Table Of Content

12 FINITISTIC DIMENSION THROUGH INFINITE PROJECTIVE DIMENSION. FRANÇOISHUARD,MARCELOLANZILOTTA,OCTAVIOMENDOZA 7 0 0 Abstract. WeshowthatanartinalgebraΛhavingatmostthreeradicallay- 2 ersofinfiniteprojectivedimensionhasfinitefinitisticdimension,generalizing p the known result for algebras with vanishing radical cube. We also give an e equivalence between the finiteness of fin.dim.Λ and the finiteness of a given S classofΛ-modulesofinfiniteprojectivedimension. 6 2 1. Introduction. ] T LetΛbe anartinalgebra,andconsidermodΛthe classoffinitely generatedleft R Λ-modules. The finitistic dimension of Λ is then defined to be . h t a fin.dim.Λ=sup{pdM : M ∈modΛ and pdM <∞}, m where pdM denotes the projective dimension of M. It was conjectured by Bass in [ the 60’s that fin.dim.Λ is always finite. Since then, this conjecture was shown to hold for many classes of algebras [2, 7, 4, 9, 10, 11]. In particular the conjecture 1 v holds for artin algebras with vanishing radical cube [2, 5, 10]. In this paper, we 3 generalizethisresulttoartinalgebrashavingatmostthreeradicallayersofinfinite 5 projective dimension. 2 4 TheoremA.If Λhasatmostthreeradicallayersofinfiniteprojectivedimension, . Λ 9 then fin.dim.Λ is finite. 0 7 We also provide a bound on the finitistic dimension of Λ under the preceding 0 hypothesis. In order to achieve our goal, we use the Ψ function of Igusa and : v Todorov [7] and introduce the notion of infinite-layer length which counts in an i efficientmanner the number of (notnecessarilyradical)layersofinfinite projective X dimensionofamodule. OurnextresultshowsthatthefinitisticdimensionofΛisin r a somesensecontrolledbythe syzygiesofthe simple Λ-modulesofinfinite projective dimension. Let us denote by Ωn(M) the n-th syzygy of M in modΛ. Theorem B. The following conditions are equivalent for an artin algebra Λ. (a) The finitistic dimension of Λ is finite, (b) There exists s∈N such that {Ωs(M)}M∈K(Λ) is in addΩs+1(Σ), (c)Thereexistss∈Nsuchthat{Ωs(M)}M∈K(Λ) isoffiniterepresentationtype. whereΣisthe directsumofallisomorphismclassesofsimpleΛ-modulesofinfinite projective dimension, and K (Λ) is the class formed by taking all indecomposable M in modΛ satisfying the following two properties: (a) pdM =∞ and (b) M is a summand of radL for some L∈modΛ of finite projective dimension. 1 2000Mathematics SubjectClassification: 16G70, 16G20,16E10 2 Keywordsandphrases: finitisticdimension,compositionfactors. 1 2 F.HUARD-M.LANZILOTTA-O.MENDOZA 2. Notations and definitions. For an artin algebra Λ and a Λ-module M, we denote by topM and socM the top and socle of M respectively. Given a class C of objects in modΛ, the projective dimension of C is pdC := sup{pdM : M ∈ C} if the class C is not empty, otherwise it is zero. Moreover, the finitistic dimension of the class C is fin.dim.C :=pd{M ∈C : pdM <∞}. We let S∞ be the finite set consisting of allisomorphismclasses ofsimple Λ-modules of infinite projective dimension. Simi- larly,wedenotebyS<∞ thefinitesetconsistingofallisomorphismclassesofsimple Λ-modules of finite projective dimension. Then α:=pdS<∞ is