Kato, Kiriko

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Stable Module Theory with Kernels

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Mathematical Journal of OkayamaUniversityVolume 43, Issue 1 2001 Article 5JANUARY 2001Stable Module Theory with KernelsKiriko Kato∗∗Osaka Women’s UniversityCopyright c©2001 by the authors. Mathematical Journal of Okayama University is produced byThe Berkeley Electronic Press (bepress). http://escholarship.lib.okayama-u.ac.jp/mjouMath. J. Okayama Univ. 43(2001), 31{41STABLE MODULE THEORY WITH KERNELSKiriko KATO1. IntroductionAuslander and Bridger introduced the notion of projective stabilizationmodR of a category of ¯nite modules. The category modR is known tobe non-abelian. But realistically, modR is almost abelian. It fails to beabelian because of the lack of kernel and cokernel. In fact, each morphismhas a pseudo-kernel and a pseudo-cokernel (see x3). On the other hand, apseudo-kernel of a monomorphism does not necessarily vanish. In this paperwe focus on how modR is similar or dissimilar to an abelian category (x4).What is a monomorphism? Which object makes monomorphisms split?One reason for similarity is that modR is closely related to the homotopycategory of complexes. We discuss the functor from modR to homotopycategory (x2). The method we use already produced important results inrepresentation theory on commutative rings [2], [5].Throughout the paper, R is a commutative semiperfect ring, equivalentlya ¯nite direct sum of local rings; that is, each ¯nite module has a projec-tive cover (see [4] for semiperfect rings). The category of ¯nitely generatedR-modules is denoted by modR, and the category of ¯nite projective R-modules is denoted by projR. For an abelian category A, K(A) stands forthe homotopy category of complexes where a complex is denoted asF ² : ¢ ¢ ¢ ! Fn¡1 dFn¡1¡¡¡¡! Fn dFn¡¡! Fn+1 ! ¢ ¢ ¢ :A morphism in K(A) is a homotopy equivalence class of chain maps. Adegree-shifting T is an autofunctor on K(modR);(TF )n = Fn+1; dTF n = dF n+1;¿·nF ², ¿¸nF ² are truncations;¿·nF ² : ¢ ¢ ¢ ! Fn¡2 ! Fn¡1 ! Fn ! 0! 0! ¢ ¢ ¢ ;¿¸nF ² : ¢ ¢ ¢ ! 0! 0! Fn ! Fn+1 ! Fn+2 ! ¢ ¢ ¢ ;and F ¤² is the cocomplex such as F ¤n = (Fn)¤, dF¤n = (dFn¡1)¤ where ¤ meansHomR( ; R). The projective stabilization modR is de¯ned as follows:² Each object of modR is an object of modR.² For A;B 2 modR, a set of morphisms from A to B isHomR(A;B)=P(A;B);311Kato: Stable Module Theory with KernelsProduced by The Berkeley Electronic Press, 200132 K. KATOwhere P(A;B) := ff 2 HomR(A;B) j f factors through some pro-jective moduleg. Each element is denoted as f = f mod P(A;B). IfA;B 2 modR are isomorphic in modR, we write A st»= B.For an R-module M , de¯ne a transpose TrM of M to be Cok ±¤ whereP±¡! Q ! M ! 