A construction of tilting modules associated with a simple module

JOURNAL

OF ALGEBRA

145, 224-230 (1992)

A Construction of Tilting Modules with a Simple Module

Associated

HISAAKI FUJITA Institute of Malhematics, University of Tsukuba, Tsukuba-shi, Ibaraki 305, Japan Communicated by Kent R. Fulller Received March 5, 1990

Let R be a ring with identity and r a positive integer. A right R-module T is a tilting module (of projective dimension
...

+P,+PO+T+O,

where P,, .... P, are finitely generated and projective, (2) Ext’,( T, T) = 0 for all integers 1 d i < r, and that (3) there is an exact sequence of right R-modules 0 + R + T, + T, + . . . -+ T,y+ 0, where T,, .... T, are in add(T,), i.e., direct summands of finite direct sums of copies of T (see Miyashita [S]). If T is a tilting module and proj.dim(T,) = r, then note that s = Tcodim(R,)=r (see [4, 61) and we call Tan r-tilting module. Associated with certain simple modules over artin algebras, constructions of l-tilting modules are given in Auslander, Platzeck, and Reiten [ 1] and Brenner and Butler [2]. In [S, Sect. 41, Miyashita considered an analogue of them in the case of r-tilting modules. In this note we introduce a general construction of an r-tilting module associated with a certain simple module over a semiperfect Noetherian ring and we give an example to show that our construction covers the previous one properly. In what follows, let R be a basic semiperfect Noetherian ring, J its Jacobson radical, R = R/J, and e a primitive idempotent of R. The functor Hom,( , R) is denoted by ( )*. We write homomorphisms on the opposite side of the scalars. We shall prove the following 224 0021-8693/92 $3.00 Copyright 0 1992 by Academic Press, Inc. All rights of reproduction in any form reserved.

225

TILTING MODULES

THEOREM. Assume that (a) Exti(&?, R.?)=0 and (b) Extk(&, R)=O for 0 6 i < r. Then for each i (1 d i 6 r), there exists a right R-module Yi such that the endomorphism ring of Y, is local and that (1 - e) R @ Yi is an i-tilting module.

We begin with a general result, from which we can afford the desired Yls of the theorem. Let S be an arbitrary ring with identity, A4 a maximal ideal of S such that S/M is Artinian, and U a simple left S-module such that MU = 0. For a positive integer r, consider a set of short exact sequences 0 + K,$+ QiA L, + 0 (0
1. For any left S-module X, the following

statements are

equivalent.

(1) Ext’,(U,X)=Ofor

Ofi
* fi (2) t&Q,,-Q++ Homd , Xl.

. ..-%QT'

zs exact, where ( )* is the functor

Proof We first show the assertion when r = 1. Since Cok f, E QO/L,, Kerf: g (QO/L1)*. Apply ( )* to 0 + K,,/L, + QO/L, -+ U+ 0. Then we obtain an exact sequence 0 + U* -+ (QO/L,)* + (K,/L,)*. Since S/M is simple Artinian, KO/L1 is isomorphic to a direct sum of copies of U. Thus U* = 0 if and only if 0 -+ Q$ -% Q: is exact. Let r > 1. By induction on r, it is sufficient to show that if Exti. U, X) = 0 for 0 d i < r then Ext’,( U, X) = 0 if and only if QT- 1& Q,? ‘:+I + QF+r is exact. Apply ( )* to the following commutative diagram with exact column and rows Q r+l fi+ I I r

&

4 0Since (KJL,,

KrIL,

I-

QJL,

1-

L,-0 II L,-

0.

,)* = 0, L,* g (Q,/Lr+ ,)*, so that

isexact.Hence,asf,*=g,*~j,*~h,*_,,Qf-,zQ,* “+‘bQF+, is exact if and only ifj,* 0h,*_ , is an epimorphism. Sincej,* is a monomorphism, this is equivalent to Extk(K,- ,/L,, X) = 0 and Extt(L,- r, X) = 0. On the other

226

HISAAKI

FUJITA

hand, note that Extk( U, X) E Exti- ‘(KO, X). From the short exact sequence 0 + L1 ---)K,, -+ K,/L, -+ 0, we obtain an exact sequence O=Ext’,-‘(K,/L,,

X)-,Ext’,-‘(K,,,

X)+Ext>-‘(L,,

X)+Ext;(K,/L,,

X).

