aa r X i v : . [ m a t h . A C ] N ov ON THE RANK OF MULTI-GRADEDDIFFERENTIAL MODULES
JUSTIN W. DEVRIES
Abstract. A Z d -graded differential R -module is a Z d -graded R -module D with a morphism δ : D → D such that δ = 0. For R = k [ x , . . . , x d ], thispaper establishes a lower bound on the rank of such a differential module whenthe underlying R -module is free. We define the Betti number of a differentialmodule and use it to show that when the homology H ( D ) = ker δ/ im δ of D isnon-zero and finite dimensional over k then there is an inequality rank R D ≥ d . Introduction
Let k be a field and set R = k [ x , . . . , x d ]. A differential R -module D is an R -module with a square-zero homomorphism δ : D → D called the differential . Thehomology of D is defined in the usual way: H ( D ) = ker δ/ im δ . Differential moduleshave played an important role in the work of Avramov, Buchweitz, Iyengar, andMiller on the homology of finite free complexes [1, 2]. In this context, differentialmodules arise naturally when working with DG-modules: some constructions withdesirable properties do not respect the grading but do preserve the differential (see[2, 10] for some instances of this phenomenon).Motivated by a conjecture of Avramov, Buchweitz and Iyengar, this paper isconcerned with establishing bounds on the ranks of Z d -graded differential modules.They conjecture: Conjecture 1.1 ([1, Conjecture 5.3]) . Let R be a regular local ring of dimension d , and F a differential R -module admitting a finite free flag. If H ( F ) has non-zerofinite length, then rank R F ≥ d . In this conjecture, a free flag on a differential module is a filtration compatiblewith the differential that provides appropriate lifting properties in the category ofdifferential modules (see Definition 2.7). In their paper they show that Conjec-ture 1.1 is true for d ≤ d = 3 when R is a unique factorization domain [1,Theorem 5.2].The main theorem (Theorem 6.4) of this paper establishes a lower bound onthe rank of a differential module by finding a lower bound on the Betti number ofa differential module. A consequence of this result is the following theorem thatpartially answers Conjecture 1.1 in the multi-graded case.
Theorem 1.2.
Let F be a finitely generated Z d -graded differential R -module withdifferential δ : F → F that is homogeneous of degree zero, such that F is free as an R -module. If H ( F ) has non-zero finite length then rank R F ≥ d . This result is new even for complexes of R -modules. Given a complex of Z d -graded free R -modules F = . . . / / F / / F / / F / / . . . the module L i F i with differential δ = L i ∂ i forms a differential module. When H ( F ) has non-zero finite length as an R -module then Theorem 1.2 implies that(1.1) X i rank R F i ≥ d . This inequality is already known when F is a resolution—i.e. H ( F ) is concentratedin a single homological degree—from the work of Charalambous and Santoni onthe Buchsbaum-Eisenbud-Horrocks problem [6, 11]. Recall that for a Z -gradedpolynomial ring, the Buchsbaum-Eisenbud-Horrocks problem is to show that allnon-zero finite length R -modules M satisfy β i ( M ) ≥ (cid:0) di (cid:1) , where β i ( M ) is the i -thBetti number of M [3, 9]. Summing the binomial coefficients gives (1.1) when F is afree resolution of a non-zero finite length Z d -graded module M . However, when F isnot acyclic it is not clear how to establish (1.1) without using differential modules.Working with differential modules provides an advantage in that it simultane-ously treats the case of free resolutions and free complexes with homology spreadamong several homological degrees, as well as other contexts. One such applica-tion arises in the conjectures of Carlsson and Halperin concerning a lower boundon the rank of DG-modules with non-zero finite length homology [4, 8]. For thisconnection between differential modules and DG-modules see [1, § § § §
4. Section 5 adapts an inequality of Santoni [11] to differentialmodules. The main result, Theorem 6.4, is proved in § Differential modules
