aa r X i v : . [ m a t h . A C ] F e b MATRIX FACTORIZATIONS WITH MORE THAN TWO FACTORS
TIM TRIBONE
Abstract.
Given an element f in a regular local ring, we study matrix factorizations of f with d ≥ ϕ , ϕ , . . . , ϕ d ) such that their productis f times an identity matrix of the appropriate size. Several well known properties of matrixfactorizations with 2 factors extend to the case of arbitrarily many factors. For instance, we showthat the stable category of matrix factorizations with d ≥ Introduction
Let S be a regular local ring and f a non-zero non-unit in S . A matrix factorization of f isa pair of n × n matrices ( ϕ, ψ ) such that ϕψ = f · I n , where I n is the identity matrix of size n . The correspondence given by Eisenbud [Eis80, Corollary 6.3] between matrix factorizationsand maximal Cohen-Macaulay modules has made these objects an important tool for studyinghypersurface rings. We consider a natural generalization in which the factorizations have two ormore factors. In other words, we consider tuples of n × n matrices ( ϕ , ϕ , . . . , ϕ d ), for some d ≥ ϕ ϕ · · · ϕ d = f · I n .Several authors, including [Chi78, BHS88, HUB91, BES17] and also [Ber06, BL07], have studiedthe existence of matrix factorizations with more than 2 factors, though, both sets of authors studyspecial cases of the definition we present. The first set of authors focus their attention mainly onmatrices which have only linear entries; factorizations of this type correspond to Ulrich modules overthe hypersurface defined by f . The second set of authors tackle a different question. Specifically,Bergman asked in [Ber06] whether the adjoint (the matrix of cofactors) of the generic n × n matrixfactors into a product of two non-invertible n × n matrices.Our approach is to study the category of matrix factorizations of f with d ≥ dS ( f ). In Section 2 we examine the basic structure of MF dS ( f ). We show thatMF dS ( f ) is a Frobenius category and therefore its stable category is naturally triangulated. In thecase d = 2, the stable category we present coincides with the one originally studied by Eisenbud.In particular, it is equivalent to the stable category of maximal Cohen-Macaulay R = S/ ( f )-modules. In Sections 3 and 4 we extend results of Kn¨orrer and Solberg which give module theoreticdescriptions of MF dS ( f ) for all d ≥
2. Specifically, Theorem 4.4 extends [Kn¨o87, Proposition 2.1] byidentifying the category of matrix factorizations with d ≥ d -fold branched cover of R . Theequivalence of categories given by Kn¨orrer, which we generalize, is an important initial step inthe classification of simple hypersurface singularities. In Section 5, we investigate the suspensionfunctor on the stable category of matrix factorizations with d ≥ S ( f ) and MF dS ( f ) for d > Definitions and Background
In this section we collect the definitions, notation, and conventions that we will use throughout.
Definition 1.1.
Let S be a regular local ring, f a non-zero non-unit in S , and d ≥ matrix factorization of f with d factors is a d -tuple of homomorphisms between finitely generatedfree S -modules of the same rank, ( ϕ : F → F , ϕ : F → F , . . . , ϕ d : F → F d ), such that ϕ ϕ · · · ϕ d = f · F . Depending on the context, we may omit the free S -modules in the notation and simply write( ϕ , ϕ , . . . , ϕ d ). If the free S -modules F , . . . , F d are of rank n , we say ( ϕ , ϕ , . . . , ϕ d ) is of size n .As we will see, there is an inherent cyclic nature to matrix factorizations with d factors. It willtherefore be convenient to adopt the following notation conventions. Notation 1.2.
