Maximal Cohen-Macaulay Modules over Two Cubic Hypersurface ... · Maximal Cohen-Macaulay Modules...

Maximal Cohen-Macaulay Modulesover Two Cubic Hypersurface

Singularities

Corina Baciu

1.Gutachter: Prof.Dr. G. Pfister2.Gutachter: Prof.Dr. D. Popescu

Vollzug der Promotion: 03.08.2005

Vom Fachbereich Mathematik der Universitat Kaiserslautern zur Verleihung desakademischen Grades Doktor der Naturwissenschaften

(Doctor rerum naturalium, Dr.rer.nat.)

genehmigte Dissertation.

D 386

Contents

I Simple node singularity 11

1 Rank one, graded, MCM modules over the hypersurfaceR = k[y1, y2, y3]/(y3

1 + y21y3 − y2

2y3) 131.1 Two–generated graded MCM R–modules . . . . . . . . . . . . 131.2 Three-generated graded MCM R-modules . . . . . . . . . . . 17

2 Rank two, graded, MCM modules over the hypersurfaceR = k[y1, y2, y3]/(y3

1 + y21y3 − y2

2y3) 272.1 The classification of rank two locally free MCM R–modules . 272.2 The rank two, stable vector bundles on ProjR and their matrix

factorizations . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.3 The classification of non–locally free, rank 2, MCM R–modules 40

2.3.1 6–generated . . . . . . . . . . . . . . . . . . . . . . . . 412.3.2 5–generated modules . . . . . . . . . . . . . . . . . . . 472.3.3 4–generated modules . . . . . . . . . . . . . . . . . . . 54

II Fermat surface 63

3 Rank one, graded, MCM modules over the hypersurfaceR = k[x1, x2, x3, x4]/(x3

1 + x32 + x3

3 + x34) 65

4 Rank two, orientable, MCM modules over the hypersurfaceR = k[x1, x2, x3, x4]/(x3

1 + x32 + x3

3 + x34) 69

4.1 Orientable 6–generated MCM modules . . . . . . . . . . . . . 734.2 Orientable 4–generated MCM modules . . . . . . . . . . . . . 88

5 Rank two, non – orientable, MCM modules over the hyper-surface R = k[x1, x2, x3, x4]/(x3

1 + x32 + x3

3 + x34) 97

5.1 Non–orientable, rank 2, 4–generated MCM modules . . . . . . 995.2 Non–orientable, rank 2, 5–generated MCM modules . . . . . . 105

Introduction

The present work is concerned with the classification of graded maximalCohen-Macaulay modules (shortly MCM modules ) over the affine cone of twovery special cubic hypersurfaces: the simple node singularity (Newtonknot)and the Fermat surface.

The category of MCM modules has been intensively studied (mostly in80’s) and a long series of interesting results appeared. We note here the oneof Greuel-Knorrer ([GK]), Buchweitz-Greuel-Schreyer ([BGS]), Auslander-Reiten ([AR]), Solberg ([S1], [S2]). D.Eisenbud and J.Herzog classified (see[EH]) the homogeneous Cohen–Macaulay rings of finite representation type(that has only a finite number of isomorphism classes of indecomposablegraded MCM modules, up to degree shiftings). They are isomorphic to oneof the following rings:

1. k[x1, x2, ..., xn],

2. k[x]/(xm) for some m > 0,

3. k[x1, x2, ..., xn]/(x21 + x2

2 + ... + x2n),

4. k[x1, x2]/(x21x2 + x3

2),

5. k[x1, x2, x3]/(x1x2, x2x3, x3x1),

6. the scroll of type (m) for some m,

7. the scroll of type (2, 1), and

8. k[x1, x2, ..., x6]/I, I being the ideal generated by all the 2× 2-minors ofthe symmetric matrix

x1 x2 x3

x2 x4 x5

x3 x5 x6

Geometrically, the above rings can be described as:

1

• Arbitrary dimension: Pn and smooth quadric hypersurfaces

• Dimension 0: 3 or fewer distinct points in P2

• Dimension 1: A curve of degree n in Pn. ( rational normal curves )

• Dimension 2: The rational normal scroll S(1, 2) in P4 andthe Veronese embedding of P2 in P5.

The graded MCM modules over a graded algebra S = ⊕n≥0

Sn generated

by S1, with k = S0 algebraically closed field, correspond, by sheafification, toaCM sheaves on the projective cone Y = ProjS (Theorem of Grothendieck,see for ex. [KL]). We say that over a projective variety Y of dimensionn, with OY (1) a very ample line bundle, a bundle E on Y is ArithmeticallyCohen–Macaulay (shortly aCM) if the graded module Γ∗(E) = ⊕

k∈ZΓ(Y, E(k))

is graded MCM module (over the affine cone over Y ) and Hp(Y, E(t)) = 0for p 6= 0, n and for all t ∈ Z.

In particular, if Y is a smooth curve, the aCM sheaves are the vectorbundles. So, in this case, the problem of the classification of graded MCMmodules is equivalent to the classification of the vector bundles on the pro-jective cone.

In 2001, G. - M. Greuel and Yu. Drozd (see [DG]) have proved that thecategory of vector bundles over a reduced curve is:

1) finite, for a configuration of projective lines of the type

C

CCCCCC

CCCCCCC

CCCCCCC

2) tame, for

(a) smooth elliptic curve

2

(b) a configuration of projective lines of the type

AAAA

AA

(c) simple node

3) wild, otherwise.

For projective varieties of dimension 1, the problem of classification ofaCM coherent sheaves is quasi-closed by Drozd and Greuel in the above citedpaper: long time ago, in 1957, Atiyah (see [At]) has described the categoryof vector bundles over an elliptic curve and the same done Drozd and Greuelfor the other curves with tame category of vector bundles: the simple nodesingularity and the configuration of projective lines

AAAA

AA

In the case of the simple node singularity the vector bundles are describedas B(d, m, λ) where:• λ ∈ k∗, continuous parameter• m natural number• d = d1, .., dr cycle of integers, d1 + ..+dr is the degree of the vector bundleand rm is the rank (see also [B]).

As we can look to MCM modules from two points of view, the algebraicone and the geometric one, if the results from both sides are combined, onecan get very useful information. An example, in this sense, is the paper ’Max-imal Cohen-Macaulay Modules over the Cone of an Elliptic Curve’ (Journalof Algebra, 253, 209-236, 2002) of R. Laza, G. Pfister and D. Popescu, where,using the classification of the vector bundles over a smooth elliptic curvemade by Atiyah, there are explicitly described the graded MCM modulesover R = k[y1, y2, y3]/〈y3

1 + y32 + y3

3〉.They give canonical normal forms for the matrix factorizations of all gradedreflexive R–modules of rank 1 (see Section 3 in [LPP]) and show effectively

3

how one can produce the indecomposable graded reflexive R–modules ofgrater rank using Singular (see Section 5 in [LPP]).

Is it possible to do the same in the case of the simple node, using theabove description of vector bundles?

The problem seems to be more difficult, since there is one singularity. So,the MCM modules on the affine cone of the simple node (that has a non–isolated singularity) correspond to the coherent aCM sheaves on the simplenode, not all locally free. The classification of Drozd and Greuel can provideus at most the locally free MCM modules.

In the first part of the present work, combining methods of commuta-tive algebra and computer algebra with tools from algebraic geometry, wedescribe explicitly:• all rank one graded MCM modules;• all rank two, indecomposable, graded MCM modules• all rank two locally free MCM modules corresponding to stable vector bun-dles.

The modules are described using the technique of matrix factorization,introduced by Eisenbud. We remind here the basics.

Let k be an algebraically closed field of characteristic zero and S the poly-nomial ring S = k[x1, ..., xn]. Consider R = S/f the hypersurface ring givenby some irreducible homogeneous polynomial f ∈ S of degree d and M agraded MCM module over R. As an S–module, M is Cohen-Macaulay mod-ule, so, by Auslander-Buchsbaum formula, it has a graded minimal resolutionof the form

0 −→j=n⊕j=1

S(βj)ϕ−→

j=n⊕j=1

S(αj) −→M −→ 0,

with n is the minimal number of generators of M and ϕ the multiplicationwith a square n × n matrix A with homogeneous entries. Because of theminimality of the resolution, the entries of A can be either zero or of strictlypositive degree.Tensoring by R we obtain an infinite, graded, 2-periodic resolution of M overR, of the form

...ϕ−→

j=n⊕j=1

R(αj − d)ψ−→

j=n⊕j=1

R(βj)ϕ−→

j=n⊕j=1

R(αj) −→M −→ 0.

4

Again, the map ψ is the multiplication by a square n× n matrix A′ withhomogeneous entries. In the paper ’Homological algebra with applicationto group representations’(1980), D.Eisenbud has proved the remarkable factthat the pair of matrices (A,A′) form a matrix factorization of the polyno-mial f . This means, AA′ = A′A = fIdn.

The matrix factorizations proved to be a very useful tool for the study ofMCM modules over hypersurface rings.If (A,A′) is a matrix factorization of the polynomial f and ϕ a graded mapdefined by A, Cokerϕ is a graded MCM module over R = S/f .

Of course, two matrix factorizations (A,A′) and (B,B′) give isomorphicmodules if and only if there exist two invertible matrices U, V such thatB = UAV,B′ = V A′U .

The rank of the module CokerA is precisely the integer r such thatdetA = f r. It follows immediately that the minimal number of genera-tors of a MCM R–module of rank r is smaller equal dr, with d the degree ofthe polynomial f .

The module CokerA is decomposable if and only if the matrix A decom-poses, after some linear transformations.

In the second part of this work it is proved that, over the affine cone ofa smooth hypersurface, a rank two graded MCM module is orientable if andonly if it has a skew symmetric matrix factorization.

In the subsection 2.2, I prove that, over the affine cone of curves of arith-metic genus 1, the matrix factorizations give information on the stability ofthe sheafification of graded MCM modules.

Another very useful tool in the classification of MCM modules is givenby the extensions of MCM modules. Over hypersurface rings, it is veryinteresting the following question:If L and F are two graded MCM modules with the matrix factorizations(A,A′), respectively (B,B ′), how are looking like the matrix factorizations ofthe module E, if there exists a graded exact sequence:

0→ L→ E → F → 0 ?

We answer to this question in the next theorem.

5

Theorem 0.1. Let S = k[x1, ..., xn] and R = S/f a hypersurface ring definedby an irreducible homogeneous polynomial f . Consider L, F two graded MCMR–modules with the matrix factorizations (A,A′), respectively (B,B ′) and the

graded extension 0→ Lα−→ E

β−→ F → 0.Then E is a graded MCM R–module and has a matrix factorization (M,M ′),

M of the type M =

(A D0 B

), D a matrix with homogeneous entries in S

such that A′ ·D ·B′ = 0 in R.

Proof. Denote s = µ(L), t = µ(F ), r1 =rank(L), r2 =rank(F ).Consider the following graded diagram:

0 −→j=s⊕j=1R(α′j)

i−−−→j=s+t⊕j=1

R(α′j)π−−−→

j=s+t⊕

j=s+1R(α′j) −−−→ 0

yAy

yB

0 −→j=s⊕j=1R(αj)

i−−−→j=s+t⊕j=1

R(αj)π−−−→

j=s+t⊕

j=s+1R(αj) −−−→ 0

ypAyδ

ypB

0 −→ CokerAα−−−→ E

β−−−→ CokerB −−−→ 0

where, pA, pB are the natural projections, π is the projection on the last t

variables and i the natural inclusion. Let ϕB :j=s+t⊕

j=s+1R(αj)→ E be a graded

map such that β ϕB = pB and consider ϕA = αpA. Define the graded mapδ as the sum of ϕA and ϕB. Then, δ makes the above diagram commutative.Using Snake Lemma, we get the graded exact sequence

0 −→ ImAi−→ Ker δ

π−→ ImB −→ 0

and the surjectivity of δ. Therefore, Ker δ is a graded MCM R–module of

rank s+ t− r1 − r2 such that E 'j=s+t⊕j=1

R(αj)/Ker δ.

So, there exists (M1,M′1) a graded matrix factorization of f such that E '

CokerM ′1 and such that the graded sequence

0 −→ ImAi−→ ImM ′

1π−→ ImB −→ 0

is exact.

6

Since Im

(A0

)⊂ ImM ′

1, there exist two invertible matrices U and V

such that UM ′1V =

(A D1

0 B1

). Denote M ′′ = UM ′

1V . Then CokerM ′′ is

a graded MCM module isomorphic to E. Thus, there exists the matrix B ′1with homogeneous entries such that (B,B ′1) is a matrix factorization of fand A′D1B

′1 = 0 in R.

We have therefore a graded commutative diagram:

0 −→ ImAi−−−→ ImM ′′ π−−−→ ImB1 −−−→ 0yid

yo

0 −→ ImAi−−−→ ImM ′

1π−−−→ ImB −−−→ 0

and an inclusion ImB1 → ImB of two graded MCM modules of the samerank. This means that they are isomorphic, and therefore, there exist twoinvertible matrices U1 and V1 such that B = U1B1V1. Denote D = D1V1 and

M =

(A D0 B

). Then CokerM ' CokerM ′′ ' E. Evidently, A′DB′ = 0 in

R.

As mentioned before, in the first part of this work, I describe matrix fac-torizations of all the isomorphism classes of the rank one and two graded,indecomposable MCM modules over the hypersurface R = k[y1, y2, y3]/(y3

1 +y2

1y3 − y22y3), that is the affine cone over the simple node singularity.

In the first section, it is proven that, up to shiftings, there are threefamilies, parametrized by the points of the curve, of isomorphism classes ofgraded, rank 1, MCM modules; two of them contain 2–minimally generatedmodules, the other contains 3–minimally generated modules.

I identify also the graded maps, that, after sheafification, produce the linebundles of degree 0, 1 or -1.

In the second subsection there are described the rank two, graded, inde-composable MCM R–modules. In 2.1, I construct an algorithm to produceall rank two locally free MCM modules. In this section, the main role isplayed by the classification of vector bundles on the simple node, especially

7

by the property of B(0, 2, 1) and B((0, 1), 1, λ) to be uniquely determinedthrough the extensions:

0 −→ OY −→ B(0, 2, 1) −→ OY −→ 0,

and0 −→ OY −→ B((0, 1), 1, λ) −→ B(1, 1,−λ) −→ 0.

As a consequence of the above results, in 2.2, I give matrix factorizationsof all rank two graded MCM R–modules with stable sheafification on thesimple node.In the subsection 2.3 I describe explicitly all the isomorphism classes of in-decomposable, non–locally free graded MCM modules of rank two: there aretwo families parametrized by the regular points of the curve and 22 countablefamilies.

Remark:It is known that a graded hypersurface has countable CM–representationtype iff it is isomorphic to A∞, that is: k[x1, x2, ..., xn]/(x2

2 + ...+ x2n).

But what about Cohen–Macaulay algebras of bigger dimension ?For rings (respectively projective varieties) of bigger dimension, that are notof finite representation type, one can hope only to classify families of MCMmodules (resp. aCM bundles) of some low rank. On Fano threefolds withPic(X) ∼= Z and on hypersurfaces of degree up to 6 and dimension at least3, there are obtained a lot of results regarding the rank two vector bundles,in papers like [AC], [Ma1], [Ma2], [CM1], [CM2], [CM3], [AF], [F1].

In this work, in the second part, we classify all graded, rank two, inde-composable MCM modules over the affine cone of the Fermat surface, thatis k[x1, x2, x3, x4]/(x3

1 + x32 + x3

3 + x34).

The graded rank one MCM modules over this algebra were classified in [EP].There are a finite number, and they correspond to 27 lines, 27 conics respec-tively 72 twisted cubic curves on the surface. For the completion of the work,I present this classification in the third section.In the fourth section we classify the orientable rank two MCM modules. Themain role here is the description of the orientable modules (over normal rings)

8

with the help of codimension two Gorenstein ideals, realized by Herzog andKuhl in [HK].In the case of 6–generated orientable modules, we use the fact that they haveskew symmetric matrix factorizations. We prove and use a very interestingproperty of the Fermat surface: a 6–generated orientable bundle, when re-stricted to the hyperplane section curve, splits into two line bundles on thecurve.( In fact, it is known that the Fermat surface is a very special cubicsurface. On it, three lines can meet at a point, while this does not happen ona general nonsingular cubic surface.) Classifying the 6–generated orientable,rank two, MCM modules, we obtain a 5 dimensional moduli space of iso-morphism classes of indecomposable modules whose restriction to V (f, x4)splits into two non-isomorphic modules and a 2-dimensional moduli space ofisomorphism classes of indecomposable modules whose restriction to V (f, x4)splits into two isomorphic modules.In the last section, we classify the non–orientable rank two MCM modules.We use a similar idea as in the case of orientable, 4–generated modules, butin this case, the ideal is not any more Cohen–Macaulay.Recently, Daniele Faenzzi, has obtained some very interesting results on themoduli spaces of aCM vector bundles on a cubic surface in P3 ( see [F2]).Part of them will be used in this work, in order to get some informations onthe stability of the orientable modules constructed by us.

I would like to mention that, through all the work, a constant support forthe computations is given by the computer algebra system SINGULAR.(see[GPS]) Part of the procedures that I used can be found in [LPP].

There are several interesting issues that leaded me to these computationsand would direct the next research:

1. The MCM modules over a non-isolated singularity;

2. The form of the matrix factorizations corresponding to stable vectorbundles

3. The behavior of non-orientable modules over a general cubic surface

4. The classification of indecomposable, rank 2, MCM modules over ageneral cubic surface

In the present work we have constructed examples and tools useful in solvingthe above problems.

9

I specify that some results from the first part of this work were publishedin [Ba1] and [Ba2], and were presented at the NATO Advanced ResearchWorkshop on Computational Commutative and Non–Commutative Alge-braic Geometry, Chisinau, June 2004, the Workshop on Cohen–MacaulayRings and Related Structures, Constanta, April, 2005, and seminars at theUniversity Kaiserslautern. The results from the second part are going to bepublished in Journal of Algebra, and they were presented to a seminar at theUniversity Essen.

Acknowledgments : I would like to express my gratitude to all thepeople that supported me during my work, especially to Prof. G. Pfister andProf. D. Popescu, that introduced me to the world of MCM modules and,with friendly and useful advises directed me through it. I am thanksfull toDr. V. Vuletescu, Dr. I. Bourban and Dr. D. Faenzzi for fruitful discussionson this theme. I would like to mention that, during this work I was supportedby the DFG-Schwerpunkt ”Globale Methoden in der komplexen Geometrie”.

10

Part I

Simple node singularity

In this part of the work, we classify the rank one and rank two, gradedMCM over the affine cone of the simple node singularity. We give the matrixfactorizations of the rank two stable bundles. Furthermore, we classify thenon–locally free, rank one and rank two MCM modules.

11

1 Rank one, graded, MCM modules over the

hypersurface R = k[y1, y2, y3]/(y31 + y2

1y3− y22y3)

It is known that for any two graded R-modules M and N , M ⊗OX N ∼=M ⊗R N and ˜HomR(M,N) ∼= HomOX (M, N). If M and N are maximal

Cohen-Macaulay R-modules, then M ∼= N if and only if M ∼= N .In the classification of the vector bundles on the simple node Y = ProjR

from [B], the line bundles of degree d ∈ Z are B(d, 1, λ) with λ ∈ k∗ (λ runover all regular points of the curve Y ).The tensor product of two line bundles is given by:

B(d, 1, λ)⊗ B(d′, 1, λ′) = B(d+ d′, 1, λ · λ′).

In this section are classified the rank one graded MCM R–modules. Asmentioned in the Introduction, their minimal number of generators is smallerequal to e(R), the multiplicity of the ring. Therefore, they are two or threeminimally generated.

1.1 Two–generated graded MCM R–modules

Let s = (0 : 0 : 1) be the unique singular point of the curve V (f) ⊂ P2k and

denote V (f)reg = V (f)\s.Then V (f)reg = (λ1 : λ2 : 1), λ3

1 + λ21 − λ2

2 = 0, λ1 6= 0 ∪ (0 : 1 : 0).For any λ = (λ1 : λ2 : 1) in V (f) denote:

ϕλ =

(y1 − λ1y3 y2y3 + λ2y

23

y2 − λ2y3 y21 + (λ1 + 1)y1y3 + (λ2

1 + λ1)y23

),

ψλ =

(y2

1 + (λ1 + 1)y1y3 + (λ21 + λ1)y2

3 −(y2y3 + λ2y23)

−(y2 − λ2y3) y1 − λ1y3

).

13

If λ = (0 : 1 : 0) let be:

ϕλ =

(y1 + y3 y2

2

y3 y21

), ψλ =

(y2

1 −y22

−y3 y1 + y3

).

We consider the following graded maps defined by the above matrices:

1. ψλ : R(−2)2 −→ R ⊕R(−1), λ ∈ V (f)reg;

2. ϕλ : R(−2)⊕R(−3) −→ R(−1)2, λ ∈ V (f)reg.

Theorem 1.1. For all λ ∈ V (f), (ϕλ, ψλ) is a matrix factorization of f andthe sets of graded MCM R–modules:

M−1 = Cokerϕλ |λ ∈ V (f)reg , M1 = Cokerψλ |λ ∈ V (f)reg and

M = Cokerϕs,Cokerψshave the following properties:1) Every two-generated non-free graded MCM R-module is isomorphic, up toshifting, with one of the modules from M−1 ∪M1 ∪M.2) Every two different R-modules from M−1 ∪M1 ∪M are not isomorphic.3) All the modules from M−1 ∪M1 ∪M have rank 1.4) The modules from M1 are the syzygies and also the duals of the modulesfrom M−1.

Proof. Clearly ϕλψλ = ψλϕλ = f · 12 for any λ ∈ V (f).

1) Let M be a two-generated non-free graded MCM R-module and con-sider (ϕ, ψ) a graded reduced matrix factorization of it. So ϕψ = ψϕ = f ·12

and detϕ · detψ = f 2. Since f is irreducible, we may consider detϕ =detψ = f . Because ψ is the adjoint of ϕ, it is sufficient to find

ϕ =

(ϕ11 ϕ12

ϕ21 ϕ22

)

such that detϕ = f and ϕ11 and ϕ21 are two linearly independent linearforms. So, applying some elementary transformations on the matrix ϕ, wemay suppose that:

ϕ11 = y1 − λ1y3 and ϕ21 = y2 − λ2y3, λ1, λ2 ∈ kor

ϕ11 = y1 − λy2 and ϕ21 = y3, λ ∈ k.

14

Let us consider the first case, when

ϕ =

(y1 − λ1y3 ϕ12

y2 − λ2y3 ϕ22

)

with ϕ12, ϕ22 two-forms.Notice that (detϕ)(λ1, λ2, 1) = 0. Therefore λ = (λ1 : λ2 : 1) is a point onthe curve V (f). We want to show that ϕ ∼ ϕλ.For this, consider the product ψλ · ϕ that has the form

ψλ · ϕ =

(f g0 f

)

with g = ( y21 + (λ1 + 1)y1y3 + (λ2

1 + λ1)y23 ) · ϕ12 − ( y2y3 + λ2y

23 ) · ϕ22. Since

g · (y1 − λ1y3) = ϕ12 · f − (y2y3 + λ2y23) · detϕ = f · (ϕ12 − y2y3 − λ2y

23) and

f is irreducible, we can write g = f · g1 with g1 ∈ k[y1, y2, y3]. Therefore, wehave

ψλϕ = f (

1 g1

0 1

).

Multiplying at left with ϕλ, we obtain

f ϕ = f ϕλ (

1 g1

0 1

),

that means,

ϕ = ϕλ (

1 g1

0 1

).

This equality induce the equivalence between ϕ and ϕλ.The second case (ϕ11 = y1 − λy2 and ϕ21 = y3;λ ∈ k) can be treated

exactly as above, replacing ψλ with ψλ0 , where λ0 denotes the point (0:1:0).

2) Because of the degrees of the entries, no module fromM1∪Coker ψsis isomorphic with a module from M−1 ∪ Cokerϕs.Since any two equivalent matrices have the same fitting ideals, for the rest,it is enough to consider the following fitting ideals:

• The modules from M−1 ∪M1:Fitt1(ϕλ) = Fitt1(ψλ) = 〈y1 − λ1y3, y2 − λ2y3, y

23〉, λ = (λ1 : λ2 : 1) ∈ V (f)

Fitt1(ϕλ0) = Fitt1(ψλ0) = 〈y1, y3, y22〉, λ0 = (0 : 1 : 0).

• The modules from M: Fitt1(ϕs) = Fitt1(ψs) = 〈y1, y2〉.

15

3) Follows from Corollary 6.4, [Ei1].

4) By construction, the modules of M1 are the syzygies of the modulesof M−1. Since

ϕtλ =

(0 1−1 0

)ψλ

(0 −11 0

)

Cokerψλ ∼= (Cokerϕλ)∨.

Theorem 1.2. Let Y be the projective cone over R, that is the simple nodesingularity.

1. The coherent sheaves associated to the modules from M1 give all theisomorphism classes of line bundles of degree 1 over Y ;

2. The coherent sheaves associated to the modules from M−1 give all theisomorphism classes of line bundles of degree -1 over Y ;

3. The coherent sheaves associated to the modules from M are not locallyfree.

Proof. 3) The only singular point of V (f) is (0 : 0 : 1). To prove thatCokerψs is non–locally free module, by [TJP](1.3.8), it is sufficient to provethat Fitt1(ψs)R〈y1,y2〉 6= R〈y1 ,y2〉.We can easily see that Fitt1(ψs)R〈y1,y2〉 = 〈y1, y2〉R〈y1,y2〉, so, Cokerψs andalso its dual, Cokerϕs are non–locally free modules.

1) Any line bundle of degree one on Y has the form OY (P ), with P regularpoint of Y . Following the proof of Theorem 3.8 from ([LPP]), we obtain thatthe graded MCM R-module corresponding to OY (P ) is a module from M1,for any regular point P of Y .

2) The modules of M−1 are the duals of the modules from M1, so, theyinduce the line bundles of degree -1, that are B(−1, 1, λ), λ ∈ k∗.

16

1.2 Three-generated graded MCM R-modules

For any λ = (λ1 : λ2 : 1) in V (f) let be:

αλ =

0 y1 − λ1y3 y2 − λ2y3

y1 y2 + λ2y3 (λ21 + λ1)y3

y3 0 −y1 − (λ1 + 1)y3

and βλ the adjoint of αλ.We define also the graded maps αλ : R(−2)3 −→ R(−1)3, given by the

matrices αλ.Using the same notations as in the previous section (s = (0 : 0 : 1), λ0 =

(0 : 1 : 0), V (f)reg = V (f)\s), we have the following:

Theorem 1.3. For all λ = (λ1 : λ2 : 1) ∈ V (f), (αλ, βλ) is a matrix factor-ization of f and the set of three-generated graded MCM R-modules

M0 = Cokerαλ |λ ∈ V (f)reg\λ0

has the following properties:1) All the modules from M0 have rank 1.2) Every two different modules from M0 are not isomorphic.3) Every three–generated, rank 1, graded MCM R-module is isomorphic withone of the modules from M0 or to Cokerαs.