Throughout, k is a field, R = k [ x , . . . , x d ] is the standard Z d -graded polynomialring and m = ( x , . . . , x d ). To be specific, the grading on R is such that the degreedeg( x i ) ∈ Z d of variable x i is (0 , . . . , , , , . . . ,
0) with the 1 appearing in the i -thcoordinate. For m ∈ Z d , we write m i to denote the i -th coordinate. The group Z d is equipped with the coordinate-wise partial order: a ≤ b if and only if a i ≤ b i forall i .Recall that a Z d -graded module M over R is an R -module that has a decom-position L n ∈ Z d M n as abelian groups such that multiplication by an element of R of degree m takes M n to M m + n . An R -linear map φ between Z d -graded modules M and N is a morphism if φ ( M m ) ⊆ N m . In particular, a complex of Z d -gradedmodules is required to have morphisms for its differentials. N THE RANK OF MULTI-GRADED DIFFERENTIAL MODULES 3
For n ∈ Z d the shifted (or twisted) module M ( n ) is defined to be M m + n indegree m for each m ∈ Z d with the same R -module structure as M . Given amorphism φ : M → N the shifted morphism M ( n ) → N ( n ) defined by x φ ( x )is denoted φ ( n ).We will primarily work with Z d -graded modules and Z d -graded differential mod-ules, so definitions will be given only in that context for simplicity; see [1, 5, 7] fordetails concerning arbitrary differential modules. Definition 2.1. A Z d -graded differential R -module with differential degree d ∈ Z d is a Z d -graded R -module D with a morphism δ : D → D ( d ) such that thecomposition D ( − d ) δ ( − d ) / / D δ / / D ( d )is zero. We say that D has differential δ .The homology of a differential module D is the Z d -graded R -module H ( D ) = ker δ/ im( δ ( − d )) . Any Z d -graded R -module, in particular H ( D ), will be considered as a differentialmodule with zero differential.A morphism φ : D → E between Z d -graded differential modules is a morphismof Z d -graded modules satisfying δ E ◦ φ = φ ◦ δ D . In particular, for there to be anon-zero morphism, D and E must have the same differential degree.In the usual way, a morphism φ : D → E induces a map in homology H ( φ ) : H ( D ) → H ( E ). If H ( φ ) is an isomorphism we say that φ is a quasi-isomorphism and write D ≃ E or φ : D ≃ −→ E . Given an exact sequence of differential modules0 / / D / / D / / D / / . . . / / H ( D ) / / H ( D ) / / H ( D ) / / H ( D ) / / H ( D ) / / . . . This long exact sequence is often written as a triangle:(2.1) H ( D ) / / H ( D ) z z ttttttttt H ( D ) d d JJJJJJJJJ
See [5, Chap. IV §
1] for a proof.Bounds on the rank of a differential module will be obtained by comparing therank and the Betti number. To define the Betti number we will need a notion ofa tensor product of differential modules. However adapting the usual definition ofthe tensor product between complexes fails to produce a differential module whenapplied to two differential modules. To work around this we recall the constructionof a tensor product of a complex and a differential module, along with some of itsproperties [1, § Definition 2.2.
For a complex C of Z d -graded R -modules and a Z d -graded differ-ential R -module D , define a Z d -graded differential module C ⊠ R D by setting C ⊠ R D = M n ∈ Z ( C n ⊗ R D ) , JUSTIN W. DEVRIES with differential defined by δ C ⊠ R D ( c ⊗ d ) = ∂ C ( c ) ⊗ d + ( − n c ⊗ δ D ( d ) , for c ⊗ d ∈ C n ⊗ R D .We will need the following facts concerning this product. Proposition 2.3 ([1, 1.9.3]) . Let X and Y be complexes and let D be a differentialmodule. Then there is a natural isomorphism of differential modules: ( X ⊗ R Y ) ⊠ R D = X ⊠ R ( Y ⊠ R D ) . Proposition 2.4 ([1, Proposition 1.10]) . Let X and Y be bounded below complexesof flat R -modules, i.e. X i = Y i = 0 for sufficiently small i . Then (1) the functor X ⊠ R − preserves exact sequences andquasi-isomorphisms, (2) a quasi-isomorphism φ : X → Y induces a quasi-isomorphism φ ⊠ R D : X ⊠ R D → Y ⊠ R D for all differential R -modules D . Using this tensor product, we can define the Tor functor between R -modules anddifferential R -modules, and hence define a Betti number. Definition 2.5.