0 is a projective presentation of M . The transpose ofM is uniquely determined as an object of modR. If f 2 HomR(M;N),then f induces a map TrN ! TrM , which represents a morphism Tr f 2HomR(TrN;TrM). Hence Tr is an autofunctor on modR.2. A functor to the homotopy categoryLet L be a full subcategory of K(modR) de¯ned asL = fF ² 2 K(projR) j H i(F ) = 0 (i < 0); Hj(F ¤² ) = 0 (j ¸ 0)g:Lemma 2.1. For a morphism f² in L, f² = 0 in K(modR) if and only ifH0(¿·0f) = 0 in modR.Proof. Let f² : A² ! B² be a chain map with A²; B² 2 L such thatH0(¿·0f²) = 0. Then there exists g 2 HomR(H0(¿·0A²); B0) that satis¯esH0(¿·0f²) = ½±g where ½ : B0 ! H0(¿·0B²) is the natural projective cover.We get chain maps ½² 2 HomR(B0; ¿·0B²) and g² 2 HomR(¿·0A²; B0) suchas H0(½²) = ½ and H0(g²) = g. From the assumption, ¿·0f² is homotopicto ½² ± g², which impliesf i = hi+1 ± dAi + dBi¡1 ± hiwith some hi+1 : Ai+1 ! Bi for i · ¡1. Similarly, since H0(¿·0f²)¤ = 0,we havef j = hj+1 ± dAj + dBj¡1 ± hjwith some hj : Aj ! Bj¡1 for j ¸ 2. Therefore as a morphism in L, we mayassume f i = 0 (i 6= 0; 1). Moreover, we may assume f i = 0 (i 6= 1); sincedA¡1¤ ± f0¤ = 0, we get s1 : A1 ! B0 with f0 = s1 ± dA0. Finally, to seef² = 0, observe dA0¤±f1¤ = 0, then we get u2 : A2 ! B1 with f1 = u2±dA1.Since dA1¤ ± u2¤ ± dB1¤ = f1¤ ± dB1¤ = 0, there exits a map u3 : A3 ! B2such thatdB1 ± u2 + u3 ± dA2 = 0:Thus we obtain a homotopy map u : A² ! T¡1B² which shows that f² ishomotopic to zero.The \only if" part comes from a more general result Lemma 2.2. ¤Lemma 2.2 ([5]). Let f² be a chain map between two projective complexes.If f² is homotopic to zero, then Hn(¿·nf²) = 0 for every n 2 Z.2Mathematical Journal of Okayama University, Vol. 43 [2001], Iss. 1, Art. 5http://escholarship.lib.okayama-u.ac.jp/mjou/vol43/iss1/5STABLE MODULE THEORY WITH KERNELS 33For the proof of Lemma 2.2, the argument in [5, p. 246] completely worksso we omit the proof here.Lemma 2.1 is a key lemma and we obtain the following results as corol-laries.Proposition 2.3. For A 2 modR, there exists FA² 2 L that satis¯esH0(¿·0FA²)st»= A:Such an FA² is uniquely determined by A up to isomorphisms. We ¯x thenotation FA² and call this a standard resolution of A.Proof. First take a projective resolution PA² of A:¢ ¢ ¢ ! PA¡2 ! PA¡1 ! PA0 ! A! 0;and then a projective resolution PTrA² of TrA = Cok dFA¡1¤ as0Ã TrAÃ PA¡1¤ Ã PA0¤ Ã PTrA¡2¤ Ã ¢ ¢ ¢ :De¯ne a complex FA² asFAi =(PAi (i · ¡1);PTrA¡1¡i¤ (i ¸ 0); dFAi =(dPAi (i · ¡1);dPTrA¡2¡i¤ (i ¸ 0):We easily see FA² 2 L andH0(¿·0FA²)st»= A:Suppose both FA² and F 0A² have this property. Adding some trivial com-plex P ² of projective modulesP ² : ¢ ¢ ¢ ! 0! P 0 = P 1 ! 0! ¢ ¢ ¢if necessary, we may assume that H0(¿·0FA²) »= H0(¿·0F 0A²) »= A in modR.Then there are chain maps '² : ¿·0F 0A² ! ¿·0FA² and °² : ¿·0FA² !¿·0F 0A² such that '² ± °² = 1¿·0F 0A² and °² ± '² = 1¿·0FA² . As ¿¸¡1FA¤²and ¿¸¡1F 0A¤² are acyclic, H¡1(¿¸¡1'¤²) induces a chain map ¿¸¡1F 0A¤² !