finite. Further, we denote by [M :S∞] the number of(not necessarilydistinct) compositionfactors of M of infinite projective dimension. Note that if all the composition factors of M belong to S<∞ or M =0, then we have [M :S∞]=0 and pdM ≤α. We now recallthe definition andmain properties of the function Ψ of Igusa and Todorov [7]. Let K denote the quotient of the free abelian group generated by all the symbols [M], where M ∈modΛ, by the subgroup generated by symbols of the form: (a) [C]−[A]−[B] if C ≃ A⊕B, and (b) [P] if P is projective. Then K is thefreeZ-module generatedbytheiso-classesofindecomposablefinitely generated non-projective Λ-modules. In [7], K. Igusa and G. Todorov define the function Ψ:modΛ→N as follows. The syzygy induces a Z-endomorphism on K that will also be denoted by Ω. That is, we have a Z-homomorphism Ω : K → K where Ω[M] := [ΩM]. For a given Λ-module M, we denote by < M > the Z-submodule of K generated by the indecomposable direct summands of M. Since Z is a Noetherian ring, Fitting’s lemmaimpliesthatthereisanintegernsuchthatΩ:Ωm <M >→Ωm+1 <M > is anisomorphismfor all m≥n; hence there exists a smallest non-negativeinteger Φ(M) such that Ω : Ωm < M > → Ωm+1 < M > is an isomorphism for all m≥Φ(M).LetC bethesetwhoseelementsarethedirectsummandsofΩΦ(M)M. M Then we set: Ψ(M):=Φ(M)+fin.dim.C . M The following result is due to K. Igusa and G. Todorov. Proposition 2.1. [7] The function Ψ : modΛ → N satisfies the following properties. (a) If pd M is finite then Ψ(M)=pd M. On the other hand, Ψ(M)=0 if M is indecomposable and pd M =∞, (b) Ψ(M)=Ψ(N) if add M =add N, (c) Ψ(M)≤Ψ(M ⊕N), (d) Ψ(M ⊕P)=Ψ(M) for any projective Λ-module P, (e) If 0→A→B →C →0 is an exact sequence in modΛ and pd C is finite then pd C ≤Ψ(A⊕B)+1. Y. Wang proved the following useful inequality, which is a direct consequence of 2.1 (e). Lemma 2.2. [8] If 0→A→B →C →0 is an exact sequence in modΛ and pd B is finite then pd B ≤2+Ψ(ΩA⊕Ω2C). We will also need the following result which appears in our previous paper [6]. Proposition 2.3. [6] For all M in modΛ, Ψ(M)≤1+Ψ(ΩM) FINITISTIC DIMENSION THROUGH INFINITE PROJECTIVE DIMENSION. 3 For a subcategory C of modΛ, we set Ψdim.C :=sup{Ψ(M) : M ∈C}. Since clearly Ψdim.C < ∞ implies fin.dim.C < ∞, in view of the finitistic dimension conjecture, it would be interesting to study this new invariant. 3. The functors Q and S In this section we introduce endofunctors Q and S on modΛ that associate to a Λ-moduleM aquotientQ(M)andasubmoduleS(M)ofM withthepropertiesthat thesocleofQ(M)andthetopofS(M)bothlieinaddS∞ whenever[M :S∞]6=0. Thesefunctorswillbeusedinthedefinitionoftheinfinite-layerlengthofamodule. Throughout this section, we let α=pdS<∞. Given any class C of Λ-modules, we will consider the full subcategory F(C) of modΛ whose objects are the C-filtered Λ-modules. That is, M ∈F(C) if there is afinitechain0=M ⊆M ⊆···⊆M =M ofsubmodulesofM withm≥0and 0 1 m such that each quotient M /M is isomorphic to some object in C. For example, i i−1 an object M ∈modΛ is filtered by S<∞ if and only if [M :S∞]=0. Lemma 3.1. Let M be a Λ-module. Then the set of all submodules of M filtered by S<∞ admits a unique maximal element that will be denoted by K(M). Proof. Note first that this class is not empty since [0:S∞]=0. Now let K and 1 K be two submodules of M filtered by S<∞ that are maximal with this property. 