Proof. Clearly αλβλ = βλαλ = f · 13 for any λ = (λ1 : λ2 : 1) ∈ V (f).

1) Since det (αλ) = f , by Corollary 6.4 ([Ei1]), Cokerαλ has rank 1.

2) Suppose that there exist two invertible matrices, U and V , with entriesin k, such that Uαλ = αξV for λ, ξ ∈ V (f). With the help of computer (weuse Singular[GPS]) we obtain that λ = ξ:

LIB "matrix.lib";

option(redSB);

ring r=0,(y(1..3),u(1..9),v(1..9),a,b,c,d),(c,dp);

ideal I=a3+a2-b2,c3+c2-d2;

qring Q=std(I);

17

Define the matrix A = αλ, for λ = (a : b : 1) and B = αξ, for ξ = (c : d : 1).

matrix A[3][3]= 0, y(1)-a*y(3), y(2)-b*y(3),

y(1), y(2)+b*y(3), (a2+a)*y(3),

y(3), 0, -y(1)-(1+a)*y(3);

matrix B=subst(A,a,c,b,d);

Investigation the condition Uαλ = αξV :

matrix U[3][3]=u(1..9);

matrix V[3][3]=v(1..9);

matrix C=U*A-B*V;

ideal I=flatten(C);

ideal J=ideal(det(U)-1);

J=J+transpose(coeffs(I,y(1)))[2];



ideal L=std(J);

L;

The first two entries of the ideal L are:

L[1]=b-d

L[2]=a-c

Therefore a = c and b = d, that means λ = ξ.

3) Let be M a three-generated, rank one, graded MCM R-module and(ϕ, ψ) the corresponding graded reduced matrix factorization. We can sup-pose detϕ=f and detψ = f 2. Since f ∈ 〈y1, y3〉, by [Ei2], ϕ has generalizedzeros. Thus after some elementary transformations,

ϕ =

0 ϕ1 ϕ2

ϕ3 a bϕ4 c d

ϕ1, ϕ2, ϕ3, ϕ4 and a, b, c, d linear forms, ϕ1, ϕ2, ϕ3, ϕ4 linearly indepen-dent.As f ∈ 〈ϕ1, ϕ2〉 ∩ 〈ϕ3, ϕ4〉, we can suppose that ϕ1 and ϕ3 have non-zerocoefficient of y1. So, we can choose ϕi, i = 1, 4 as follows:

ϕ1 = y1 − λ1y3, ϕ2 = y2 − λ2y3 or ϕ1 = y1 − λy2, ϕ2 = y3

ϕ3 = y1 − ξ1y2, ϕ4 = y2 − ξ2y3 or ϕ3 = y1 − ξy2, ϕ4 = y3.

18

Since detϕ = f , the points (λ1 : λ2 : 1), (ξ1 : ξ2 : 1), (λ : 1 : 0), (ξ : 1 : 0) layin V (f). Therefore λ = ξ = 0.For any λ = (λ1 : λ2 : 1) in V (f), we write ϕ1λ = y1 − λ1y3, ϕ2λ = y2 − λ2y3

and for λ = (0 : 1 : 0) we write ϕ1λ = y1, ϕ2λ = y3.Then ϕ has the form:

ϕ =

0 ϕ1λ ϕ2λ

ϕ1ξ a bϕ2ξ c d

with a, b, c, d linear forms.

Notice that, since f /∈ 〈y21, y1y3, y

23〉, ϕ can not have the form

0 y1 y3

y1 a by3 c d

,

therefore, it is not possible that λ = ξ = (0 : 1 : 0).To finish the proof, we need two helping results:

Lemma 1.4. Let M be a three-generated, rank one, graded MCM R-moduleand (ϕ, ψ) a matrix factorization of M , ϕ having the above form. Then thereexists λ′ ∈ V (f)\(0 : 1 : 0), a′, b′, c′, d′ linear forms such that the matrix

ϕ′ =

0 ϕ1λ′ ϕ2λ′

y1 a′ b′

y3 c′ d′

together with its adjoint matrix ψ′ form another matrix factorization (ϕ′, ψ′)of M .

Proof. We have to prove that after some elementary transformation the ma-trix ϕ will become ϕ′, that means, there exist two invertible 3 × 3 matricesU and V , with entries in k, such that Uϕ′ = ϕV . For this, it is sufficient toprove that there exist two invertible 3 × 3 matrices U, V such that the first

column of U−1ϕV is

0y1

y3

.

Considering U = (uij)1≤i,j≤3 and V = (vij)1≤i,j≤3, the above condition leadto the following system of equations:

ϕ1λv21 + ϕ2λv31 = y1u12 + y3u13

ϕ1ξv11 + av21 + bv31 = y1u22 + y3u23

ϕ2ξv11 + cv21 + dv31 = y1u32 + y3u33.

19

In particular, ϕ(0, 1, 0) ·

v11

v21

v31

=

000

.

Since det (ϕ(0, 1, 0)) = f(0, 1, 0) = 0, we may choose a non-zero solution(v11, v21, v31) which can be completed to an invertible matrix V and suchthat also the corresponding (u12, u13, u22, u23, u32, u33) can be completed toan invertible matrix U .

Lemma 1.5. Let M be a three-generated, rank one, graded MCM R-moduleand (ϕ, ψ) a matrix factorization of M , ϕ having the form:

ϕ =

0 ϕ1λ ϕ2λ

y1 a by3 c d

where a, b, c, d are linear forms. Then (αλ, βλ) is another matrix factorizationof M .

Proof.Observation:1. Subtracting the first column, multiplied with some nonzero constant, fromthe second, respectively the third column, one can annhilate the coefficientof the variable y1 at the entries a, b. Similarly, subtracting the first line fromthe last one, one can ”kill” y1 also in c.

2. More, since λ 6= (0 : 1 : 0) we can consider that b is in 〈y3〉k .Indeed, if b = b2y2 + b3y3, we subtract the first line multiplied with b2 fromthe second line (to ”kill” y2 in b) and then add the first column multipliedwith b2 to the second column (to ”kill” the new y1b2 in a).Therefore, instead of b we can write by3 with b ∈ k.

Consider the following polynomials (2-minors of αλ and ϕ):

γ =

∣∣∣∣y1 by3

y3 d

, δ =

∣∣∣∣y1 ay3 c

∣∣∣∣, γ =

∣∣∣∣y1 (λ2

1 + λ1)y3

y3 −y1 − (λ1 + 1)y3

∣∣∣∣

and δ =

∣∣∣∣y1 y2 + λ2y3

y3 0

∣∣∣∣.

Since detϕ = detαλ = f , we have the following equality:ϕ1λ(γ − γ) = ϕ2λ(δ − δ). (∗)

20

So ϕ1λ | δ − δ. But δ − δ = −c(y1 − λ1y3) − y3(y2 + λ2y3 + λ1c − a) anda, c ∈ 〈y2, y3〉k. Therefore, a = y2 +λ2y3 +λ1c and δ− δ = −cϕ1λ. Replacingδ−δ in (∗) we get γ−γ = −c(y2−λ2y3). But γ−γ = y1(−y1−(λ1+1)y3)−d)−y2

3(λ21 +λ1− b) and c ∈ 〈y2, y3〉k. Therefore d = −y1− (λ1 + 1)y3, b = λ2

1 +λ1

and c = 0. This shows that ϕ ∼ αλ.

Using Lemma 1.4 and Lemma 1.5 the proof of the theorem 1.3 is finished.

Theorem 1.6. Let Y be the projective cone over R.

1. The coherent sheaves associated to the modules from M0 give all theisomorphism classes of line bundles of degree 0 over Y ;

2. The coherent sheaf associated to Cokerαs is not locally free. Togetherwith the sheafifications of the modules fromM they give all the isomor-phism classes of rank one, non-locally free aCM sheaves.

Proof. By computing Fitt2(αλ)R〈y1,y2〉 we find that Cokerαλ is locally free ifand only if λ is a regular point of Y (we use again [TJP, Prop 1.3.8]). Inthe previous subsection we have proved that the modules fromM−1 andM1

have degree –1, respectively 1. So, after some shiftings, they give all linebundles of degree 3k − 1 and 3k + 1, with k ∈ Z. Therefore, the remainingrank one graded MCM R–modules (the one from M0), define the line bun-dles of degree 3k.1) We prove first that the sheafification of Cokerαξ has degree 0, whereξ = (−1 : 0 : 1).Let us compute (Cokerαξ ⊗ Cokerαξ)

∨∨.

setring S;

matrix M[3][3]= 0, y(1)-y(3), y(2),

y(1), y(2), 0,

y(3), 0, -y(1);

tensorCM(M,M);

_[1,1]=0

This means that Cokerαξ is a self dual module, so the matrix αξ correspondsto B(0, 1,−1).Consider the graded map αξ : R(−2)3 −→ R(−1)3, that should have thedegree a multiple of 3, let it be 3t.

21

From the graded exact sequence

0 −→ (Cokerαξ)∨ −→ R(1)3 −→ R(2)3 −→ Cokerαtξ −→ 0

we see that the degree of Cokerαtξ is 9 − 3t. But there exists a gradedisomorphism between Cokerαξ and Cokerαtξ ⊗R(−3). So, (9− 3t)− 9 = 3t,that means t = 0.2) Let us now consider Cokerαλ, with λ = (a : b : 1) a regular point of thesimple node. We know its sheafification has degree of the form 3t.Consider also the module Cokerψξ, with the corresponding graded mapψξ : R(−2)2 −→ R ⊕ R(−1). As we have seen in the previous subsection,this module has the degree 1. The line bundle corresponding to Cokerψξhas the form B(1, 1, µ) and the one corresponding to Cokerαλ has the formB(3t, 1, µ′).By Serre duality, for any two vector bundles F , E on the simple node,

Hom(F , E) = Ext(E ,F) = H1(E∨ ⊗F).

Therefore, Ext1(B(1, 1, µ),B(3t, 1, µ′)) = H1(B(−1, 1, µ−1) ⊗ B(3t, 1, µ′)) =H1(B(3t − 1, 1, µ−1µ′)). From [B], dimk(H

1(B(3t − 1, 1, µ−1µ′))) = 1 if andonly if 3t− 1 = −1, so, if and only if t = 0.We prove that dimk(Ext1(Cokerψξ,Cokerαλ) = 1, for any λ = (a : b : 1)regular point of the simple node. With this we are done.According to the theorem 0.1, a module M with a graded extension

0 −→ Cokerαλ −→M −→ Cokerψξ −→ 0,

is the cokernel of a graded map T : R5(−2) −→ R3(−1)⊕R⊕R(−1), given

by a square 5× 5 matrix of the form

(αλ D0 ψξ

).

D is a 3 × 2 matrix of the form

d1 d2

d3 d4

d5 d6

with linear entries, such that

βλDφξ = 0 in R.We make some linear transformations on T , in order to obtain a simple formof D. First of all, we can eliminate the variable y1 in all entries of D, bysubtracting one of the first three columns multiplied with some constants.By subtracting the last line from the first one we eliminate the variable y2

in the entry d1. We add in this way y1 to the entry d2, but we can ”kill” itusing the second column. In the same way, using the last line and the third

22

column of T , we eliminate y2 also in the entry d5. By subtracting the firstcolumn from the last one, we eliminate the variable y3 in d6. We ”kill” thenew y1 in the entry d4 using the last line.Let us now study the relation βλDφξ = 0 in R. For simplicity, we use forthis the computer.

The procedure condext returns the ideal given by the coefficients of y1

in the entries of a matrix, after it reduces the entries to the polynomialy3

1 + y21y3 − y2

2y3.This procedure will be used also later, in the subsection 2.3.

LIB"matrix.lib";

LIB"homolog.lib";

LIB"linalg.lib";

proc condext(matrix G)

ideal g=flatten(G);

matrix V;ideal P;

int k,j;

P=0;

for(j=1;j<=size(G);j++)

g[j]=reduce(g[j],std(y(1)^3+y(1)^2*y(3)-y(2)^2*y(3)));

V=coef(g[j],y(1));

for(k=1;k<=1/2*size(V);k++)

P=P+V[2,k];

;

P=interred(P);

return(P);

We define the ring R and the matrices ψξ, φξ, respectively αλ.

ring R=0,(y(1..3),d(1..6),a,b),(c,dp(3),dp(6),dp(2));

ideal i=y(1)^3+y(1)^2*y(3)-y(2)^2*y(3),a3+a2-b2;

qring S=std(i);

matrix psi[2][2]= y(1)^2,-y(2)*y(3),

23

-y(2), y(1)+y(3);

matrix phi[2][2]=y(1)+y(3), y(2)*y(3),

y(2), y(1)^2;

matrix A[3][3]=0,y(1)-a*y(3), y(2)-b*y(3),

y(1),y(2)+b*y(3), (a2+a)*y(3),

y(3), 0,-y(1)-(a+1)*y(3);

We define the matrix D and put the condition βλDφξ = 0.

matrix Aa=adjoint(A);

matrix D[3][2]=d(1..6);

matrix G=Aa*D*phi;

ideal P;

P=condext(G);

P[4]=P[4]/y(3)^2;

P[5]=P[5]/y(3)^2;

P;

The ideal of the coefficients is:

P[1]=y(2)*d(6)-y(3)*d(3)-y(3)*d(5)*a

P[2]=y(2)*d(2)+y(2)*d(5)+y(3)*d(1)*a+y(3)*d(1)-y(3)*d(5)*b

P[3]=y(2)*d(1)-y(2)*d(4)+y(3)*d(1)*b-y(3)*d(3)-y(3)*d(5)*a^2-

-y(3)*d(5)*a

P[4]=d(1)*b-d(2)*a^2-d(4)*b-d(5)*a^2-d(6)*b

P[5]=d(1)*a+d(1)-d(2)*b-d(4)*a-d(4)-d(5)*b-d(6)*a-d(6)

From P[1], we obtain y3|d6, but we have already eliminated the variabley3 in d6, so we can suppose that d6 = 0. Under this condition, P[1] impliesthat d3 = −ad5.From P[2] we obtain that y2|d1(a+1)−d5b, but d1 and d5 have no y2. There-fore, d1(a + 1) − d5b = d2 + d5 = 0. Since a 6= 0, P[5] implies that d4 = d1

and P[4] implies d5 = d1b/a2. We know that d1 has the form d1 = a1y3, with

a1 ∈ k.Since we need that the matrix T do not decomposes (nonzero extension),a1/a

2 can not be zero. Dividing the last two columns by a1/a2, and multi-

plying the last two lines with a1, we obtain the matrix:

24

T =

0 y1 − ay3 y2 − by3 a2y3 −by3

y1 y2 + by3 (a2 + a)y3 −by3 a2y3

y3 0 −y1 − (a + 1)y3 by3 00 0 0 y2

1 −y2y3

0 0 0 −y2 y1 + y3

.

One can see easily that this matrix do not decomposes. (There are no ma-trices U, V such that D = αλU + V ψξ)This proves that dimk(Ext1(Cokerψξ,Cokerαλ) = 1, and with this we aredone.

Proposition 1.7. Every three-generated, rank 2, indecomposable, gradedMCM R-module is isomorphic with one of the modules Coker βλ, λ = (λ1 :λ2 : 1) ∈ V (f).

Proof. Let M be a three-generated, rank two, indecomposable, graded MCMR-module and (ϕ, ψ) a matrix factorization of M . Then Cokerψ is a three-generated, rank one, graded MCM R-module. Therefore, it is isomorphic toone of the modules from M0 ∪ Cokerαs and Cokerϕ is isomorphic to oneof the modules from Coker βλ|λ = (λ1 : λ2 : 1), λ ∈ V (f).

25

We present a short overview over the results concerning the rank onegraded MCM R–modules:

The rank one graded MCM R-modules corresponding to the line bundleson Y are given by the cokernel of some graded maps defined by the followingmatrices, with (λ1 : λ2 : 1) a regular point on Y :

(y1 − λ1y3 y2y3 + λ2y

23

y2 − λ2y3 y21 + (λ1 + 1)y1y3 + (λ2

1 + λ1)y23

),

(y1 + y3 y2

2

y3 y21

),

(y2

1 + (λ1 + 1)y1y3 + (λ21 + λ1)y2

3 − y2y3 − λ2y23

−(y2 − λ2y3) y1 − λ1y3

),

(y2

1 −y22

−y3 y1 + y3

),

0 y1 − λ1y3 y2 − λ2y3

y1 y2 + λ2y3 (λ21 + λ1)y3

y3 0 −y1 − (λ1 + 1)y3

.

The rank one graded MCM R-modules corresponding to the non-locallyfree aCM sheaves are given by the following set of matrices:

(y1 y2y3

y2 y21 + y1y3

),

(y2

1 + y1y3 − y2y3

−y2 y1

),

0 y1 y2

y1 y2 0y3 0 −y1 − y3

.

26

2 Rank two, graded, MCM modules over the

hypersurface R = k[y1, y2, y3]/(y31 + y2

1y3− y22y3)

In the following there are described all isomorphism classes of rank two, in-decomposable MCM modules over R = k[y1, y2, y3]/(f), f = y3

1 +y21y3−y2

2y3,using their matrix factorizations.In the case of locally free MCM modules, it is used the classification of vectorbundles over the projective cone Y = Proj R, from [B].The main ”ingredient” is the existence of two extensions that determineuniquely the rank 2 vector bundles B(0, 2, 1) and B((0, 1), 1, λ); very impor-tant is also the Singular procedure for computing the tensor product oftwo locally free MCM modules ( see [LPP] ).Furthermore, it is given a characterization of the MCM modules with stablesheafification, over a projective curve with arithmetic genus 1, using a matrixfactorization of it. I construct also the matrix factorizations correspondingto stable vector bundles of rank two over the simple node Y .In the last part of this section, there are computed matrix factorizations ofthe non–locally free, rank two, MCM modules over R, using Proposition 0.1.They correspond to the rank two, non–locally free aCM sheaves on the sim-ple node Y . There are two families parametrized by the regular points of thecurve Y = V (f) and 22 countable families.

2.1 The classification of rank two locally free MCM R–modules

According to the classification of the vector bundles on the simple nodesingularity from [B], there are two types of rank two, indecomposable, vectorbundles on ProjR:• B(a, 2, λ), with a ∈ Z and λ ∈ k∗•B(d, 1, λ), with d a 2-cycle with entries in Z and λ ∈ k∗(d = (a, b), a 6= b).

To generate the first type of rank two vector bundles it is sufficient toknow the bundle B(0, 2, 1) and the line bundles, because:

27

B(a, 2, λ) ∼= B(a, 1, λ)⊗ B(0, 2, 1).

The fact that the bundle B(0, 2, 1) is uniquely determined by the exact se-quence

0 −→ OY −→ B(0, 2, 1) −→ OY −→ 0, (1)

provide a way to determine the graded MCM R-module corresponding toit. Using the matrix factorizations of the rank 1 MCM nodules given inthe previous section and a Singular–procedure that computes the tensorproduct of two locally free MCM modules, one can easily construct matrixfactorizations for the vector bundles B(a, 2, λ).

The second type of rank two vector bundles can be generated by thebundles B((0, n), 1, λ) and the line bundles, using the tensor product formula:B((a, b), 1, λ)⊗ B(c, 1, µ) ∼= B((a + c, b+ c), 1, λµ2).

For d = (a, b) and e = (c, d) two 2-cycles with entries in Z, we have:

B(d, 1, λ)⊗ B(e, 1, µ) ∼= B(f1, 1, λ · µ)⊕ B(f2, 1, λ · µ),

where f1 = (a+c, b+d) and f2 = (a+d, b+c). If fi = (α, α)(i = 1 or 2), thenB(fi, 1, λ · µ) splits as: B(fi, 1, λ · µ) = B(α, 1,

√λ · µ)⊕ B(α, 1,−√λ · µ).

Therefore, inductively, we can obtain all B((0, n), 1, λ), n ∈ N∗, if weknow the bundles B((0, 1), 1, λ). By duality, (B(d, 1, λ)∨ ∼= B(−d, 1, λ−1))we obtain also B((0, n), 1, λ) with n negative integer.The bundles B((0, 1), 1, λ) are uniquely determined by the existence of theexact sequences

0 −→ OY −→ B((0, 1), 1, λ) −→ B(1, 1,−λ) −→ 0. (2)

Using this, I compute the graded MCM R-modules corresponding to them.So, inductively, one can obtain all rank two graded indecomposable locallyfree MCM R-modules.

In the sequel we determine the module M2 corresponding to B(0, 2, 1).

Lemma 2.1. Let be ρ =

y21 + y1y3 − y2 − y3 0−y2y3 y1 0 − y3

0 0 y1 y2

,

28

ψ =

y1 y2 y3 0y2y3 y2

1 + y1y3 0 y3

0 0 y21 + y1y3 − y2

0 0 −y2y3 y1

,

γ =(

0 0 y2y3 y21 + y1y3

)and ϕ =

(ργ

). Then (ψ, ϕ) is a matrix

factorization of Ω1R(m), where m is the unique graded maximal ideal of R,

m = 〈y1, y2, y3〉. More, the following exact sequence

ψ−→ R(−3)⊕R(−2)3 ρ−→ R(−1)3 (y1,y2,y3)−−−−−→ m −→ 0 (3)

is a graded minimal free resolution of m. In particular, Ω1R(m) has no free

summands.

Proof. Clearly ϕψ = ψϕ = f · 14 and the above sequence is a complex.

Let u1, u2, u3 ∈ R such that

3∑

i=1

yiui = 0. We show that u =

u1

u2

u3

is an

element of Imρ. Subtracting multiples of the second and third columns ofρ from u we may suppose that u1 depends only on y1. As the maps aregraded, we may suppose that u is graded, so u1 = ays1, a ∈ k, s ∈ N. If a 6= 0,

s+ 1 ≥ 3, so, subtracting from u multiples of

y21

−y2y3

y21

∈ Imρ, we reduce

to the case u1 = 0. Then y2u2 + y3u3 = 0 and, since y2, y3 is a regularsequence in R, u is a multiple of the fourth column of ρ.To show that Ker ρ ⊂ Imψ, it is enough to prove that Ker ρ ⊂ Kerϕ.We denote with ρ3 the third row of ρ and choose ν an element of Ker ρ. Sincey1γ = y2y3ρ3 and ρ3ν = 0, y1(γν) = 0. So γν = 0 and ν is an element ofKerϕ.Because no entry of ϕ or ψ is unite, Ω1

R(m) has no free summands.

Proposition 2.2. There exists a graded exact sequence:

0 −→ Ri−→ Ω2

R(m)⊗R(3)π−→ m −→ 0 (4)

and Ω2R(m)⊗R(3) corresponds to the bundle B(0, 2, 1).

29

Proof. 1) We prove the existence of the exact sequence (4). Define the map

i : R −→ Ω2R(m) ⊗ R(3) by i(1) =

0y3

−y2

y1

(the fourth column of ψ) and

let π : Ω2R(m)⊗R(3) −→ m be the projection on the first component. Since

Ω2R(m) ⊗ R(3) = Imψ ⊗ R(3) ⊂ R ⊕ R(1)3, i and π are graded morphisms.

Clearly, i is injective, π is surjective and Im i ⊂ Ker π. We prove thatKer π ⊂ Imi.

Let v = ψ ·

abcd

be an element in Ker π. Then

abc

∈ Im ρ.

Denote by ψ′ the 4×3 matrix obtained from ψ by eliminating the last column.

Then v = ψ′ ·

abc

+ d

0y3

−y2

y1

. So we have to prove that the columns

of ψ′ρ are in Imi.

We write ψ′ρ =

0 0 0 00 0 y3 · (−y2y3) y3 · (−y2

1 − y1y3)0 0 −y2 · (−y2y3) − y2 · (−y2

1 − y22)

0 0 y1 · (−y2y3) y1 · (−y21 − y2

2)

.

Therefore, v ∈ R ·

0y3

−y2

y1

= Imi.

2) Let M2 be the graded R-module Ω2R(m) ⊗ R(3). We prove that it is

indecomposable.If it would decompose, it would be isomorphic to Coker θ, with θ of the form(A 00 B

), A and B quadratic matrices with determinant equal to f . By the

Corollary 6.4 from [Ei1], A and B define rank 1, graded MCM R-modules, sothey are equivalent to one of ϕλ, ψλ|λ ∈ V (f).(see the previous sections)Since θ ∼ ϕ, their fitting ideals are equal, so Fitt2(θ) = Fitt2(ϕ) = m2.

But the elements of degree 2 from Fitt2(θ) are given just by lA1 · lB1 ,lA1 · lB2 , lA2 · lB1 , lA2 · lB2 where lA1 , l

A2 respectively lB1 , l

B2 are the entries of A and

B of degree 1. The ideal generated by them is not m2 since m2 is minimally

30

generated by 6 elements. Therefore, M2 is indecomposable.

3) To prove that M2 is locally free, it is sufficient to notice thatFitt2(ϕ)R〈y1,y2〉 = R〈y1,y2〉 and Fitt3(ϕ) = 0 (see Proposition 1.3.8, [TJP]).

4) From the exact sequence (4) we get the following exact sequence ofvector bundles on Y = ProjR:

0 −→ OY −→ M2 −→ OY −→ 0.

M2 is an indecomposable vector bundle of rank 2, so it is isomorphic toB(0, 2, 1).

Using this result we obtain the following theorem:

Theorem 2.3. The rank two vector bundles of the type B(a, 2, λ), with λ ∈ k∗and a integer, are sheafification of the R-modules (M2 ⊗ L)∨∨ ⊗ R(k), withL in M0 ∪M1 ∪M−1 and k ∈ Z. More, the modules corresponding to

B(0, 2, λ), λ 6= 1, have 6 generators;B(−1, 2, λ), have 4 generators;B(1, 2, λ), have 4 generators.

Proof. The first statement is a direct consequence of the previous proposi-tion, if we consider the following tensor product:B(a, 2, λ) ∼= B(a, 1, λ)⊗ B(0, 2, 1).Therefore the modules corresponding to the bundles B(0, 2, λ) are (M2 ⊗Cokerαλ)

∨∨ and we can compute their matrix factorizations using the com-puter. We use the procedures that compute the reflexive hull and the tensorproduct in the category of locally free Cohen-Macaulay modules from [LPP].

