coordinate number n i + 2 and therefore a s is in the coordinate number n i + 1 + s = j , s = j + i n 1 and consequently s + 1 = j + i n . This shows that

ker ( e 3 M γ ( β α γ ) n i e 1 M ) = { ( 0 , 0 , ( 0 , 0 , , a 1 , , a j ) t ) | a r k , r = 1 , , j } .

We compute the image for

d 5 × : ( e 3 M ( β α γ ) i e 3 M ) ,

we have

d 5 × ( 0 , 0 , ( b 1 , b 2 , , b j ) t ) = ( 0 , 0 , ( 0 , 0 , , 0 , b 1 , b 2 , , b j i ) t )

where b 1 is the i t h coordinate number i + 1 and hence

( e 3 M ( β α γ ) i e 3 M ) = { ( 0 , 0 , ( b 1 , , b j i ) t ) | b t k , t = 1 , , j i } .

We therefore have

dim k E x t Λ ( ( S 1 , , S 3 ) t , M ) = i + 1 ( j + i n ) = n + 1 j for j n + 1

dim k E x t Λ t ( 3 i , 3 j + 2 ) = n ( j + 1 ) for t = 1 , 2 , 3 , n j for t = 4 and n + 1 j for t = 5 . We therefore have n = j + 1 , n = j and n = j 1 .

It follows from the computations of the forth and fifth Ext-groups that the condition E x t Λ n ( M , N ) = 0 for n 0 E x t Λ n ( N , M ) = 0 for n 0 holds.

We now discuss the Ext-groups of the Nakayama algebras with the projectives of length 3 n + 2 using the combination ( 3 i + 1,3 j + 1 ) , where 3 i + 1 and 3 j + 1 are also modules of length 3 i + 1 and 3 j + 1 respectively.

Let Γ be a path with the relations β α γ γ β α , γ β α α γ β and α γ β β α γ where the length of each relation is 3 n + 2, n is a positive integer. Let

Λ = k Γ / β α γ γ β α , γ β α α γ β , α γ β β α γ .

The projectives of the above path algebra are as follows:

P 1 = ( S 1 , S 2 , , S 1 , S 2 ) t , P 2 = ( S 2 , S 3 , , S 2 , S 3 ) t , P 3 = ( S 3 , S 1 , , S 3 , S 1 ) t .

The above projectives P i , i = 1 , 2 , 3 each has length n + 2 . The minimal projective resolution of module ( S 1 , , S 3 ) i of length 3 i is given as;

Q 6 d 6 Q 5 d 5 Q 4 d 4 Q 3 d 3 Q 2 d 2 Q 1 d 1 Q 0 d 0 N 0.

where N is the module of length 3 i , Q 6 i = Q 6 i + 1 = P 1 , Q 6 i + 2 = Q 6 i + 3 = P 3 and Q 6 i + 4 = Q 6 i + 5 = P 2 for i 0 and d 6 i + 1 is a multiplication by ( γ β α ) i , d 6 i + 2 is a multiplication by β α ( γ β α ) n i , d 6 i + 3 is a multiplication by ( β α γ ) i , d 6 i + 4 is a multiplication by α γ ( β α γ ) n i , d 6 i + 5 is a multiplication by ( α γ β ) i and d 6 i is multiplication by γ β ( α γ β ) n i .

The combinations have been reduced to four class because in the above resolution, we see that the module of length 3 i is in the same group with the module of length 3 i + 2 . The two modules therefore have the same properties with respect to the condition E x t Λ n ( M , N ) = 0 for n 0 E x t Λ n ( N , M ) = 0 for n 0 .

From the above resolution we have,

Λ e 1 d 7 Λ e 2 d 6 Λ e 2 d 5 Λ e 3 d 4 Λ e 3 d 3 Λ e 1 d 2 Λ e 1 d 1 N d 0 0.

The p d N = since the resolution is periodic. The period is 6. The truncation of the above resolution is given as:

P Λ e 1 d 6 Λ e 2 d 5 Λ e 2 d 4 Λ e 3 d 3 Λ e 3 d 2 Λ e 1 d 1 Λ e 1 d 0 0.

Applying H o m ( , M ) where M is the module of length 3 j , we have

0 d 0 × H o m Λ ( Λ e 1 , M ) d 1 × H o m Λ ( Λ e 1 , M ) d 2 × H o m Λ ( Λ e 3 , M ) d 3 × H o m Λ ( Λ e 3 , M ) d 4 × H o m Λ ( Λ e 2 , M ) d 5 × H o m Λ ( λ e 2 , M ) d 6 × H o m Λ ( Λ e 1 , M ) ,

where H o m Λ ( Λ e 1 , M ) e 1 M , H o m Λ ( Λ e 2 , M ) e 2 M and H o m Λ ( λ e 3 , M ) e 3 M .