2 Since K +K is a quotient of K ⊕K ∈ F(S<∞), we get that K +K is also 1 2 1 2 1 2 filtered by S<∞. Hence K =K +K =K . (cid:3) 1 1 2 2 Let Q(M) be the quotient M/K(M), that is we have the following exact sequence: 0 → K(M) i→M M → Q(M) → 0. Given a morphism f : M → N of Λ-modules, we have the following diagram: 0 // K(M) iM //M //Q(M) // 0 (cid:31) (cid:31) (cid:31) (cid:31) K(f) f Q(f) (cid:31) (cid:31) 0 //K((cid:15)(cid:15)N) iN //N(cid:15)(cid:15) // Q(N(cid:15)(cid:15) ) //0 Now fi (K(M)) is a quotient of K(M)∈F(S<∞) therefore it is a submodule of M N filtered by S<∞. It then follows fromthe maximality of K(N) that fi factors M uniquely through i , giving the map K(f). Passing to the cokernels, we obtain a N map Q(f) : Q(M) → Q(N) that makes the above diagram commute. It is then straightforwardtoverifythatK andQ,asdefinedabove,areindeedfunctorswhose main properties are listed below. Note that pdK(M)≤α. Proposition 3.2. The functors K, Q : modΛ → modΛ defined above have the following properties. (a) Q(M)=0 if and only if M ∈F(S<∞). (b) If socM ∈addS∞ then Q(M)=M, (c) pdM <∞ if and only if pdQ(M)<∞, (d) If [M :S∞]6=0 then socQ(M)∈addS∞, (e) Ωα+1M ⊕P ∼=Ωα+1Q(M)⊕P′ for some projective Λ-modules P and P′, 4 F.HUARD-M.LANZILOTTA-O.MENDOZA (f) pdM ≤max{pdQ(M),α}, (g) If f : M → N is a monomorphism (epimorphism), then Q(f) : Q(M) → Q(N) is a monomorphism (epimorphism), (h) Q2 =Q, K2 =K and KQ=0=QK. Proof. Statements (c), (e) and (f) are easily verified using the exact sequence 0→K(M)→M →Q(M)→0 and the fact that pdK(M)≤pdS<∞ =α<∞. (a) By definition, we have the equivalences 0 = Q(M) ⇐⇒ K(M) = M ⇐⇒ [M :S∞]=0. (b) If K(M)6= 0, then 0 6= socK(M)⊆ socM ∈ addS∞, a contradiction since K(M)∈F(S<∞). Thus K(M)=0 and Q(M)=M. (d) Since [M : S∞] 6= 0, we have from (a) that Q(M) 6= 0. Assume that socQ(M) admits a simple summand S of finite projective dimension, and consider the following commutative diagram i 0 // K(M) // E //S //0 f (cid:15)(cid:15) (cid:15)(cid:15) 0 // K(M) // M //Q(M) //0 wheretheupperexactsequenceisobtainedbypullback. Applyingthesnakelemma, we infer that f is a monomorphism. Moreover, E is filtered by S<∞ since both K(M) and S are so. The maximality of K(M) then implies that i is an isomorphism, thus S =0, a contradiction. (g) Let f : M → N be a morphism of Λ-modules. Consider the following exact and commutative diagram 0 // K(M) // M //Q(M) //0 K(f) f Q(f) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) 0 //K(N) // N // Q(N) //0 If f :M →N is an epimorphism, then from the diagram above, we see that Q(f) is also an epimorphism. Suppose that f : M → N is a monomorphism. If M ∈ F(S<∞), then by (a) Q(M)=0; and hence Q(f) is a monomorphism. Assume now that [M :S∞]6= 0. It then follows from (d) that socQ(M) ∈ addS∞. Applying the snake lemma to the diagram above, we get a monomorphism from X := KerQ(f) to Y := CokerK(f). IfX 6=0,then06=socX ⊆socQ(M)∈addS∞. Also,socX ⊆socY, a contradiction since Y ∈F(S<∞). So X =0 and Q(f) is a monomorphism. (h)Bytakingthepull-backtothecanonicalmorphismsM →Q(M)←KQ(M), we get the following exact and commutative diagram FINITISTIC DIMENSION THROUGH INFINITE PROJECTIVE DIMENSION. 