For a differential R -module D and an R -module M setTor R ( M, D ) = H ( P ⊠ R D )where P is a free resolution of M . This is well-defined as different choices of freeresolution produce quasi-isomorphic differential modules by Proposition 2.4. Definition 2.6.
We define the
Betti number β R ( D ) of a differential R -module D to be β R ( D ) = rank k Tor R ( k, D ) . The connection between ranks of differential modules and Betti numbers is pro-vided by free flags, a notion of a free resolution for differential modules [1, § Definition 2.7. A free flag on a differential module F is a family { F n } n ∈ Z of R -submodules such that(1) F n = 0 for n < F n ⊆ F n +1 for all n ,(3) δ F ( F n ) ⊆ F n − for all n ,(4) S n ∈ Z F n = F ,(5) F n /F n − is a free R -module for all n .A differential module F with a free flag resolves D if there is a quasi-isomorphism F ≃ / / D .With differential modules that admit a free flag providing a resolution of a dif-ferential module, the Tor functor is balanced, which gives the connection betweenthe rank and Betti number of a differential module. Proposition 2.8 ([1, Proposition 2.4]) . Let F be a differential module with a freeflag. Then the functor − ⊠ R F preserves exact sequences and quasi-isomorphisms. N THE RANK OF MULTI-GRADED DIFFERENTIAL MODULES 5
Lemma 2.9.
Let P be a free resolution of a module M and let F be a free flagresolving a differential module D . Then H ( P ⊠ R D ) is isomorphic to H ( M ⊠ R F ) as R -modules.Proof. Let ε : P → M and η : F → D be quasi-isomorphisms. Then there aremorphisms P ⊠ R D P ⊠ R F P ⊠ R η o o ε ⊠ R F / / M ⊠ R F. By Proposition 2.4 and Proposition 2.8 these are quasi-isomorphisms. (cid:3)
Theorem 2.10.
Let F be differential module admitting a free flag. Then β R ( F ) ≤ rank R F. Proof.
By Lemma 2.9, β R ( F ) = rank k Tor R ( k, F ) = rank k H ( k ⊠ R F ) . Since k is an R -module, k ⊠ R F = k ⊗ R F . Since H ( k ⊗ R F ) is a subquotient of k ⊗ R F , we have rank k H ( k ⊠ R F ) ≤ rank k k ⊗ R F = rank R F. (cid:3) Remark . When δ ( F ) ⊆ m F we have β R ( F ) = rank R F as the differential of k ⊠ R F is zero. In general, the inequality can be strict; see Example 4.3.We finish this section by collecting a few properties of the Tor functor for uselater. Lemma 2.12.
Consider an exact sequence of differential R -modules / / D α / / D β / / D / / . For each R -module M there is an exact commutative diagram: Tor R ( M, D ) α / / Tor R ( M, D ) β w w nnnnnnnnnnnn Tor R ( M, D ) γ g g PPPPPPPPPPPP
Proof.
Take a free resolution P of the module M . By Proposition 2.4 the sequenceof differential modules remains exact after applying P ⊠ R − :0 / / P ⊠ R D P ⊠ α / / P ⊠ R D P ⊠ β / / P ⊠ R D / / . The exact triangle (2.1) coming from this exact sequence is the desired one. (cid:3)
Lemma 2.13.
Let R → S be a homomorphism of rings where S is flat over R . Let M be an S -module and D a differential R -module. Then Tor S ( M, S ⊠ R D ) ∼ = Tor R ( M, D ) . Proof.