The letter d will always be an integer indicating the number of factors in a matrixfactorization. When d is clear from context, all indices are taken modulo d unless otherwise specified.More specifically, let i = j ∈ Z d and let A , A , . . . , A d be symbols indexed over Z d . Let ˜ i and ˜ j be integer representatives of i, j within the range 0 < ˜ i, ˜ j ≤ d . The notation A i A i +1 · · · A j will betaken to mean ( A i A i +1 · · · A j − A j if ˜ i ≤ ˜ jA i A i +1 · · · A d A · · · A j − A j if ˜ i ≥ ˜ j . We follow a similar convention for indexing a decreasing list of symbols over Z d .A non-zero finitely generated module M over a local ring A is called maximal Cohen-Macaulay (MCM) if depth A ( M ) = dim A (the Krull dimension of A ). Our first observation is that matrixfactorizations of f with d factors encode MCM modules over the hypersurface ring R = S/ ( f ), andthat Definition 1.1 is more symmetric than it seems. Lemma 1.3.
Let S be a regular local ring and f a non-zero non-unit in S . Let ( ϕ : F → F , ϕ : F → F , . . . , ϕ d : F → F d ) be a matrix factorization of f with d ≥ factors. For any k ∈ Z d ,(i) ϕ k ϕ k +1 · · · ϕ k − = f · F k , and(ii) if cok ϕ k = 0 , then cok ϕ k is an MCM R -module.Proof. (i) We proceed by induction on d ≥
2. For the case d = 2, we simply need to show that ϕψ = f · F implies ψϕ = f · G . This fact is well known but we include the argumenthere for clarity. Suppose ( ϕ : G → F, ψ : F → G ) is a matrix factorization with 2 factors.Since f is a non-zero element in the domain S , it follows that both ϕ and ψ are injective.Canceling ϕ on the left of the equation ϕψϕ = f · ϕ = ϕ · f , we find ψϕ = f · G .Now, assume d > d factors. Let k ∈ Z d and notice that, by viewing the composition ϕ k ϕ k +1 : F k +2 → F k asa single homomorphism, the ( d − ϕ , ϕ , . . . , ϕ k − , ϕ k ϕ k +1 , ϕ k +2 , . . . , ϕ d )is a matrix factorization of f with d − ϕ k ϕ k +1 · · · ϕ k − = f · F k .(ii) Let k ∈ Z d . By (i), we have that ϕ k ϕ k +1 · · · ϕ k − = f · F k . In particular, f · cok ϕ k = 0,that is, cok ϕ k is an R -module. Also, as in (i), the homomorphism ϕ k is injective since f ∈ S is non-zero. Thus, we have a short exact sequence0 F k +1 F k cok ϕ k , ϕ k hich implies that pd S (cok ϕ k ) ≤
1. By the Auslander-Buchsbaum formula, we have thatdepth(cok ϕ k ) = dim( S ) − pd S (cok ϕ k ) ≥ dim( S ) − R ) . That is, cok ϕ k is an MCM R -module. (cid:3) Definition 1.4.
Let S be a regular local ring and f a non-zero non-unit in S . Let X = ( ϕ : F → F , . . . , ϕ d : F → F d ) and X ′ = ( ϕ ′ : F ′ → F ′ , . . . , ϕ ′ d : F ′ → F ′ d ) be matrix factorizations of f .(i) A morphism of matrix factorizations between X and X ′ is a d -tuple of S -module homo-morphisms, α = ( α , α , . . . , α d ), making each square of the following diagram commute: F F d · · · F F F ′ F ′ d · · · F ′ F ′ . α ϕ d α d ϕ d − ϕ α ϕ α ϕ ′ d ϕ ′ d − ϕ ′ ϕ ′ Composition of morphisms is defined component-wise, that is, if α = ( α , . . . , α d ) : X → X ′′ and β = ( β , . . . , β d ) : X ′ → X are morphisms of matrix factorizations, then α ◦ β =( α β , α β , . . . , α d β d ) : X ′ → X ′′ . The matrix factorizations X and X ′ are isomorphic ifthere exists a morphism α = ( α , . . . , α d ) : X → X ′ such that α k is an isomorphism foreach k ∈ Z d .(ii) Let MF dS ( f ) denote the category of matrix factorizations of f with d factors. The additivestructure on MF dS ( f ) is given by the direct sum of X and X ′ X ⊕ X ′ := ( ϕ ⊕ ϕ ′ : F ⊕ F ′ → F ⊕ F ′ , . . . , ϕ d ⊕ ϕ ′ d : F ⊕ F ′ → F d ⊕ F ′ d ) . (iii) Motivated by Lemma 1.3, we define functors T j : MF dS ( f ) → MF dS ( f ), j ∈ Z d , by T j ( ϕ , ϕ , . . . , ϕ d ) = ( ϕ j +1 , ϕ j +2 , . . . , ϕ j − , ϕ j )and T j ( α , α , . . . , α d ) = ( α j +1 , α j +2 , . . . , α j − , α j )for any ( α , α , . . . , α d ) ∈ Hom MF dS ( f ) ( X, X ′ ). We refer to T = T as the shift functor onMF dS ( f ). Definition 1.5.