5 0 0 j (cid:15)(cid:15) (cid:15)(cid:15) 0 // K(M) // E //KQ(M) // 0 (cid:15)(cid:15) (cid:15)(cid:15) 0 // K(M) //M //Q(M) // 0 (cid:15)(cid:15) (cid:15)(cid:15) Q2(M) Q2(M) (cid:15)(cid:15) (cid:15)(cid:15) 0 0 Hence, by 3.1 and the fact that E ∈ F(S<∞), we conclude that j is an isomorphism so that KQ(M) = 0 and Q(M) = Q2(M). On the other hand, since K(M)∈F(S<∞), we get from 3.2(a) that QK(M)=0 and K2(M)=K(M). (cid:3) We now proceed to give the construction of the functor S which is dual to Q. We omit the proofs since they are essentially the same as those we previously did. In what follows, a quotient of M is an epimorphism g : M → C; moreover, if g : M → C and g : M → C are two quotients of M, we say that g is greater 1 1 2 2 1 than or equal to g if there is an epimorphism h : C → C such that hg = g . 2 1 2 1 2 Finally, we say that a quotient g :M →C is filtered by S<∞ in case C is so. Lemma 3.3. Let M be a Λ-module. Then the set of all quotients of M filtered by S<∞ admits a unique (up to isomorphism) maximal element that will be denoted by p :M →C(M). M Considertheexactsequence0→S(M)→M p→M C(M)→0. Givenamorphism of Λ-modules f :M →N, we have the following diagram: 0 //S(M) //M pM //C(M) // 0 (cid:31) (cid:31) (cid:31) (cid:31) S(f) f C(f) (cid:31) (cid:31) 0 //S(N(cid:15)(cid:15) ) // N(cid:15)(cid:15) pN //C(N(cid:15)(cid:15) ) //0 Now p f(M) is a submodule of C(N) ∈ F(S<∞), therefore M → p f(M) is a N N quotient of M filtered by S<∞. It then follows from the maximality of p :M → M C that p f factors uniquely through p giving us the map C(f). By passing to M N M kernels,weobtainthemapS(f)thatmakestheabovediagramcommute. Itisnow straightforward to verify that S and C are functors. We then have the following properties of such functors, the proof of which is dual to 3.2. Proposition 3.4. The functors S, C : modΛ → modΛ, defined above, have the following properties. (a) S(M)=0 if and only if M ∈F(S<∞), (b) If topM ∈addS∞, then S(M)=M, 6 F.HUARD-M.LANZILOTTA-O.MENDOZA (c) pdM <∞ if and only if pdS(M)<∞, (d) If [M :S∞]6=0 then topS(M)∈addS∞, (e) ΩαM ⊕P ∼=ΩαS(M)⊕P′ for some projective Λ-modules P and P′, (f) pdM ≤max{pdS(M),α}, (g) If f : M → N is a monomorphism (epimorphism), then S(f) : S(M) → S(N) is a monomorphism (epimorphism), (h) C2 =C, S2 =S and CS =0=SC. Note that it can also be shown that QS and SQ are naturally isomorphic functors. 