Let P be a free resolution of M . Then using Proposition 2.3 one gets:Tor S ( M, S ⊠ R D ) = H ( P ⊠ S ( S ⊠ R D )) ∼ = H (( P ⊗ S S ) ⊠ R D ) ∼ = H ( P ⊠ R D )= Tor R ( M, D ) . (cid:3) JUSTIN W. DEVRIES Compression
Every complex of R -modules produces a differential module by forming its com-pression . This construction allows results about differential modules to be trans-lated to results about complexes of modules. In fact, the differential modules pro-duced by compressing always have differential degree so it is sufficient to restrictto differential modules with differential degree if one is interested in establishingresults about complexes. Note that not every differential module of differentialdegree arises this way (see Example 3.3). Construction 3.1 ([1, 1.3]) . If C is a complex of Z d -graded R -modules, then its compression is the Z d -graded differential module C ∆ = M n ∈ Z C n with differential δ C ∆ = L n ∈ Z ∂ Cn .We have deg( δ C ∆ ) = because the differentials of the complex C are requiredto have degree zero. By the definition of δ C ∆ , we have H ( C ∆ ) = L n ∈ Z H n ( C ).When the complex C is bounded below and consists of free R -modules then thecompression has a free flag. Indeed, suppose C i = 0 for i sufficiently small. Thensetting F n = L i ≤ n C i forms a free flag.Computing the Betti number of a compression is a straight-forward applicationof Theorem 2.10 and Remark 2.11. Lemma 3.2.
Let C be a bounded below complex of free modules that is minimal inthe sense that ∂ Cn ( C n ) ⊆ m C n − . Then β ( C ∆ ) = X i rank R C i . When C is a minimal free resolution of a module M we have β ( C ∆ ) = X i β i ( M ) , where β i ( M ) is the usual Betti number of M .Proof. Since C is a bounded below complex of free modules, C ∆ has a free flag.We have δ ( C ∆ ) = M n ∈ Z ∂ n ( C n ) ⊆ M n ∈ Z m C n − = m C ∆ , so by Remark 2.11 we have β ( C ∆ ) = rank R C ∆ = X i rank R C i . When C is a minimal free resolution of M we have rank R C i = β i ( M ), whichcompletes the proof. (cid:3) Obviously differential modules with non-zero differential degree do not comefrom compressing a complex, but the following example shows that there are alsodifferential modules with differential degree zero that are not compressions of acomplex.
N THE RANK OF MULTI-GRADED DIFFERENTIAL MODULES 7
Example 3.3.
Let R = k [ x, y ] and let F = R (0 , ⊕ R ( − , ⊕ R (0 , − ⊕ R ( − , − F as column vectors, define a differential δ by left-multiplicationby the matrix x y xy − y x . This is a differential module with deg δ = . Represented diagrammatically thishas the form of a Koszul complex on x, y modified by adding an additional map: R ( − , − (cid:2) − yx (cid:3) / / xy ) ) R ( − , ⊕ R (0 , − [ x y ] / / R (0 , / / . Reading the diagram from right to left produces a free flag:0 ⊂ R (0 , ⊂ R (0 , ⊕ R ( − , ⊕ R (0 , − ⊂ R (0 , ⊕ R ( − , ⊕ R (0 , − ⊕ R ( − , −
1) = F. To calculate H ( F ), consider the first differential submodule of the flag F = R (0 , H ( F ) = R (0 , H ( F/F ) = ( R ( − , ⊕ R (0 , − /R ( − y ⊕ x ) . From the short exact sequence0 / / F / / F / / F/F / / . . . / / H ( F ) / / H ( F/F ) α / / H ( F ) β / / H ( F ) / / . . . where the map α is given by the matrix (cid:2) x y (cid:3) . Since α is injective, β must be asurjection, giving H ( F ) = H ( F ) / im α = R/ ( x, y ) = k. To compute the Betti number, note that δ ( F ) ⊆ m F , so we have β R ( F ) =rank R F = 4 by Remark 2.11.4. Non-positive differential degree
Every differential R -module with a free flag is free as an R -module, but not con-versely (see Example 4.4). Even when a differential module admits a free flag theremay be no way to “minimize,” unlike finite free complexes that can be decomposedinto an acyclic complex and a minimal complex C with ∂ ( C ) ⊆ m C (see Example4.3). Restricting to the case of a differential module D with deg δ D ≤ we canavoid both of these difficulties. Theorem 4.1.