Let S be a regular local ring with maximal ideal n and let f be a non-zero non-unitin S . Set R = S/ ( f ) and fix d ≥ R ) denote the category of maximal Cohen-Macaulay R -modules.(ii) An R -module M is stable if it has no direct summands isomorphic to R .(iii) For an R -module M , let syz R ( M ) denote the reduced first syzygy of M .(iv) A matrix factorization X = ( ϕ , ϕ , . . . , ϕ d ) is called stable if cok ϕ k is a stable R -modulefor all k ∈ Z d .(v) A homomorphism between free S -modules ϕ : G → F is called minimal if Im ϕ ⊆ n F .(vi) A matrix factorization ( ϕ , ϕ , . . . , ϕ d ) ∈ MF dS ( f ) is called reduced if ϕ k is minimal for all k ∈ Z d .(vii) A non-zero matrix factorization X ∈ MF dS ( f ) is indecomposable if X ∼ = X ′ ⊕ X ′′ implies X ′ = 0 or X ′′ = 0.Eisenbud showed in [Eis80] that there is a one-to-one correspondence between reduced matrixfactorizations (with 2 factors) and stable MCM R -modules. This correspondence can be realizedin the form of a stable equivalence between the categories MF S ( f ) and MCM( R ). We record twoconsequences that will be needed later. Proposition 1.6. [Eis80]
Let S be a regular local ring, f a non-zero non-unit in S , and set R = S/ ( f ) . i) For any MCM R -module M , there exists a matrix factorization ( ϕ, ψ ) ∈ MF S ( f ) such that ϕ is minimal and cok ϕ ∼ = M .(ii) A matrix factorization ( ϕ, ψ ) ∈ MF S ( f ) is reduced if and only if it is stable. In this case, syz R (cok ϕ ) ∼ = cok ψ and syz R (cok ψ ) ∼ = cok ϕ . We will see in Section 5 that only one direction of (ii) holds when d >
2. Finally, we stateanother observation, also made by Eisenbud, that will help us identify matrix factorizations.
Lemma 1.7. [Eis80, Corollary 5.4]
Let S be a regular local ring and f a non-zero non-unit in S .Suppose A : G → F and B : F → G are homomorphisms of finitely generated free S -modules suchthat AB = f · F . Then, rank F = rank G .Proof. Once again, since f is a non-zerodivisor, the maps A and B are injective. Since f · cok A = 0,cok A is a torsion S -module. From the short exact sequence0 G F cok A A it follows that rank S F = rank S G . (cid:3)
2. MF dS ( f ) is a Frobenius Category In this section we show that there is a natural choice of an exact structure on the categoryMF dS ( f ) which induces the structure of a triangulated category on the stable category MF dS ( f )defined below. We start by recalling the axioms that define an exact category. The axioms anddefinitions below follow the presentation given in [B¨u10] and we refer the reader to this paper formore information on exact categories.Let A be an additive category. A pair of composable morphisms A ′ A A ′′ i p is calleda kernel-cokernel pair if i is a kernel of p and p is a cokernel of i . Given a collection of kernel-cokernel pairs, E , we call a morphism i : A ′ → A an admissible monomorphism if there exists amorphism p : A → A ′′ such that A ′ A A ′′ i p is an element of E . Dually, a morphism p : A → A ′′ is an admissible epimorphism is there exists a morphism i : A ′ → A such that theircomposition is in E . We will indicate admissible monomorphisms and admissible epimorphisms bythe arrows and ։ respectively.An exact structure on A is a class E of kernel-cokernel pairs which is closed under isomorphismsand such that the following axioms hold:(E0) The identity morphism 1 X is an admissible monomorphism for all X ∈ A .