LIB"matrix.lib";

option(redSB);

proc reflexivHull(matrix M)

module N=mres(transpose(M),3)[3];

N=prune(transpose(N));

return(matrix(N));

31

proc tensorCM(matrix Phi, matrix Psi)

int s=nrows(Phi);

int q=nrows(Psi);

matrix A=tensor(unitmat(s),Psi);

matrix B=tensor(Phi,unitmat(q));

matrix R=concat(A,B);

return(reflexivHull(R));

proc M2(ideal I)

matrix A=syz(transpose(mres(I,3)[3]));

return(transpose(A));

ring R=0,(y(1..3)),(c,dp);

ideal i=y(1)^3+y(1)^2*y(3)-y(2)^2*y(3);

qring S=std(i);

ideal I=maxideal(1);

matrix C=M2(I);

We define the matrix αλ, for λ = (a : b : 1).

ring R1=0,(y(1..3),a,b),(c,dp);

ideal I=y(1)^3+y(1)^2*y(3)-y(2)^2*y(3),a3+a2-b2;

qring S1=std(I);

matrix A[3][3]= 0, y(1)-a*y(3), y(2)-b*y(3),

y(1), y(2)+b*y(3), (a2+a)*y(3),

y(3), 0, -y(1)-(a+1)*y(3);

matrix C=imap(S,C);

We compute the dimension of the matrix corresponding to (M2⊗Cokerαλ)∨∨.

nrows(tensorCM(C,A));

6

32

The same, the modules corresponding to the bundles B(−1, 2, λ) are (M2 ⊗Cokerϕλ)

∨∨. We compute the size of the matrix corresponding to this mod-ule.

matrix B[2][2]=y(1)-a*y(3), y(2)*y(3)+b*y(3)^2,

y(2)-b*y(3),y(1)^2+(a+1)*y(1)*y(3)+(a2+a)*y(3)^2;

nrows(tensorCM(C,B));

4

B= y(1)+y(3),y(2)^2,

y(3), y(1)^2;

nrows(tensorCM(C,B));

4

By duality, it follows also the last statement.

Let ξ = (−1 : 0 : 1) and λ0 = (0 : 1 : 0) two regular points on the curveY = ProjR. We consider the graded maps ψξ : R(−2)2 −→ R ⊕R(−1) andαλ0 : R(−2)3 −→ R(−1)3, given by the matrices with the same name. (seethe previous section)

Lemma 2.4. If we denote by µ0 the regular point on the nodal curve Y such

that ˜Cokerψξ ∼= B(1, 1, µ0), then OY (1) = B(3, 1,−µ30) and the line bundle

B(1, 1,−µ0) is given by Cokerψλ0 .

Proof. Since the degree of Cokerψλ0 is 1 (see 1.2 ) we can denote ˜Cokerψλ0

by B(1, 1, θ).Let us compute, the tensor product (Cokerαξ ⊗ Cokerψξ)

∨∨

matrix N[2][2]= y(1)^2, -y(2)*y(3),

-y(2), y(1)-y(3);

matrix L=tensorCM(M,N);

print(L);

y(1), -y(3),

y(2)^2, -y(1)^2+y(1)*y(3)

It is easy to see that, after some elementary transformations, the matrix Lbecomes ψλ0 , so we have obtained that (Cokerαξ⊗Coker ψξ)

∨∨ ∼= Cokerψλ0 .By sheafification, it becomes B(0, 1,−1)⊗B(1, 1, θ) ∼= B(1, 1,−µ0), thereforeθ = µ0.

Let us now compute the reflexive hull of Cokerψλ0⊗Cokerψλ0⊗Cokerψλ0 .

33

matrix L[2][2]= y(1)^2, -(y(1)^2+y(2)^2),

-y(3), y(1);

tensorCM(L,tensorCM(L,L));

_[1,1]=0

This means that (Cokerψλ0 ⊗ Cokerψλ0 ⊗ Cokerψλ0)∨ ∼= R, so the linebundle OY (1) is B(3, 1,−µ3

0).

In a similar way, we get the following relations:

Proposition 2.5. Let be λ0 = (0 : 1 : 0). For any λ = (λ1 : λ2 : 1) ∈ V (f),we have:

Cokerα(λ1:λ2:1)∼= (Cokerϕλ0 ⊗ Cokerψ(λ1 :λ2:1))

∨∨

Coker α(λ1:−λ2:1)∼= (Coker α(λ1:λ2:1))

t ∼= (Cokerψλ0 ⊗ Coker ϕ(λ1:λ2:1))∨∨.

For the rest of this section, we fix the notations ξ, λ0 and µ0 as in theprevious lemma.

Let us now determine the graded MCM modules corresponding to thebundles B((0, 1), 1, λ) with λ ∈ k∗. Consider the module Ω1

R(M2) = Cokerψ,whereψ : R(−4)3 ⊕ R(−3) −→ R(−3)⊕ R(−2)3 is given as in lemma 2.1.

Lemma 2.6. Consider λ a regular point on the curve.

1. (Ω1R(M2)⊗ Cokerαλ)

∨∨ has

4 generators, if λ = (1 : 0 : 1);3 generators, otherwise;

2. (Ω1R(M2)⊗ Cokerϕλ)

∨∨ has 5 generators;

3. (Ω1R(M2)⊗ Cokerψλ)∨∨ has 5 generators;

Proof. We define the module Ω1R(M2) by:

matrix D=transpose(syz(C));

and use the procedure tensorCM as before.

34

We compute the matrix corresponding to the MCM module(Ω1

R(M2)⊗ Cokerϕξ)∨∨.

matrix D=transpose(syz(C));

matrix B[2][2]=y(1)+y(3), y(2)*y(3),

y(2), y(1)^2;

matrix A=tensorCM(D,B);

After some linear transformations, the matrix A it becomes:

A =

y1 0

0 0

0 −y3

αξ

0 ψξ

.

Remark: This matrix is linear equivalent to the matrix T obtained in theproof of Theorem 1.6, for a = −1 and b = 0.

Proposition 2.7. 1. The graded module corresponding to B((0, 1), 1, µ0)is (Ω1

R(M2)⊗ Cokerϕξ)∨∨ ⊗R(2).

2. Ω1R(M2) = B((−4,−5), 1, µ−9

0 ).

Proof. 1) We have the following graded commutative diagram:

0 −−−→ R(−2)3 i−−−→ R(−2)5 π−−−→ R(−2)2 −−−→ 0yαξyA

yψξ

0 −−−→ R(−1)3 i−−−→ R(−1)3 ⊕ R⊕R(−1)π−−−→ R ⊕R(−1) −−−→ 0,

where π is the projection on the last two components and i is defined as

i(

abc

) =

abc00

.

35

Using the Snake–Lemma and the surjectivity of the map KerAπ−→

Kerψξ, we obtain the graded exact sequence

0 −→ Cokerαξ −→ CokerA −→ Cokerψξ −→ 0.

By sheafification, it becomes

0 −→ B(0, 1,−1) −→ B −→ B(1, 1, µ0) −→ 0,

where B is CokerA . We tensorise it with the locally free sheaf B(0, 1,−1)and we get:

0 −→ OY −→ B ⊗ B(0, 1,−1) −→ B(1, 1,−µ0) −→ 0.

Since B((0, 1), 1, µ0) is uniquely determined by the existence of the exactsequence

0 −→ OY −→ B((0, 1), 1, µ0) −→ B(1, 1,−µ0) −→ 0,

the vector bundle B ⊗ B(0, 1,−1) is isomorphic to B((0, 1), 1, µ0). But thismeans that B is isomorphic to B((0, 1), 1, µ0).

Consider the map ϕξ : R(−2) ⊕ R(−3) −→ R(−1)2 as in the subsection1.1, such that the degree of Cokerϕξ is -1.From the exact sequence (3) we see that deg(Ω1

R(M2))=−9. Therefore theR-module corresponding to B((0, 1), 1, µ0) is (Ω1

R(M2)⊗Coker ϕξ)∨∨⊗R(2).

It is the cokernel of the graded map A : R(−2)5 −→ R(−1)3 ⊕ R ⊕ R(−1),defined by the matrix A (see above).

2) The previous statement implies that

Ω1R(M2) = B((0, 1), 1, µ0) ⊗ ˜(Cokerϕξ)∨ ⊗ OY (−2).(we consider the graded

maps given as above)

Using 1.1 and 1.2 we obtain that ˜(Cokerϕξ)∨ = ˜Cokerψξ = B(1, 1, µ0).

Therefore Ω1R(M2) = B((0, 1), 1, µ0)⊗ B(1, 1, µ0)⊗ B(−6, 1, µ−6

0 ) =B((0, 1), 1, µ0)⊗ B(−5, 1, µ−5

0 ) = B((−5,−4), 1, µ−90 ).

2.2 The rank two, stable vector bundles on ProjR and their matrix factorizations

Let Y ⊂ P2 be a rational curve. For a locally free sheaf on Y , E , we definedeg E to be deg E = χ(E) + (pa(Y )− 1)rank (E).

36

The vector bundle E is called stable if for any torsion free quotient F of E ,

deg Erank E <

deg Frank F .

A graded MCM module over the affine cone of Y is called stable if the itssheafification is a stable vector bundle on Y .

Denote S = k[y1, y2, y3] and let F be the homogeneous polynomial defin-ing the curve Y .Let M be a graded, indecomposable, locally free MCM module over the affinecone of Y and let µ be the minimal number of generators of M.Consider (A,A′) a graded matrix factorization of F corresponding to M .Set LA the vector space

LA = D ∈ Mµ×µ(S)|A′DA′ = 0 modulo(F )

and define the following equivalence on it: two matrices D and D′ from LAare equivalent (D ∼ D′) iff there exist two quadratic matrices U and V suchthat D −D′ = UA− AV .Denote SA = LA/ ∼.

Remark 2.8. If (B,B′) is another matrix factorization of M , the vectorspaces SA and SB are isomorphic.

Indeed, if U and V are the invertible matrices such that B = UAV , weconstruct the vector spaces isomorphism:

θ : SB −→ SA,

θ(D) = U−1DV −1.

Theorem 2.9. Let Y ⊂ P2 be a projective curve with arithmetic genus 1 andlet M be a graded, indecomposable, locally free MCM module over the affinecone of Y . The following statements are equivalent:

1. M is a stable module;

2. dim SA = 1 for (A,A′) a matrix factorization of M ;

3. dim SA = 1 for all (A,A′) matrix factorizations of M .

37

Proof. A.Caldararu has proved that a vector bundle E on a curve with arith-metic genus 1 is stable if and only if is simple, that means Ext1(E , E)=1(see a proof in [B]) The proof of the theorem follows now immediately from0.1 and the previous remark.

In his PhD work (see [B]), I.Burban has proved:

Proposition 2.10. A bundle B(d, m, λ) on the simple node Y is stable ifand only if the following conditions are fulfilled:1. m = 12. |di − dj| ≤ 1 for any 1 ≤ i, j ≤ r3. no possible differences d− d[t]contains a subsequence of the form 10...01,where d[t] = dt+1...drd1...dt−1dt is a shifting of d.

Therefore, the rank two stable vector bundles on Y are of the typeB((a, a+1), 1, λ) with a integer and λ ∈ k∗. Since

B((a, b), 1, λ)⊗ B(c, 1, µ) ∼= B((a + c, b+ c), 1, λµ2)

(a, b, c integers, λ, µ ∈ k∗), if we succeed to compute the module correspond-ing to one of the rank two stable bundles, we can generate all the other bytensoring it with the modules corresponding to line bundles.

From Proposition 2.7 we see that Ω1R(M2) is a stable vector bundle. More,

using Lemma 2.6 we can say that Ω1R(M2), up to shifting, is the only rank

two stable R–module that is is minimally 4–generated. (Notice that its dualis just the module itself shifted with 3) We choose this vector bundle to bethe one generating together with the line bundles all the other stable vectorbundles.

Proposition 2.11. There exists only one indecomposable, stable, rank twoMCM R-module that is even minimally generated. It is 4–generated, andby tensoring it with rank one MCM modules, it generates all other stablemodules of rank two. More, all 3-generated, rank 2, indecomposable, gradedMCM R-modules that are locally free, have stable sheafification.

Proof. It follows immediately from Lemma 2.6 and Proposition 1.7. Fordetails, see [Ba2].

Proposition 2.12. The isomorphism classes of rank 2, indecomposable, sta-ble vector bundles are given by the following MCM R-modules:

38

1. Cokerψ, for ψ =

y1 y2 y3 0y2y3 y2

1 + y1y3 0 y3

0 0 y21 + y1y3 − y2

0 0 −y2y3 y1

,

2. Coker βλ, λ ∈ V (f)reg\(0 : 1 : 0), that are 3–generated;

3. Coker Sλ, where

Sλ =

y1(1 + 2a) + y3(a + a2) −y2 + by3

0 0

y2 − by3 0

α(a:−b:1)

0 ψ(a:−b:1)

for λ = (a : b : 1) ∈ V (f)reg

or

Sλ =

α(0:0:1)

0 ψ(0:0:1)

y3 0

0 y3

0 0

for λ = (0 : 1 : 0);

4. Coker Stλ, λ ∈ V (f)reg.

Proof. Using the procedure ”tensorCM” we compute the matrix correspond-ing to (Ω1

R(M2)⊗ ϕλ) for all regular points λ on Y .

ring R=0,(y(1..3)),dp;

ideal i=y(1)^3+y(1)^2*y(3)-y(2)^2*y(3);

qring S=std(i);

ideal m=maxideal(1);

matrix D=mres(m,3)[3];

ring R1=0,(y(1..3),a,b),dp;

ideal I=y(1)^3+y(1)^2*y(3)-y(2)^2*y(3),a3+a2-b2;

39

qring S1=std(I);

matrix D=imap(S,D);

matrix M[2][2]= y(1)-a*y(3), y(2)*y(3)+b*y(3)^2,

y(2)-b*y(3), y(1)^2+(a+1)*y(1)*y(3)+(a2+a)*y(3)^2;

matrix B=tensorCM(D,M);

B=permcol(B,5,4); B=permcol(B,2,4); B=addrow(B,1,a+1,4);

B=addcol(B,1,a+1,3); B=permrow(B,1,3); B=permrow(B,2,4);

B=addrow(B,1,-y(3),5); B=permrow(B,4,5); B=multrow(B,3,-1);

B=multcol(B,1,-1); B=multcol(B,2,-1); B=multrow(B,5,-1);

B=multrow(B,4,-1); B=addcol(B,1,-b,4); B=addcol(B,1,-b,4);

print(B);

This gives the matrices Sλ. Using Lemma 2.6 we are done.

2.3 The classification of non–locally free, rank 2, MCM R–modules

Remark 2.13. It is known that on a smooth curve any vector bundle of rankr ≥ 2, say E , fits in an extension

0→ L→ E → F → 0

where F is a vector bundle of rank r − 1 and L is a line bundle.Over an isolated curve singularity, any coherent sheaf C of rank r has anextension of type

0→ C1 → C → C2 → 0

where C1 and C2 are coherent sheaves of rank 1, respectively r − 1. It canhappen that C1 and C2 are non–locally free but C is vector bundle.

The theorem 0.1 describe the extensions of graded MCM modules overa hypersurface ring. This description gives us an algorithm to compute ma-trix factorizations of the non–locally free, rank two, indecomposable, gradedMCM modules over the ring R.Since their minimal number of generators is smaller equal to 2e(R), theycan be 3, 4, 5 or 6–minimally generated. The one that are three minimallygenerated, are isomorphic, up to shiftings, to Coker βs, as it was proved inProposition 1.7.

40

2.3.1 6–generated

We will describe in details only the 6-generated ones, the other cases, 4 and5–generated are explained roughly.

For each m ∈ Z, m ≥ 1, define the matrix:

δm =

0 y1 y2 0 ym3 −ym3y1 y2 y1 0 ym3 −ym3y3 0 −y1 0 0 ym30 0 0 0 y1 y2

0 0 0 y1 y2 y1

0 0 0 y3 0 −y1

.

Theorem 2.14. There are countably many isomorphism classes of graded,indecomposable, rank two, non–locally free MCM R-modules that are mini-mally 6–generated.They are cokernel of graded maps defined by the matrices

δm, δtm|m ∈ Z, m ≥ 1.

Proof. Let M be a graded, indecomposable, rank two, non–locally free MCMR-modules with µ(M) = 6. Then, up to a shifting, M fits in a gradedextension of the type

0→ Cokerαs →M → Cokerαλ ⊗R(k)→ 0 (5)

or in one of the type

0→ Cokerαλ ⊗ R(k)→M → Cokerαs → 0 (6)

with k ∈ Z, λ = (a : b : 1) ∈ V (f).In the above graded exact sequences we consider, as in the previous sec-

tion, the graded maps αλ : R(−2)3 −→ R(−1)3, for which Cokerαλ hasdegree 0. So Cokerαλ ⊗ R(k) has degree 3k.

The modules M with an extension of type (6) are duals of some modulesfrom the first extension. Therefore, it is enough to prove the statement forthe modules with an extension of type (5).

By Theorem 0.1, the module M has a matrix factorization (S, S ′), with S

a matrix of the form S =

(αs D0 αλ

). The matrix D has homogeneous entries

and it fulfills the condition βs ·D · βλ = 0 mod (f).

41

The corresponding graded map S, is defined asS : R(−2)3 ⊕ R(k − 2)3 −→ R(−1)3 ⊕ R(k − 1)3, so, the matrix D shouldhave homogeneous entries of degree 1 − k. If k ≥ 2, the extension splits, ifk = 1 the module M decomposes. We need, therefore, to consider only thenegative shiftings of Cokerαλ.

Denote the entries of D with d1, ..., d9, so that D =(d1 d2 d3d4 d5 d6d7 d8 d9

)and denote

m = 1− k. The matrix S has the form:

S =

0 y1 y2 d1 d2 d3

y1 y2 0 d4 d5 d6

y3 0 −y1 − y3 d7 d8 d9

0 0 0 0 y1 − ay3 y2 − by3

0 0 0 y1 y2 + by3 (a2 + a)y3

0 0 0 y3 0 −y1 − (a+ 1)y3

.

We make some linear transformations, in order to simplify the matrix S.(In this way we do not change the module Coker S)By subtracting the first 3 columns from the last three, (with a correspondingmultiplication factor) we can ”kill” the variable y1 in all entries of D.We kill y2 in d1 by subtracting again the third column from the fourth one;the new appeared y1 in d7 disappear if we subtract the line 5 from the thirdone.Similarly, we can kill also y3 in d1, using the line 6 and column 2. So we cansuppose that d1 = 0.In the same way, we eliminate y2 and y3 from d4, using the column 2 andline 5, respectively line 6 and column 1. So we can suppose also that d4 = 0.Using the first column and the line 5, one can eliminate y3 in d7. Thereforewe consider d7 = a7y

m2 , with a7 a constant. Using the first column and the

line 4, one can eliminate y3 also in d8. So we can consider that d8 = a8ym2 , a8

constant.The same, using the second column and the line 4, we eliminate y2 in d5 andwe write d5 = a5y

m3 , a5 ∈ K.

We check now the condition βs ·D · βλ = 0 to get more informations onthe entries of D.

LIB"matrix.lib";

LIB"homolog.lib";

42

LIB"linalg.lib";

proc condext(matrix G)

ideal g=flatten(G);

matrix V;ideal P;

int k,j;

P=0;

for(j=1;j<=size(G);j++)

g[j]=reduce(g[j],std(y(1)^3+y(1)^2*y(3)-y(2)^2*y(3)));

V=coef(g[j],y(1));

for(k=1;k<=1/2*size(V);k++)

P=P+V[2,k];

;

P=interred(P);

return(P);


ideal i=y(1)^3+y(1)^2*y(3)-y(2)^2*y(3),a3+a2-b2;

qring S=std(i);

matrix A[3][3]=0,y(1)-a*y(3), y(2)-b*y(3),

y(1),y(2)+b*y(3), (a2+a)*y(3),

y(3), 0,-y(1)-(a+1)*y(3);

matrix B=subst(A,a,0,b,0);

matrix Aa=adjoint(A);

matrix Ba=subst(Aa,a,0,b,0);

matrix D[3][3]=d(1..9);

D[1,1]=0;D[2,1]=0;

matrix G=Ba*D*Aa;

ideal P;

P=condext(G);

P;

P[1]=y(2)*y(3)*d(8)+y(3)^2*d(6)+y(3)^2*d(7)*a^2+y(3)^2*d(7)*a-

43

y(3)^2*d(8)*b+y(3)^2*d(9)*a+y(3)^2*d(9)

P[2]=y(2)*y(3)*d(7)*b-y(2)*y(3)*d(8)*a-y(3)^2*d(2)*a^2-

y(3)^2*d(2)*a+y(3)^2*d(3)*b+y(3)^2*d(5)*b-y(3)^2*d(6)*a

P[3]=y(2)*y(3)*d(7)*a-y(3)^2*d(2)*b+y(3)^2*d(3)*a+

y(3)^2*d(5)*a+y(3)^2*d(7)*a*b-y(3)^2*d(8)*a^2+

y(3)^2*d(9)*b

P[4]=y(2)*y(3)*d(2)-y(2)*y(3)*d(9)-y(3)^2*d(2)*b+y(3)^2*d(3)*a

P[5]=y(2)^2*d(8)+y(2)*y(3)*d(2)*a+y(2)*y(3)*d(2)+

y(2)*y(3)*d(6)-y(3)^2*d(5)*a^2-y(3)^2*d(5)*a+

y(3)^2*d(6)*b

P[6]=y(2)^2*d(7)+y(2)*y(3)*d(3)+y(2)*y(3)*d(5)-y(3)^2*d(5)*b+

y(3)^2*d(6)*a

P[7]=y(2)^2*d(2)-y(2)^2*d(9)-y(2)*y(3)*d(2)*b+y(2)*y(3)*d(3)*a

Since no y3 appear in d7 and d8, from the conditions P[6] and P[5] we getthat d7 = d8 = 0. After we substitute in the ideal P, d7 and d8 with 0, wedivide with y3 or y2 where is possible.

P=subst(P,d(7),0,d(8),0);

P[1]=P[1]/y(3)^2; P[2]=P[2]/y(3)^2; P[3]=P[3]/y(3)^2;

P[4]=P[4]/y(3); P[5]=P[5]/y(3); P[6]=P[6]/y(3); P[7]=P[7]/y(2);

P=interred(P);

P;

P[1]=d(6)+d(9)*a+d(9)

P[2]=d(2)*b-d(3)*a-d(5)*a-d(9)*b

P[3]=d(2)*a^2+d(2)*a-d(3)*b-d(5)*b+d(6)*a

P[4]=y(2)*d(3)+y(2)*d(5)-y(3)*d(5)*b+y(3)*d(6)*a

P[5]=y(2)*d(2)-y(2)*d(9)-y(3)*d(2)*b+y(3)*d(3)*a

We apply some more linear transformations to the matrix S.Notice that, if we denote the third column of S with c3 and first column withc1, −c3 + (1− a)c1 has on the third position y1 − ay3. So, if we eliminate y2

from d9 subtracting the fourth line from the third one, we can make againzero on the position [3,5] subtracting −c3 + (1 − a)c1 from the column 5.We destroy the new y1 from the position of d5 with the line 4, but it stillremains there a term g · y3, with g a polynomial of degree −k. If k = 0, g is

44

a constant, so we do not need to make any other transformations. If k ≥ 1,we eliminate so possible y2 from the new d5 using the second column and theline 4. So, at the end of this transformations, if we denote d9 = a9y

m3 , a9

constant and we notice that d6 = −d9(a+ 1) (see P[1]) , S is looking like:

S =

0 y1 y2 0 d2 d3

y1 y2 0 0 a5ym3 −(a+ 1)a9y

m3

y3 0 −y1 − y3 0 0 a9ym3

0 0 0 0 y1 − ay3 y2 − by3

0 0 0 y1 y2 + by3 (a2 + a)y3

0 0 0 y3 0 −y1 − (a+ 1)y3

.

From the condition P[4], we obtain that d3 is divisible with ym3 . But d3

is a homogeneous polynomial of degree m, so d3 = a3ym3 , a3 constant. The

condition P[4] becomes y3(−a2a9−aa9−a5b)+y2(a3 +a5) = 0. So, a3 = −a5

and a(a + 1)a9 = −a5b. With this information, the ideal P becomes:

P=subst(P,d(3),-d(5));

P;

P[1]=d(6)+d(9)*a+d(9)

P[2]=d(2)*b-d(9)*b

P[3]=d(2)*a^2+d(2)*a+d(6)*a

P[4]=-y(3)*d(5)*b+y(3)*d(6)*a

P[5]=y(2)*d(2)-y(2)*d(9)-y(3)*d(2)*b-y(3)*d(5)*a

From the condition P[5], in a similar way as above, one get d2 = a2ym3 ,

with a2 constant and y3(aa5 + ba2) + y2(a9− a2) = 0. Therefore, a2 = a9 andaa5 + ba9 = 0. We write the new form of S:

S =

0 y1 y2 0 a9ym3 −a5y

m3

y1 y2 0 0 a5ym3 −(a + 1)a9y

m3

y3 0 −y1 − y3 0 0 a9ym3

0 0 0 0 y1 − ay3 y2 − by3

0 0 0 y1 y2 + by3 (a2 + a)y3

0 0 0 y3 0 −y1 − (a+ 1)y3

with aa5 + ba9 = 0.We want that the module Coker S is not locally free.

This is equivalent to the condition Fitt2(S)R〈y1,y2〉 6= R〈y1,y2〉. (see [TJP])

45

(we denote with Fitt2 the ideal generated by all 4–minors of the matrix)To check this, we substitute, in the matrix S, the variables y1 and y2 with 0and the variable y3 with 1. The fitting ideal of the new matrix is zero if andonly if Coker S is non–locally free module.

matrix A1=subst(A,y(1),0,y(2),0,y(3),1);

matrix B1=subst(B,y(1),0,y(2),0,y(3),1);

matrix D1[3][3]=0,d(9), -d(5),

0,d(5),-(a+1)*d(9),

0, 0, d(9);

matrix G[6][3]=D1,A1;

matrix S1[6][6]=concat(B1,G);

ideal F=fitting(S1,2);

F;

F[1]=d(5)*b+d(9)*a^2+d(9)*a

F[2]=d(5)*a+d(9)*b

F[3]=d(5)^2-d(9)^2*a-d(9)^2

If a 6= 0, d5 = −d9b/a. Since CokerS do not decomposes, a9 6= 0 and canbe chosen to be a. We obtain in this way the matrix

S =

0 y1 y2 0 aym3 bym3y1 y2 0 0 −bym3 −(a2 + a)ym3y3 0 −y1 − y3 0 0 aym30 0 0 0 y1 − ay3 y2 − by3

0 0 0 y1 y2 + by3 (a2 + a)y3

0 0 0 y3 0 −y1 − (a+ 1)y3

.

After some linear transformations, this matrix becomes:

S =

0 y1 y2 0 ym−13 (ay3 − y1) ym−1

3 (by3 − y2)y1 y2 0 0 ym−1

3 (−by3 − y2) −(a2 + a)ym3y3 0 −y1 − y3 0 0 ym−1

3 ((a+ 1)y3 + y1)0 0 0 0 y1 − ay3 y2 − by3

0 0 0 y1 y2 + by3 (a2 + a)y3

0 0 0 y3 0 −y1 − (a+ 1)y3

.

46

As one can easily see, this matrix decomposes, after some evident lineartransformations.

If a = b = 0, d5 = d9 or d5 = −d9 (from F[3]). As before, a9 6= 0 and canbe chosen to be 1. With this choice, the matrix S becomes δm or δtm.

Let us prove now the indecomposability.Suppose δm decomposes.Then, there exist the invertible matrices U and V , and there exist ν1, ν2 ∈V (f) such that: δm · U = V ·

(αν1 00 αν2

).

Write U =(U1 U2U3 U4

)and V =

(V1 V2V3 V4

).