Let F be a finitely generated Z d -graded differential R -module with deg δ F ≤ that is free as an R -module. Then F has a free flag and there is asubmodule F ′ that is a direct summand in the category of Z d -graded differential R -modules such that (1) F ′ has a free flag, JUSTIN W. DEVRIES (2) δ ( F ′ ) ⊆ m F ′ , (3) H ( F ′ ) = H ( F ) .Remark . The hypothesis that deg δ ≤ is necessary. See Examples 4.3 and 4.4. Proof.
We induce on rank R F : if rank R F = 1 then the differential of F is multipli-cation by an element of R . Since R is a domain, this element must be zero; hence F = F is a free flag. As δ ( F ) = 0 we conclude that δ ( F ) ⊆ m F as well.Now suppose rank R F >
1. If δ F ( F ) m F then there is some homogeneous basiselement e with δ F ( e ) m F . We first show that e and δ ( e ) are linearly independent.Let e, v , . . . , v n be a basis for F and write(4.1) δ ( e ) = re + s v + · · · + s n v n . Suppose that ae + bδ ( e ) = 0 with a, b ∈ R . In terms of the basis we have( a + br ) e + bs v + · · · + bs n v n = 0 . As this is a basis we conclude that a + br = 0 and bs i = 0 for all i . Since δ ( e ) m F one of r, s , . . . , s n is not in m . If s i m for some i we conclude that b = 0 andtherefore e and δ ( e ) are linearly independent. If s i ∈ m for all i then we have r m .Condensing (4.1) gives δ ( e ) = re + v with v ∈ m F and re m F . Because δ is adifferential, 0 = δ ( e ) = r e + rv + δ ( v ) . We have δ ( v ) ∈ m F and rv ∈ m F since v ∈ m F , and therefore r e ∈ m F . But r isa unit, so re ∈ m F , a contradiction.So e and δ F ( e ) are linearly independent. By Nakayama’s lemma we can take { e, δ F ( e ) } to be part of a basis of F . Let G = Re ⊕ Rδ F ( e ). Then G is a differentialsub-module. So we have an exact sequence of differential modules:(4.2) 0 / / G / / F / / F/G / / . Since H ( G ) = 0, the long exact sequence in homology coming from (4.2) shows that H ( F/G ) = H ( F ). The module F/G is a free summand of F since G is generatedby basis elements of F . So by induction F/G has a free flag { G n } n ∈ Z . Setting F = Rδ F ( e ) ,F = Rδ F ( e ) ⊕ Re,F n = Rδ F ( e ) ⊕ Re ⊕ G n − , n ≥ F . The induction hypothesis also shows that F/G has a directsummand F ′ with a free flag such that δ ( F ′ ) ⊆ m F ′ and such that H ( F ′ ) = H ( F/G ) = H ( F ). This completes the proof when δ F ( F ) m F .Now suppose that δ F ( F ) ⊆ m F . In this case it suffices to show that F has a freeflag. Let e , . . . , e n be a homogeneous basis for F . Let e be a minimal element of { deg( e ) , . . . , deg( e n ) } under the partial order on Z d . Set G = M deg( e i )= e Re i . Then δ F ( G ) ⊆ G since deg( δ F ( e i )) ≤ deg( e i ) for all i as the degree of δ F is non-positive in each coordinate. So G is a differential sub-module.We claim that δ F | G = 0. When deg δ F < , we have δ F | G = 0 as deg( δ F ( e i )) < deg( e i ) and all the generators e i of G have the same degree. When deg δ F = the N THE RANK OF MULTI-GRADED DIFFERENTIAL MODULES 9 matrix representing δ F | G has entries in k since all generators of G are in the samedegree. So δ F | G = 0, otherwise there would be an element of δ F ( G ) that is not in m F , contrary to assumption.Since δ F | G = 0 we get δ F ( F ) = 0 by setting F = G . As F is generatedby basis elements of F , the quotient F/F is a free R -module, so the inductionhypothesis produces a free flag { G n } n ∈ Z for F/F . Setting F n = F ⊕ G n − for n ≥ F n = 0 for n < F . (cid:3) The next example illustrates many of the difficulties in dealing with differentialmodules with non-zero differential degree. It provides an obstruction to extendingTheorem 6.4 and Lemma 4.1 to differential modules with deg δ > . By [1, Theo-rem 5.2], a differential module over k [ x, y ] with a free flag must have rank at least4. From this we conclude that the following example also shows that Lemma 4.1cannot be extended to differential modules with deg δ > as no summand can havea free flag. Example 4.3.