Using the assumption, we obtain Ext;-i(L,, X) g Ext;;‘(K,, X) g ExtLp2(L2, X)r ... z Exti(L, _ i, X). It is now easy to conclude the assertion. The following lemma is well known in noncommutative Noetherian ring theory (cf., e.g., [3, Chap. ll]), but we give a proof for the readers convenience. LEMMA 2. Let S be a right Noetherian ring with identity, M and N maximal ideals of S such that S/M and S/N are Artinian, and let U and V be simple right S-modules such that UM = 0 and VN = 0. Then Extk. V, U) = 0 if and only if M n N = NM.

Proof:

Consider the short exact sequence of right S-modules 0 --) N/NM + SfNM + SJN -+ 0.

(e)

Since S/(Mn N) is semisimple, (e) splits if and only if Mn N = NM. If Exti( V, U) = 0 then (e) splits. If Mn N= NM then any extension of U by V splits, so that Exti( V, U) = 0. Construction of Yi. Put M= ann,(i&?). Since R is semiperfect, R/M is simple Artinian. Put L, = R?, Q, = Re, and K, = Je. For i 2 1, inductively, take a projective cover gi: Qj -+ Li of Li and put K,= put L,=MK,-,, Ker gi. Let fi: Qi- Qi- i be the map g,jihi_, where ji: Li-+ K,- i and hi-11 Ki-1 -+ Qip i are inclusion maps. Then we put Y, = Cok f T. Remark. To use the minimal projective resolution of Et? seemed to be natural to construct an r-tilting module (see [.5, Theorem 4.31). This is, however, not always available. EXAMPLE. Let D be a local Dedekind domain with a unique maximal ideal ~0. Let n be the tiled D-order

D

rt2D

DDDD D

7cD

D

D

n2D

XD

7cD D

227

TILTING MODULES

Put k = D/nD. Let R be the k-algebra A/n/l and let ei (ER) be the image of the matrix unit eii of A for 1 6 i< 5. We denote by D( ) the duality functor Hom,( , k). Then the minimal projective resolution of D(Re4)R is given by O-+e,R-+e,R@e,R+e,R@e,R +e,R@e,R+D(Re,)+O.

Note that D( R)R is isomorphic to (1 - e4)R @ D( Re,). Hence for 0 6 i < 2, Ext,(D(R), F,R) = 0. Note also that all maximal ideals of R are idempotent. Therefore R satisfies (a) (by Lemma 2) and (b) of the theorem for e = e4. But, RZ, appears in the top of the kernel of the projective cover of L, = Je,, while the Y, is isomorphic to D(Re,) and (1 - e,)R @ Y, is a 3-tilting module. The next lemma may be essentially known (cf., e.g., [S, Proposition 4.2]), but we give a sketch of a proof. LEMMA 3. Let X be a finitely generated left R-module which does not have projective direct summands and satisfies Exti(X, R) = 0. Let f: P + X be a projective cover of X and K = Ker f Then RK has no projective direct summands and M,(K) E M,(X).

Proof If K has a projective direct summand Q #O then the inclusion map Q + P splits, which contradicts that K is small in P. Let CIE End,(X). Then, since P is projective, there exist /?E End,(P) and y E End,(K) such that the following diagram is commutative.

1

O-K-P-X-O

1

/

1

If CIfactors through a projective module (hence P) then so does y. This defines a map from M,(X) to m,(K). Since Exti(X, R) = 0, this map is one-to-one and onto, so that it becomes a ring isomorphism. In order to show that the endomorphism rings of Y, are local, we need the following LEMMA 4. Assume that Extk(&, RZ) = 0 and Hom,(Rt?, R) = 0. Let g,: P, + X be a projective cover of a finitely generated left R-module X, K,, = Ker g,, and L, = MK, and let g, : P, + L, be a projective cover of L,. Letf:P,+P,bethemapP,-%L,qK,,~POand Y=Cokf*.

228

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FUJITA

(1) Assume that X is torsionless and that P, has no direct summands isomorphic to Re. Then tf RX is non-projective with a local endomorphism ring then so is Y,. (2) Moreover, assume that ExtL(&?, R) = 0 and Exti(X, R) = 0. Then tf RX is non-projective with a local endomorphism ring then so is RL, . Proof Put K = KO/L1 and X’ = P,/L, ., Since L, c K, c rad( P,,), /?:P, + X’ is a projective cover. Hence P, .f P, + X’ -+ 0 is a minimal projective presentation of X’ and Y = Cok f* = Tr(X’). Assume that X’ is projective. Then MK, = L, = 0 and so K, is isomorphic to a finite direct sum of copies of RF, so that (K,,)* = 0. Since K, c P,, K, = 0. Hence X is projective.

(1) It is sufficient to show that End,(X’) z End,(X). Consider the short exact sequence O-K-X’*X-0.