We discuss the Ext-groups of the Nakayama algebras with projectives of length 3 n + 2 by considering the case: ( 3 i + 1,3 j + 1 ) . 3 i + 1 is the module ( S 1 , , S 1 ) i + 1 of length 3 i + 1 and 3 j + 1 is the module ( S 1 , , S 1 ) i + 1 of length 3 j + 1 .

The minimal projective resolution of the module ( S 1 , , S 1 ) i + 1 of length 3 i + 1 is given as;

Q 6 d 6 Q 5 d 5 Q 4 d 4 Q 3 d 3 Q 2 d 2 Q 1 d 1 Q 0 d 0 A 0

where Q 6 i = Q 6 i + 3 = P 1 , Q 6 i + 1 = Q 6 i + 4 = P 2 and Q 6 i + 2 = Q 6 i + 5 = P 3 for i 0 , d 6 i + 1 is a multiplication by α ( γ β α ) i + 1 , d 6 i + 2 is a multiplication by β ( α γ β ) n i , d 6 i + 3 is a multiplication by γ ( β α γ ) i + 1 , d 6 i + 4 is a multiplication by α ( γ β α ) n i , d 6 i + 5 is a multiplication by β ( α γ β ) i + 1 and d 6 i + 6 is a multiplication by γ ( β α γ ) n i and A is the module of length 3 i + 1 .

Considering the above resolution, we have

Λ e 1 d 6 Λ e 3 d 5 Λ e 2 d 4 Λ e 1 d 3 Λ e 3 d 2 Λ e 2 d 1 Λ e 1 d 0 A 0.

The p d A = since the resolution is periodic. The period ic 6. The truncation of the above resolution is given as;

P : Λ e 1 d 6 Λ e 3 d 5 Λ e 2 d 4 Λ e 1 d 3 Λ e 3 d 2 Λ e 2 d 1 Λ e 1 d 0 0.

Applying H o m Λ ( , M ) , where M is in the module ( S 1 , , S j ) j of length 3 j , we have

0 d 0 × H o m Λ ( Λ e 1 , M ) d 1 × H o m Λ ( Λ e 2 , M ) d 2 × H o m Λ ( Λ e 3 , M ) d 3 × H o m Λ ( Λ e 1 , M ) d 4 × H o m Λ ( Λ e 2 , M ) d 5 × H o m Λ ( λ e 3 , M ) d 6 × H o m Λ ( Λ e 1 , M ) ,

where H o m Λ ( Λ e 1 , M ) e 1 M , H o m Λ ( Λ e 2 , M ) e 2 M and H o m Λ ( λ e 3 , M ) e 3 M .

We compute the Ext-groups for ( 3 i + 1,3 j + 1 ) . The first Ext-group is;

E x t Λ 1 ( ( S 1 , , S 1 ) i + 1 , M ) = ker ( e 2 M β ( α γ β ) n i e 3 M ) / ( e 1 M α ( γ β α ) i + 1 e 2 M ) ,

where M is the module of length 3 j + 1 . We compute the kernel for

d 2 × : ( e 2 M β ( α γ β ) n i e 3 M ) .

We have

d 2 × ( 0 , ( a 1 , a 2 , , a s , , a j ) t , 0 ) = ( 0 , 0 , ( 0 , 0 , , 0 , a 1 , , a s ) t ) ,

where a 1 is in the coordinate number n i + 1 and therefore a s is in the coordinate number n i + s = j , s = j + i n and consequently, s + 1 = j + i + 1 n . This shows that

ker ( e 2 M β ( α γ β ) n i e 3 M ) = { ( 0 , ( 0 , 0 , , 0 , a s + 1 , , a j ) t , 0 ) | a r k , r = s + 1 , , j } .

We compute the image

d 1 × : ( e 1 M α ( γ β α ) i + 1 e 2 M ) .

We have

d 1 × ( ( b 1 , b 2 , , b j + 1 ) t , 0 , 0 ) = ( 0 , ( 0 , 0 , , 0 , b 1 , , b j ( i + 1 ) ) t , 0 ) ,

where b 1 is in the coordinate number i + 2 . Hence

( e 1 M α ( γ β α ) i + 1 e 2 M ) = { ( 0 , ( 0 , 0 , , 0 , b 1 , , b j ( i + 1 ) ) t , 0 ) | b t k , t = 1 , , j ( i + 1 ) } .