Let R = k [ x, y ] and let F = R (0 , ⊕ R (0 , ⊕ R (1 , ⊕ R (1 , δ = x y
10 0 0 − y x . This is a differential module with differential degree (1 , R (1 , (cid:2) − yx (cid:3) / / ( ( R (0 , ⊕ R (1 , [ x y ] / / R (0 , / / . As in Example 3.3, reading the diagram from right to left gives a free flag. Thesame computation from Example 3.3 shows that H ( F ) = k . As F has a free flag,we can compute β R ( F ) by rank k H ( k ⊠ R F ). Applying k ⊠ R − to (4.3) we have thevector space k (suppressing the grading) with differential given by the diagram: k / / ! ! k / / k / / . The homology is k , so β R ( F ) = 2.This final example shows that a differential module that is free as an R -moduleneed not have a free flag; thus Lemma 4.1 cannot be strengthened to apply todifferential modules with deg δ > . Example 4.4.
Let F be as in Example 4.3. Let e be the basis element in degree( − , −
1) and set G = Re ⊕ Rδ F ( e ). Then calculation shows that F/G is thedifferential module D = R (0 , ⊕ R (1 ,
0) with δ = (cid:20) xy − y x − xy (cid:21) . This is a differential module with deg δ = (1 , H ( G ) = 0, an exact sequenceargument shows that the map F → F/G is a quasi-isomorphism; hence H ( D ) = H ( F ) = k . As F admits a free flag, it is a resolution of D . So we have β R ( D ) = β R ( F ) = 2.The differential module D itself cannot have a free flag since rank R D = 2 < High-low decompositions
The main tool, Theorem 5.5, we use for finding a bound on the Betti numbercomes from an inequality of Santoni [11] formulated to apply to differential modules.The modified statements and proofs are provided in this section for completeness.The essential idea is to use information about the “top” and “bottom” degree partsto derive information about the entire module. The meaning of “top” and “bottom”is made precise by a high-low decomposition , Definition 5.4.For this section let S be an arbitrary commutative ring, and let C be a classof differential S -modules which is closed under taking submodules and quotients.Take λ to be a superadditive function from C to an ordered commutative monoidsuch that λ ( C ) ≥ C ∈ C . Recall that λ is superadditive if an exact sequence0 / / A / / B / / C / / C gives an inequality λ ( B ) ≥ λ ( A ) + λ ( C ) . Example 5.1.
For our purposes, S will be a Z d -graded polynomial ring over afield, C will be the collection of differential S -modules with non-zero homology infinitely many degrees and λ will be the length of the homology module. Lemma 5.2.
Let B be a differential S -module and suppose we have the followingcommutative diagrams in C : A ψ A (cid:15) (cid:15) (cid:15) (cid:15) ι / / B ψ B (cid:15) (cid:15) B ′ ε ′ / / / / φ B (cid:15) (cid:15) C ′ (cid:127) _ φ C (cid:15) (cid:15) A ′′ (cid:31) (cid:127) ι ′′ / / B ′′ B ε / / C Then the following inequalities hold: λ (im ι ) ≥ λ (im ι ′′ ) ,λ (im ε ) ≥ λ (im ε ′ ) . Furthermore, if ει = 0 then λ ( B ) ≥ λ (im ι ′′ ) + λ (im ε ′ ) . Proof.