Let j − 1 be the degree of the entries of the matrices U1 and V1 and j ′ − 1,the degree of the entries of U2 and V2. Then, the entries of U3 and V3 shouldhave degree j −m, the one of U4 and V4 should have degree j ′ −m.Since det U = det V = 1, if m ≥ 2, U3 = 0 (and V3 = 0) or U4 = 0 (andV4 = 0). If m = 1, since αsU3 = V3αν1 , we get U3 = V3 = 0 or ν1 = sand U3 = V3 = tId, t 6= 0. In both cases, (also for m ≥ 2), we obtain thatν1 = ν2 = s and that

0 ym3 −ym30 ym3 −ym30 0 ym3

= V2αs − αsU2,

that is impossible. (in the right hand-side, the entry [1,2] is in the ideal〈y1, y2〉, so, it can not be −ym3 ). Therefore, for any m, the matrices δm andδtm are indecomposable.With a similar proof, one can show that there not exist two invertible matricesU and V such that Uδm = δtmV ; more, because of degree reason, it is clearthat for two different m1 andm2, δm1 and δm2 do not give isomorphic modules.This complete the proof of the theorem.

2.3.2 5–generated modules

Denote v a root of -1.For all m ∈ Z, m ≥ 1, we define the matrices:

47

αmψ1 =

0 y1 y2 ym3 −vym3y1 y2 0 vym3 ym3y3 0 −y1 − y3 vym3 00 0 0 y2

1 + y1y3 − y2y3

0 0 0 −y2 y1

and

αmψ2 =

0 y1 y2 ym3 vym3y1 y2 0 −vym3 ym3y3 0 −y1 − y3 −vym3 00 0 0 y2

1 + y1y3 − y2y3

0 0 0 −y2 y1

,

αmϕ1 =

0 y1 y2 ym3 ym+13

y1 y2 0 −ym3 −ym+13

y3 0 −y1 − y3 0 ym+13

0 0 0 y1 y2y3

0 0 0 y2 y21 + y1y3

and

αmϕ2 =

0 y1 y2 −ym3 ym+13

y1 y2 0 −ym3 ym+13

y3 0 −y1 − y3 0 −ym+13

0 0 0 y1 y2y3

0 0 0 y2 y21 + y1y3

.

Define also:

αψ0 =

0 y1 y2 0 0y1 y2 0 y2 −y2

y3 0 −y1 − y3 0 y2

0 0 0 y21 −y2

2

0 0 0 −y3 y1 + y3

.

Theorem 2.15. There are countably many isomorphism classes of graded,indecomposable, rank two, non–locally free MCM R-modules that are mini-mally 5–generated. They are given by the matrices:

αmψ1, αmψ2, α

mϕ1, α

mϕ2, αψ0|m ∈ Z, m ≥ 1

and their transpositions.

48

Proof. Let M be a graded, indecomposable, rank two, non–locally free MCMR-modules with µ(M) = 5. Then, up to a shifting, M or M∨ fits in one ofthe following graded extensions:

0→ Cokerαs →M → Cokerψλ ⊗R(k)→ 0 (7)

0→ Cokerαλ →M → Cokerψs ⊗R(k)→ 0 (8)

0→ Cokerαs →M → Cokerϕλ ⊗R(k)→ 0 (9)

0→ Cokerαλ →M → Cokerϕs ⊗R(k)→ 0 (10)

with k ∈ Z, λ ∈ V (f).As before, we consider the graded maps αλ : R(−2)3 −→ R(−1)3,

ψλ : R(−2)2 −→ R ⊕R(−1), ϕλ : R(−2)⊕ R(−3) −→ R(−1)2.• Consider first the extension (7) (in the case of the extension (8) the

proof and results are identical).As before, consider (S, S ′) a matrix factorization of M , with S a matrix

of the form S =(αs D0 ψλ

). The matrix D has homogeneous entries and it

fulfills, in the ring R, the condition βs ·D · ϕλ = 0.Since the corresponding graded map S, is defined as S : R(−2)3 ⊕ R(k −2)2 −→ R(−1)3 ⊕R(k)⊕R(k − 1), the matrix D should have homogeneousentries of degree 1− k. So, if k ≥ 2, the extension splits, if k = 1 the moduleM decomposes. We need, therefore, to consider only the negative shiftingsof Cokerψλ.

Denote the entries of D with d1, ..., d6, so that D =(d1 d2d3 d4d5 d6

)and let

m = 1− k.As in the previous proof, we make some linear transformations to elim-

inate the variable y1 from all di, i = 1, ..., 6 and the variable y2 from d1, d3

and d5.The extension condition, βs ·D ·ϕλ = 0, implies that y2 do not appear in

d2 and d4 and that d6 = 0. We obtain this making the necessary calculationswith Singular:


ideal i=y(1)^3+y(1)^2*y(3)-y(2)^2*y(3),a3+a2-b2;

qring S=std(i);

matrix psi[2][2]=

y(1)^2+(a+1)*y(1)*y(3)+(a2+a)*y(3)^2,-(y(2)*y(3)+b*y(3)^2),

-(y(2)-b*y(3)), y(1)-a*y(3);

49

matrix phi[2][2]=

y(1)-a*y(3), y(2)*y(3)+b*y(3)^2,

y(2)-b*y(3), y(1)^2+(a+1)*y(1)*y(3)+(a2+a)*y(3)^2;

matrix A[3][3]=0,y(1)-a*y(3), y(2)-b*y(3),

y(1),y(2)+b*y(3), (a2+a)*y(3),

y(3), 0,-y(1)-(a+1)*y(3);

matrix B=subst(A,a,0,b,0); matrix Ba=adjoint(B);

matrix D[3][2]=d(1..6);

matrix G=Ba*D*phi;

ideal P;

P=condext(G);

P;

P[1]=y(2)*d(6)-y(3)*d(3)-y(3)*d(5)*a-y(3)*d(5)-y(3)*d(6)*b

P[2]=y(2)*d(2)+y(2)*d(5)-y(3)*d(1)*a-y(3)*d(2)*b

P[3]=y(2)*d(1)-y(2)*d(4)+y(3)*d(3)*a+y(3)*d(4)*b

P[4]=y(3)^2*d(1)*b+y(3)^2*d(2)*a^2+y(3)^2*d(2)*a-y(3)^2*

d(4)*b+y(3)^2*d(5)*a^2+y(3)^2*d(5)*a+y(3)^2*d(6)*a*b

P[5]=y(3)^2*d(1)*a+y(3)^2*d(2)*b-y(3)^2*d(4)*a+y(3)^2*d(5)*b+

y(3)^2*d(6)*a^2

From P[1], we find d6 = a6ym3 , with a6 a constant and (y2− y3b)a6 = y3(a3 +

a5 + aa5). Therefore, a6 = 0 and a3 = a5 + aa5. The same, from P[2], wehave d2 = a2y

m3 , a2 constant. More, a2 = −a5 and aa1 = a5b. From P[3], we

obtain d4 = a4ym3 and a4 = a1, a4 ∈ K. So the matrix S is looking like:

S =

0 y1 y2 a1ym3 −a5y

m3

y1 y2 0 (a5 + aa5)ym3 a1ym3

y3 0 −y1 − y3 a5ym3 0

0 0 0 y21 + (a + 1)y1y3 + (a2 + a)y2

3 − y2y3 − by23

0 0 0 −y2 + by3 y1 − ay3

with aa1 − ba5 = 0.We impose now that the module Coker S is non-locally free. For this, weshould have Fitt2(S) ·R〈y1,y2〉 6= R〈y1,y2〉.

50

matrix A1=subst(psi,y(1),0,y(2),0,y(3),1);


matrix D1[3][2]=d(1),-d(5),d(5)*(-a+1),

d(1), d(5), 0;




F;

F[1]=d(5)*b

F[2]=d(5)*a

F[3]=d(1)*b-d(5)*a^2-d(5)*a

F[4]=d(1)*a-d(5)*b

F[5]=d(1)^2-d(5)^2*a+d(5)^2

As before, the indecomposability of Coker S oblige to the conditions a = b =0 and a2

1 + a25 = 0. Therefore, S becomes αmψ1 or αmψ2. If λ = (0 : 1 : 0), with

very similar computations, we obtain only one indecomposable extension,Cokerαψ0 . The indecomposability follows as in the previous theorem.• Consider now the extension (9)(the proof and results in the last case

are identical with this one). In this case, the corresponding graded map S isdefined as S : R(−2)3 ⊕R(k − 2)⊕ R(k − 3) −→ R(−1)3 ⊕R(k − 1)2.On the first column, the matrix D should have homogeneous entries of degree1− k and on the second column of degree 2− k. So, if k ≥ 3, the extensionsplits, if k = 2 the module M decomposes. In the case k = 1, the first columnof D should be 0. Writing the condition of extension, we obtain that alsothe second column annihilates, so the module Coker S decomposes.Consider now k ≤ 0.

Denote the entries of D with d1, ..., d6, so that D =(d1 d2d3 d4d5 d6

)and let m = 1−k.

We make again some linear transformations to eliminate the variable y1 fromall d1, d2, d3, d5 and d6 and the variable y2 from d5. More, we eliminate y2

1

from d4 in case that there exists. We write d4 = y1d7 + d8. Also, in casem ≥ 1, we can eliminate y2y3 from d2 and d6. Subtracting the first columnfrom the the fourth one (multiplied with the right polynomial) we eliminatealso y3 in d5, so we can consider d5 = 0. ( the new appeared y1 in d3 is killedusing the fourth line) Using the second column one can eliminate y1y2 fromd4, so d7 = a7y

m3 , with a7 constant.

51

The extension condition βs ·D · ϕλ = 0 gives:

ring R1=0,(y(1..3),d(1..8),a,b),(c,dp(3),dp(8),dp(2));

ideal i=y(1)^3+y(1)^2*y(3)-y(2)^2*y(3),a3+a2-b2;

qring S1=std(i);

matrix Ba=imap(S,Ba);

matrix psi=imap(S,psi);

matrix D[3][2]=d(1), d(2),

d(3),y(1)*d(7)+d(8),

0, d(6);

matrix G=Ba*D*psi;P=interred(P);

ideal P;

P=condext(G);

P[1]=P[1]/y(3); P[2]=P[2]/y(3); P[3]=P[3]/y(3);

P[4]=P[4]/y(3); P[7]=P[7]/y(3)^2 ;

P=subst(P,d(8),y(3)*d(7)*a+y(3)*d(7)-d(6)*a-d(6));

P=interred(P);

P;

P[1]=y(3)*d(1)*b+y(3)*d(3)*a+y(3)*d(7)*b+d(2)*a-d(6)*b

P[2]=y(3)*d(1)*a^2+y(3)*d(1)*a+y(3)*d(3)*b+y(3)*d(7)*a^2+

y(3)*d(7)*a+d(2)*b-d(6)*a^2-d(6)*a

P[3]=y(2)*d(7)*a+y(2)*d(7)+y(3)*d(1)*a*b+y(3)*d(1)*b+

y(3)*d(3)*a^2+y(3)*d(3)*a+y(3)*d(7)*a*b+y(3)*d(7)*b+

d(2)*a^2+d(2)*a-d(6)*a*b-d(6)*b

P[4]=y(2)*y(3)*d(3)+y(3)^2*d(3)*b+y(3)^2*d(7)*a^2+

y(3)^2*d(7)*a+y(2)*d(2)-y(3)*d(6)*a^2-y(3)*d(6)*a

P[5]=y(2)*y(3)*d(1)+y(2)*y(3)*d(7)+y(3)^2*d(1)*b-y(2)*d(6)+

y(3)*d(2)*a

So, d4 = y1d7 + y3(a + 1)(d7 − d6) and y3|d2 ( see P[4] ). But we haveeliminated y2y3 from d2, so d2 = a2y

m+13 , where a2 is constant from K.

The same, P[5]implies that d6 = a6ym+13 , a6 constant. Taking in account

the form of d7, P[5] shows that also d1 = a1ym1 , with a1 constant. More,

a1 + a7 − a6 = ba1 + aa2. P[4] gives that also d3 = a3ym3 , a3 constant such

that a3 = −a2. With this information, P becomes:

52

P=subst(P,d(6),y(3)*(d(1)+d(7)),d(2),-y(3)*d(3));

P=interred(P);

P;

P[1]=y(2)*d(7)*a+y(2)*d(7)

P[2]=y(3)^2*d(1)*b-y(3)^2*d(3)*a

P[3]=y(3)^2*d(1)*a^2+y(3)^2*d(1)*a-y(3)^2*d(3)*b

Therefore, d7 = 0 and a1b− a3a = 0.The matrix S is looking like:

S =

0 y1 y2 a1ym3 a2y

m+13

y1 y2 0 −a2ym3 −a1y

m+13

y3 0 −y1 − y3 0 a1ym+13

0 0 0 y1 − ay3 y2y3 + by23

0 0 0 y2 − by3 y21 + (a + 1)y1y3 + (a2 + a)y2

3

.

We impose now that the module is non-locally free. For this, we shouldhave Fitt2(S)R〈y1,y2〉 6= R〈y1 ,y2〉.

matrix A1=subst(phi,y(1),0,y(2),0,y(3),1);


matrix D1[3][2]=d(1),d(2),-d(2),

-d(1), 0, d(1);




F;

F[1]=d(2)*b^2

F[2]=d(2)*a*b

F[3]=d(2)*a^2-d(2)*b^2

F[4]=d(1)*b+d(2)*a

F[5]=d(1)*a+d(2)*b

F[6]=d(1)^2-d(2)^2

If d2 = 0 then d1 should be also zero, so the matrix decomposes. If d2 6= 0 ,then a = b = 0 and d1 = d2 or d1 = −d2. More, we can choose d2 = 1. So,

53

the matrix S is equivalent to αmϕ1 or αmϕ1.With similar computations, one can prove that there are no indecomposableextensions of this type, if λ = (0 : 1 : 0).

The indecomposability of the matrices αmϕ1 and αmϕ1 follows as in the the-orem 2.14.

2.3.3 4–generated modules

For all λ = (a : b : 1) ∈ V (f), with a 6= 0 we define the matrix:

ϕψλ =

y1 y2y3 −by3 ay3

y2 y21 + y1y3 ay1 + (a2 + a)y3 −by3

0 0 y21 + (a+ 1)y1y3 + (a2 + a)y2

3 −y2y3 − by23

0 0 −y2 + by3 y1 − ay3

.

If λ = (0 : 1 : 0) let be:

ϕψλ =

y1 y2y3 0 y2

y2 y21 + y1y3 y1 0

0 0 y21 − y2

2

0 0 −y3 y1 + y3

.

For all m ∈ Z, m ≥ 1 we define:

ϕmψ1 =

y1 y2y3 ym3 ym3y2 y2

1 + y1y3 y1ym−13 + ym3 −ym3

0 0 y21 + y1y3 −y2y3

0 0 −y2 y1

and

ϕmψ2 =

y1 y2y3 −ym3 ym3y2 y2

1 + y1y3 y1ym−13 + ym3 ym3

0 0 y21 + y1y3 −y2y3

0 0 −y2 y1

,

ψmϕ1 =

y21 + y1y3 −y2y3 ym+1

3 y1ym+13 + ym+2

3

−y2 y1 ym3 ym+13

0 0 y1 y2y3

0 0 y2 y21 + y2y3

and

54

ψmϕ2 =

y21 + y1y3 −y2y3 ym+1

3 −y1ym+13 − ym+2

3

−y2 y1 −ym3 ym+13

0 0 y1 y2y3

0 0 y2 y21 + y2y3

,

ϕmϕ1 =

y1 y2y3 ym3 −ym+13

y2 y21 + y1y3 ym3 −y1y

m3 − ym+1

3

0 0 y1 y2y3

0 0 y2 y21 + y1y3

and

ϕmϕ2 =

y1 y2y3 −ym3 −ym+13

y2 y21 + y1y3 ym3 y1y

m3 + ym+1

3

0 0 y1 y2y3

0 0 y2 y21 + y1y3

.

Theorem 2.16. There are two families of isomorphism classes of graded, in-decomposable, rank two, non–locally free MCM R-modules that are minimally4–generated. One family is countable, given by

ϕmψ1, ϕmψ2, ψ

mϕ1, ψ

mϕ2, ϕ

mϕ1, ϕ

mϕ2|m ∈ Z, m ≥ 1

and their transpositions. The second family is parametrized by the curveV (f):

ϕψλ, ϕtψλ|λ ∈ V (f)reg.

Proof. Let M be a graded, indecomposable, rank two, non–locally free MCMR-modules with µ(M) = 4. With a similar argumentation as in previoustheorems, it is sufficient to consider that, up to a shifting, M or M∨ fits inone of the following graded extensions:

0→ Cokerϕs →M → Cokerψλ ⊗ R(k)→ 0 (11)

0→ Cokerψs →M → Cokerϕλ ⊗ R(k)→ 0 (12)

0→ Cokerϕs →M → Cokerϕλ ⊗R(k)→ 0 (13)

with k ∈ Z, λ ∈ V (f).As before, we consider the graded maps ψλ : R(−2)2 −→ R⊕ R(−1),

ϕλ : R(−2)⊕R(−3) −→ R(−1)2.

55

• Suppose M has an extension of type (11).Therefore, moduleM has a matrix factorization (S, S ′), with S =

(ϕs D0 ψλ

).

The matrix D has homogeneous entries of degree m = 1− k.So, if k ≥ 2, the extension splits, if k = 1 the module M decomposes. Weneed, therefore, to consider only the negative shiftings of Cokerψλ. Denotethe entries of D with d1, ..., d4, so that D =

(d1 d2d3 d4

).

By linear transformations, we can eliminate the variable y1 in d1, d2, d4 andy2 in d4. In case that m ≥ 2, we can eliminate also y2y3 in d2 and y2

1 in d3.We write d3 = y1d5 + d6 and d4 = a4y

m3 .

The extension condition βs ·D · ϕλ = 0 mod (f) :

setring R;

matrix p=subst(psi,a,0,b,0);

matrix D[2][2]= d(1),d(2),

y(1)*d(5)+d(6),d(4);

matrix G=p*D*phi;

ideal P;

P=condext(G);

P=subst(P,d(6),y(3)*d(5)*a+y(3)*d(5));

P=interred(P);

P;

P[1]=y(3)^2*d(5)*b-y(3)*d(1)*a-y(3)*d(2)*b+y(3)*d(4)*a

P[2]=y(3)^2*d(5)*a^2+y(3)^2*d(5)*a+y(2)*d(1)-y(2)*d(4)+

y(3)*d(4)*b

P[3]=y(2)*y(3)*d(5)-y(2)*d(2)+y(3)*d(1)*a+y(3)*d(2)*b

P[4]=y(2)*y(3)*d(1)-y(2)*y(3)*d(4)+y(3)^2*d(1)*b+

y(3)^2*d(2)*a^2+y(3)^2*d(2)*a

We find out that d3 = d5(y1 + (a + 1)y3). From P[3] we see that y3|d2, andsince y2y3 is eliminated from d2, we can write d2 = a2y

m3 . Now, using also

P[4], we notice that also d1 has the form d1 = a1ym3 . And again, from P[3],

d5 = a5ym3 . Further more, a5 = a2, a4 = a1 and aa1 + ba2 = 0.



matrix B1=subst(phi,a,0,b,0,y(1),0,y(2),0,y(3),1);

matrix D1[2][2]= d(1),d(2),

56

d(2)*(a+1),d(1);




F;

F[1]=d(1)*b+d(2)*a^2+d(2)*a

F[2]=d(1)*a+d(2)*b

F[3]=d(1)^2-d(2)^2*a-d(2)^2

If a 6= 0, then a2 should be nonzero (otherwise the matrix decomposes), andwe can choose it to be equal to a. Then a1 = b. But, in case m ≥ 2, the newobtained matrix S , after some simple linear transformations, decomposes.In case m = 1 the matrix is:

ϕψλ =

y1 y2y3 −by3 ay3

y2 y21 + y1y3 ay1 + (a2 + a)y3 −by3

0 0 y21 + (a+ 1)y1y3 + (a2 + a)y2

3 − y2y3 − by23

0 0 −y2 + by3 y1 − ay3

.

In case a = b = 0, a1 = a2 or a1 = −a2. Choosing a2 = 1, we get thematrices ϕmψ1 and ϕmψ2.

If λ = (0 : 1 : 0), with similar calculation, we get only one indecomposableextension:

S =

y1 y2y3 0 y2

y2 y21 + y1y3 y1 0

0 0 y21 − y2

2

0 0 −y3 y1 + y3

.

As one can easily see, the module CokerS is non–locally free. Indecom-posability follows as in theorem 2.14.• Consider now that M has an extension of type (12), that means,

0→ Cokerψs →M → Cokerϕλ ⊗ R(k)→ 0.

In this case, the corresponding matrix D has homogeneous entry of degreem = 1−k on the position [2,1], entries of degree m+ 1 on the main diagonaland of degree m + 2 on the position [1,2]. So, if k ≥ 3, the extension splits.If k = 2 the only nonzero entry of D is on position [1,3] and has degree 1. Ifk = 1 the entry [2,1] should be zero. Writing the condition of extension, oneget easily that D = 0, so S decomposes.Let us consider now only the negative shiftings of Cokerϕλ. Denote theentries of D with d1, ..., d4, so that D =

(d1 d2d3 d4

).

57

By linear transformations, we can eliminate the variable y1 in d1, d3, d4 andy2 in d4. In case that m ≥ 0, we can eliminate also y2

1 in d2. We writed2 = y1d5 + d6 and d4 = a4y

m3 . More, in case that d5 has y2y3 we eliminate

it using the third line.The extension condition βs ·D · ϕλ = 0 means:

matrix p=subst(phi,a,0,b,0);

matrix D[2][2]=d(1),y(1)*d(5)+d(6),

d(3), d(4);

matrix G=p*D*psi;

ideal P;

P=condext(G);

P=subst(P,d(6),y(3)*d(5)*a+y(3)*d(5));

P=interred(P);

P;

P[1]=y(3)^2*d(3)*b-y(3)*d(1)*a+y(3)*d(4)*a-y(3)*d(5)*b

P[2]=y(3)^2*d(3)*a^2+y(3)^2*d(3)*a+y(2)*d(1)-y(2)*d(4)+

y(3)*d(4)*b

P[3]=y(2)*y(3)*d(3)-y(2)*d(5)+y(3)*d(1)*a+y(3)*d(5)*b

P[4]=y(2)*y(3)*d(1)-y(2)*y(3)*d(4)+y(3)^2*d(1)*b+

y(3)^2*d(5)*a^2+y(3)^2*d(5)*a

We find out that d2 = y1d5+d5y3(a+1). From P[3] we see that y3|d5, and sincey2y3 is eliminated from d5, we can write d5 = a5y

m+13 . Since d4 = a4y

m+13 ,

from P[4], we get that also d1 has the form d1 = a1ym+13 . And, therefore,

from P[1], d3 = a3ym3 . Furthermore, a4 = a1, a5 = a3 and (ba3 + aa1 = 0




matrix D1[2][2]=d(1),d(3)*(a+1),

d(3), d(1);

matrix G[4][2]=D1,B1;

matrix S1[4][4]=concat(A1,G);


F;

58

F[1]=d(1)*b-d(3)*a^2-d(3)*a

F[2]=d(1)*a-d(3)*b

F[3]=d(1)^2-d(3)^2*a-d(3)^2

If a 6= 0, a3 should be nonzero, and we can choose it to be a. So a1 = a4 = −band a3 = a5 = a. The matrix S has the form:

S =

y21 + y1y3 −y2y3 −bym+1

3 ay1ym+13 − (a2 + a)ym+2

3

−y2 y1 aym3 −bym+13

0 0 y1 − ay3 y2y3 + by23

0 0 y2 − by3 y21 + (a+ 1)y2y3 + (a2 + a)y2

3

.

As in other cases, if m ≥ 1, this matrix decomposes, after some lineartransformations.

If a = b = 0, we need a1 = a3 or a1 = −a3. We choose a1 = 1 and inthis way we get the matrices ψmϕ1 and ψmϕ2. If λ = (0 : 1 : 0), one get noindecomposable extensions of this type.• We consider now the last case, when M has an extension of type (13).

Let (S, S ′) be a matrix factorization of the module M such that S =(ϕs D0 ϕλ

).

The matrix D has homogeneous entries of degree m = 1 − k on the firstcolumn, and of degree m + 1 on the second. So, if k ≥ 2, the extensionsplits. If k = 1 the first column of D should be zero. Writing the conditionof extension, one get easily that D = 0, so S decomposes.Let us consider now only the negative shiftings of Cokerϕλ. Denote theentries of D with d1, ..., d4, so that D =

(d1 d2d3 d4

).

By linear transformations, we can eliminate the variable y1 in d1, d2, d3 andy2 in d1. In case that m ≥ 1, we can eliminate also y2y3 in d2 and y2

1 in d4.We write d4 = y1d5 + d6 and d1 = a1y

m3 .

The extension condition βs ·D · ϕλ = 0 gives :

matrix p=subst(psi,a,0,b,0);

matrix D[2][2]=d(1),d(2),d(3),y(1)*d(5)+d(6);

matrix G=p*D*psi;

ideal P;

P=condext(G);

P=subst(P,d(6),y(3)*d(5)*a+y(3)*d(5));

P=interred(P);

P;

59

P[1]=y(2)*d(1)+y(2)*d(5)-y(3)*d(3)*a-y(3)*d(5)*b

P[2]=y(3)^2*d(1)*b+y(3)^2*d(3)*a+y(3)^2*d(5)*b+

y(3)*d(2)*a

P[3]=y(3)^2*d(1)*a^2+y(3)^2*d(1)*a+y(3)^2*d(3)*b+

y(3)^2*d(5)*a^2+y(3)^2*d(5)*a+y(3)*d(2)*b

P[4]=y(2)*y(3)*d(3)-y(3)^2*d(1)*a^2-y(3)^2*d(1)*a+

y(2)*d(2)-y(3)*d(2)*b

We find out that d3 = y1d5 + d5y3(a + 1). From P[4] we see that y3|d2, andsince y2y3 is eliminated from d2, we can write d2 = a2y

m+13 . Since d1 = a1y

m3 ,

from the same condition, we get that also d3 has the form d3 = a3ym3 . And,

therefore, from P[1], d5 = a5ym3 . Furthermore, a5 = −a1, a2 = −a3 and

aa3 − ba1 = 0We impose now that the module is non-locally free. For this, we should

have Fitt2(S)R〈y1,y2〉 6= R〈y1 ,y2〉.

matrix A1=subst(phi,y(1),0,y(2),0,y(3),1);


matrix D1[2][2]=d(1),-d(3), d(3),-d(1)*(a+1);




F;

F[1]=d(1)*b-d(3)*a

F[2]=d(1)*a^2+d(1)*a-d(3)*b

F[3]=d(1)^2*a+d(1)^2-d(3)^2

If a 6= 0, a1 should be nonzero, and we can choose it to be a. So a1 = −a5 = aand a3 = −a2 = b. The matrix S has the form:

S =

y1 y2y3 aym3 −bym+13

y2 y21 + y1y3 bym3 −ay1y

m3 − (a2 + a)ym+1

3

0 0 y1 − ay3 y2y3 + by23

0 0 y2 − by3 y21 + (a+ 1)y2y3 + (a2 + a)y2

3

.