We have dim k E x t Λ 1 ( ( S 1 , , S 1 ) t , M ) = i + 2 ( j + i + 1 n ) = n + 1 j for j n + 1 . Next, we compute the second Ext-group.

E x t Λ 2 ( ( S 1 , , S 1 ) t , M ) = ker ( e 3 M γ ( β α γ ) i + 1 e 1 M ) / ( e 2 M β ( α γ β ) n i e 3 M ) .

We compute the kernel for

d 3 × : ( e 3 M γ ( β α γ ) i + 1 e 1 M ) .

We have

d 3 × ( 0 , 0 , ( a 1 , a 2 , , a j ) t ) = ( ( 0 , 0 , , 0 , a 1 , , a j ( i + 1 ) ) t , 0 , 0 ) ,

where a 1 is in the coordinate number i + 2 . This shows that

ker ( e 3 M γ ( β α γ ) i + 1 e 1 M ) = { ( 0 , 0 , ( 0 , 0 , , 0 , a 1 , , a j ) t ) | a r k , r = 1 , , j } ,

where a 1 is in the coordinate number j i . We compute the image for

d 2 × : ( e 2 M β ( α γ β ) n i e 3 M ) .

We have

d 2 × ( 0 , ( b 1 , b 2 , , b s , , b j ) t , 0 ) = ( 0 , 0 , ( 0 , 0 , , 0 , b 1 , , b s ) t ) ,

where b 1 is in the coordinate number n i + 1 and therefore b s is the coordinate number n i + s = j and s = j + i n . This implies that

( e 2 M β ( α γ β ) n i e 3 M ) = { ( 0 , 0 , ( 0 , 0 , , 0 , b 1 , , b j ) t ) | b t k , t = 1 , , j } ,

where b 1 is in the coordinate number n i + 1 . We have

dim k E x t Λ 2 ( ( S 1 , , S 1 ) t , M ) = n i + 1 ( j i ) = n + 1 j

for j n + 1 . We compute the third Ext-group.

E x t Λ 3 ( ( S 1 , , S 1 ) t , M ) = ker ( e 1 M α ( γ β α ) n i e 2 M ) / ( e 3 M γ ( β α γ ) i + 1 e 1 M ) .

We compute the kernel for

d 4 × : ( e 1 M α ( γ β α ) n i e 2 M ) .

We have

d 4 × ( ( a 1 , a 2 , , a s , , a j + 1 ) t , 0 , 0 ) = ( 0 , ( 0 , 0 , , 0 , a 1 , , a s ) t , 0 ) ,

where a 1 is in the coordinate number n i + 1 and therefore a s is in the coordinate number n i + s = j + 1 , s = j + i + 1 n and consequently, s + 1 = j + i + 2 n . Hence

ker ( e 1 M α ( γ β α ) n i e 2 M ) = { ( ( 0 , 0 , , 0 , a s + 1 , , a j + 1 ) t , 0 , 0 ) | a r k , r = s + 1 , , j + 1 } .

We compute the image for

d 3 × : ( e 3 M γ ( β α γ ) i + 1 e 1 M ) .

We have

d 3 × ( 0 , 0 , ( b 1 , b 2 , , b j ) t ) = ( ( 0 , 0 , , 0 , b 1 , , b j i ) t , 0 , 0 ) ,

where b 1 is in the coordinate number i + 3 . This implies that

( e 3 M γ ( β α γ ) i + 1 e 1 M ) = { ( ( 0 , 0 , , 0 , b 1 , , b j i ) t , 0 , 0 ) | b t k , t = 1 , , j i } .

We therefore have dim k E x t Λ 3 ( ( S 1 , , S 1 ) t , M ) i + 3 ( j + i + 2 n ) = n + 1 j for j n + 1 . We discuss the fourth Ext-group.

E x t Λ 4 ( ( S 1 , , S 1 ) t , M ) = ker ( e 2 M β ( α γ β ) i + 1 e 3 M ) / ( e 1 M α ( γ β α ) n i e 2 M ) .

We compute the kernel for

d 5 × : ( e 2 M β ( α γ β ) i + 1 e 3 M ) .