¿¸¡1FA¤². With this map for the positive part, '² can be extended to a chainmap f² : FA² ! F 0A² such that ¿·0f² = '². Similarly we get a chain mapg² : F 0A² ! FA² such that ¿·0g² = °². It is easy to see H0(¿·0(f² ± g²)) =1A and H0(¿·0(g² ± f²)) = 1A. From Lemma 2.1, we have f² ± g² = 1F 0A²and g² ± f² = 1FA² . ¤Proposition 2.4. For f 2 HomR(A;B), there existsf² 2 HomK(modR)(FA²; FB²)3Kato: Stable Module Theory with KernelsProduced by The Berkeley Electronic Press, 200134 K. KATOthat satis¯es H0(¿·0f²) = f . Such an f² is uniquely determined by f up toisomorphisms, so we use the notation f² to describe a chain map with thisproperty for given f .Proof. As in the proof of Proposition 2.3, we obtain a chain map f². Unique-ness follows from Lemma 2.1. ¤Since the operation H0¿·0 commutes with composition, the next lemmais an immediate corollary of Proposition 2.4.Lemma 2.5. For f 2 HomR(A;B) and g 2 HomR(B;C), we havef² ± g² = (f ± g)²:To sum up, we construct a functor.Theorem 2.6. The mapping A 7! FA² gives a functor from modR toK(modR), and this gives a category equivalence between modR and L.Every short exact sequence of modules induces that of projective reso-lutions. But it does not necessarily induces an exact sequence of standardresolutions.Lemma 2.7. A short exact sequence0! A! B ! C ! 0in modR induces a short exact sequence of chain complexes0! FA² ! FB² ! FC² ! 0if and only if 0! C¤ ! B¤ ! A¤ ! 0 is also exact.Proof. If 0 ! FA² ! FB² ! FC² ! 0 is exact, so is 0 ! ¿¸1FC¤² !¿¸1FB¤² ! ¿¸1FA¤² ! 0, which induces an exact sequence of homology:0! C¤ ! B¤ ! A¤ ! 0:With no assumption, we have a diagram with exact rows:0 // ¿·0FA² //²²¿·0FB² //²²¿·0FC² //²²00 // A // B // C // 0:(2.1)If 0! C¤ ! B¤ ! A¤ ! 0 is exact, similarly we get a diagram with exactrows:0 // ¿¸1FC¤² //²²¿¸1FB¤² //²²¿¸1FA¤² //²²00 // C¤ // B¤ // A¤ // 0:(2.2)4Mathematical Journal of Okayama University, Vol. 43 [2001], Iss. 1, Art. 5http://escholarship.lib.okayama-u.ac.jp/mjou/vol43/iss1/5STABLE MODULE THEORY WITH KERNELS 35Applying HomR( ; R) to (2.2) and connecting the dualized diagram to(2.1), we get a desired exact sequence 0! FA² ! FB² ! FC² ! 0. ¤3. Pseudo-kernels and pseudo-cokernelsFor A;B 2 modR, put A² = FA², B² = FB². For f 2 HomR(A;B),consider the chain map f² : A² ! B² with H0(¿·0f²) = f . Putting C² =C(f²)², we get a triangle(3.1) T¡1C² n²¡! A² f²¡! B² c²¡! C²:In general, C² does not belong to L any more but it satis¯es the following:H i(C²) = 0 (i < ¡1); Hj(C¤²) = 0 (j > ¡1):De¯nition and Lemma 3.1. As objects of modR, Ker f := H¡1(¿·¡1C²)and Cok f := H0(¿·0C²) are uniquely determined by f .Proof. Lemma 2.1 guarantees that C² is uniquely determined in K(projR).Together with Lemma 2.2, we know that Hn(¿·nC²) are also uniquely de-termined by f . ¤Putnf := H0(¿·0n²) : Ker f ! A; cf := H0(¿·0c²) : B ! Cok f:The triangle (3.1) gives an exact sequence of the following form:(3.2) 0! Ker fnfqf¡¡¡! A© P (f ½)¡¡¡! B ! 0with some projective module P . In fact, Ker f is characterized with thisproperty:Proposition 3.2. If an R-linear map ½0 : P 0 ! B from a projective moduleP 0 makes ~f 0 : A© P 0 (f ½0)¡¡¡! B a surjective mapping, then Ker ~f 0 st»= Ker f .Proof. It is easy to show that both of the composites P 0 ½0¡! B ! Cok fand P½¡! B ! Cok f are projective covers of Cok f . There exist t 2HomR(PB; P 0) and u 2 HomR(PB; A) such that ½ ¡ ½0 ± t = f ± u. If tis not an epimorphism, add some s : Q ! P with Q 2 projR to makeP ©Q (t s)¡¡¡! P 0 surjective. From the diagramA© P ©Q (f ½ s±½0) // BA© P (f ½) //(1 0 00 1 0)OOB;5Kato: Stable Module Theory with KernelsProduced by The Berkeley Electronic Press, 200136 K. KATOwe get Ker fst»= Ker(f ½ s ± ½0). Also we have a diagramA© P ©Q (f ½ s±½0) //(1 u 00 t s) ²²BA© P 0 (f ½0) // B;hence Ker ~f 0st»= Ker(f ½ s ± ½0). ¤Lemma 3.3. With notation as above, we have the following :1) f ± nf = 0.2) If x 2 HomR(X;A) satis¯es f ± x = 0, there existshx 2 HomR(X;Ker f)such that x = nf ± hx.The proof is straightforward from the de¯nition. Strictly speaking, Ker fis not the kernel of f . Because it lacks the uniqueness of hx in 2) ofLemma 3.3. (See Example 3.4).Example 3.4. Let R = k[[x; y; z]]=(x2 ¡ yz), A = R=(yz) and B =R=(yz; y2; z2). Let f : A! B be the natural map induced from the inclusion(yz) ½ (yz; y2; z2). Since f is surjective, Ker f st»= Ker f »= R=(z) © R=(y),and the sequence 0 ! Ker f nf¡! A f¡! B ! 0 is exact. Put X = Tr k andlet u 2 HomR(X;Ker f) be as follows:0 // Rxyz//(10) ²²R3 //(0 0 10 0 0)²²X //u²²00 // R2z 00 y// R2 // Ker f // 0:Easily we get nf ± u = 0A = uf ± 0K where 0A = 0 2 HomR(X;A) and0K = 0 2 HomR(X;Ker f). Also we have u 6= 0K from this diagram. ¤Dually, (Cok f; cf ) satis¯es the following, which comes from the observa-tionCok f = TrKer Tr f; cf = TrnTr f :Lemma 3.5. 1) cf ± f = 0.2) If y 2 HomR(B; Y ) satis¯es y ± f = 0, there existsey 2 HomR(Cok f; Y )such that y = ey ± cf .6Mathematical Journal of Okayama University, Vol. 43 [2001], Iss. 1, Art. 5http://escholarship.lib.okayama-u.ac.jp/mjou/vol43/iss1/5STABLE MODULE THEORY WITH KERNELS 37Two modules Ker f and Ker f are not always stably isomorphic. But weget the following.Lemma 3.6. 1) There is an exact sequence 0 ! L ! M ! N ! 0such that Lst»= Ker f , Mst»= Ker f and Nst»= ­1R(Cok f).2) There is an exact sequence 0 ! L0 ! M 0 ! N 0 ! 0 such thatM 0st»= Cok f , N 0st»= Cok f and ­1R(L0) is the surjective image ofKer f .Proof. 1) The claim easily follows from the following diagram:0²²0²²0²²0 // Ker f //²²A //(10) ²²Im f //²²00 // Ker f //²²A© P (f ½) //(01) ²²B //²²00 // ­1R(Cok f) //²²P //²²Cok f //²²0:0 0 02) Dualizing (3.2) with R, we get a map ·f : Afjf¡¡¡! B©Q with some pro-jective module Q such that Cok ·fst»= Cok f . And consider the commutativediagram:0²²Q²²0 // Im ·f //²²B ©Q //²²Cok ·f //²²00 // Im f // B //²²Cok f // 0;0where the middle column is a split exact sequence. If we put L0 the kernel ofepimorphism Cok ·f ! Cok f , then ­1R(L0) is the kernel of the natural mapA=Ker ·f »= Im ·f ! Im f »= A=Ker f:7Kato: Stable Module Theory with KernelsProduced by The Berkeley Electronic Press, 200138 K. KATOTherefore ­1R(L0) »= Ker f=Ker ·f . ¤Corollary 3.7. 