60

One can notice immediately, that after some simple linear transforma-tions, S decomposes.If a = b = 0, we need a1 = a3 or a1 = −a3. We can choose a3 = 1. In thiscase, the matrix S becomes one of ϕmϕ1 or ϕmϕ2.If λ = (0 : 1 : 0), one get no indecomposable extensions of this type. The in-decomposability of the matrices ψmϕ1, ψmϕ2, ϕmϕ1 and ϕmϕ2 follows as in theorem2.14.

61

Part II

Fermat surface

In this part of the work we would classify the rank two, graded MCMmodules on the affine cone of the Fermat surface. In this purpose, we treatseparately the orientable and non–orientable modules. With the help of someresults from [F2], we give also the matrix factorizations of orientable, 6– and4–generated stable modules. In the first section we remind the classificationof rank one MCM modules, that can be found in [EP] and [F2].

63

3 Rank one, graded, MCM modules over the

hypersurface

R = k[x1, x2, x3, x4]/(x31 + x3

2 + x33 + x3

4)

The aCM line bundles on a smooth cubic surface in P3 are classified in [F2]by the following theorem:

Theorem 3.1. Let X = V (F ) be a smooth hypersurface in P = P3 definedby the cubic form F and let L be a normalized aCM line bundle on Y. Thenthe minimal graded free resolution on L takes one of the following forms:

0→ OP(−3)F=f(L)−−−−→ OP → L→ 0 (14)

0→ O2P(−2)

f(L)−−→ OP(−1)⊕OP → L → 0 (15)

0→ OP(−1)⊕OP(−2)f(L)−−→ O2

P → L → 0 (16)

0→ O3P(−1)

f(L)−−→ O3P → L→ 0, (17)

where det(f(L)) = F .The line bundles with resolutions of type (15), (16) respectively (17) are iso-morphic to O(L), O(C), respectively O(T ), where L is a line on X, C aconic on X, and T a twisted cubic on X.

From this theorem one can conclude that there are 27+27+72 isomor-phism classes of graded MCM modules of rank one, over a hypersurface ringwith smooth projective cone, K[x1, x2, x3, x4]/f , with f an irreducible gradedpolynomial of degree 3. More, 54 of them are minimally 2–generated, andthe other 72 are minimally 3–generated.

In the case of the Fermat surface we have the exact description of thismodules (their matrix factorizations), that can be found in [EP].Let R = K[x1, x2, x3, x4]/f , f = x3

1 + x32 + x3

3 + x34.

For every a, b ∈ K such that a3 = b3 = −1 and for any σ = (i j s) permutationof the set 2, 3, 4 with i < j, we denote

ϕσ(a, b) =

(x1 − axs −(x2

i + bxixj + b2x2j)

xi − bxj x21 + ax1xs + a2x2

s

)

65

and

ψσ(a, b) =

(x2

1 + ax1xs + a2x2s x2

i + bxixj + b2x2j

−(xi − bxj) x1 − axs

).

Theorem 3.2. (ϕσ(a, b), ψσ(a, b)) is a matrix factorization off = x3

1 + x32 + x3

3 + x34, for all σ, a, b as above. The sets:

M = Cokerϕσ(a, b) | a, b, σ

andN = Cokerψσ(a, b) | a, b, σ

have the following properties:

1. Every two generated, non–free, graded MCM module is isomorphic withone of the modules of M∪N .

2. Every two different graded MCM modules fromM∪N are not isomor-phic.

3. All modules from M∪N have rank one.

From Theorem 3.1 one can see that the matrices ϕσ(a, b) correspond tothe 27 quadrics on the surface and the matrices ψσ(a, b) correspond to the27 lines on the surface.

To describe the 3–generated rank one MCM modules, we make the fol-lowing notations.For a, b, c, d, ε ∈ K such that a3 = b3 = c3 = d3 = −1, ε3 = 1, ε 6= 1, andbcd = εa, we set

α(b, c, d, ε) =

0 x1 − ax4 x2 − bx3

x1 − cx2 −b2x3 − abc2ε2x4 b2c2x3 − abcε2x4

x3 − dx4 c2x2 + bc2x3 + acx4 −x1 − cx2 − ax4

and

β(b, c, d, ε) = α(b, c, d, ε)t,

that is, the transpose of α(b, c, d, ε). Then each of the matrices α(b, c, d, ε)and β(b, c, d, ε) forms with its adjoint, α(b, c, d, ε)∗, respectively β(b, c, d, ε)∗,a matrix factorization of f .

66

For a, b, c ∈ K, distinct roots of −1, and ε as above, we set

η(a, b, c, ε) =

0 x1 + x2 x3 − ax4

x1 + εx2 −x3 + cx4 0x3 − bx4 0 −x1 − ε2x2

and

ϑ(a, b, c) =

0 x1 + x3 x2 − ax4

x1 − a2bx3 −x2 + cx4 0x2 − bx4 0 −x1 + ab2x3

.

The matrices η(a, b, c, ε) and ϑ(a, b, c) form with their adjoint, η(a, b, c, ε)∗,respectively ϑ(a, b, c)∗, matrix factorizations of f .

Theorem 3.3 ((3.4) in [EP]). Let

P = Cokerα(b, c, d, ε), Coker β(b, c, d, ε) | b, c, d, ε ∈ K,b3 = c3 = d3 = −1, bcd = εa, ε3 = 1, ε 6= 1

and

Q = Coker η(a, b, c, ε), Coker ϑ(a, b, c) | ε3 = 1, ε 6= 1

and (a, b, c) is a permutation of the roots of − 1.Then the sets P,Q of rank 1, 3–generated, MCM graded R–modules have thefollowing properties:

(i) Every 3–generated, rank 1, indecomposable, graded MCM R–module isisomorphic with one module from P ∪ Q.

(ii) If M = Cokerα(b, c, d, ε) (or M = Coker β(b, c, d, ε)) belongs to P andN ∈ P, then N ' M if and only if N = Cokerα(bε, cε, dε, ε2) (orN = Coker β(bε, cε, dε, ε2)).

(iii) Any two different modules from Q are not isomorphic.

(iv) Any module of Q is not isomorphic with some module of P.

Notice that the 3–generated modules from P ∪ Q correspond to the 72twisted cubic curves on the Fermat surface.The map M 7→ Ω1

R(M) is a bijection between the 3–generated, indecompos-able, graded, MCM R–modules of rank two and the 3–generated, indecom-posable, graded, MCM R–modules of rank 1. Thus, from the above theoremwe obtain the description of the rank 2, 3–generated, indecomposable, gradedMCM R–modules.

67

Theorem 3.4. Let

P∗ = Cokerα(b, c, d, ε)∗, Coker β(b, c, d, ε)∗ | b, c, d, ε ∈ K,b3 = c3 = d3 = −1, bcd = εa, ε3 = 1, ε 6= 1

and

Q∗ = Coker η(a, b, c, ε)∗, Coker ϑ(a, b, c)∗ | ε3 = 1, ε 6= 1

and (a, b, c) is a permutation of the roots of − 1 .

Then the sets P∗,Q∗ of rank 2, 3–generated, MCM graded R–modules havethe following properties:

(i) Every 3–generated, rank 2, indecomposable, graded MCM R–module isisomorphic with one module from P∗ ∪Q∗.

(ii) If M = Cokerα(b, c, d, ε)∗ (or M = Coker β(b, c, d, ε)∗) belongs to P∗and N ∈ P∗, then N ' M if and only if N = Cokerα(bε, cε, dε, ε2)∗

(or N = Coker β(bε, cε, dε, ε2)∗).

(iii) Any two different modules from Q∗ are not isomorphic.

(iv) Any module of Q∗ is not isomorphic with some module of P∗.

Corollary 3.5. There are 72 isomorphism classes of rank 2, indecomposable,graded MCM modules over R with three generators.

68

4 Rank two, orientable, MCM modules over

the hypersurface

R = k[x1, x2, x3, x4]/(x31 + x3

2 + x33 + x3

4)

Let R = k[x1, .., xn]/f be a normal hypersurface ring. A torsion free R-module M of rank m is orientable if and only if detM = 0, where detM(called determinant of M) is the corresponding class of the bidual (∧mM)∗∗

in Cl(R).J. Herzog and M. Kuhl have proved ([HK]) that any graded, indecomposable,orientable MCM R–module of rank 2 and can be written as M ∼= Ω2

R(I),where I is a graded Gorenstein ideal of codimension 2. Two such modulesare isomorphic if and only if their corresponding ideals are evenly linked. (Weremind that two ideals I, J of R are said to be (directly) linked with respectto a regular sequence y = y1, ...yg in I ∩ J if (y) : J = I and (y) : I = J .The ideals I and J are evenly linked if there exists a sequence of idealsI = I0, I1, ..., Il = J such that l is even and It is directly linked to It+1 for alli = 0, l − 1. )In the same paper it is proved that any graded, indecomposable, orientableMCM R–module of rank 2 has an even number of generators and, more, if2s is the minimal number of generators of M , than the minimal number ofgenerators of the corresponding Gorenstein ideal I is 2s− 1. (!s ≤ deg(f)).Using this characterization we will describe in the following subsection, therank two, graded, indecomposable, 4-generated orientable MCM over thecubic hypersurface ring k[x1, x2, x3, x4]/(f), f = x3

1 + x32 + x3

3 + x34. Notice

that over a cubic hypersurface an orientable MCM module is 4–generated or6–generated. To describe the 6–generated one, we will use another method,mainly, we prove that they have a skew symmetric matrix factorization.To find all such matrix factorizations, we restrict first the module to thehyperplane section (x4 = 0).

We use an interesting property of the Fermat surface that the rank two6–generated orientable bundles, when restricting to the hyperplane sectioncurve splits into two line bundles on the curve (a smooth elliptic curve). Weprove in the following that an orientable MCM R–module of rank two has askew symmetric matrix factorization.

69

Let ϕ = (aij)1≤i,j≤2s be a generic skew symmetric matrix, that is

aii = 0, aij = −aji, for all i, j = 1, . . . , 2s.

Thendet(ϕ) = pf(ϕ)2,

where pf(ϕ) denotes the Pfaffian of ϕ (see [Bo1, §5, no. 2] or [BH, (3.4)]).Like determinants, Pfaffians can be developed along a row. Set ϕij the matrix

obtained from ϕ by deleting the ith and jth rows and columns. Then, forall i = 1, . . . , 2s,

pf(ϕ) =2s∑

j=1j 6=i

(−1)i+jσ(i, j)aij pf(ϕij), (18)

where σ(i, j) denotes sign(j − i). Multiplying (18) by pf(ϕ), we obtain

det(ϕ) =

2s∑

j=1

aijbij, (19)

for bij = (−1)i+jσ(i, j) pf(ϕij) pf(ϕ) when i 6= j and bii = 0. Since ϕ is ageneric matrix we see from (19) that bij is exactly the algebraic complementof aij and so the transpose matrix B of (bij) is the adjoint matrix of ϕ. Set

ψ =1

pf(ϕ)B.

Thenϕψ = ψϕ = pf(ϕ) Id2s,

as it is stated also in [[JP], §3].

Theorem 4.1. Let S = K[x1, ..., xn] and f ∈ S a polynomial of degree ddefining a smooth irreducible hypersurface in Pn−1. The cokernel of a homo-geneous skew symmetric matrix over S of order 2s ≤ 2d, with determinantf 2 defines a graded MCM R–module M of rank 2, where R = S/f .Conversely, each graded indecomposable orientable MCM R–module M ofrank 2 is the cokernel of a map given by a skew symmetric homogeneousmatrix ϕ over S, whose determinant is f 2.

70

Proof. Let ϕ a homogeneous skew symmetric matrix over S of order 2s,s ≤ d, with detϕ = f 2. Let ψ be given for ϕ as above, that is the (i, j) entryof ψ is σ(i, j) pf(ϕij). Since pf(ϕ) = f , we get

ϕψ = ψϕ = f · Id2s .

Therefore (ϕ, ψ) is a matrix factorization of f which defines a graded MCMR–module of rank 2.To prove the converse, consider M a graded indecomposable orientable MCMR–module M of rank 2. According to Herzog and Kuhl [HK], M must havean even minimal number of generators and it is isomorphic to the secondsyzygy over R of a Gorenstein ideal I ⊂ R of codimension 2. More, if 2sis the minimal number of generators of M , than the minimal number ofgenerators of I is 2s− 1. Using the Buchsbaum–Eisenbud Theorem (see e.g.[BH], (3,4)) there exists an exact sequence

0 −→ Sd3−→ S2s−1 d2−→ S2s−1 d1−→ S (20)

such that d2 is a skew symmetric homogeneous matrix, d3 is the dual of d1,d3 = dt1, and

d1 =(

pf((d2)1

),− pf

((d2)2

), . . . , pf

((d2

)2s−1

)),

where (d2)i denotes the (2s− 2)× (2s− 2) skew symmetric matrix obtained

by deleting the ith row and column of d2. Also I = J/(f), with J = Im d1.Since f ∈ J there exists v : S −→ S2s−1 such that d1v = f . It is easy to seefrom (20) that I has the following minimal resolution over S :

0 −→ S

(d30

)−−−→ S2s (d2 ,v)−−−→ S2s−1 d1−→ I −→ 0.

As in [Ei1], since fI = 0, there exists a map h : S2s−1 → S2s such that(d2, v)h = −f · Id2s−1 and we obtain the following exact sequence

R2s

(h|d3

0

)−−−−→ R2s (d2,v)−−−→ R2s−1 d1−→ I −→ 0. (21)

On the other hand, ϕ =(

d2 v−vt 0

)is a skew symmetric homogeneous matrix of

order 2s. Let ψ defined as above. By construction, ψ has the form(

c dt1−d1 0

)

71

and so (d2, v)(

c−d1

)= −f · Id2s−1 = pf(ϕ) · Id2s−1. Taking h =

(c−d1

)above,

we obtain from (21) the following exact sequence:

R2s ϕ−→ R2s ψ−→ R2s (d2 ,v)−−−→ R2s−1 d1−→ I −→ 0,

which givesCokerϕ ∼= Imψ = Ω2

R(I).

We have pf(ϕ) = −f and so det(ϕ) = f 2. Notice that all maps appeared inthe proof are graded, but we omitted the necessary shiftings. In [BEPP] onecan find a complete proof for the particular case d = 3, n = 4, s = 3.

The geometric version of this theorem (regarding vector bundles on smoothhypersurfaces) is the following:

If ε ∈ −1, 1, we say that a matrix A is ε–symmetric if At = εA.Correspondingly we have a notion of ε-symmetric duality on a vector bundle.

Theorem 4.2. Let Y = V(FY ) be a smooth hypersurface of degree d in Pn

and let F be a aCM rank r vector bundle on Y . Then the minimal gradedfree resolution of the sheaf F , extended to zero on Pn, takes the form:

0 −→ Syz(F)f(F)−−→ Gen(F)

p(F)−−→ F −→ 0

with Gen(F) = ⊕si=1OPn(ai), Syz(F) = ⊕si=1OPn(bi) and det (f(F)) = F r.Moreover, suppose that there exists an ε–symmetric duality κ : F ⊗ F →OY (d+ t). Then we have a natural isomorphism Syz(F) ∼= Gen(F)∗(t),and f(F) = εf(F)t. In particular, if F is a rank two vector bundle, thematrix f(F) is skew symmetric if and only if ∧2(F) ∼= OY (t), for somet ∈ Z.

Since we shall use it in this section, we recall from [LPP] the descriptionof the rank 1, three-generated, non–free, graded MCM R3–modules, for R3 =k[x1, x2, x3]/[(x3

1 + x32 + x3

3].First we recall the notations. Let P0 = [−1 : 0 : 1] ∈ V (f3). For eachλ = [λ1 : λ2 : 1] ∈ V (f3), λ 6= P0, we set

αλ =

0 x1 − λ1x3 x2 − λ2x3

x1 + x3 −x2 − λ2x3 −wx3

x2 wx3 (1− λ1)x3 − x1

,

72

where w =λ2

2

λ1+1and, if λ = [λ1 : 1 : 0] ∈ V (f3), we set

αλ =

0 x1 − λ1x2 x3

x1 + x3 −λ1x1 λ1x1 + λ21x2

x2 x3 − x1 −x1

.

Let βλ the adjoint matrix of αλ.

Theorem 4.3 ((3.7) in [LPP]). (αλ, βλ) is a matrix factorization for allλ ∈ V (f3), λ 6= P0, and the set of 3–generated MCM graded R3–modules,

M0 = Cokerαλ | λ ∈ V (f3), λ 6= P0

has the following properties:

(i) All the modules from M0 have rank 1.

(ii) Each two different modules from M0 are not isomorphic.

(iii) Every 3–generated, rank 1, non–free, graded MCM R3–module is iso-morphic with one module from M0.

4.1 Orientable 6–generated MCM modules

Let S = K[x1, x2, x3, x4], and R = S/(f), f = x31 + x3

2 + x33 + x3

4.We have proved that a non–free graded orientable 6–generated MCM

R–module corresponds to a skew symmetric homogeneous matrix over S oforder 6, whose determinant is f 2.

Let Λ be such a matrix. Notice that Λ has linear entries and the matrixΛ := Λ|x4=0, obtained from Λ by restricting the entries to x4 = 0, is ahomogeneous matrix over S3 = K[x1, x2, x3], whose determinant is f 2

3 , wheref3 = x3

1 + x32 + x3

3. Therefore, CokerΛ defines a graded rank 2, 6–generatedMCM over R3 = S3/(f3). These modules were explicitly described in [LPP].

Lemma 4.4. Let M be a non–free graded orientable 6–generated MCM mod-ule over R. Then the restriction of M to the curve defined by f = x4 = 0splits into a direct sum of a 3–generated MCM of rank 1 and its dual.

73

Especially, there exists λ ∈ V (f3) r P0 and a skew symmetric matrixΓ ∈ M6×6(K), such that M is the cokernel of a map given by the matrix

Λ = x4 · Γ +(

0 −αtλαλ 0

).

(The same notations as in [LPP] and in Introduction.)

Proof. Let Λ1 be a skew symmetric homogeneous matrix over S, correspond-ing to M , and denote Λ1 = Λ1|x4=0. Suppose that the MCM S3–module cor-responding to Λ1 is indecomposable. Then we can generate it as described inTheorem 4.2 and Lemma 5.4 from [LPP]. Denote with D the matrix whichwe obtain by this means.

Since D ∼ Λ1, and Λ1 is skew symmetric, there exist two invertible ma-trices U, V ∈ M6×6(K) such that U · D · V + (U · D · V )t = 0. Therefore,there exists T ∈ M6×6(K) an invertible matrix such that T ·D+ (TD)t = 0.(Take T = (V t)−1 · U .)

With the help of Singular, we find that, in fact, there is no invertiblematrix T such that T ·D is skew symmetric. Therefore, the module corre-sponding to Λ1 should decompose.

First, we generate the matrix D.

LIB"matrix.lib";

option(redSB);

proc reflexivHull(matrix M)

module N=mres(transpose(M),3)[3];

N=prune(transpose(N));

return(matrix(N));

proc tensorCM(matrix Phi, matrix Psi)

int s=nrows(Phi);

int q=nrows(Psi);

matrix A=tensor(unitmat(s),Psi);

matrix B=tensor(Phi,unitmat(q));

matrix R=concat(A,B,U);

return(reflexivHull(R));

74

proc M2(ideal I)

matrix A=syz(transpose(mres(I,3)[3]));

return(transpose(A));

ring R=0,(x(1..3)),(c,dp);

qring S=std(x(1)^3+x(2)^3+x(3)^3);

ideal I=maxideal(1);

matrix C=M2(I);

ring R1=(0,a),(x(1..3),e,b),lp;

ideal I=x(1)^3+x(2)^3+x(3)^3,(a-1)^3+b3+1,e*b+a2-3*a+3,e*a-b2;

qring S1=std(I);

matrix B[3][3]= 0, x(1)-(a-1)*x(3), x(2)-b*x(3),

x(1)+x(3), -x(2)-x(3)*b, -x(3)*e,

x(2), x(3)*e, -x (1)+(-a+2)*x(3);

matrix C=imap(S,C); matrix D=tensorCM(C,B);

We check the existence of the invertible matrix T, such that T ·D is skewsymmetric.

ring R2=0,(x(1..3),a,e,b,t(1..36)),dp;

ideal I=x(1)^3+x(2)^3+x(3)^3,(a-1)^3+b3+1,e*b+a2-3*a+3,e*a-b2;

qring S2=std(I);

matrix D=imap(S1,D);

matrix T[6][6]=t(1..36);

matrix A=T*D+transpose(T*D); ideal I=flatten(A);

ideal I1=transpose(coeffs(I,x(1)))[2];



ideal J=I1+I2+I3+ideal(det(T)-1);

ideal L=std(J);

L;

L[1]=1

Therefore, there does not exist such invertible matrix T .

75

This means, that after some linear transformations, Λ1 decomposes intotwo matrices of order three and rank 1 with determinant f3 = x3

1 + x32 + x3

3,which correspond to two points λ1, λ2 in V (f3)r P0, P0 = [−1 : 0 : 1]. Letus denote them by A and B. We can consider A = αλ1 , B = αλ2 .

Since Λ1 is skew symmetric, there exists an invertible matrix U ∈ M6×6(K)

such that U · ( A 00 B ) is skew symmetric. Consider U =

(U1 U2U3 U4

). The skew

symmetry condition gives the following equalities:

U1 · A+ (U1 · A)t = 0U4 ·B + (U4 ·B)t = 0U2 ·B + At · U t

3 = 0U3 ·A + Bt · U t

2 = 0

So U1 · αλ1 and U4 · αλ2 are skew symmetric, so they have only zeros on themain diagonal. Since the entries of the second and third line and column ofαλ1 and αλ2 are linearly independent, we easily obtain that U1 = U4 = 0.Therefore, U2 and U3 are invertible matrices and B = −U−1

2 · At · U t3.

We have obtained Λ1 ∼(αλ1

0

0 αtλ1

)∼(

0 −αtλ1αλ1

0

).

Therefore, there exists Γ ∈ M6×6(K) skew symmetric, and λ ∈ V (f3) rP0 such that Λ1 ∼ Λ = x4 · Γ +

(0 −αtλαλ 0

). We can write Γ =

(Γ1 −Γt2Γ2 Γ3

),

Γi ∈ M3×3(K), i = 1, 2, 3, Γ1 and Γ3 skew symmetric.

Remark 4.5 (Notation). For any λ = [a : b : c] ∈ V (f3)rP0 there existsa unique point in V (f3) r P0 which we denote as λt, such that αtλ ∼ αλt .We find λt = [c : b : a].1

For λ = [a : b : 1] we denote with Uλ and Vλ two invertible matrices suchthat Uλ · αtλ = αλt · Vλ.

If a 6= 0, then we can take Uλ =

(b2 b(a+1) −(a+1)2

−(a+1)2 b2 −b(a+1)

b(a+1) (a+1)2 b2

)and Vλ = U t

λ.

If a = 0, then we can take Uλ =

(−b2 −b 1−2b 1 b2

2b2 2b 1

)and Vλ =

(1 −2b 2b2

−b −b2 −1−b −b2 2

).

Notice that λt = λ for λ = [1 : b : 1] ∈ V (f3), and λ 6= λt for all otherλ ∈ V (f3)r P0.

Remark 4.6. For any λ = [1 : b : 0] ∈ V (f3) r P0 and any Λ = x4 ·Γ +

(0 −αtλαλ 0

)skew symmetric with det Λ = f 2, we have λt = [0 : b : 1] in

1If λ corresponds to the 3–generated rank 1 MCM N , then λt corresponds to its dualN∨.

76

V (f3) r P0 and Λ′ = x4Γ′ +(

0 −αtλt

αλt 0

)skew symmetric with det Λ′ = f 2

such that Λ ∼ Λ′.

Indeed, take Λ′ = U ·Λ · U t where U =(

0 T1T2 0

), T1 = 1√

3

(b b2 00 1 −b22 0 1

)and

T2 = 1√3

(−1 b b

2 b2 b2

−2b2 1 −2

).

Therefore, Coker Λ and Coker Λ′ define two isomorphic MCM modules.This is the reason why we may only consider the case λ = [a : b : 1] ∈V (f3)r P0, from now on.

Remark 4.7. Consider λ = [a : b : 1] ∈ V (f3) r P0 and Λ = x4 · Γ +(0 −αtλαλ 0

)as in Lemma 4.4. Then there exists Λ = x4 · Γ +

(αλ 00 αλt

)with

det Λ = f 2 such that Λ ∼ Λ.

Indeed, consider Λ =(

0 Id−Uλ 0

)· Λ ·

(Id 00 V −1

λ

).

We obtain Γ =(

Γ2 Γ3·V −1λ

−Uλ·Γ1 Uλ·Γt2·V −1λ

).

Lemma 4.8. Consider Λ = x4 · Γ +(

0 −αtλαλ 0

)as above. Then the MCM

module M corresponding to Λ is indecomposable if and only if Γ1 6= 0 orΓ3 6= 0.

Proof. Suppose M is indecomposable. If Γ1 = Γ3 = 0, then ( 0 IdId 0 ) · Λ =(

x4·Γ2+αλ 00 −x4·Γt2−αtλ

), so Λ decomposes after some linear transformation.

This contradicts the indecomposability of M = Coker Λ, so we musthave Γ1 6= 0 or Γ3 6= 0.

Now, let us suppose Γ1 6= 0 or Γ3 6= 0 and prove that M is indecompos-able.

Suppose M decomposes. Then there exists a matrix(T1 00 T2

)equivalent to

Λ with T1, T2 two matrices of order three and rank 1, with detT1 = detT2 = fand T1|x4=0 = αλ1 , T2|x4=0 = αλ2 , where λ1, λ2 ∈ V (f3)r P0.

Since Λ is skew symmetric, after some linear transformations,(αλ1

00 αλ2

)

should also become skew symmetric. As we saw in the proof of Lemma 4.4,this gives αλ2 ∼ αtλ1

, so λ2 = λt1.Using Remark 4.6, there exist U, V ∈ M6×6(K) invertible matrices such

that U · Λ · V =(T1 00 T2

)= x4 ·

(N1 00 N2

)+(αλ1

00 α

λt1

).

77

Therefore,

U ·(αλ 00 αλt

)=

(αλ1 00 αλt1

)·V −1 (1)

U ·(

Γ2 Γ3 · V −1λ

−Uλ · Γ1 Uλ · Γt2 · V −1λ

)=

(N1 00 N2

)·V −1 (2) .

Let us consider U =(U1 U2U3 U4

)and V −1 =

(V1 V2V3 V4

)with Ui, Vi ∈ M3×3(K),

i = 1, . . . , 4.The first system of equations gives:

U1 · αλ = αλ1 · V1

U2 · αλt = αλ1 · V2

U3 · αλ = αλt1 · V3

U4 · αλt = αλt1 · V4

By comparing the coefficients of x1, x2, x3 on the left–hand side and right–hand side of the above equalities, we obtain easily:

Ui = Vi = Ki · Id3 with Ki ∈ K, i = 1, . . . , 4 .

Moreover, if λ 6= λ1, then K1 = K4 = 0 and if λ 6= λt1, then K2 = K3 = 0.Since U is invertible, we have λ = λ1 or λ = λt1.