4. The infinite-layer length of a module. In this section, we introduce the invariant ℓℓ∞(M) for a Λ-module M. Our goal is to count the number of ”layers” of infinite projective dimension of M. A natural way to proceed would be to consider radical layers. Recall that a radical layerofamoduleM isasemisimplemoduleoftheformradiM/radi+1M forsome 0≤i <ℓℓ(M), where ℓℓ(M) is the Loewy length of M. It then seems reasonable to define the ”infinite-layer length” ℓ∞(M) of a module M as follows. Definition 4.1. For any M ∈ modΛ, we set ℓ∞(M) to be the number of radical layers of M that have infinite projective dimension. As natural as it seems, this definition has some flaws. Firstly, if one proceeds duallyanddefinesℓ (M)usingsoclelayers,itisnottrueingeneralthatℓ∞(M)= ∞ ℓ (M). Also this length does not behave well with submodules. That is, if K is a ∞ submodule of M, we do not always have that ℓ∞(K)≤ℓ∞(M). We willnowdefine another”infinite-layerlength” thatsatisfiesthe abovestated properties and is better than ℓ∞ in the sense that ℓℓ∞(M) ≤ ℓ∞(M). As an analogy, the Loewey length of a module M can be defined to be the smallest nonnegativeintegerisuchthatradiM =0. Inourcase,sinceweareonlyinterested inlayersofinfiniteprojectivedimension,weusethefunctorS to”level”ourmodule prior to taking the radical. This guarantees that at each step, a layer consisting of a maximal number of simples of infinite projective dimension is removed. Given a module M, we start by calculating S(M) = M0. Unless S(M) = 0, the top of M0 lies in addS∞. We then take the radical of M0, and let M1 := S(rad(M0)). Iterating the process, we let Mi+1 :=S(rad(Mi)) for all i≥0. This procedure leads us to consider the following additive functor F :=S◦rad :modΛ→modΛ, whereS isthefunctordefinedintheprevioussectionandrad istheradicalfunctor. Note that F0 is the identity functor; and so, by using the functor F, it is now easy to see that Mi =Fi(S(M)) for all i≥0. Definition 4.2. Let Λ be an artin algebra. For a finitely generated Λ-module M, we define the infinite-layer length of M to be ℓℓ∞(M):=min{i≥0 : Fi(S(M))=0}. FINITISTIC DIMENSION THROUGH INFINITE PROJECTIVE DIMENSION. 7 Example 4.3. Consider the bound quiver algebra Λ=kQ/I where Q is given by α (cid:18)(cid:18) >> •1 |µ|||1||||||||µ|||2|>> BBBBBβBBB 5• •2 ρ4•OO oo •3~~|γ|~~||1||||||||γ|||2| δ and I is generated by the set of paths {α3,αβ,ρµ α,µ β,γ δ−γ δ} where i=1,2. i i 1 2 Then S∞ ={S(1),S(4)} and the projective Λ-modules are 1 2 == (cid:2)(cid:2) == 1 2 3 3 3 5 ::: ::: (cid:4)(cid:4)(cid:4) (cid:4)(cid:4)(cid:4) ::: P(1)= 1 3 3 P(2)= 4 P(3)= 4 P(5)= 1 1 (cid:4)(cid:4)(cid:4) 4 5 5 1 1 (cid:2)(cid:2) == == 5 1 1 1 1 1 1 (cid:2)(cid:2) == 1 1 and P(4)=radP(3). In order to compute ℓℓ∞(P(1)), we need to find the smallest nonnegative integer i such that Fi(S(P(1))) = 0. Note that S(P(1)) = P(1) since topP(1) = S(1) ∈ addS∞. Thus FS(P(1)) = F(P(1)) = Srad(P(1)) is the following module: 4 FS(P(1))= 1 ⊕ 5 (cid:2)(cid:2)(cid:2) ::: 1 1 1 Continuing this process, we obtain F2S(P(1)) = (S(1))3 and F3S(P(1)) = 0, thus ℓℓ∞(P(1)) = 3. It is easy to see that among the six radical layers of P(1) only the fifth has finite projective dimension, therefore ℓ∞(P(1)) = 5. Note that P(1) has four socle layers of infinite projective dimension, giving ℓ (P(1)) = 4. ∞ Further computations show that ℓℓ∞(P(2)) = ℓℓ∞(P(3)) = ℓℓ∞(P(4)) = 2 and that ℓℓ∞(P(5))=3. Lemma 4.4. Consider the functor Fi :modΛ→modΛ for i≥0. Then (a) Fi and Fi◦S preserve monomorphisms and epimorphisms, (b) FiS(M)⊆Fi(M)⊆S◦radi(M) for all M ∈modΛ and i≥1. Proof. (a) From 3.4 (g), we know that S preserves both monomorphisms and epimorphisms. This is also true of the functor rad. Consequently, for all non negativeintegers i,we havethat Fi and Fi◦S also preservesmonomorphisms and epimorphisms. 