1) Ker fst»= Ker f if f is an epimorphism.2) Cok fst»= Cok f if f is a split monomorphism.Notations. For a given homomorphism f : A ! B in modR, put A² =FA², B² = FB² and C² = C(f²)². Set K² = FKer f ² and L² = FCok f ².Chain maps nf ² 2 HomK(modR)(K²; A²) and cf ² 2 HomK(modR)(B²; L²) areinduced from nf and cf . Since f² ± nf ² = 0 and cf ² ± f² = 0, there exist"² 2 HomK(modR)(K²; T¡1C²) and ±² 2 HomK(modR)(C²; L²) such thatnf² = n² ± "²; and cf ² = ±² ± c²:Notice thatC("²)i = 0 (i · ¡1); and C(±²)j = 0 (j ¸ ¡1)because "² and ±² induce 1Ker f and 1Cok f .K²nf²//"²²²A²T¡1C²n² // A²f² // B²c² // C²±²²²B²cf²// L²:4. Monomorphisms, epimorphisms, and split morphismsIf Ker f = 0, then f is injective. But the vanishing of Ker f is nota necessary condition for f to be injective; let A;B be two modules withpdB ¸ 2. Let f be a split monomorphism A ! A © B. Obviously nf = 0but Ker fst»= ­1R(B) is not projective. We investigate what is an injectivemorphism in modR.Proposition 4.1. With notations as in x3, the following are equivalent.1) f is a monomorphism in modR.2) Ext1R(f;¡) : Ext1R(B;¡)! Ext1R(A;¡) is surjective.3) nf = 0.4) There exists '² 2 HomK(modR)(K²; T¡1B²) that makes the diagramcommutative:K²"² //'² %%KKKKKKKKT¡1C²T¡1B²T¡1c²OO8Mathematical Journal of Okayama University, Vol. 43 [2001], Iss. 1, Art. 5http://escholarship.lib.okayama-u.ac.jp/mjou/vol43/iss1/5STABLE MODULE THEORY WITH KERNELS 395) ¿·¡1c² is a split epimorphism in K(modR).6) ­1R(f) is a split monomorphism and Ker f »= Cok ­1R(f) in modR.Proof. 1) , 3). A morphism f is called a monomorphism if and only iff±x = 0 always implies x = 0, which is equivalent to nf = 0 from Lemma 3.3.1) , 2). An exact sequence0! Ker fnfqf¡¡¡! A© P (f ½)¡¡¡! B ! 0induces a long exact sequence¢ ¢ ¢ ! Ext1R(B; )Ext1R(f; )¡¡¡¡¡¡! Ext1R(A; )Ext1R(nf ; )¡¡¡¡¡¡¡! Ext1R(Ker f; )! ¢ ¢ ¢ :So Ext1R(f; ) is surjective if and only if Ext1R(nf ; ) is zero, which is equivalentto the condition nf = 0 from [1] (1.44).3) , 4). Lemma 2.1 shows that nf = 0 if and only if nf ² = n² ±"² = 0, that is, some '² : K² ! T¡1B² exists and "² = T¡1c² ± '² sinceT¡1B² T¡1c²¡¡¡¡! T¡1C² n²¡! A² ! B² is a triangle.4)) 5). Applying ¿·0 to the diagram in 4), we get 5) since ¿·0(T¡1c²) =T¡1(¿·¡1c²) and ¿·0"² is the identity.5) ) 6). Put X² = C(¿·¡1c²). Then a triangleT¡1X² ! ¿·¡1B²¿·¡1c²¡¡¡¡! ¿·¡1C² ! X²induces a split exact sequence0! H¡2(X²) !¡! H¡1(¿·¡1B²)! H¡1(¿·¡1C²):By de¯nition, H¡1(¿·¡1B²)st»= B and H¡1(¿·¡1C²)st»= Ker f . We claimthat H¡2(X²)st»= A and via this isomorphism, !st»= ­1R(f). Since B¡1 !C¡1 = X¡1 is surjective, so is dX¡2, which implies H¡2(X²)st»= Cok dX¡3.Moreover, Cok dX¡3 = Cok dC(c)¡3st»= Cok dA¡2 = ­1R(A) as ¿·¡2X² =¿·¡2C(c)² and C(c)² »= A².6) ) 4). With no assumption, we have the diagramF­1R(A)² ­1R(f)²//®²²²F­1R(B)² v² //¯²²²C(­1R(f))²°²²²t²))SSSSSSSSSK²"²uukkkkkkkkkkkT¡1A²T¡1f² // T¡1B²T¡1c² // T¡1C²9Kato: Stable Module Theory with KernelsProduced by The Berkeley Electronic Press, 200140 K. KATOwhere ®², ¯² are canonical maps induced by 1­1R(A) and 1­1R(B), which induce°². The map Cok ­1R(f) ! Ker f induces t². Now if we assume thecondition 6), there exists a chain map s² : C(­1R(f))² ! F­1R(B)² such thatv²±s² = 1C(­1R(f))² and t² = 1K² . Hence "² = °² = °²±v²±s² = T¡1c²±¯²±s²so we get the chain map '² = ¯² ± s². ¤If Ext1R(B;R) = 0, then H¡1(C¤²) = 0, which implies "² = 1; K² =T¡1C². Thus we have the next lemma:Lemma 4.2. The following are equivalent for B 2 modR.1) In modR, every monomorphism to B splits.2) Ext1R(B;R) = 0.Proof. 