For the first inequality, there is a surjection im ι ։ im ψ B ι , so λ (im ι ) ≥ λ (im ψ B ι ) = λ (im ι ′′ ψ A ) . Because ψ A is surjective there is also a surjection im ι ′′ ψ A ։ im ι ′′ . This gives thedesired inequality, λ (im ι ) ≥ λ (im ι ′′ ).For the second inequality, there is an inclusion im ε ′ ֒ → im ε since φ C is injective.By superadditivity, λ (im ε ) ≥ λ (im ε ′ ).For the final inequality, note that ει = 0 implies that im ι ⊆ ker ε . The exactsequence 0 / / ker ε / / B / / im ε / / , N THE RANK OF MULTI-GRADED DIFFERENTIAL MODULES 11 then implies λ ( B ) ≥ λ (im ε ) + λ (im ι ) ≥ λ (im ε ′ ) + λ (im ι ′′ ) using the first twoinequalities. (cid:3) Lemma 5.3.
Let D be a differential S [ x ] -module and consider it as a differential S -module via the inclusion S ֒ → S [ x ] . Viewing S [ x ] ⊠ S D as a S [ x ] -module via theaction r ( s ⊗ d ) = ( rs ) ⊗ d , there is a sequence of differential S [ x ] -modules / / S [ x ] ⊠ S D σ / / S [ x ] ⊠ S D ε / / D / / , with σ (1 ⊗ d ) = x ⊗ d − ⊗ xd and ε ( a ⊗ d ) = ad . This sequence is exact andfunctorial in D . The map σ is given by multiplication by x if and only if xD = 0 .Proof. It is straight-forward to check that σ and ε are morphisms and that thesequence is exact and functorial. Evidently σ is multiplication by x when xD = 0.The exactness of the sequence shows that the converse holds. (cid:3) The following definition and theorem are the differential module versions of San-toni’s results for R -modules [11]. Definition 5.4.
A differential S [ x ]-module D admits a high-low decomposition ifthere are non-zero differential S [ x ]-modules D h and D ℓ each annihilated by x andthere are morphisms of differential S [ x ]-modules D h (cid:31) (cid:127) / / D and D / / / / D ℓ that split in the category of differential S -modules. Theorem 5.5.
Let K be an S [ x ] -module such that xK = 0 , and assume C is closedunder Tor S [ x ] ( K, − ) . Let D ∈ C be a differential module which admits a high-lowdecomposition. Then λ (Tor S [ x ] ( K, D )) ≥ λ (Tor S ( K, D ℓ )) + λ (Tor S ( K, D h )) . Proof.
Applying the functoriality of Lemma 5.3 to the high-low decomposition D h (cid:31) (cid:127) / / D and D / / / / D ℓ gives two exact commutative diagrams:0 (cid:15) (cid:15) (cid:15) (cid:15) (cid:15) (cid:15) / / S [ x ] ⊠ S D h σ ′ / / (cid:15) (cid:15) S [ x ] ⊠ S D h ε ′ / / (cid:15) (cid:15) D h / / (cid:15) (cid:15) / / S [ x ] ⊠ S D σ / / S [ x ] ⊠ S D ε / / D / / / / S [ x ] ⊠ S D σ / / (cid:15) (cid:15) S [ x ] ⊠ S D ε / / (cid:15) (cid:15) D / / (cid:15) (cid:15) / / S [ x ] ⊠ S D ℓ σ ′′ / / (cid:15) (cid:15) S [ x ] ⊠ S D ℓ ε ′′ / / (cid:15) (cid:15) D ℓ / / (cid:15) (cid:15)
00 0 0In both diagrams the first two columns are split exact over S [ x ] due to the high-lowdecomposition. Because D h and D ℓ are annihilated by x , Lemma 5.3 implies that σ ′ and σ ′′ are multiplication by x . The S [ x ]-action on Tor S [ x ] ( K, − ) is via K and xK = 0, so after applying Tor S [ x ] ( K, − ) and using Lemma 2.13 the maps σ ′ and σ ′′ become zero, leaving0 (cid:15) (cid:15) (cid:15) (cid:15) / / Tor S ( K, D h ) ε ′ / / (cid:15) (cid:15) Tor S [ x ] ( K, D h ) γ ′ / / (cid:15) (cid:15) ( † ) Tor S ( K, D h ) / / (cid:15) (cid:15) S ( K, D ) σ / / Tor S ( K, D ) ε / / Tor S [ x ] ( K, D ) γ / / Tor S ( K, D ) σ / / Tor S ( K, D )andTor S ( K, D ) σ / / Tor S ( K, D ) ε / / (cid:15) (cid:15) ( ‡ ) Tor S [ x ] ( K, D ) γ / / (cid:15) (cid:15) Tor S ( K, D ) σ / / (cid:15) (cid:15) Tor S ( K, D )0 / / Tor S ( K, D ℓ ) ε ′′ / / (cid:15) (cid:15) Tor S [ x ] ( K, D ℓ ) γ ′′ / / Tor S ( K, D ℓ ) / / (cid:15) (cid:15)
00 0Using Lemma 5.2 on the commutative squares ( † ) and ( ‡ ) gives the desired inequal-ity. (cid:3) Lower bound on the Betti number
In order to apply the results for high-low decompositions we need to establishsome results on the existence of high-low decompositions D h and D ℓ with H ( D h ) =0 and H ( D ℓ ) = 0.Recall that m i denotes the i -th coordinate of a d -tuple m ∈ Z d . Definition 6.1.