(E0 op ) The identity morphism 1 X is an admissible epimorphism for all X ∈ A .(E1) Admissible monomorphisms are closed under composition.(E1 op ) Admissible epimorphisms are closed under composition.(E2) The push-out of an admissible monomorphism X Y and an arbitrary morphism X → X ′ exists and induces an admissible monomorphism as in the diagram X YX ′ Y ′ . (E2 op ) The pull-back of an admissible epimorphism X ′′ ։ Y ′ and an arbitrary morphism Y → Y ′ exists and induces an admissible epimorphism as in the diagram X YX ′ Y ′ . iven an additive category A and a class E satisfying these axioms, the pair ( A , E ) is called an exact category .Let S be a regular local ring, f ∈ S a non-zero non-unit, and set R = S/ ( f ). Fix an integer d ≥ X = ( ϕ : F → F , . . . , ϕ d : F → F d ) , X ′ = ( ϕ ′ : F ′ → F ′ , . . . , ϕ ′ d : F ′ → F ′ d ) , and X ′′ = ( ϕ ′′ : F ′′ → F ′′ , . . . , ϕ ′′ d : F ′′ → F ′′ d ) be matrix factorizations in MF dS ( f ). Definition 2.1.
Suppose we have a pair of morphisms α = ( α , . . . , α d ) : X → X ′′ and β =( β , . . . , β d ) : X ′ → X in MF dS ( f ). Then, the composition X ′ X X ′′ β α is called a short exact sequence of matrix factorizations if the sequence0 F ′ k F k F ′′ k β k α k is a short exact sequence of free S -modules for each k ∈ Z d . Lemma 2.2.
A short exact sequence of matrix factorizations is a kernel-cokernel pair in MF dS ( f ) .Proof. Let X ′ X X ′′ β α be a short exact sequence of matrix factorizations. First weshow that β is the kernel of α . Certainly, we have that αβ = 0. Suppose g : Y → X is anothermorphism such that αg = 0, where Y = ( ψ : G → G , . . . , ψ d : G → G d ). Let k ∈ Z d . We havethe following diagram of free S -modules G k F ′ k F k F ′′ k ˜ g k g k β k α k where the bottom row is exact. Since β k is the kernel of α k , there exists a unique S -homomorphism˜ g k : G k → F ′ k such that β k ˜ g k = g k . It suffices to show that ˜ g = (˜ g , ˜ g , . . . , ˜ g d ) : Y → X ′ is amorphism of matrix factorizations since each ˜ g k , k ∈ Z d , is uniquely determined. That is, we needto show that the diagram G k +1 G k F ′ k +1 F ′ k ˜ g k +1 ψ k ˜ g k ϕ ′ k commutes for all k ∈ Z d . Note that g k ψ k = ϕ k g k +1 and ϕ k β k +1 = β k ϕ ′ k since g and β are morphismsin MF dS ( f ). Then, β k ˜ g k ψ k = g k ψ k = ϕ k g k +1 = ϕ k β k +1 ˜ g k +1 = β k ϕ ′ k ˜ g k +1 and since β k is injective,we can cancel it on the left to conclude that ˜ g k ψ k = ϕ ′ k ˜ g k +1 as desired. Hence, ˜ g is the uniquemorphism such that β ◦ ˜ g = ( β ˜ g , . . . , β d ˜ g d )= ( g , . . . , g d )= g. This completes the proof that β is a kernel of α . The proof that α is a cokernel of β is similar. (cid:3) Let E d denote the class of short exact sequences of matrix factorizations in MF dS ( f ). The firstfour axioms of an exact category are satisfied by the pair (MF dS ( f ) , E d ) directly from the definitions.The axioms [E2] and [E2 op ] also hold, which we will show below. Before we do, we need to knowmore about the form of the admissible morphisms in (MF dS ( f ) , E d ). Lemma 2.3.