We know that αλ1 = T1|x4=0 where T1 is a matrix of order three overS = K[x1, x2, x3, x4] of rank 1 and with determinant f . So Coker T1 is agraded 3–generated rank 1 MCM R–module. In [EP], all the isomorphismclasses of such modules are given explicitly. We obtain αλ1 ∼ α|x4=0 orαλ1 ∼ αt|x4=0 or αλ1 ∼ η|x4=0 or αλ1 ∼ ν|x4=0.

Using Singular we check that none of the above matrices is equivalentto α[1:`:1], therefore, λ1 6= λt1.

LIB"matrix.lib";

option(redSB);

ring r=0,(x(1..3),l,a,b,c,d,e,v(1..9),u(1..9)),dp;

ideal I=x(1)^3+x(2)^3+x(3)^3,

l^3+2,

a3+1,b3+1,c3+1,d3+1,e2+e+1,bcd-e*a;

qring s=std(I);

proc isomorf(matrix X,matrix Y)

78

matrix U[3][3]=u(1..9);

matrix V[3][3]=v(1..9);

matrix C=U*X-Y*V;

ideal I=flatten(C);




ideal J=I1+I2+I3+ideal(det(U)-1,det(V)-1);

ideal L=std(J);

return(L);

We write the matrix A corresponding to the point (1:l:1).

matrix A[3][3]=0, x(1)-x(3), x(2)-l*x(3),

x(1)+x(3), -x(2)-l*x(3), -1/2*l^2*x(3),

x(2), 1/2*l^2*x(3), -x(1);

We write the matrices corresponding to the rank 1 3–generated MCMmodules, restricted to the hyperplane (x(4) = 0).

matrix alpha[3][3]=0, x(1), -x(3)*b+x(2),

-x(2)*c+x(1), -x(3)*b^2, x(3)*b^2*c^2,

x(3), x(3)*b*c^2+x(2)*c^2, -x(2)*c-x(1);

matrix alphat=transpose(alpha);

matrix eta[3][3]=0,x(1)+x(2), x(3),

x(1)+e*x(2), -x(3), 0,

x(3), 0, -x(1)-e^2*x(2);

matrix nu[3][3]=0, x(1)+x(3), x(2),

x(1)-a^2*b*x(3), -x(2), 0,

x(2), 0, -x(1)+a*b^2*x(3);

isomorf(alpha,A); L[1]=1

isomorf(alphat,A); L[1]=1

79

isomorf(eta,A); L[1]=1

isomorf(teta,A); L[1]=1

If λ = λ1 6= λt1 as a solution of the system (1), we obtain: U = V =(K1·Id 0

0 K4·Id), K1 ·K4 6= 0.

Replacing U and V in (2), we get:

K1 · Γ3 · Vλ = 0

K4 · Uλ · Γ1 = 0.

Since K1 6= 0, K4 6= 0 and Uλ, Vλ are invertible matrices, Γ1 and Γ3 shouldbe 0–matrices, which is a contradiction to our hypothesis.

If λ = λt1 6= λ1, we obtain, as a solution of (1): U = V =(

0 K2·IdK3 Id 0

),

K2 ·K3 6= 0.

Replacing U and V in (2), we obtain:

K2 · Uλ · Γ1 = 0

K3 · Γ3 · Vλ = 0.

Therefore, we must have again Γ1 = Γ3 = 0.

For each λ = [a : b : 1] ∈ V (f3) r P0, we define a family of skewsymmetric homogeneous indecomposable matrices of order six over S =K[x1, x2, x3, x4] with determinant f 2:

Mλ :=

Λ(λ,Γ) = x4 · Γ+

(0 −αtλαλ 0

), det Λ(λ,Γ) = f 2,

Γ =

(Γ1 −Γt2Γ2 Γ3

),

Γ1,Γ3 skew symmetric, Γ1 6= 0or Γ3 6= 0

.

Notice that, as in the proof of Lemma 4.8, if Λ(λ,Γ) ∼ Λ(λ′,Γ′), then λ′ = λ orλ′ = λt.

Lemma 4.9. Let λ = [a : b : 1] ∈ V (f3)r P0 with a 6= 1.

1. Inside the family Mλ, two matrices, Λ and Λ′, are equivalent if and

only if there exists k ∈ K∗ such that Λ′ = Uk · Λ · U tk, Uk =

(k Id 0

0 1k

Id

).

This condition means:

Γ′2 = Γ2

Γ′1 = k2 · Γ1

Γ′3 = 1k2 · Γ3.

2. A matrix Λ from Mλ is equivalent to a matrix Λ′ from Mλ′ , λ′ 6= λ

if and only if λ′ = [1 : b : a] and Λ′ = Uk · Λ · U tk, where k ∈ K∗ and

Uk =(

0 k·U−1λ

− 1kUλ 0

).

80

Proof. We assume a 6= 0. The case a = 0 is treated similarly. Two matrices,Λ = Λ(λ,Γ) and Λ′ = Λ(λ′,Γ′), are equivalent if and only if Λ and Λ

′are

equivalent (see Remark 4.7).

If U and V are two invertible matrices such that U ·Λ = Λ′ · V , as in the

proof of Lemma 4.8, we obtain

U = V =

(K1 Id K2 IdK3 Id K4 Id

)with K1 = K4 = 0 if λ 6= λ′ and

K2 = K3 = 0 if λ′ 6= λt .

Since U · Λ = Λ′ · V , we have:

(0 Id−Uλ′ 0

)−1

·U ·(

0 Id−Uλ 0

)·Λ·(

Id 00 V −1

λ

)·U−1·

(Id 00 V −1

λ′

)−1

= Λ′.

(∗)

1. If λ = λ′ then λ′ 6= λt, so U =(K1 Id 0

0 K4 Id

)with K1 6= 0, K4 6= 0. So

(∗) implies:(K4·Id 0

0 K1 Id

)· Λ ·

(1K1

Id 0

0 1K4

Id

)= Λ′. For k =

√K4

K1and

Uk =(k Id 0

0 1k·Id

)we have Λ′ = Uk · Λ · U t

k.

2. If λ′ = λt then λ′ 6= λ, so U =(

0 K2 IdK3 Id 0

), K2 6= 0, K3 6= 0. Replacing

U in (∗) we obtain:

Λ′ =

(0 −K3U

−1λt

−K2Uλ 0

)· Λ ·

(0 1

K3Vλt

1K2V −1λ 0

).

Since a 6= 0 and a 6= 1, Vλ = U tλ, λt =

[1a

: 1b

: 1], Uλt = 1

a2 · Uλ,Vλt = 1

a2Utλ (see Remark 4.5).

So Λ′ =(

0 −K3a2U−1λ

−K2Uλ 0

)· Λ ·

(0 1

K3· 1a2 ·Utλ

1K2

(U−1λ )t 0

)=(

0 kU−1λ

− 1kUλ 0

)·

Λ ·(

0 kU−1λ

− 1kUλ 0

)t, where k2 = −a2 · K3

K2.

In a similar way, we can prove the following lemma:

Lemma 4.10. Let λ = [1 : b : 1] ∈ V (f3)r P0.

81

1. Inside the familyMλ, two matrices Λ and Λ′ are equivalent if and onlyif Λ′ = T · Λ · T t, where

T =(K4·Id K3·U−1

λK2·Uλ K1·Id

), K1, K2, K3, K4 ∈ K such that K1K4−K2K3 = 1 .

2. No λ ∈ V (f3) r P0, [1 : b : 1] exists, such that a matrix from Mλ isequivalent to a matrix from M[1:b:1].

Now let us see “how large” the familyMλ is for a given λ in V (f3)rP0.For Λ = Λ(λ,Γ) in Mλ, we denote:

Γ1 =

0 a1 a2

−a1 0 a3

−a2 −a3 0

, Γ2 =

a7 a8 a9

a10 a11 a12

a13 a14 a15

, Γ3 =

0 a4 a5

−a4 0 a6

−a5 −a6 0

.

The condition det Λ = f 2 provides 10 equations in the above 15 parameters.Six of these equations are linear in the entries of Γ2 and form a linear systemof dimension three.

1. If b = 0 the solution of this system is:

a7 = −a12 · (a2 + 1)

a8 = a10 = a15 = 0

a9 = a11 − a13

a14 = a2 · a12 .

2. If b 6= 0, the system has the following solution:

a8 = − ba+1

a7 + a15

a9 = − a−1b(a+1)

a7 − a2

b2· a15

a10 = ba+1· a7

a12 = − a2+3(a+1)2 · a7 + b

a+1a11 + 1−a

b(a+1)a15

a13 = a−1b(a+1)

· a7 + a11 + a2

b2· a15

a14 = 2(1−a)(a+1)2 · a7 − b

a+1· a11 + a−1

b(a+1)· a15 .

82

The other four equations are linear in the entries of Γ1 with coefficientsin K[a4, . . . , a15] and have dimension five:

LIB"matrix.lib";

option(redSB);

ring r=0,(x(4),x(1),x(2),x(3),e,a,b,a(1..15)),dp;

ideal ii=a3+b3+1,e*b+a2-a+1,e*a+e-b2;

qring s=std(ii);

matrix B[10][1];

B[1,1]=x(4)*a(1);

B[2,1]=x(4)*a(2);

B[3,1]=-x(4)*a(7);

B[4,1]=-x(4)*a(10)-(x(1)+x(3));

B[5,1]=x(4)*a(3);

B[6,1]=-x(4)*a(8)-(x(1)-a*x(3));

B[7,1]=-x(4)*a(11)+x(2)+b*x(3);

B[8,1]=-x(4)*a(9)-x(2)+b*x(3);

B[9,1]=-x(4)*a(12)+e*x(3);

B[10,1]=x(4)*a(4);

matrix V[1][5];

V[1,1]=-x(4)*a(13)-x(2);

V[1,2]=-x(4)*a(14)-e*x(3);

V[1,3]=-x(4)*a(15)+x(1)+(a-1)*x(3);

V[1,4]=x(4)*a(5); V[1,5]=x(4)*a(6);

poly p1=B[5,1]*B[10,1]-B[6,1]*B[9,1]+B[7,1]*B[8,1];

poly p2=B[2,1]*B[10,1]-B[3,1]*B[9,1]+B[4,1]*B[8,1];

poly p3=B[1,1]*B[10,1]-B[3,1]*B[7,1]+B[4,1]*B[6,1];

poly p4=B[1,1]*B[9,1]-B[2,1]*B[7,1]+B[4,1]*B[5,1];

poly p5=B[1,1]*B[8,1]-B[2,1]*B[6,1]+B[3,1]*B[5,1];

poly g=V[1,1]*p1-V[1,2]*p2+V[1,3]*p3-V[1,4]*p4+V[1,5]*p5;

poly f=x(4)^3+x(1)^3+x(2)^3+x(3)^3; g=g-f;

For our skew symmetric matrix the condition g = f is equivalent to(det Λ = f 2).

matrix H=coef(g,x(4)*x(1)*x(2)*x(3));

83

for(int j=1;j<=13;j++)

H[1,j]=0;

ideal I=H; I=interred(I);

I[1]=a(9)-a(11)+a(13) I[2]=a(8)+a(10)-a(15) I[3]=a(7)+a(12)+a(14)

I[4]=a*a(10)-e*a(11)+b*a(12)+2*e*a(13)+2*b*a(14)-2*a*a(15)+a(10)

+a(15)

I[5]=2*e*a(10)+2*b*a(11)-2*a*a(12)-b*a(13)-a*a(14)-e*a(15)

+a(12)+2*a(14)

I[6]=a(3)*a(4)-a(2)*a(5)+a(1)*a(6)+a(11)^2+a(10)*a(12)-a(11)*a(13)

+a(13)^2-a(10)*a(14)-2*a(12)*a(15)-a(14)*a(15)

I[7]=a(1)*a(4)+a(3)*a(5)+a(2)*a(6)-a(10)^2+a(11)*a(12)+a(12)*a(13)

+2*a(11)*a(14)-a(13)*a(14)+a(10)*a(15)-a(15)^2

I[8]=2*e^2*a(12)+2*a*b*a(12)-3*b^2*a(13)+2*e^2*a(14)-a*b*a(14)

-3*e*b*a(15)-6*e*a(11)-b*a(12)+12*e*a(13)+2*b*a(14)-6*a*a(15)

I[9]=a(3)*a(5)*a(10)-a(2)*a(6)*a(10)-a(2)*a(5)*a(11)-a(1)*a(6)*a(11)

+a(1)*a(5)*a(12)+a(3)*a(6)*a(12)-a(2)*a(5)*a(13)

+2*a(1)*a(6)*a(13)+a(13)^3+a(2)*a(4)*a(14)+a(3)*a(6)*a(14)

+a(10)*a(11)*a(14)+a(12)^2*a(14)-2*a(10)*a(13)*a(14)

+a(12)*a(14)^2+a(3)*a(5)*a(15)+2*a(2)*a(6)*a(15)

+a(11)*a(14)*a(15)-2*a(13)*a(14)*a(15)-a(15)^3-1

I[10]=2*e*a(2)*a(4)-2*e*a(1)*a(5)+2*b*a(2)*a(5)-2*a*a(3)*a(5)

+2*b*a(1)*a(6)-4*a*a(2)*a(6)-2*b*a(11)^2+2*a*a(11)*a(12)

+2*e*a(12)^2+5*b*a(11)*a(13)-4*a*a(12)*a(13)-2*b*a(13)^2

-2*b*a(10)*a(14)-a*a(11)*a(14)+2*a*a(13)*a(14)-2*e*a(14)^2

+3*e*a(11)*a(15)-6*e*a(13)*a(15)-2*b*a(14)*a(15)+6*a*a(15)^2

+4*a(3)*a(5)+2*a(2)*a(6)-a(11)*a(12)+2*a(12)*a(13)

+2*a(11)*a(14)-4*a(13)*a(14)-6*a(15)^2

ideal J=I[1],I[2],I[3],I[4],I[5],I[8];

This is the ideal generated by the linear equations in the entries of (Γ2).We compute its dimension.

ring r1=0,(a(7..15),e,a,b),(c,dp(9),dp(3));

ideal I=a3+b3+1,e*b+a2-a+1,e*a+e-b2;

qring s1=std(I);

ideal J=imap(s,J);

ideal JJ=std(J);

84

dim(JJ);

4

Since we have defined e as a variable, the dimension of our ideal is, in fact,3. We compute now the dimension of the ideal generated by the remaining4 equations, I[6], I[7], I[9], I[10], in the quotient ring given by the previousideal.

ideal J1=I[6],I[7],I[9],I[10];

ring r2=0,(a(1..15),e,a,b),(c,dp(15),dp(3));

ideal J=imap(s1,J);

J=J+(a3+b3+1,e*b+a2-a+1,e*a+e-b2);

ideal JJ=std(J);

qring s2=JJ;

ideal J1=imap(s,J1);

ideal JJ1=std(J1);

dim(JJ1);

6

As before, in fact, the ideal has dimension is 5.Let us summarize the results.Let M be an indecomposable graded rank 2, 6–generated MCM and M

the restriction of M to the elliptic curve on our surface defined by f = x4 = 0.Then M ∼= Nλ⊕N∨λ for a suitable 3–generated rank 1 MCM Nλ = coker(αλ),λ ∈ V (f, x4)r[−1 : 0 : 1 : 0] ∼= V (f3)r[−1 : 0 : 1] =: C. If λ = [a : b : c]and λt := [c : b : a], thenN∨λ

∼= Nλt, in particular, there exist skew–symmetric3×3–matrices Γ1,Γ3 with constant entries not being zero simultaneously and

a 3× 3–matrix Γ2 such that M = coker(Λ) for Λ = x4

(Γ1 −Γt2Γ2 Γ3

)+(

0 −αtλαλ 0

),

Γ1 =(

0 a1 a2−a1 0 a3−a2 −a3 0

), Γ3 =

(0 a4 a5

−a4 0 a6−a5 −a6 0

), Γ2 =

(a7 a8 a9a10 a11 a12a13 a14 a15

)and det(Λ) = f 2.

Let A15 be the 15–dimensional affine space with the coordinates (a1, . . . , a15)

and G be the subgroup of Sl2(K) generated by the matrices gk =(

0 k− 1k

0

),

k ∈ K r 0. Consider the action of G on A15: G × A15 → A15, (gk, a) →a′ = (k2a1, k

2a2, k2a3,

1k2a4,

1k2a5,

1k2a6, a7, .., a15). Denote A = A15/G.

A point (λ; a) ∈ C × A corresponds to a matrix Λ = x4

(Γ1 −Γt2Γ2 Γ3

)+

(0 −αtλαλ 0

). The group G acts on C×A in the following way: let (λ; a) ∈ C×A

correspond to Λ = x4

(Γ1 −Γt2Γ2 Γ3

)+(

0 −αtλαλ 0

), then gk(λ; a) = (λt; b), where

(λt; b) corresponds to gk

(Uλ 0

0 U−1λ

)Λ(Utλ 0

0 Ut−1λ

)gtk, Uλ defined in Remark 4.5.

85

Let M ⊆ C × A be the G–invariant closed subset defined by det(Λ) = f 2.Let π : M → C be the canonical projection. If λ = [1 : b : 1] withb3 = −2, then Sl2(K) acts on π−1(λ) via the representation Sl2(K) →(

K1 Id K2U−1λ

K3Uλ K4 Id

), K1K4 −K2K3 = 1

,(K1 K2K3 K4

)7→(K1 Id K2U

−1λ

K3Uλ K4 Id

).

Theorem 4.11. 1. Every indecomposable graded rank 2, 6–generated MCMis represented by a point in M.

2. Mr π−1([1 : b : 1] | b3 = −2)/G is the moduli space of isomorphismclasses of indecomposable graded rank 2, 6–generated MCM M such thatthe restriction to V (f, x4), M ∼= Nλ ⊕ N∨λ for Nλ being not self–dual.This moduli space is 5–dimensional.

3. Sl2(K) acts on π−1([1 : b : 1] | b3 = −2) and π−1([1 : b : 1] | b3 =−2)/Sl2(K) is the moduli space of isomorphism classes of indecom-posable graded rank 2, 6–generated MCM M such that the restrictionto V (f, x4), M ∼= Nλ ⊕Nλ for Nλ being self–dual.

Remark 4.12. It is well known that the ideal defining 5 general points inP3K (this means any four from them are not on a hyperplane) is Goren-

stein. Restricting to the 5 general points on the surface V (f) we get a familyof Gorenstein ideals whose isomorphism classes of 2-syzygies over R (theyare indecomposable, graded, rank 2, 6-generated MCM modules) form a 5-parameter family (see [Mi], [IK]). Here we give an example.Let [1 : 0 : 0 : −1], [1 : 0 : −1 : 0], [1 : −1 : 0 : 0], [1 : −u : 0 : 0], [1 : −u : 1 : −u],u2 + u+ 1 = 0 be 5 points in general position on V (f) and I the ideal definedby these points in R. I is generated by the following quadratic forms:x2x4 +ux3x4, −ux2x3 +ux3x4, x1x4 +x2

4− (1−u)x3x4, u(x1 +x3)x3 +2x3x4,−x3x4 − x2

1 + ux1x2 − u2x22 + x2

3 + x24.

The second syzygy of I over R is the cokernel of a skew symmetric matrix Adefined byA[1, 1] = A[2, 2] = A[3, 3] = A[4, 4] = A[5, 5] = A[6, 6] = 0,A[1, 2] = (−3u− 2)x3 + (2u− 1)x4 = −A[2, 1],A[1, 3] = −ux1 + (−2u+ 1)x2 + (u+ 1)x3 + ux4 = −A[3, 1],A[1, 4] = (u− 2)x1 − x2 + (−3u− 4)x3 + (2u− 1)x4 = −A[4, 1],A[1, 5] = (u+ 1)x3 − ux4 = −A[5, 1],A[1, 6] = −ux1 + (u+ 1)x2 + (1/7u+ 3/7)x3 + (−3/7u− 2/7)x4 = −A[6, 1],A[2, 3] = (u− 2)x1 − x2 + x3 + (−u+ 2)x4 = −A[3, 2],A[2, 4] = (3u+ 2)x1 + (2u+ 3)x2 + 4ux3 + x4 = −A[4, 2],A[2, 5] = (−3u− 1)x3 + (u− 2)x4 = −A[5, 2],

86

A[2, 6] = (−u− 2)x1 + (−u+ 1)x2 + (−u− 1)x3 + ux4 = −A[6, 2],A[3, 4] = −3x3 = −A[4, 3],A[3, 5] = (u+ 1)x3 = −A[5, 3],A[3, 6] = (−6/7u− 4/7)x3 + x4 = −A[6, 3],A[4, 5] = (−3u− 1)x3 = −A[5, 4],A[4, 6] = −ux3 + ux4 = −A[6, 4],A[5, 6] = −x1 − ux2 = −A[6, 5].

This matrix is equivalent to Λ(λ,Γ) = x4 · Γ +

(0 −αtλαλ 0

)for

λ = (0 : u+ 1 : 1) and Γ given by:

a1 = −43u− 2

3, a2 = −u− 1, a3 = 2

3u+ 1

3,

a4 = −14u+ 1, a5 = −u− 1

2, a6 = −3

2u+ 3

4,

a7 = −12, a8 = −1

2u, a9 = 1

2u,

a10 = −12u− 1

2, a11 = u+ 3

2, a12 = u+ 1,

a13 = 12u+ 3

2, a14 = −u− 1

2, a15 = −u− 1

2.

Similar problem is treated in [F2], where D. Faenzzi describe the modulispaces of orientable aCM bundles on a smooth cubic surface in P3. He givesalso some informations on the stability of this bundles.

Theorem 4.13. Let X be a smooth cubic surface in P3 and let E be anindecomposable aCM rank two vector bundle on X. Then the minimal gradedfree resolution of E takes one of the following forms:

(A) 0→ O6(−1)→ O6 → 0;

(B) 0→ O4(−1)⊕O(−2)→ O5 → 0;

(C) 0→ O5(−2)→ O ⊕O4(−1)→ 0;

(D) 0→ O2(−1)⊕O3(−2)→ O4 → 0;

(E) 0→ O(−1)⊕O3(−2)→ O3 ⊕O(−1)→ 0;

(F) 0→ O4(−2)→ O2 ⊕O2(−1)→ 0;

(G) 0→ O3(−2)⊕O(−3)→ O ⊕O3(−1)→ 0;

(H) 0→ O3(−2)→ O3 → 0.

87

Proposition 4.14. Let E be an indecomposable rank 2 aCM bundle on Xhaving resolution (A). Then c1(E) ∼= T1 + T2, c2(E) ∼= T1 · T2, where T1 andT2 are twisted cubics contained in X with 3 ≤ T1 · T2 ≤ 5. Furthermore, thematrix f(E) is skew–symmetric if and only if T1 · T2 = 5.

Proposition 4.15. Let E be indecomposable aCM of rank two, with f(E) isskew–symmetric and resolution (A).The general E is stable and MCMs(2; c1(E), c2(E)) (the moduli space of sta-ble rank two aCM bundles with c1 and c2 as Chern classes) is smooth ofdimension 5.

Using this results one can give the matrix factorizations of the rank two,orientable stable modules. We remind, that by stable module, we understandMCM module with stable sheafification on the projective cone.

Proposition 4.16. The matrices from

⋃

λ∈V (f)

Mλ

describe all rank two, orientable MCM modules with stable sheafification.

4.2 Orientable 4–generated MCM modules

LetM be a graded, indecomposable, orientable, 4–generated MCMR–moduleof rank 2. We remind that there exists I graded 3–generated Gorensteinideal in R, of codimension 2, such that M ∼= Ω2

R(I). We can write thenI as I = J/(f), with J ⊂ S = k[x1, x2, x3, x4] a graded, 3–generated idealcontaining f , f ∈ mJ by [HK]. Let α1, α2, α3 be a minimal system of ho-mogeneous generators of J . Since dimS/J = dimR/I = 1, it follows thatα1, α2, α3 is a regular system of elements in S.

Let a, b ∈ K with a3 = b3 = −1 and σ = (i j s) be a permutation of theset 2, 3, 4 with i < j. Set

wσ1 = x1 − axs, wσ2 = xi − bxj ,vσ1 = x2

1 + ax1xs + a2xs, vσ2 = x2i + bxixj + b2x2

j .

88

Then we havef = wσ1vσ1 + wσ2vσ2.

For any λ point of the surface V (f) ⊂ P3, we set three pairs of polyno-mials, (piλ, qiλ), 1 ≥ i ≥ 3 such that

f =

3∑

i=1

piλqiλ.

If λ = [λ1 : λ2 : λ3 : 1], set

piλ = xi − λix4, qiλ = x2i + λixix4 + λ2

ix24, for 1 ≤ i ≤ 3.

If λ = [λ1 : λ2 : 1 : 0], set

piλ = xi − λix3, qiλ = x2i + λixix3 + λ2

ix23, for 1 ≤ i ≤ 2

p3λ = x4, q3λ = x24.

If λ = [λ1 : 1 : 0 : 0] ∈ V (f), we set

p1λ = x1 − λ1x2, q1λ = x21 + λ1x1x2 + λ2

1x22,

p2λ = x3, q2λ = x23,

p3λ = x4, q3λ = x24.

Since f ∈ (α1, α2, α3) and eventually we are interested in Ω2R(α1, α2, α3),

we may suppose that either αi is in the set piλ, qiλ for each 1 ≤ i ≤ 3, orαi is in the set wσi, vσi for each 1 ≤ i ≤ 2 and β = α3 is a regular elementin R/(α1, α2).

Lemma 4.17. Let M be a graded, indecomposable, 4–generated MCM R–module of rank 2. Then M is isomorphic to one of the following modules:

1. Ω2R(p1λ, p2λ, p3λ) or Ω2

R(q1λ, q2λ, q3λ), for some λ ∈ V (f),

2. Ω2R(wσ1, vσ2, β) or Ω2

R(wσ1, wσ2, β) for some a, b, σ and β as above.

Proof. For any λ ∈ V (f) set

Iλ = (p1λ, p2λ, p3λ)

and

ϕλ =

0 p3λ −p2λ −q1λ

−p3λ 0 −p1λ q2λ

p2λ p1λ 0 q3λ

q1λ −q2λ −q3λ 0

, ψλ =

0 −q3λ q2λ p1λ

q3λ 0 q1λ −p2λ

−q2λ −q1λ 0 −p3λ

−p1λ p2λ p3λ 0

.

89

There exists the following graded exact sequence:

R3(−5)⊕ R(−6)ϕλ−→ R4(−4)

ψλ−→ R3(−2)⊕R(−3)A−→ R3(−1)

τ−→ Iλ −→ 0,

where τ = (−p1λ, p2λ, p3λ) and A is given by the first three rows of ϕλ. Thus,Ω2(Iλ) ∼= Coker(ϕλ) and (ϕλ, ψλ) is a matrix factorization of the polynomialf , defining the module Ω2(Iλ). The ideals Iλ and (q1λ, q2λ, p3λ) belong to thesame even linkage class since

Iλ ∼ (q1λ, p2λ, p3λ) ∼ (q1λ, q2λ, p3λ).