8 F.HUARD-M.LANZILOTTA-O.MENDOZA (b) For all i ≥ 0, we have that FiS(M) ⊆ Fi(M) since S(M) ⊆ M and Fi preservesmonomorphisms. Thelastinclusionwillbeprovedbyinductiononi≥1. Firstly, we assert that (∗) rad ◦S(X)⊆rad(X) for any X ∈modΛ. Indeed, since S(X) ⊆ X, then (∗) follows from the fact that the functor rad preserves monomorphism. We now proceed to the proof of the last inclusion. If i=1 there is nothing to prove. So assume that i>1. Then by hypothesis we have that Fi−1(M)⊆S◦radi−1(M). Hence, by (a) and (∗), we get Fi(M)⊆F ◦S◦radi−1(M)=S◦rad ◦S(radi−1(M))⊆S◦radi(M). (cid:3) The next result shows that ℓℓ∞ behaves naturally with monomorphisms, epimorphisms and direct sums. Proposition 4.5. Let Λ be an artin algebra and L, M and N be Λ-modules. (a) If f :L→M is a monomorphism, then ℓℓ∞(L)≤ℓℓ∞(M), (b) If g :M →N is an epimorphism, then ℓℓ∞(N)≤ℓℓ∞(M), (c) ℓℓ∞(M ⊕N)=max(ℓℓ∞(M), ℓℓ∞(N)), (d) ℓℓ∞(M)≤ℓℓ∞( Λ). Λ Proof. (a) Let f : L → M be a monomorphism and assume that ℓℓ∞(M) = m. Then Fm(S(M))= 0; and so by 4.4(a) we get Fm(S(L)) =0. Thus ℓℓ∞(L)≤m, proving (a). A similar argument holds for (b). (c) Since the functor FiS is aditive, we have that FiS(M ⊕N)≃FiS(M)⊕FiS(N), therefore max(ℓℓ∞(M), ℓℓ∞(N)) ≤ ℓℓ∞(M ⊕N). The reverse inequality follows from the fact that Fi0S(X)=0 implies FiS(X)=0 for all i≥i0. (d) Using that M is a quotient of Λn for some natural number n, we obtain Λ from (b) that ℓℓ∞(M)≤ℓℓ∞( Λn); thus (d) follows from (c). (cid:3) Λ Proposition 4.6. Let Λ be an artin algebra and M a Λ-module. (a) ℓℓ∞(M)=0 if and only if M ∈F(S<∞), (b) ℓℓ∞(S(M))=ℓℓ∞(M), (c) If topM lies in addS∞ then ℓℓ∞(radM)=ℓℓ∞(M)−1. Proof. (a)LetM ∈F(S<∞).Thenby3.4(a)wehaveS(M)=0,thusF0(S(M))= 0 and ℓℓ∞(M) = 0. Conversely, if ℓℓ∞(M) = 0 then S(M) = 0 and hence M ∈ F(S<∞). (b) This follows from the definition and the fact that S2(M)=S(M). (c) It follows from (a) that ℓℓ∞(M)> 0 and from 3.4(b) that S(M)= M. For alli≥1,wehaveFi(S(M))=Fi(M)=(S◦rad)i−1S(radM)=Fi−1(S(radM)). Therefore ℓℓ∞(radM)=ℓℓ∞(M)−1. (cid:3) FINITISTIC DIMENSION THROUGH INFINITE PROJECTIVE DIMENSION. 9 We now proceed to show that the function ℓℓ∞ behaves well with quotient of socles. This will allow us to show that ℓℓ∞ and its dual ℓℓ , which has yet to be ∞ defined, are the same invariants. Proposition 4.7. Let Λ be an artin algebra and M a Λ-module. If socM ∈ addS∞ then ℓℓ∞(M/socM)=ℓℓ∞(M)−1. Proof. Suppose that socM ∈ addS∞ and let m = ℓℓ∞(M) > 0. We claim that N := Fm−1(S(M)) ⊆ socM. Indeed, from 4.4(a), we have 0 6= N ⊆ M and from the definition of ℓℓ∞, we have S(rad(N)) = Fm(S(M)) = 0. In particular, 0 6= socN ⊆ socM ∈ addS∞. Now if N is not semisimple, then radN 6= 0. Therefore,we get from3.4(a) that S(rad(N))6=0 since 06=soc(radN)⊆socN ∈ addS∞. This is a contradiction, hence N is semisimple and N = socN, proving that N ⊆socM. Consider