Let γ = ( γ , . . . , γ d ) : X → X ′′ be a morphism of matrix factorizations. γ is an admissible epimorphism if and only if the S -homomorphisms γ , . . . , γ d are surjec-tions.(2) γ is an admissible monomorphism if and only if the S -homomorphisms γ , . . . , γ d are splitinjections.Proof. We prove only (2) as the proof of (1) is similar. Suppose γ is an admissible monomor-phism. Then, there exists an admissible epimorphism π = ( π , π , . . . , π d ) : X → X ′′ such that X ′ X X ′′ γ π is a short exact sequence of matrix factorizations. In particular,0 F ′ k F k F ′′ k γ k π k is a short exact sequence of S -modules for each k ∈ Z d . Since F ′′ k is free, this sequence is split andtherefore γ k is a split injection.Conversely, suppose the homomorphisms γ , . . . , γ d are each split injections. For k ∈ Z d , set F ′′ k := cok γ k and π k : F k → F ′′ k the natural projection map. Notice that F ′′ k is a free S -module ofrank equal to rank F k − rank F ′ k . Now, for each k ∈ Z d , there exists a map ϕ ′′ k : F k +1 → F k suchthat the following diagram with split exact rows commutes:0 F ′ F F ′′ F ′ d F d F ′′ d F ′ F F ′′ F ′ F F ′′ . ϕ ′ d γ π ϕ d ϕ ′′ d γ d ϕ ′ d − π d ϕ d − ϕ ′′ d − ϕ ′ ϕ ϕ ′′ γ ϕ ′ π ϕ ϕ ′′ γ π In particular, there exists t : F ′′ → F such that π t = 1 F ′′ . The splitting allows us to computethe composition along the right most column: ϕ ′′ ϕ ′′ · · · ϕ ′′ d = π ϕ ϕ · · · ϕ d t = f · π t = f · F ′′ . Since the free modules F ′′ , F ′′ , . . . , F ′′ d are all of the same rank, we have that X ′′ = ( ϕ ′′ : F ′′ → F ′′ , . . . , ϕ ′′ d : F ′′ → F ′′ d ) ∈ MF dS ( f ) and X ′ X X ′′ γ ( π ,...,π d ) is a short exact sequence of matrix factorizations. (cid:3) Lemma 2.3 indicates that not every monomorphism of matrix factorizations is an admissiblemonomorphism. The simplest example of this arises when d = 2. Example 2.4.
Suppose ( ϕ : G → F, ψ : F → G ) ∈ MF S ( f ) with cok ψ = 0. Then, the tuple( ψ, G ) forms a morphism between the matrix factorizations ( ϕ, ψ ) → ( f · G , G ). This morphismis a monomorphism, in the sense that it can be cancelled on the left, but it is not admissible sincethe cokernel of ψ is not a free S -module. he same is true of epimorphisms, that is, there are epimorphisms that are not admissible.For ( ϕ, ψ ) ∈ MF S ( f ) with cok ψ = 0, the tuple (1 F , ψ ) forms a morphism between the matrixfactorizations ( f · F , F ) → ( ϕ, ψ ). If ( a, b ) ◦ (1 F , ψ ) = ( a ′ , b ′ ) ◦ (1 F , ψ ) for some morphisms( a, b ) , ( a ′ , b ′ ), then a = a ′ and bψ = b ′ ψ . We can pre-compose both sides of the second equationwith ϕ to get b · f = b ′ · f , hence b = b ′ . So, (1 F , ψ ) can be cancelled on the right but is notadmissible epimorphism since ψ is not surjective.Actually, further inspection of these examples shows they are both monomorphisms and bothepimorphisms but neither is admissible of either type. In particular, neither is an isomorphism. InAbelian category, a monomorphism which is also an epimorphism must be an isomorphism. Similarexamples can be constructed for all d > dS ( f ) is notAbelian for any d ≥ Proposition 2.5.