( For the first link we consider the regular sequence p1λq1λ, p2λ, p3λ and forthe second one the sequence q1λ, p2λq2λ, p3λ.) Similarly, one can see thatIλ is evenly linked with the ideals (q1λ, p2λ, q3λ) and (p1λ, q2λ, q3λ). By [HK,Theorem 2.1], we obtain that

Coker(ϕλ) ∼= Ω2R(Iλ) ∼= Ω2

R(q1λ, q2λ, p3λ) ∼= Ω2R(q1λ, p2λ, q3λ)

∼= Ω2R(p1λ, q2λ, q3λ).

Analogously, we see that

Coker(ψλ) ∼= Ω2R(q1λ, q2λ, q3λ) ∼= Ω2

R(p1λ, p2λ, q3λ) ∼= Ω2R(p1λ, q2λ, p3λ)∼= Ω2

R(q1λ, p2λ, p3λ).

Thus, the case when αi is one of the forms piλ, qiλ gives (1).Now let σ, a, b as above and β ∈ S which is regular on R/(wσ1, vσ2). Set

Iσβ(a, b) = (wσ1, vσ2, β)

and

ϕσβ(a, b) =

0 wσ1 −vσ2 0−wσ1 0 −β wσ2

vσ2 β 0 vσ1

0 −wσ2 −vσ1 0

,

ψσβ(a, b) =

0 −vσ1 wσ2 βvσ1 0 0 −vσ2

−wσ2 0 0 −wσ1

−β vσ2 wσ1 0

.

We have the following exact sequence:

R4 ϕσβ(a,b)−−−−−→ R4 ψσβ(a,b)−−−−−→ R4 B−→ R3 τ ′−→ Iσβ(a, b) −→ 0 ,

90

where τ ′ = (−β, vσ2, wσ1) and B is the matrix given by the first three rowsof ϕσβ(a, b). Thus,

Ω2R

(Iσβ(a, b)

) ∼= Coker(ϕσβ(a, b)

).

As above, we see that

Ω2R

(Iσβ(a, b)

) ∼= Ω2R(wσ2, vσ1, β)

andΩ2R(wσ1, wσ2, β) ∼= Ω2

R(vσ1, vσ2, β) ∼= Coker(ψσβ(a, b)

).

Thus, the case when αi is one of the forms wσi, vσi for i ≤ 2 gives (2).

LetM =

Coker(ϕλ),Coker(ψλ) | λ ∈ V (f)

.

For a, b, σ as above, set

ϕσ(a, b) = ϕσ,xjxs(a, b), ψσ(a, b) = ψσ,xjxs(a, b),

that is, β = xjxs. Let

P =

Coker(ϕσ(a, b)

),Coker(ψσ(a, b)) | a, b, σ as above

.

Theorem 4.18. The set M∪Pcontains only non–isomorphic, indecompos-able, graded, orientable, 4–generated MCM R–modules of rank 2 and everyindecomposable, graded, orientable, 4–generated MCM R–module of rank 2is isomorphic with one module of M∪P.

Proof. Applying Lemma 4.17, we must show in the case (2) that β can betaken xjxs. Since vσ1−wσ1(x1 +2axs) = 3a2x2

s, adding in ϕσβ(a, b) multiplesof the last row to the second one and multiples of the first column to the thirdone, we may suppose the entry (2, 3) of the form γ+xsδ, with γ, δ dependingonly on xj, xi. These transformations modify the entries (2, 2), (3, 3) whichare now possibly non–zero. Adding similar multiples of the last column tothe second one and multiples of the first row to the third one, we obtainϕσ,β(a, b) of the same type as before but with β = γ + xsδ. We may reduceto consider δ 6∈ K. Indeed, if δ ∈ K, then, acting on the rows and columns ofϕσβ(a, b), we obtain that M = Coker(ϕσβ(a, b)) is decomposable or belongsto the set M. Now let δ be not constant. Similarly, adding in ϕσβ(a, b)multiples of the first row to the second one and multiples of the last columnto the third one we may suppose that the entry (2, 3) has the form εxjxs

91

with ε ∈ K. These transformations modify the entries (2, 2), (3, 3). Aftersimilar transformations, we obtain ϕσβ(a, b) of the same type as before butwith β = εxjxs. If ε = 0 we see that ϕσβ(a, b) is a direct sum of two 2× 2–matrices, which contradicts the indecomposability of M = Coker

(ϕσβ(a, b)

).

So ε 6= 0. Divide the second and the third column of ϕσβ(a, b) with ε, andmultiply the first and the last row of ϕσβ(a, b) with ε. We reduce to the caseε = 1, that is β = xjxs.

Now we show that two different modules fromM∪P are not isomorphic.Note that the Fitting ideals of ϕλ (respectively ψλ) modulo (x1, . . . , x4)2

have the form (p1λ, p2λ, p3λ) and the Fitting ideals of ϕσ(a, b) (respectivelyψσ(a, b)) modulo (x1, . . . , x4)2 have the form (wσ1, wσ2) and these ideals areall different. Thus,

Coker(ϕλ) | λ ∈ V (f) ∪ Coker

(ϕσ(a, b)

)| σ, a, b as above

contains only non–isomorphic modules (similarly for ψ’s). It follows that, ifN,P ∈ M∪P are isomorphic and different, then N ' Ω1

R(P ).If N = Coker(ϕλ), for λ ∈ V (f), then this is not possible since the

ideals (p1λ, p2λ, p3λ) and (q1λ, q2λ, q3λ) are not in the same even linkage class.Indeed, by the proof of (1) in Lemma 4.17, (p1λ, p2λ, p3λ) is evenly linked with(q1λ, q2λ, p3λ) and this last ideal is obviously directly linked with (q1λ, q2λ, q3λ).If N = Coker

(ϕσ(a, b)

)for some σ, a, b, and N ' Ω1

R(N), then the ideals(wσ1, vσ2, xjxs) and (wσ1, wσ2, xjxs) are evenly linked. But these ideals aredirectly linked by the regular sequence wσ1, vσ2wσ2, xjxs, contradiction!

It remains to show that M∪ P contains only indecomposable modules.If N ∈ M, let us say N = Coker(ϕλ) for λ = [λ1 : λ2 : λ3 : 1], we see thatN/x4N is exactly the module corresponding to the matrix

0 x3 −x2 −x21

−x3 0 −x1 x22

x2 x1 0 x23

x21 −x2

2 −x23 0

whose cokernel is the special module M2 (see [LPP] for the special module ofrank 2 which corresponds to the special bundle from Atiyah classification).Thus, N/x4N is indecomposable and, by Nakayama’s Lemma, N is indecom-posable. Now let N ∈ P, N = Coker

(ψσ(a, b)

). By the permutation of the

92

rows and the columns of ψσ(a, b), we may suppose that it has the form:

wσ1 −vσ2 xjxs 0wσ2 vσ1 0 xjxs0 0 vσ1 vσ2

0 0 −wσ2 wσ1

.

Suppose N is decomposable. Then ψσ(a, b) is equivalent with a direct sumof two matrices of order 2 say A1, A2. Let B1, B2 be the submatrices of theψσ(a, b) given by the first two lines and columns, respectively the last twolines and columns. Certainly A1, A2, B1, B2 define some maximal Cohen-Macaulay modules of rank one N1, N2, T1, T2, and due to the particular formof ψσ(a, b) we have the following exact sequence

0→ T1 → N1 ⊕N2 = N → T2 → 0.

Note that ψσ(a, b) is modulo xj or xs the sum of B1, B2. Thus Ti/xjTi ∼=Ni/xjNi for i = 1, 2 and similarly for xs. Since we have the whole descriptionof rank one maximal Cohen-Macaulay modules we can see that Ai is equiv-alent with Bi modulo xj and modulo xs only when Ai is equivalent with Bi.Thus Ti ∼= Ni for i = 1, 2 and so N ∼= T1⊕T2. By a subtle result of Miyata ([Mi] ) this happens only if the above exact sequence splits. This means thatthere exist two matrices A,B of order two such that

xjxs · Id2 =

(wσ1 −vσ2

wσ2 vσ1

)A +B

(vσ1 vσ2

−wσ2 wσ1

),

which is impossible.

Proposition 4.19. The aCM bundles corresponding to

Remarks 4.20. 1. There exists a bijection between

P1 =

Coker(ϕσ(a, b)

)| σ, a, b

and the 2–generated, non–free, MCM R–modules, which remind usof Atiyah’s classification. Thus, P1 contains 27 modules correspondingto 27 lines and 27 pencils of conics of V (f).Similarly, P2 =

Coker

(ψσ(a, b)

)| σ, a, b

contains 27 modules.

2. M is a kind of “blowing up” of M2,Ω1R(M2) from [LPP] (see the proof

of Theorem 4.18). Note also that M consists of two classes of modulesparameterized by the points of V (f), which is also in Atiyah’s idea.

93

3. The matrices ϕ defining the modules of M∪P are skew symmetric asour Theorem 4.1 predicted.

Lemma 4.21. (With the notations from Theorem 4.13)

1. The aCM bundles corresponding to the modules

Cokerψλ,Cokerψσ(a, b) | λ ∈ V (f), σ, a, b

have a minimal resolution of type (E);

2. The aCM bundles corresponding to the MCM modules

Cokerϕλ,Cokerϕσ(a, b) | λ ∈ V (f), σ, a, b

have a minimal resolution of type (G);

3. The modules

Cokerϕσ(a, b),Cokerψσ(a, b) | λ, σ, a, b

are extensionsof rank one MCM modules, that correspond to 27 lines (conics).

4. For β linear form,

Cokerϕσβ(a, b),Cokerψσβ(a, b) | λ ∈ V (f), σ, a, b

are also extensions, but they are isomorphic to modules from M.

Proof. The results follow straightforeward, by simple computations.For 3), we write ψσ(a, b) as in the proof of indecomposability (Theorem 4.18).We get that the corresponding aCM bundle, E fits in one extension of thetype:

0→ OX(C)→ E → OX(L)→ 0,

where C is a conic and L a line on the surface X = ProjR. Similarly, ifF denotes the corresponding aCM bundle of some ϕσ(a, b), then we have anextension of type:

0→ OX(L+H)→ F → OX(C)→ 0,

for H = OX(1).

Theorem 4.22. 1. For any λ ∈ X, X = ProjR, the module Cokerψλ,corresponds to a stable aCM bundle. More, all stable, indecomposableaCM bundles with a resolution of type (E) have a matrix factorizationof type ψλ.

94

2. The aCM bundles corresponding to the modules

Cokerϕλ,Cokerϕσ(a, b) | λ ∈ V (f), σ, a, b

are semistable.

The theorem follows from the previous lemma, using some actual resultsof D. Faenzzi:

Proposition 4.23. ( see [F2] ) Let E be an indecomposable rank two aCMbundle on a smooth cubic hypersurface X in P3.

1. If E has a minimal resolution of type (E), then c1(E) ≡ H and c2(E) =2. ( H = OX(1))

2. If it has a resolution of type (G), then c1(E) = 0, c2(E) = 1.

3. If E is a general bundle with resolution of type (E), then E is stable.The moduli space of stable rank 2 aCM bundles with c1(E) ≡ H andc2(E) = 2, denoted MCMs(2;H, 2) is birational to X.

4. If E has resolution of type (G), then E is semistable with h0(End(E)) =2 and the moduli space of semistable rank 2 aCM bundles with c1(E) =0 and c2(E) = 1, denoted MCMss(2; 0, 1) is birational to X, while .MCMs(2; 0, 1) is empty.

95

5 Rank two, non – orientable, MCM mod-

ules over the hypersurface R = k[x1, x2, x3, x4]/(x31+

x32 + x3

3 + x34)

Let M be a graded non–orientable, rank 2, MCM R–module, without freedirect summands. We should like to express M as a 2–syzygy of an ideal I,M ∼= Ω2

R(I), with µ(M) = µ(I)+1 (this is known in orientable case by [HK],see here Section 3).

The following proposition can be found in [Bo1, Korollar 2].

Proposition 5.1. Let (A,m) be a Noetherian normal local domain withdimA ≥ 2 and N a finite torsion–free A–module. Then there exists a finitefree submodule F ⊂ N such that N/F is isomorphic with an ideal of A andthe canonical map F/mF → N/mN is injective.

Applying Proposition 5.1, we obtain the following exact sequence:

0→ R −→M −→ I −→ 0 (22)

for an ideal I ⊂ R, which induces an exact sequence

0 −→ K = R/m −→M/mM −→ I/mI −→ 0.

Thus µ(M) = µ(I) + 1.As we know in the orientable case to obtain MCM R–modules of rank

2 we must choose I such that Ext1R(I, R) is a cyclic R–module or, more

precisely, such that R/I is Gorenstein. In the non–orientable case one canalso show that Ext1

R(I, R) must be a cyclic R–module, but this is not veryhelpful since it is hard to check this condition for arbitrary I. Below we shallstate an easier condition.

Let J ⊂ S = K[X1, . . . , X4] be an ideal such that f ∈ mJ and I = J/(f).

Lemma 5.2. Let

0 −→ Ss3d3−→ Ss2

d2−→ Ss1d1−→ J −→ 0

be a minimal free S–resolution of an ideal J with depth S/J = 1.

If rank Ω2R

(J/(f)

)= 2 and µ

(Ω2R

(J/(f)

))= µ(I) + 1 then s1 = s2 ≤ 5

and s3 = 1.

97

Proof. As in the proof of Theorem 4.1, we obtain a minimal free resolutionof I = J/(f) over S in the following way:

Let v : S → Ss1 be an S–linear map such that jd1v = f IdS, where j :

J → S is the inclusion. Let d1 be the composite map Ss1d1−→ J → J/(f) = I.

Then the following sequence

0 −→ Ss3

(d30

)−→ Ss2+1 (d2,v)−−−→ Ss1

d1−→ I −→ 0

is exact and forms a minimal free resolution of I over S. Since

f · Ss1 ⊂ Im(d2, v),

there exists an S–linear map h : Ss1 → Ss2+1 such that

(d2, v)h = f IdSs1

and we obtain the following exact sequence

Rs3+s1

„h

∣∣d30

«

−−−−→ Rs2+1 (d2 ,v)−−−→ Rs1 d1−→ I −→ 0,

which is part of a minimal free R–resolution of I. Thus, M = Ω2R(I) is

the image of the first map above and so s1 + s3 = s2 + 1 = s1 + 1 becauseµ(M) = µ

(Ω1R(M)

)= µ(I)+1 by hypothesis. It follows that s3 = 1, s1 = s2.

As µ(M) ≤ 3 rankRM = 6 we obtain s1 ≤ 5.

Let detN be the corresponding class of the bidual (∧nN)∗∗, n = rankN ,in Cl(R) for a torsion free R–module N . Since det is an additive function,we obtain det(M) = 0 if and only if det(I) = 0. Thus, M is non–orientableif and only if I is non–orientable, that is, codim(J) ≤ 1 for all ideals J ⊂R isomorphic with I, according to [HK]. Since M has rank 2, we obtaincodim(I) = 1. Thus, dimR/I = 2 and, from (22), we obtain depthR/I = 1,that is, R/I is not Cohen–Macaulay. Also from (22) we obtain Ω2

R(M) 'Ω2R(I) and so M ' Ω2

R(I).

Proposition 5.3. Each graded, non–orientable, rank 2, s–generated MCMR–module is the second syzygy Ω2

R(I) of an (s − 1)–generated graded idealI ⊂ R with depthR/I = 1 and dimR/I = 2.

98

5.1 Non–orientable, rank 2, 4–generated MCM modules

As in Section 3, let u, a, b ∈ K, with

a3 = b3 = −1, u2 + u+ 1 = 0,

σ = (i j s) be a permutation of the set 2, 3, 4 with i < j and set

wσ1 = x1 − axs, wσ2 = xi − bxj,vσ1 = x2

1 + ax1xs + a2x2s, vσ2 = x2

i + bxixj + b2x2j .

We havevσ1 = v′σ1v

′′σ1, vσ2 = v′σ2v

′′σ2

forv′σ1 = x1 − uaxs, v′′σ1 = x1 + (1 + u)axs,v′σ2 = xi − ubxj, v′′σ2 = xi + (1 + u)bxj.

SetI1σ(a, b, u) = (xsv

′σ2, vσ2, wσ1),

I2σ(a, b, u) = (xjv′′σ1, vσ1, wσ2),

I3σ(a, b, u) = (xsv′′σ2, vσ2, vσ1),

I4σ(a, b, u) = (xjv′σ1, vσ1, vσ2),

I5σ(a, b, u) = (xsv′′σ2, vσ2, wσ1),

I6σ(a, b, u) = (xjv′σ1, vσ1, wσ2),

I7σ(a, b, u) = (xsv′σ2, vσ2, vσ1),

I8σ(a, b, u) = (xjv′′σ1, vσ2, vσ1)

Set

ϕ1σ(a, b, u) =

wσ1 −wσ2 0 xsvσ2 vσ1 xsv

′σ2 0

0 0 wσ1 −v′′σ2

0 0 −wσ2v′σ2 −vσ1

,

ϕ2σ(a, b, u) =

wσ1 −wσ2 0 −xjvσ2 vσ1 xjv

′′σ1 0

0 0 wσ2 −v′σ1

0 0 −wσ1v′′σ1 −vσ2

,

99

ψ3σ(a, b, u) =

wσ1 −wσ2 −xs 0vσ2 vσ1 0 −xsv′′σ2

0 0 −v′σ2 wσ1

0 0 −vσ1 −wσ2v′′σ2

,

ψ4σ(a, b, u) =

wσ1 −wσ2 xj 0vσ1 vσ2 0 −xjv′σ1

0 0 −v′′σ1 wσ2

0 0 −vσ2 −wσ1v′σ1

,

ϕ3σ(a, b, u) =

vσ1 wσ2 0 −xs−vσ2 wσ1 xsv

′′σ2 0

0 0 −wσ2v′′σ2 −wσ1

0 0 vσ1 −v′σ2

,

ϕ4σ(a, b, u) =

vσ1 wσ2 xjv′σ1 0

−vσ2 wσ1 0 −xj0 0 −wσ1v

′σ1 −wσ2

0 0 vσ2 0− v′′σ1

,

ψ1σ(a, b, u) =

vσ1 wσ2 0 xs−vσ2 wσ1 −xsv′σ2 0

0 0 vσ1 −v′′σ2

0 0 −wσ2v′σ2 −wσ1

,

ψ2σ(a, b, u) =

vσ1 wσ2 0 xj−vσ2 wσ1 −xjv′′σ1 0

0 0 vσ2 −v′σ1

0 0 −wσ1v′′σ1 −wσ2

.

Theorem 5.4. 1. For each 1 ≤ t ≤ 4, the pair(ϕtσ(a, b, u), ψtσ(a, b, u)

)

forms a matrix factorization of Ω2R

(Itσ(a, b, u)

).

2. The set

N =

Coker(ϕtσ(a, b, u)

), Coker

(ψtσ(a, b, u)

)| 1 ≤ t ≤ 4, σ, a, b, u

contains only graded, indecomposable, non–orientable, 4–generated MCMR–modules of rank 2.

3. Every indecomposable, graded, non–orientable, 4–generated MCM mod-ule over R of rank 2 is isomorphic with one module of N .

100

4. The modules of N are pairwise isomorphic. In particular, there exist216 isomorphism classes of indecomposable, graded, non–orientable, 4–generated MCM modules over R of rank 2.

Proof. (1) It is easy to check that

ϕtσ(a, b, u) · ψtσ(a, b, u) = f · Id4

and the following sequence is exact:

R(−6)4 ϕ1σ(a,b,u)−−−−−→ R(−5)2 ⊗R(−4)2 ψ1σ(a,b,u)−−−−−→ R(−3)4 A1−→ R(−2)2 ⊗R(−1)

−→ I1σ(a, b, u) −→ 0,

where A1 is the 3 × 4−matrix formed by the first three rows of ϕ1σ(a, b, u).Thus, (1) holds for t = 1, the other cases being similar.

(2) Clearly I1σ(a, b, u) ⊂ (v′σ2, wσ1) and so dimR/I1σ(a, b, u) = 2. As xsis zero–divisor in R/I1σ(a, b, u) we see that depthR/I1σ(a, b, u) = 1 and, byProposition 5.3, Ω2

R(I) is non–orientable, 4–generated of rank 2. Note thatthe module Coker

(ϕ1σ(a, b, u)

), as in the last part of the proof of Theorem

4.18, is indecomposable because there exist no two matrices A,B of ordertwo such that

(0 xs

xsv′σ2 0

)=

(wσ1 −wσ2

vσ2 vσ1

)A +B

(wσ1 −v′′σ2

−wσ2v′σ2 −vσ1

).

Similarly, the cases t > 1 follows.(3) Now letM be an indecomposable, graded, non–orientable, 4–generated

MCM R–module of rank 2. By Proposition 5.3, there exists a graded idealI ⊂ R with dimR/I = 2, depthR/I = 1, which is 3–generated and such thatM ' Ω2

R(I). Then I = J/(f) with J ⊂ S = K[x1, x2, x3, x4] is a 3–generatedideal containing f . We have still f ∈ mJ , though we are not in the orientablecase (see [EP1] for details). Let α1, α2, α3 be a minimal system of homoge-neous generators of J . If f does not belong to the ideal generated by twoαt, then, as in Section 3, f =

∑3t=1 ptqt and, after a renumbering, we may

suppose that αt is necessarily either pt or qt, for all 1 ≤ t ≤ 3. Then α1, α2, α3

is a regular system of elements in S and so R/I = S/J is Cohen–Macaulaywhich is false.

Thus, we may suppose f ∈ (α1, α2). Then there exist a, b ∈ K witha3 = b3 = −1, and σ = (i j s) a permutation of the set σ = 2, 3, 4, i < j,such that αt is necessarily either wσt or vσt, for t = 1, 2. If α1 = wσ1, α2 =

101

wσ2, then R/(α1, α2) is a domain and α1, α2, α3 must be a regular system ofelements in S and so, again, R/I = S/J is Cohen–Macaulay, contradiction!

We have the following cases:Case I: α1 = wσ1

Then α2 must be vσ2 and we have

(α1, α2) = (v′σ2, wσ1) ∩ (v′′σ2, wσ1).

It follows that a zero–divisor of R/(α1, α2) must be either in (v′σ2, wσ1) orin (v′′σ2, wσ1). As we know, α3 is a zero–divisor in R/(α1, α2) and so α3 ∈(v′σ2, wσ1) or α3 ∈ (v′′σ2, wσ1).

I(a) Supposeα3 ∈ (v′σ2, wσ1).

Subtracting from α3 a multiple of wσ1, we may take α3 = v′σ2β for a form βof S. Note that the matrices

ϕ =

0 wσ1 −v′′σ2 0−wσ1 0 −β wσ2

vσ2 βv′σ2 0 vσ1

0 −wσ2v′σ2 −vσ1 0

,

ψ =

0 −vσ1 wσ2 βvσ1 0 0 −v′′σ2

−wσ2v′σ2 0 0 −wσ1

−βv′σ2 vσ2 wσ1 0

give the following exact sequence:

−→ R4 ϕ−→ R4 ψ−→ R4 B1−→ R3 −→ I −→ 0,

where B1 is given by the first three rows of ϕ. Thus. (ϕ, ψ) is a matrixfactorization of Ω2

R(I) ' M . Adding in ϕ multiples of the first row to thesecond one and adding multiples of the fourth column to the third one,we may suppose that the entry (2, 3) of ϕ depends only on x1, xs. Thesetransformations modify also the entries (2, 2) and (3, 3), which are now notzero. Adding similar multiples of the first column to the second one and ofthe fourth row to the third one, we obtain ϕ of the same type as before butwith β depending only on x1, xs. Since vσ1 − wσ1(x1 + 2axs) = 3ax2

s, addingin ϕ multiples of the first column to the third one and multiples of the fourthrow to the second row, we may suppose that the entry (2, 3) has the form

102

λxs for some λ ∈ K. These transformations modify also the entries (3, 3)and (2, 2), which are now not zero. Adding similar multiples of the first rowto the third one and of the fourth column to the second column, we obtain ϕof the same type as before but with β = λxs. If λ = 0, then, clearly, ϕ is thedirect sum of two 2–matrices which contradicts that M is indecomposable.So. λ 6= 0. Now we divide the second and the third column of ϕ by λ andmultiply the first and the fourth row by λ. The new ϕ is as before but withλ = 1, that is ϕ = ϕ1σ(a, b, u).

I(b) Supposeα3 ∈ (v′′σ2, wσ1).

Then we may take α3 = v′′σ2β, for a form β. With a similar proof as above,we obtain M ' Coker

(ψ3σ(a, b, u)

).

Case II: α2 = wσ2.Then α1 = vσ1. It follows that (α1, α2) = (v′σ1, wσ2)∩ (v′′σ1, wσ2). We have

the following two subcases:II(a) α3 ∈ (v′σ1, wσ2). We may suppose α3 = v′σ1β, for a form β and we

obtain that M ' Coker(ψ4σ(a, b, u)

).

II(b) α3 ∈ (v′′σ1, wσ2). In this subcase we may take α3 = v′′σ1β, for a formβ and we obtain that M ' Coker

(ϕ2σ(a, b, u)

).

Case III: α1 = vσ1, α2 = vσ2.Then (α1, α2) = (v′σ1, v

′σ2)∩(v′σ1, v

′′σ2)∩(v′′σ1, v

′σ2)∩(v′′σ1, v

′′σ2). We proceed as

in the above cases, taking α3 from one prime ideal of the above decompositionof(α1, α2)

), let us say α3 ∈ (v′σ1, v

′σ2), that is α3 = v′σ1β + v′σ2γ for some

β, γ ∈ S. Suppose that one cannot reduce the problem to the case β = 0 orγ = 0, this implies, for example, that v′σ1 does not divide γ and v′σ2 does notdivide β. Then Ω1

S

((α1, α2, α3)

)⊂ S3 contains the columns of the following

matrix

vσ2 α3 0 v′′σ2β−vσ1 0 α3 v′′σ1γ

0 −vσ1 −vσ2 −v′′σ1v′′σ2

and we can see that µ(Ω1S

((α1, α2, α3)

)≥ 4, which contradicts Lemma 5.2.

Thus we may suppose, let us say α3 = v′σ1β, where β is not a multipleof v′′σ1. Now we may proceed as in the above cases and we obtain, in order,M ' Coker

(ϕ4σ(a, b, u)

), M ' Coker

(ϕ3σ(a, b, u)

), M ' Coker

(ψ1σ(a, b, u)

),

and M ' Coker(ψ2σ(a, b, u)

).

(4) The matrices

ϕtσ(a, b, u), ψtσ(a, b, u), 1 ≤ t ≤ 4, σ, a, b, u,

103

are equivalent in pairs. Namely:

ϕ1σ(a, b, u) ∼ ψ3σ(a, b, u2), ϕ2σ(a, b, u) ∼ ψ4σ(a, b, u2),

ϕ3σ(a, b, u) ∼ ψ1σ(a, b, u2), ϕ4σ(a, b, u) ∼ ψ2σ(a, b, u2).