(e)

Apply Hom,( , X) to (e). Then since RX is torsionless and Hom,(&, R) = 0, Hom,(cc, X): End,(X) + Hom,(X’, X) is a group isomorphism. Apply Hom,(X’, ) to (e). Then by the assumption, Hom,(X’, a): End,(X) + Hom,(X’, X) is a monomorphism. Let ge Hom,(X’, X). Then there exists h’: P, + X’ such that h’a = pg. Since (L,) h’a = (L,) fig= 0, Kr Ker a 3 (L,)h’. Since M is indempotent by Lemma 2, (L,)h’=M(L,)h’c M(Ker a)=O. Hence there is a map h: X’ + X’ such that h’ = ph. Hence /?ha = h’a = pg. Since a is onto, g = ha, so that Hom,(X’, a) is a group isomorphism. Hence Hom,(X’, a)-‘~ Hom,(a, X): End,(X) + End,(X’) becomes a ring isomorphism. (2) Apply ExtL( , R) to (e). Then by the assumption we obtain ExtL(X’, R) = 0. Hence by Lemma 3 we can conclude the assertion. Proof of Theorem. It is sufficient to show that the endomorphism ring of Y, is local and that (1 - e) R @ Y, is an r-tilting module. We divide the proof into several steps.

(1) For 1~ i < r, Qi has no direct summands isomorphic to Re. Proof M(L,/JL,)

(2)

Since A4 is idempotent by (a) and Lemma 2, ML, = L, and so = Li/JLi, so that Qi has no direct summands isomorphic to Re. For 1 Q id r, Yi is non-projective with a local endomorphism

ring. Proof. It follows from (a) that L1 = K,. Hence, when r= 1, Y, = Tr(&). Since Hom,(&!, R) = 0, R? is not projective. Hence Y, is nonprojective with a local endomorphism ring. When r = 2, it follows from

TILTING

229

MODULES

Lemma 3 that RL, is non-projective with a local endomorphism ring. Hence, by (1) and Lemma 4(l), Y, is non-projective with a local endomorphism ring. Let r > 3. As in the proof of Proposition 1, the assumption (b) implies that ExtL(L,, R) g ExtRf ‘(RF, R) = 0 for 1 d ib r- 2. Therefore, using (1) and Lemma 4 as above, we conclude the assertion. (3) The minimal projective resolution of Y, is given by 0Proof

Re*

,...-,

1;,Q:”

Q; -

y, -

0.

(P)

This follows from Proposition 1.

(4) Ext’,(Y,,(l-e)R)=Ofor Proof:

“’

l
Consider the complex (1-e)R@Q,-

... -+(l-e)R@Q,-(l-e)R@QO ‘(1 -e)R@Re+O.

(*)

Apply ( 1 - e) R @- to the short exact sequence

Since R is basic, (1 - e)R@ RF= 0. Hence we have an isomorphism (l-e)R@Li+(l-e)R@Kip, for l
for 1
ProojI Apply - @RF to the projective resolution (p) of Y,. It follows from (1) that QT @ &! = 0 for 1 < i < r, so that the assertion holds. l
and

230

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FUJITA

we obtain Tor,!( Y,, Li) z Tar,!+ i( Y,, Qi- ,/Li) 2 Tar,!+ i( Y,, Lj- i) for 1 < j < r - 2. Hence by induction on i, we conclude the assertion. (7) Ext’,(Y,, Y,)=O for 1
Consider the complex

Since Hom,(QT, Y,)g Y,@Qi for O
REFERENCES

1. M. AUSLANDER,M. I. PLATZECK,AND I. REITEN, Coxeter functors without diagrams, Trans. Amer. Math. Sot. 250 (1979), 146. 2. S. BRENNERAND M. C. BUTLER,Generalization of BernsteinCrelfand-Ponomarev reflection functors, in “Lecture Notes in Math.,” Vol. 832, pp. 103-169, Springer, Berlin, 1980. 3. K. R. GOODEARL AND R. B. WARFIELD, JR., An introduction to non-commutative Noetherian rings, in “London Math, Sot. Stud. Texts,” Vol. 16, Cambridge Univ. Press, London/New York, 1989. 4. D. HAPPEL, Triangulated categories in the representation theory of linite dimensional algebras, in “London Math. Sot. Lecture Note Ser.,” Vol. 119, Cambridge Univ. Press, London/New York, 1988. 5. Y. MIYASHITA, Tilting modules of finite projective dimension, Math. 2. 193 (1986), 113-146. 6. T. WAKAMATSU, On modules with trivial self-extensions, J. Algebra 114 (1988), 106-I 14.