The collection E d of short exact sequences of matrix factorizations in MF dS ( f ) satisfies the axioms [E2] and [E2 op ] .Proof. We will show that [E2] holds. The proof that [E2 op ] is satisfied is similar. Suppose we havea diagram in MF dS ( f )(2.1) X YX ′ β q where Y = ( ψ : G → G , . . . , ψ d : G → G d ), β = ( β , . . . , β d ), and q = ( q , . . . , q d ). Let k ∈ Z d .We may take the push-out of the injection q k and the map β k which yields the diagram0 F k G k cok q k F ′ k P k cok q k . β k q k p k α k We make the following observations from this diagram: Since the morphism q is an admissiblemonomorphism, the map q k is a split injection. Hence, cok q k is a free S -module. It follows thatthe bottom sequence also splits and so P k is free with rank S P k = rank S F ′ k + rank S G k − rank S F k .This also implies that α k is a split injection. Since k was arbitrary, this yields d free S -modules P , P , . . . , P d , each of the same rank, and d -tuples α = ( α , . . . , α d ) and p = ( p , . . . , p d ).Next, let k ∈ Z d and consider the diagram F k G k F ′ k P k P k − . β k q k p k p k − ψ k − α k − ϕ ′ k − α k χ k − There is a unique homomorphism χ k − : P k → P k − depicted above since p k − ψ k − q k = p k − q k − ϕ k − = α k − β k − ϕ k − = α k − ϕ ′ k − β k . In particular, the map χ k − is given by χ k − (( a k ′ , b k )) = α k − ϕ ′ k − ( a k ) + p k − ψ k − ( b k ) = ( ϕ ′ k − ( a k ) , ψ k − ( b k )) ∈ P k − , or any ( a k , b k ) ∈ F ′ k ⊕ G k . In other words, χ k − is the map induced by the direct sum ϕ ′ k − ⊕ ψ k − on the quotients P k → P k − . These maps link together to form a sequence of compositions P P d · · · P P . χ d χ d − χ χ From the explicit description of χ k we have that χ χ · · · χ d = f · P . Since the free S -modules P , . . . , P d are all of the same rank, it follows that Y ′ = ( χ : P → P , · · · , χ d : P → P d ) is amatrix factorization of f .It is not hard to see that α : X ′ → Y ′ and p : Y → Y ′ form morphisms of matrix factorizationsand that these morphisms render (2.1) a commutative square. As we showed above, the map α k is asplit injection for all k ∈ Z d . Hence, α is an admissible monomorphism by Lemma 2.3. To finish theproof, it suffices to check the necessary universal property which we omit as it is a straightforwardcomputation. (cid:3) Corollary 2.6.
The pair (MF dS ( f ) , E d ) is an exact category. (cid:3) With the exact structure on MF dS ( f ) fixed, we will often omit reference to E d . We proceed nowwith the main goal of this section: to show that the exact category MF dS ( f ) is a Frobenius category.First, we recall the necessary definitions which can also be found in [B¨u10].An object in an exact category ( A , E ) is called projective , respectively injective , if it satisfies theusual lifting property with respect to admissible epimorphisms, respectively admissible monomor-phisms. The pair ( A , E ) is said to have enough projectives if for every object X ∈ A , there is anadmissible epimorphism P ։ X with P projective. Dually, ( A , E ) has enough injectives if for everyobject X ∈ A , there is an admissible monomorphism X I with I injective. The exact category( A , E ) is said to be a Frobenius category if it has enough projectives, enough injectives, and theclasses of projective objects and injective objects coincide.
Definition 2.7.
For each i ∈ Z d , let P i denote the matrix factorization of size 1 whose i thcomponent is multiplication by f on S while the rest are the identity on S . In other words, P i is given by the composition S S d · · · S i +1 S i · · · S S
11 1 1 f where S k = S for each k ∈ Z d . We also set P = L i ∈ Z d P i . Lemma 2.8.