We shall prove that the matrices of the set

N ′ =ϕtσ(a, b, u), | 1 ≤ t ≤ 4, σ, a, b, u

are pairwise non–equivalent. We shall consider the matrices which are ob-tained from the matrices of N ′, reducing their entries modulo m2. If A,B ∈N ′ are equivalent, then there exist P,Q, two invertible 4× 4–matrices withthe entries in K[x1, x2, x3, x4] such that PA = BQ. Let A and B be the ma-trices obtained from A, respectively B, by reducing modulo m2 their entries.From the equality PA = BQ, we obtain that there exist two invertible scalarmatrices P , Q ∈ M4(K) such that P A = BQ. This means that the matrices

A, B are also equivalent by some scalar invertible matrices. We constructthe “reduced” matrices ϕtσ(a, b, u) and for all t. We see that the matricesϕ1σ(a, b, u), ϕ2σ(a, b, u), have the entries of the rows 2 and 4 zero and the restof the matrices have the entries of the columns 1 and 3 zero. First, we choosetwo matrices A, B, one of them with the rows 2 and 4 zero and the otherwith the columns 1 and 3 zero. Suppose that A ∼ B. It results that thereare two invertible scalar 4× 4–matrices U, V such that

A U = V B.

From this equality we obtain that the rows 2 and 4 in the matrix V B arezero. Looking at the two possibilities to choose the matrix B, we see thatthe non–zero elements of the columns 2 and 4 in B are linear independent.From the above equality we get that V is not invertible.

Hence, we could find two equivalent matrices in the set N ′ only if bothhave the rows 2 and 4 zero or the columns 1 and 3 zero. It is clear that we mayreduce the study of the equivalent matrices A = ϕ1σ(a, b, u), B = ϕ2σ(a, b, u),which have the rows 2 and 4 zero. Let U, V ∈ M4×4(K) be invertible matrices

such that A U = V B. We may transform the reduced matrices A, B suchthat the last two rows are zero. Let

A =

(A1 A2

0 0

), B =

(B1 B2

0 0

), U =

(U1 U2

U3 U4

), V =

(V1 V2

V3 V4

),

104

be the decomposition of our matrices in 2 × 2 blocks. Comparing the ele-ments in the above equality, we obtain contradiction with the fact that U isinvertible.

In the same way we check that if ϕ1σ(a, b, u) and ϕ1τ (n, p, v) are different,then they are not equivalent.

Let M(σ, a, b), M ′(τ, n, p) be two rank one MCM-modules correspondingto lines and N(σ, a, b), N ′(τ, n, p) be two rank one MCM-modules correspond-ing to conics (that is Coker

(ϕσ(a, b)

), Coker

(ψσ(a, b)

)by [EP]).

Remark 5.5. There exists an indecomposable extension in Ext1R(M(σ, a, b),

M ′(τ, n, p)) only if σ = τ . In this case, for fixed M(σ, a, b) there exists 4non–orientable MCM–modules, which are extensions E of the form

0→M ′i → E →M(σ, a, b)→ 0 .

for some M ′i , i = 1, 2 of type M ′(σ, n, p). So we have 4× 27 non–orientable

MCM–modules. Similarly, taking now extensions F of the form

0→ Ni → F → N(σ, a, b)→ 0, i = 1, 2

we obtain another 4× 27 non–orientable MCM–modules. Thus all are 216 =8× 27.

5.2 Non–orientable, rank 2, 5–generated MCM modules

As in Section 3, let u, a, b ∈ K, with

a3 = b3 = −1, u2 + u+ 1 = 0,

σ = (i j s) be a permutation of the set 2, 3, 4 with i < j and set

wσ1 = x1 − axs, wσ2 = xi − bxj,vσ1 = x2

1 + ax1xs + a2x2s, vσ2 = x2

i + bxixj + b2x2j .

We havevσ1 = v′σ1v

′′σ1, vσ2 = v′σ2v

′′σ2

105

forv′σ1 = x1 − uaxs, v′′σ1 = x1 + (1 + u)axs,v′σ2 = xi − ubxj, v′′σ2 = xi + (1 + u)bxj.

Consider the following ideals:Set

J1σ(a, b, u) = (vσ1, vσ2, v′σ1v′σ2, v

′′σ1v′′σ2),

J2σ(a, b, u) = (vσ1, vσ2, v′σ1v′σ2, v

′′σ1v′σ2),

J3σ(a, b, u) = (vσ1, vσ2, v′σ1v′σ2, v

′′σ1(v′σ2 + v′′σ2)),

J4σ(a, b, u) = (vσ1, vσ2, v′σ1v′′σ2, v

′′σ1v′σ2),

J5σ(a, b, u) = (vσ1, vσ2, v′σ1v′′σ2, v

′′σ1v′′σ2),

J6σ(a, b, u) = (vσ1, vσ2, v′σ1v′′σ2, v

′′σ1(v′σ2 + v′′σ2)),

J7σ(a, b, u) = (wσ1v′σ1, vσ2, v

′σ1v′σ2, wσ1v

′′σ2),

J8σ(a, b, u) = (wσ1v′σ1, vσ2, v

′σ1v′σ2, wσ1v

′σ2),

J9σ(a, b, u) = (wσ1v′σ1, vσ2, v

′σ1v′σ2, wσ1(v′σ2 + v′′σ2)),

J10σ(a, b, u) = (wσ1v′σ1, vσ2, v

′σ1v′′σ2, wσ1v

′σ2),

J11σ(a, b, u) = (wσ1v′σ1, vσ2, v

′σ1v′′σ2, wσ1v

′′σ2),

J12σ(a, b, u) = (wσ1v′σ1, vσ2, v

′σ1v′′σ2, wσ1(v′σ2 + v′′σ2)),

and denote by J the set of these ideals.Set:

ρ1σ(a, b, u) =

0 wσ1 −v′σ2 −v′′σ2 0v′σ1 wσ2 0 0 −v′′σ2v

′′σ1

−v′′σ2 0 v′′σ1 0 00 0 0 v′σ1 vσ2

0 0 0 −wσ2 wσ1v′′σ1

,

ω1σ(a, b, u) =

−wσ2v′′σ1 wσ1v

′′σ1 −wσ2v

′σ2 0 v′′σ2v

′′σ1

vσ1 vσ2 v′σ1v′σ2 v′′σ2v

′′σ1 0


′′σ2 wσ1v

′σ1 0 (v′′σ2)2

0 0 0 wσ1v′′σ1 −vσ2

0 0 0 wσ2 v′σ1

,

106

ρ2σ(a, b, u) =

0 wσ1 −v′σ2 0 0v′σ1 wσ2 0 0 −v′′σ1v

′σ2

−v′′σ2 0 v′′σ1 −v′′σ1 00 0 0 v′σ1 vσ2

0 0 0 −wσ2 wσ1v′′σ1

,

ω2σ(a, b, u) =


′′σ1 −wσ2v

′σ2 0 v′′σ1v

′σ2

vσ1 vσ2 v′σ1v′σ2 v′σ2v

′′σ1 0


′′σ2 wσ1v

′σ1 v′′σ1wσ1 0

0 0 0 wσ1v′′σ1 −vσ2

0 0 0 wσ2 v′σ1

,

ρ3σ(a, b, u) =

0 wσ1 −v′σ2 −v′′σ2 0v′σ1 wσ2 0 0 −v′′σ1(v′σ2 + v′′σ2)−v′′σ2 0 v′′σ1 −v′′σ1 0

0 0 0 v′σ1 vσ2

0 0 0 −wσ2 wσ1v′′σ1

,

ω3σ(a, b, u) =


′′σ1 −wσ2v

′σ2 0 v′′σ1(v′σ2 + v′′σ2)

vσ1 vσ2 v′σ1v′σ2 v′′σ1(v′σ2 + v′′σ2) 0


′′σ2 wσ1v

′σ1 v′′σ1wσ1 (v′′σ2)2

0 0 0 wσ1v′′σ1 −vσ2

0 0 0 wσ2 v′σ1

.

Replacing v′σ2 by v′′σ2 and conversely, we get other three pairs of matrices,ρiσ(a, b, u), ωiσ(a, b, u), i = 4, 5, 6. Next, replacing wσ1 by v′′σ1 and conversely,we get other three pairs of matrices, ρiσ(a, b, u), ωiσ(a, b, u), i = 7, 8, 9, and,finally, performing the both changes, we get the pairs of matricesρiσ(a, b, u), ωiσ(a, b, u), i = 10, 11, 12.

Now let us consider the following ideals:

107

T1σ(a, b, u) = (wσ2v′′σ1, vσ1, wσ2v

′σ2, xsv

′′σ1),


′σ2, xjv

′′σ1),


′σ2, (xj + xs)v

′′σ1),


′′σ2, xsv

′′σ1),


′′σ2, xjv

′′σ1),


′′σ2, (xj + xs)v

′′σ1),

T7σ(a, b, u) = (wσ2wσ1, wσ1v′σ1, wσ2v

′σ2, xswσ1),


′σ2, xjwσ1),


′σ2, (xj + xs)wσ1),


′′σ2, xswσ1),


′′σ2, xjwσ1),


′′σ2, (xj + xs)wσ1),

We denote by T the set of these ideals and set:

µ1σ(a, b, u) =

0 −v′σ1 v′σ2 0 −xswσ1 wσ2 0 xs 0−v′′σ2 0 v′′σ1 0 0

0 0 0 v′σ1 −wσ2

0 0 0 vσ2 wσ1v′′σ1

,

ν1σ(a, b, u) =

wσ2v′′σ1 vσ1 wσ2v

′σ2 −xsv′′σ1 0

−v′′σ1wσ1 vσ2 −v′σ2wσ1 0 −xswσ2v

′′σ2 v′σ1v

′′σ2 −wσ1v

′σ1 −xsv′′σ2 0

0 0 0 wσ1v′′σ1 wσ2

0 0 0 −vσ2 v′σ1

,

µ2σ(a, b, u) =

0 −v′σ1 v′σ2 0 −xjwσ1 wσ2 0 xj 0−v′′σ2 0 v′′σ1 0 0

0 0 0 v′σ1 −wσ2

0 0 0 vσ2 wσ1v′′σ1

,

108

ν2σ(a, b, u) =


′σ2 −xjv′′σ1 0

−v′′σ1wσ1 vσ2 −v′σ2wσ1 0 −xjwσ2v

′′σ2 v′σ1v

′′σ2 −wσ1v

′σ1 −xjv′′σ2 0

0 0 0 wσ1v′′σ1 wσ2

0 0 0 −vσ2 v′σ1

,

µ3σ(a, b, u) =

0 −v′σ1 v′σ2 0 −(xj + xs)wσ1 wσ2 0 xj + xs 0−v′′σ2 0 v′′σ1 0 0

0 0 0 v′σ1 −wσ2

0 0 0 vσ2 wσ1v′′σ1

,

ν3σ(a, b, u) =


′σ2 −(xj + xs)v

′′σ1 0

−v′′σ1wσ1 vσ2 −v′σ2wσ1 0 −(xj + xs)wσ2v

′′σ2 v′σ1v

′′σ2 −wσ1v

′σ1 −(xj + xs)v

′′σ2 0

0 0 0 wσ1v′′σ1 wσ2

0 0 0 −vσ2 v′σ1

,

Replacing v′′σ2 by v′σ2 and conversely, we get other three pairs of matrices,µiσ(a, b, u), νiσ(a, b, u), i = 4, 5, 6. Next, replacing wσ1 by v′′σ1 and conversely,we get other three pairs of matrices, µiσ(a, b, u), νiσ(a, b, u), i = 7, 8, 9, and,finally, performing the both changes, we get the pairs of matricesµiσ(a, b, u), νiσ(a, b, u), i = 10, 11, 12.

Clearly, the pair of matrices (ρiσ(a, b, u), ωıσ(a, b, u)) forms a matrix fac-torization of Ω2

R(Jiσ(a, b, u)/(f)) for 1 ≤ i ≤ 12, and the pair(µiσ(a, b, u), νiσ(a, b, u)) forms a matrix factorization of Ω2

R(Tiσ(a, b, u)/(f))for 1 ≤ i ≤ 12.

Lemma 5.6. Let M be a graded non–orientable, rank 2, 5–generated MCMR–module, without free direct summands. Then there exists an ideal J ∈J ∪ T such that f ∈ J and M ∼= Ω2

(J/(f)

)or M∗ ∼= Ω2

(J/(f)

), where M∗

is the dual of M . Conversely, for every J ∈ J ∪T , the module Ω2(J/(f)

)is

a non–orientable, rank two, 5–generated MCM R–module without free directsummands.

Proof. The second statement follows easily, as we already have the matrixfactorizations above of those ideals. Let M be as above. As in the beginningof Section 4 we see that M ∼= Ω2

(J/(f)

), for J an ideal of S containing f ,

109

with µ(J) = 4, dim S/J = 2, depth S/J=1 and µ(Ω1S(J)

)= 5. We may also

suppose J = (α1, α2, α3, α4) with f ∈ (α1, α2), where αt is necessarily eitherwσt or vσt for t = 1, 2 for some a, b and a certain permutation σ as above.Clearly we cannot have, simultaneously, αt = wσt because then (α1, α2) is aprime ideal and one cannot find α3, α4 zero divisors, as we need. We treatthe following cases:

Case I: α1 = wσ1

Then we have α2 = vσ2 and (α1, α2) is the intersection of the prime ideals(v′σ2, wσ1), (v′′σ2, wσ1). Since α3, α4 must be zero divisors in S/(α3, α4) we havethe following possibilities:

(I1) α3 = v′σ2β, α4 = v′σ2γ, (I2) α3 = v′′σ2β,α4 = v′′σ2γ,(I3) α3 = v′σ2β, α4 = v′′σ2γ, (I4) α3 = v′′σ2β, α4 = v′σ2γ,

for some homogeneous β, γ from m = (x1, x2, x3, x4). In the first case wesee that the relations given by the columns of the following matrix:

vσ2 α3 α4 0 0−wσ1 0 0 γ β

0 −wσ1 0 0 −v′′σ2

0 0 −wσ1 −v′′σ2 0

,

are elements in Ω1S(J) ⊂ S4. Clearly these columns are part of the minimal

system of generators of Ω1S(J) because wσ1, v

′′σ2 form a regular system in S.

The subcase (I2) is similar, this contradicts Lemma 5.2.Suppose now (I3) holds. Then the relations given by the columns of the

following matrix

vσ2 v′σ2β v′′σ2β 0 0−wσ1 0 0 β γ

0 −wσ1 0 −v′′σ2 00 0 −wσ1 0 −v′σ2

,

are part of a minimal set of generators of Ω1S(J) (note that wσ1, v

′′σ2, v

′σ2 form

a regular system in S). Contradiction! Case (I4) is similar.Case II: α1 = vσ1, α2 = vσ2

Since (α1, α2) = (v′σ1, v′σ2) ∩ (v′σ1, v

′′σ2) ∩ (v′′σ1, v

′′σ2) ∩ (v′′σ1, v

′σ2), we see that

the zero divisors of S/(α1, α2) must be in one of the prime ideals of the abovedecomposition. Suppose α3 ∈ (v′σ1, v

′σ2). If α3 = β1v

′σ1 + β2v

′σ2 then, as in

the proof of Case III of Proposition 5.4, we see that there are at least fourminimal relations between first three α. Then all α have at least five minimalrelations. Contradiction! Thus, α3 as well α4 are multiples of one v′σt, v

′′σt. So

we have the following possibilities:

110

(II1) α3 = v′σ1β, α4 = v′σ1γ, (II2) α3 = v′′σ1β, α4 = v′′σ1γ,(II3) α3 = v′σ2β, α4 = v′σ2γ, (II4) α3 = v′′σ2β, α4 = v′′σ2γ,(II5) α3 = v′σ1β, α4 = v′′σ1γ, (II6) α3 = v′σ1β, α4 = v′′σ2γ,(II7) α3 = v′σ2β, α4 = v′′σ1γ, (II8) α3 = v′σ2β, α4 = v′′σ2γ,(II9) α3 = v′σ1β, α4 = v′σ2γ, (II10) α3 = v′′σ1β, α4 = v′′σ2γ.

Subcase: α3 = v′σ1β, α4 = v′σ1γ, (vσ2v′′σ1, γ) ∼= 1, (vσ2v

′′σ1, β) ∼= 1

We see that the relations given by the columns of the following matrix

vσ2 β γ 0 0−vσ1 0 0 α3 α4

0 −v′′σ1 0 −vσ2 00 0 −v′′σ1 0 −vσ2

,

are part of a minimal system of generators of Ω1S(J), which must be false.

Indeed, it is easy to see that the last four columns are part of a minimalsystem of generators of Ω1

S(J). If the first column belongs to the modulegenerated by the last four, then there exist λ1, λ2, λ3, λ4 ∈ S such that:

vσ2 = λ1β + λ2γ,−vσ1 = λ3v

′σ1β + λ4v

′′σ1γ,

0 = λ1v′′σ1 + λ3vσ2,

0 = λ2v′′σ1 + λ4vσ2.

It follows that vσ2 | λ1 and vσ2 | λ2 and so we obtain 1 ∈ (β, γ). Contra-diction! If (vσ2v

′′σ1, β) 6∼= 1, then we are in the subcase (II5), (II6), . . . . In the

same way we treat (II2), (II3), (II4).Subcase: α3 = v′σ1β, α4 = v′′σ1γWe see that the relations given by the columns of the following matrix

vσ2 β γ 0 0−vσ1 0 0 α3 α4

0 −v′′σ1 0 −vσ2 00 0 −v′σ1 0 −vσ2

,

are elements in Ω1S(J). The columns two and three, together with the last

two columns divided by (β, vσ2), respectively (γ, vσ2), are part of a minimalsystem of generators. Since µ

(Ω1S(J)

)= 4, we see that the first column is a

linear combination of the others, as above. Thus, there exist λ1, λ2, λ3, λ4 ∈ Ssuch that:

vσ2 = λ1β + λ2γ,−vσ1 = λ3v

′σ1β/(β, vσ2) + λ4v

′′σ1γ/(γ, vσ2),

0 = λ1v′′σ1 + λ3vσ2/(β, vσ2),

0 = λ2v′σ1 + λ4vσ2/(γ, vσ2).

111

It follows that vσ2/(β, vσ2)|λ1 and vσ2/(γ, vσ2)|λ2 and so we obtain 1 ∈(β, γ), which is false, as above, if (β, vσ2) ∼= 1, (γ, vσ2) ∼= 1. Clearly β, γcannot be multiples of vσ2 because otherwise J is only 3–generated. Theanalysis of the possibilities (β, vσ2) = v′σ2 and (β, vσ2) = v′′σ2 will lead to theconclusion that J ∈ J . In this way one can discuss all the above cases.

Theorem 5.7. Let

E1 = Coker(ρiσ(a, b, u)

), Coker

(µiσ(a, b, u)

)) | σ, a, b, u, i = 1, 6,

E2 be the set of the duals of the modules from the set E1, and E = E1 ∪ E2.

1. The set E contains only indecomposable, graded, non–orientable, 5–generated MCM R–modules of rank 2.

2. Every indecomposable, graded, non–orientable, 5–generated MCM mod-ule over R of rank 2 is isomorphic with one module of E .

3. There are 648 isomorphism classes of indecomposable, graded, non–orientable MCM modules over R of rank 2, with five generators.

Proof. (1) For the proof of indecomposability we may proceed as in thelast part of the proof of Theorem 4.18. For example, let N be the mod-ule Coker

(ρ1σ(a, b, u)

)and suppose that it decomposes. Then ρ1σ(a, b, u) is

equivalent with a direct sum of two matrices: A1, of order three and A2, oforder two. Let B1, B2 be the submatrices of ρ1σ(a, b, u) given by the firstthree lines and columns, respectively the last two lines and columns. Cer-tainly A1, A2, B1, B2 define some maximal Cohen-Macaulay modules of rankone that we denote, respectively, by N1, N2, T1, T2, and due to the particularform of ρ1σ(a, b, u) we have the following exact sequence

0→ T1 → N1 ⊕N2 = N → T2 → 0.

Since ρ1σ(a, b, u) is modulo xj the sum of B1, B2, Ti/xjTi ∼= Ni/xjNi fori = 1, 2. Looking at the description of rank one maximal Cohen-Macaulaymodules we can see that Ai is equivalent with Bi modulo xj only when Ai isequivalent with Bi. Thus Ti ∼= Ni for i = 1, 2 and so N ∼= T1 ⊕ T2. By [Mi],this happens only if the above exact sequence splits, that is impossible.(2) It is enough to observe that the matrices

ρiσ(a, b, u), a, b, u, σ, i = 1, 12

112

andµiσ(a, b, u), a, b, u, σ, i = 1, 12,

are pairwise equivalent. Indeed, one may show that

ρ7σ(a, b, u) ∼ ρ4σ(a, b, u), ρ8σ(a, b, u) ∼ ρ5σ(a, b, u), ρ9σ(a, b, u) ∼ ρ6σ(a, b, u),

and

ρ10σ(a, b, u) ∼ ρ1σ(a, b, u), ρ11σ(a, b, u) ∼ ρ2σ(a, b, u), ρ12σ(a, b, u) ∼ ρ3σ(a, b, u).

One may find, in each case, a pair of some permutations matrices Ui, Vi suchthat

Uiρiσ(a, b, u) = ρ(i−3)σ(au, b, u)Vi, i = 7, 8, 9,

andUiρiσ(a, b, u) = ρ(i−9)σ(au, b, u)Vi, i = 10, 11, 12.

In a similar way one may group in pairs the matrices µiσ(a, b, u).(3) It is a laborious task to prove the that the modules of the list E arepairwise non–isomorphic. One can use the following procedure in Singular:

LIB"matrix.lib";

option(redSB);

proc isomorph5(matrix X, matrix Y)

matrix U[5][5]=u(1..25);

matrix V[5][5]=v(1..25);

matrix C=U*X-Y*V;

ideal I=flatten(C);

ideal J=det(U)-1,det(V)-1;

for (int j=1;j<=size(I);j++)

J=J+transpose(coef(I[j],x(1)*x(2)*x(3)*x(4)))[2];

ideal K=std(J);

return(K);

113

Corollary 5.8. Let

F1 = Coker(ωiσ(a, b, u)

), Coker

(νiσ(a, b, u)

)| σ, a, b, u, i = 1, 6,

F2 be the set of the duals of the modules from the set F1, and F = F1 ∪ F2.

1. The set F contains only indecomposable, graded, non–orientable, 5–generated MCM R–modules of rank 3.

2. Every indecomposable, graded, non–orientable, 5–generated MCM mod-ule over R of rank 3 is isomorphic with one module of F .

3. There are 648 isomorphism classes of indecomposable, graded, non–orientable MCM modules over R of rank 3, with 5 generators.

Proof. The map M 7→ Ω1R(M) is a bijection between the 5–generated, in-

decomposable, graded, MCM R–modules of rank 2 and the 5–generated,indecomposable, graded, MCM R–modules of rank 3.

Remark 5.9. For each 2-gen MCM module M (line or conic) there exist twonon-isomorphic 3-gen MCM modules P1, P2 and 3 non-isomorphic extensionsfor each:

0→ Pi → Eij →M → 0,

i = 1, 2, j = 1, 2, 3. So there are 6× 54 MCM of type Eij. Taking the dualswe get another 6× 54 MCM. Thus all are 648 = 12× 54.

Lemma 5.10. There exist no graded, indecomposable, non–orientable, rank2, 6–generated MCM modules.

Proof. Suppose there exist such MCM module M . Then M ∼= Ω2R

(J/(f)

)for

a certain 5–generated ideal J = (α1, α2, α3, α4, α5) of S as hinted at in thefirst part of Section 4. Then any four elements from the αt must generate anideal J ′′ in J ∪ T because, otherwise, µ(Ω1

S

(J ′′/(f)

)> 4 and so, obviously

µ(Ω1S(J/(f)) > 5. So we may suppose αt = vσt for t = 1, 2 and after some

permutations α3 = v′σ1v′′σ2. Set J ′ = (α1, α2, α3). If (J ′, α4) ∈ J . and

(J ′, α5) ∈ J then there are 4 minimal relations of (J ′, α4) and 4 minimalrelations of (J ′, α5) over S, among them at least 6 minimal relations of Jover S which contradicts Lemma 5.2. In the same way we treat the othercases.

Corollary 5.11. There exist no indecomposable, graded, non–orientable,rank 4, 6–generated MCM modules.

114

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118

Curriculum Vitae

Corina Baciu

Geboren am 13. Juli 1977 (Calarasi, Rumanien)

Ausbildung

Juli 1995 Abitur am Liceul Mihai Eminescu Calarasi, SchwerpunktePhysik und Mathematik

Okt. 1995–Juni 1999 Studium der Mathematik an der Universitat Bukarest, in einerForschungsgruppe

Okt. 1999–Aug. 2001 Diplomarbeit an der Technischen Universitat Kaiserslautern, Ar-beitsgruppe Algebraische Geometrie und Computer Algebra; Be-treuer der Diplomarbeit: Prof. Dr. G. Pfister

Okt. 2001–Aug. 2005 Doktorarbeit an der Technischen Universitat Kaiserslautern; Be-treuer: Prof. Dr. G. Pfister

Juni 2002–Juni 2005 Forderung durch den DFG–Schwerpunkt ,Globale Methoden inder komplexen Geometrie’

Veroffentlichungen

Maximal Cohen-Macaulay modules over a non–isolated hypersurface singularity, PreprintNr. 337, Technische Universitat Kaiserslautern, 2003Maximal Cohen-Macaulay modules and stable vector bundles, Proceedings of the NATOAdvanced Research Workshop on Computational Commutative and Non– CommutativeAlgebraic Geometry, Chisinau, Moldova, 6–11 June 2004Rank 2 Cohen–Macaulay modules over singularities of type x3

1 + x32 + x3

3 + x34, zusammen

mit V. Ene, G. Pfister und D. Popescu, wird erscheinen im Journal of Algebra

Curriculum Vitae

Corina Baciu

Born on July 13, 1977 (Calarasi, Romania)

Education

Sept. 1991–July 1995 Studies at Liceul Mihai Eminescu Calarasi, specializations:physics and mathematics

Oct. 1995–Juni 1999 Mathematical studies at the University of Bucharest, Romania,in a research group

Oct. 1999–Aug. 2001 Master studies at the Technical University Kaiserslautern, De-partement of Algebraic Geometriy and Computer Algebra, underthe supervision of Prof. Dr. G. Pfister

Oct. 2001–Aug. 2005 Ph. D. studies at the Technical University Kaiserslautern underthe supervision of Prof. Dr. G. Pfister

June 2002–June 2005 Research supported by the DFG–Schwerpunkt ”Globale Metho-den in der komplexen Geometrie”

Publications

Maximal Cohen-Macaulay modules over a non–isolated hypersurface singularity, PreprintNr. 337, Technische Universitat Kaiserslautern, 2003Maximal Cohen-Macaulay modules and stable vector bundles, Proceedings of the NATOAdvanced Research Workshop on Computational Commutative and Non– CommutativeAlgebraic Geometry, Chisinau, Moldova, 6–11 June 2004Rank 2 Cohen–Macaulay modules over singularities of type x3

1 + x32 + x3

3 + x34, together

with V. Ene, G. Pfister and D. Popescu, to appear in Journal of Algebra

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