We have

d 5 × ( 0 , ( a 1 , a 2 , , a j ) t , 0 ) = ( 0 , 0 , ( 0 , 0 , , 0 , a 1 , , a j i ) t ) ,

where a 1 is in the coordinate number i + 2 . Hence

ker ( e 2 M β ( α γ β ) i + 1 e 3 M ) = { ( 0 , ( 0 , 0 , , 0 , a 1 , , a j ( i + 1 ) ) t , 0 ) | a r k , r = 1 , , j ( i + 1 ) } ,

where a 1 is in the coordinate number j i . We compute the image for

d 4 × : ( e 1 M α ( γ β α ) n i e 2 M )

to have

d 4 × ( ( b 1 , b 2 , , b s , , b j ) t , 0 , 0 ) = ( 0 , ( 0 , 0 , , 0 , b 1 , , b s ) t , 0 ) ,

where b 1 is in the coordinate number n i + 1 and therefore b s is in the coordinate number n i + s = j , s = j + i n . Consequently, s + 1 = j + i + 1 n . This shows that

( e 1 M α ( γ β α ) n i e 2 M ) = { ( 0 , ( 0 , 0 , , 0 , b 1 , , b j i ) t , 0 ) | b t k , t = 1 , , j } ,

where b 1 is in the coordinate number n i + 1 . We have

dim k E x t Λ 4 ( ( S 1 , , S 1 ) t , M ) = n i + 1 ( j i ) = n + 1 j

for j n + 1 . We discuss the fifth Ext-group.

E x t Λ 5 ( ( S 1 , , S 1 ) t , M ) = ker ( e 3 M γ ( β α γ ) n i e 1 M ) / ( e 2 M β ( α γ β ) i + 1 e 3 M ) .

We compute the kernel for

d 6 × : ( e 3 M γ ( β α γ ) n i e 1 M ) .

We have

d 6 × ( 0 , 0 , ( a 1 , a 2 , , a s , , a j ) t ) = ( ( 0 , 0 , , 0 , a 1 , , a s ) t , 0 , 0 ) ,

where a 1 is in the coordinate number n i + 2 and therefore a s is in the coordinate number n i + 1 + s = j , s = j + i 1 n and consequently, s + 1 = j + i n . This shows that

ker ( e 3 M γ ( β α γ ) n i e 1 M ) = { ( 0 , 0 , ( 0 , 0 , , 0 , a s + 1 , , a j ) t ) | a r ( k , r = s + 1 , , j ) } .

We compute the image for

d 5 × : ( e 2 M β ( α γ β ) i + 1 e 3 M ) .

We have

d 5 × ( 0 , ( b 1 , b 2 , , b j ) t , 0 ) = ( 0 , 0 , ( 0 , 0 , , 0 , b 1 , , b j i ) t , 0 , 0 ) ,

where b 1 is in the coordinate number i + 2 . Hence

( e 2 M β ( α γ β ) i + 1 e 3 M ) = { ( 0 , 0 , ( 0 , 0 , , 0 , b 1 , , b j i ) t ) | b t k , t = 1 , , j } .

We therefore have

dim k E x t Λ 5 ( ( S 1 , , S 1 ) t , M ) = i + 2 ( j + i n ) = n + 2 j for j n + 2 .

dim k E x t Λ t ( 3 i + 1 , 3 j + 1 ) = n + 1 j for t = 1 , 2 , 3 and 4, and n + 2 j for t = 5 . We therefore have n + 1 = j and n + 2 = j .

It follows from the fifth Ext-group that the condition E x t Λ n ( M , N ) = 0 for n 0 E x t Λ n ( N , M ) = 0 for n 0 holds.

4. Conclusion

The study found that the Ext-groups of pairs ( M , N ) of modules over Naka- yama algebras of type ( n , n , n ) satisfy the condition E x t Λ n ( M , N ) = 0 for n 0 E x t Λ n ( N , M ) = 0 for n 0 and justified the claim with projectives of lengths 3 n + 1 and 3 n + 2 by using combinations of modules of different lengths.

Acknowledgements

We are grateful to the referee for many useful suggestions.

Cite this paper: Gyamfi, K. , Aidoo, A. and Y. B. Annor, D. (2017) Homological Properties in Relation to Nakayama Algebras. Journal of Applied Mathematics and Physics, 5, 153-167. doi: 10.4236/jamp.2017.51016.
References

[1]   Schröer, J. (2016) Nakayama Algebras, 2016.
www.math.un-bonn.de/people/Schroer/fd-atlas/FD-NakayamaAlgebras.pdf

[2]   Rotman, J.J. (2009) An Introduction to Homological Algebra. 2nd Edition, Universitex, Springer.

[3]   Auslander, M., Reiten, I. and Smalø, S. (1995) Representation Theory of Artin Algebras. Series: Cambridge Studies in Advanced Mathematics (No. 36). Cambridge University Press, Cambridge.

 
 
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