Let X ∈ MF dS ( f ) and j ∈ Z d . Then, X is projective (respectively injective) if andonly if T j ( X ) is projective (respectively injective).Proof. Suppose X is projective and j ∈ Z d . Let α = ( α , . . . , α d ) : X ′ → X ′′ be an admissiblemorphism and let p = ( p , . . . , p d ) : T j ( X ) → X ′′ be any morphism. Then, we have morphisms T − j ( α ) : T − j ( X ′ ) → T − j ( X ′′ ) and T − j ( p ) : X → T − j ( X ′′ ). The characterization of admissibleepimorphisms in Lemma 2.3 implies that T − j ( α ) is also an admissible epimorphism. Since X isprojective, there exists q = ( q , q , . . . , q d ) : X → T − j ( X ′ ) such that T − j ( α ) q = T − j ( p ). Applying T j we find that αT j ( q ) = p implying that T j ( X ) is projective. The proof of the converse and bothdirections regarding injectivity are similar. (cid:3) Lemma 2.9.
The matrix factorizations P , P , . . . , P d are both projective and injective in MF dS ( f ) .Proof. Directly from the definition we see that T j ( P i ) = P i − j for any i, j ∈ Z d . Therefore, becauseof Lemma 2.8, it suffices to show that P is both projective and injective. We shall start by showingthat P is projective. uppose α = ( α , . . . , α d ) : X → X ′′ is an admissible epimorphism and p = ( p , . . . , p d ) : P → X ′′ is an arbitrary morphism. We would like to complete the diagram(2.2) P X X ′′ pqα One component of this diagram is the following diagram of free S -modules SF F ′′ . q p α Since S is free and α is surjective, there exists a map q : S → F such that α q = p . We can usethis map to construct a morphism of matrix factorizations P → X which makes (2.2) commute.Let q = ( q , ϕ ϕ · · · ϕ d q , ϕ · · · ϕ d q , . . . , ϕ d q ). The fact that q forms a morphism P → X canbe seen in the following diagram of S -modules S S · · ·
S S SF F d · · · F F F . q ϕ d q ϕ ··· ϕ d q fϕ ϕ ··· ϕ d q q ϕ d ϕ d − ϕ ϕ ϕ Finally, αq = ( α q , α ϕ · · · ϕ d q , . . . , α d ϕ d q )= ( p , ϕ ′′ · · · ϕ ′′ d α q , · · · , ϕ ′′ d α q )= ( p , ϕ ′′ · · · ϕ ′′ d p , · · · , ϕ ′′ d p )= ( p , p , . . . , p d )= p which implies that P is projective.In order to show that P is an injective matrix factorization, let β = ( β , . . . , β d ) : X ′ X bean admissible monomorphism and a = ( a , . . . , a d ) : X ′ → P be an arbitrary morphism. We wouldlike to complete the diagram X ′ X P a βb Since β is an admissible monomorphism, each component β k is split. In particular, there exists amap t : F ′ → F such that tβ = 1 F . This splitting allows us to build the morphism b = ( a tϕ ′ · · · ϕ ′ d , a t, a tϕ ′ , . . . , a tϕ ′ · · · ϕ ′ d − ) : X → P and this morphism satisfies bβ = a . (cid:3) In Lemma 5.10, we will see that P , P , . . . , P d are the only indecomposable projective (respec-tively injective) matrix factorizations up to isomorphism.The next step is to show that MF dS ( f ) has enough projectives and enough injectives. Along theway, we construct the syzygy and cosyzygy of a matrix factorization and give explicit formulas foreach. emark 2.10. Let X ∈ MF dS ( f ). By tensoring with R = S/ ( f ), which we denote here by (cid:3) = (cid:3) ⊗ S R , one can associate to X an infinite chain of free R -modules: · · · F F F d · · · F F · · · ϕ ϕ ϕ d ϕ d − ϕ ϕ ϕ d In the case d = 2, this chain is a acyclic and, by truncating appropriately, it forms a free resolutionover R of cok ϕ (or of cok ϕ ). However, if d >
2, this chain is not acyclic. In fact, it is not evena complex. Instead, it is precisely an acyclic d - complex (see [IKM17]). With this perspective inmind, it is likely that the formulas given below can be obtained as lifted versions of the ones foundin [IKM17, Section 2]. We give explicit constructions without referencing any such lifting. Construction 2.11.