Let D be a Z d -graded differential module and let 1 ≤ i ≤ d . Wesay that D is bounded in the i -th direction if there are a, b ∈ Z such that m i [ a, b ]implies D m = 0. Remark . When D is finitely generated the condition that D is bounded in the i -th direction for all i is equivalent to the condition that rank k D < ∞ . Lemma 6.3.
Let D be a Z d -graded differential module with H ( D ) = 0 . Fix anindex ≤ i ≤ d and suppose that (deg δ D ) i = 0 . If H ( D ) is bounded in the i -thdirection then there is a Z d -graded differential module D ′ that is quasi-isomorphicto D such that D ′ has a high-low decomposition D ′ h and D ′ ℓ with H ( D ′ h ) and H ( D ′ ℓ ) both non-zero.Proof. Let a ∈ Z be the largest integer such that H ( D ) m = 0 whenever m i < a .Such an integer exists because H ( D ) is non-zero and bounded in the i -th direction.Set E = M m ∈ Z d a ≤ m i D m . N THE RANK OF MULTI-GRADED DIFFERENTIAL MODULES 13
Use induction on d . For d = 0, so that R = k , we haveTor k ( k, D ) = H ( k ⊠ k D ) ∼ = H ( D ) = 0 . So β k ( D ) ≥ d >
1. Then H ( D ) is bounded in the d -th direction by assumption.By Proposition 2.8 the Betti number is preserved under quasi-isomorphisms, so Lemma 6.3 allows us to assume that D has a high-low decomposition D h and D ℓ with H ( D h ) = 0 and H ( D ℓ ) = 0. By definition of a high-low decomposition, H ( D h )and H ( D ℓ ) are submodules of H ( D ) since the splitting happens in the category ofdifferential modules. In particular, H ( D h ) and H ( D ℓ ) are bounded in the i -thdirection for all i .Thus the induction hypothesis applies to D h and D ℓ . From Theorem 5.5 we haveinequalities: β R ( D ) ≥ β k [ x ,...,x d − ] ( D ℓ ) + β k [ x ,...,x d − ] ( D h ) ≥ d − + 2 d − = 2 d . (cid:3) Remark . Example 4.3 shows that Theorem 6.4 cannot be extended to differen-tial modules D with deg δ D > .Via Theorem 2.10 this result provides an affirmative answer to Conjecture 1.1for Z d -graded differential modules with deg δ = . Corollary 6.6. If F is a finitely generated Z d -graded differential module that isfree as an R -module such that deg δ F = and such that H ( F ) has non-zero finitelength then rank R F ≥ d . Proof.
By Lemma 4.1, F has a free flag. So Theorem 2.10 implies that β R ( F ) ≤ rank R F . Applying Theorem 6.4 gives the desired inequality. (cid:3) Acknowledgments
The author gratefully acknowledges Srikanth Iyengar for his guidance and LarsWinther Christensen for helpful comments on this paper.
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Department of Mathematics, University of Nebraska, Lincoln, NE 68588, U.S.A.
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