2)) 1). Let f : A! B be a monomorphism and let us use the samenotations as in Proposition 4.1. If Ext1R(B;R) = 0, then H¡1(C¤²) = 0,which implies T¡1C² = K², that is, "² is an isomorphism. Since nf = 0,n² = nf ² = 0 hence f² is a split monomorphism.1) ) 2). If Ext1R(B;R) 6= 0, then there exists a non-split short exactsequence0! R! A f¡! B ! 0:We see f is a monomorphism because Ker fst»= R. But f does not split. ¤Dually, we getLemma 4.3. The following are equivalent for A 2 modR.1) In modR, every epimorphism from A splits.2) Ext1R(TrA;R) = 0.Remark. The condition that ­1R(f) is a split monomorphism does not au-tomatically induce Ker fst»= Cok ­1R(f). For instance, let z 2 R be annon-zero-divisor of R. Let f be an endomorphism of R=(z2) as f = z.Then ­1R(f) is an endomorphism of R, so we have Cok ­1R(f) = 0. ButKer fst»= R=(z) is not projective.Theorem 4.4. The following are equivalent for a ring R.1) Every monomorphism in modR splits.2) Every epimorphism in modR splits.3) R is self-injective.4) Every short exact sequence 0 ! A ! B ! C ! 0 induces an exactsequence of standard resolutions 0! FA² ! FB² ! FC² ! 0.10Mathematical Journal of Okayama University, Vol. 43 [2001], Iss. 1, Art. 5http://escholarship.lib.okayama-u.ac.jp/mjou/vol43/iss1/5STABLE MODULE THEORY WITH KERNELS 415) Every short exact sequence 0 ! A ! B ! C ! 0 remains exactwhen dualized by R; 0! C¤ ! B¤ ! A¤ ! 0 is exact.Proof. The equivalence between 3) and 1) (or 2)) follows from Lemma 4.2(Lemma 4.3) respectively. We have already shown in Lemma 2.7 that 4)and 5) are equivalent. Obviously 3) implies 5), so it su±ces to prove that 5)implies 3). Let M be an arbitrary object of modR. Consider a projectivecover of M :0! ­1R(M)! P !M ! 0:If the dualized sequence remains exact, that means Ext1R(M;R) = 0. ¤AcknowledgmentI am grateful to Osamu Iyama who pointed out an error of the originalversion of this paper. I also express my gratitude to Mitsuo Hoshino whomade valuable suggestions and told that the condition 4) of Theorem 4.4 isequivalent to others.References[1] M. Auslander and M. Bridger, The stable module theory, Mem. Amer. Math. Soc.94, 1969.[2] M. Amasaki, Homogeneous prime ideals and graded modules ¯tting into long Bourbakisequences, preprint, 1999.[3] T. Kawasaki, Local cohomology modules of indecomposable surjective-Buchsbaummodules over Gorenstein local rings, J. Math. Soc. Japan 48(1996), 551{566.[4] J. Miyachi, Derived categories with applications to representations of algebras, seminarnote, 2000.[5] Y. Yoshino, Maximal Buchsbaum modules of ¯nite projective dimension, J. Algebra159(1993), 240{264.Kiriko KatoDepartment of Applied MathematicsOsaka Women's UniversitySakai, Osaka 590-0035, Japane-mail address: kiriko@appmath.osaka-wu.ac.jp(Received October 9, 2001 )11Kato: Stable Module Theory with KernelsProduced by The Berkeley Electronic Press, 2001

Stable Module Theory with Kernels

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