the projection p:M →M/socM. Then by 4.4(a) the map q =Fm−1S(p):Fm−1S(M)→Fm−1S(M/socM) is also an epimorphism and is actually the restriction of p to Fm−1S(M). We therefore have the following commutative diagram q Fm−1S(M) // Fm−1S(M/socM) i j (cid:15)(cid:15) p (cid:15)(cid:15) M //M/socM whereiandj arethe inclusions. Fromwhatprecedes,weknowthatFm−1S(M)⊆ socM,thusjq =pi=0andhenceq =0,implyingthatFm−1S(M/socM)=0and ℓℓ∞(M/socM)≤m−1. Assumenowthatℓℓ∞(M/socM)=l <m−1.Thenbyre- placingm−1forlintheabovediagram,weget0=jq =pihenceFlS(M)⊆socM. But then 0 = (S ◦rad)FlS(M) = Fl+1S(M) gives us ℓℓ∞(M) ≤ l +1 < m, a contradiction. (cid:3) We can now define the dual of ℓℓ∞. The strategy is the same as before except that we now proceed from bottom to top using the functor Q. The definition is slightly more tedious since we have to proceed with quotients by socles instead of radicals. Given a module M, we start by calculating M := Q(M). Unless 0 Q(M) = 0, the socle of M lies in addS∞. We then let M := Q(M /socM ). 0 1 0 0 Iterating the process, we let M := Q(M /socM ) for all i ≥ 0. This procedure i+1 i i leads us to consider the additive functor G:=Q/(soc ◦Q):modΛ→modΛ, where Q is the functor defined in 3.2 and soc is the socle functor. Then, by using the functor G, it is now easy to see that M =Q◦Gi(M) for all integers i≥0. i Definition 4.8. Let Λ be an artin algebra. For a finitely generated Λ-module M, we define ℓℓ (M):=min{i≥0 : Q◦Gi(M)=0}. ∞ 10 F.HUARD-M.LANZILOTTA-O.MENDOZA Example 4.9. We calculate ℓℓ (P(1)) from example 4.3. Recall that S∞ = ∞ {S(1),S(4)} hence socP(1) ∈ addS∞. Therefore QG0(P(1)) = Q(P(1)) = P(1) and QG(P(1))=Q((P(1)/socP(1)) is the following module 1 ::: QG(P(1))= 1 2 ::: 3 3 (cid:2)(cid:2) 4 Continuing this process, we get that QG2(P(1)) = S(1) and that QG3(P(1)) = 0, thus ℓℓ (P(1))=3. ∞ Proposition 4.10. Let Λ be an artin algebra and M be a finitely generated Λ- module. Then (a) ℓℓ (Q(M))=ℓℓ (M), ∞ ∞ (b) If soc(M)∈addS∞, then ℓℓ (M/socM)=ℓℓ (M)−1, ∞ ∞ (c) ℓℓ∞(D(M)) = ℓℓ (M), where D : modΛ → modΛop is the usual duality ∞ functor. Proof. (a) This follows from the definition and the fact that Q2(M)=Q(M). (b) We know from 3.2(b) that Q(M) = M. Therefore, we obtain from the definition that M = Q(Q(M)/socQ(M)) = Q(M/socM) = (M/socM) and 1 0 hence M =(M/socM) for all i≥0, giving ℓℓ (M/socM)=ℓℓ (M)−1. i+1 i ∞ ∞ (c) In order to prove (c), we need to show that D(M ) ≃ (D(M))i for all i ≥ 0 i (see in4.2). We willdo it by inductiononi.Applying the duality functor D to the exactsequence0→K(M)→M →Q(M)→0,andutilizingthefactthatDK(M) is the maximal quotient of D(M) filtered by D(S∞) together with 3.3, we get an isomorphism DQ(M)≃SD(M) which is natural on M; proving the statement for i=0. For the inductive step, we have that D(M ) = DQ(M /socM ) i+1 i i ≃ SD(M /socM ) i i ≃ Srad(DM ) i ≃ F(D(M )) i ≃ F(DQ(M )) i ≃ F(SD(M )) i = FS(D(M))i ≃ (D(M))i+1, since D(M/socM)≃rad(DM) and M =QM . (cid:3) i i We are now in position to show that ℓℓ and ℓℓ∞ coincide. ∞ Theorem 4.11. Let Λ be an artin algebra. If M is a finitely generated Λ-module, then ℓℓ∞(M)=ℓℓ (M). ∞