Let X = ( ϕ : F → F , . . . , ϕ d : F → F d ) ∈ MF dS ( f ) be of size n . Weconstruct short exact sequences K P X and
X I K ′ with P projective and I injective and give explicit formula for the resulting syzygy and the cosyzygyof X .To start, set b F k = L d − i =1 F k + i . For each k ∈ Z d , define S -homomorphisms D k : F k +1 ⊕ b F k +1 → F k ⊕ b F k and D ′ k : F k +1 ⊕ b F k +1 → F k ⊕ b F k by D k ( a k +1 , a k +2 , . . . , a k − , a k ) = ( f a k , a k +1 , . . . , a k − )and D ′ k ( a k +1 , a k +2 , . . . , a k − , a k ) = ( a k , f a k +1 , a k +2 , . . . , a k − )for all a i ∈ F i , i ∈ Z d . Set P ( X ) = ( D , D , . . . , D d ) and I ( X ) = ( D ′ , D ′ , . . . , D ′ d ). Then, the d -tuples P ( X ) and I ( X ) form matrix factorizations of f both isomorphic to L di =1 P ni .For each i, k ∈ Z d , define a homomorphism θ Xki : F i → F k given by θ Xki = ( F k i = kϕ k ϕ k +1 · · · ϕ i − ϕ i − i = k. Then, for each k ∈ Z d , define Θ Xk : b F k → F k and Ξ Xk : F k → b F k byΘ Xk ( a k +1 , a k +2 , . . . , a k − ) = X i = k θ Xki ( a i )and Ξ Xk ( a k ) = (cid:16) θ X ( k +1) k ( a k ) , θ X ( k +2) k ( a k ) , . . . , θ X ( k − k ( a k ) (cid:17) . Let k ∈ Z d and consider the following diagram:(2.3) 0 b F k +1 F k +1 ⊕ b F k +1 F k +1 b F k F k ⊕ b F k F k Ω k ǫ Xk +1 D k ρ Xk +1 ϕ k ǫ Xk ρ Xk where ρ Xk = (cid:0) F k Θ Xk (cid:1) and ǫ Xk = (cid:18) − Θ Xk b F k (cid:19) . The rows are split exact sequences of free S -modulesand one can check that right most square commutes by recalling that ϕ k θ ( k +1) i = θ ki for i = k and ϕ k θ ( k +1) k = f · F k . Thus, there is an induced map Ω k : b F k +1 → b F k as depicted. Since the rows are plit, Ω k can be computed by using the splitting, that is, Ω k = π k D k ǫ k +1 where π k : F k ⊕ b F k → b F k is projection onto b F k . In particular,Ω k ( a k +2 , a k +3 , . . . , a k − , a k ) = π k D k ǫ k +1 ( a k +2 , a k +3 , . . . , a k − , a k )= π k D k − X i = k +1 θ ( k +1) i ( a i ) , a k +2 , . . . , a k − , a k = π k f a k , − X i = k +1 θ ( k +1) i ( a i ) , a k +2 , . . . , a k − = − X i = k +1 θ ( k +1) i ( a i ) , a k +2 , . . . , a k − and therefore we can represent the components of Ω k asΩ k = − θ ( k +1)( k +2) − θ ( k +1)( k +3) − θ ( k +1)( k +4) . . . − θ ( k +1)( k − − θ ( k +1)( k ) F k +2 . . . F k +3 . . . F k +4 . . . ... ...0 0 0 . . . 0 00 0 0 · · · F k − . Since k was arbitrary, we have a d -tuple (Ω , Ω , . . . , Ω d ) which has the property thatΩ Ω · · · Ω d = π D D · · · D d ǫ = f π ǫ = f · b F . Since the free S -modules b F , b F , . . . , b F d are all of the same rank, it follows that (Ω , . . . , Ω d ) ∈ MF dS ( f ). We denote this matrix factorization by Ω MF dS ( f ) ( X ) and refer to it as the syzygy of X .Combining the diagrams (2.3) for all k ∈ Z d we have a short exact sequence(2.4) Ω MF dS ( f ) ( X ) P ( X ) X ǫ ρ where ρ = ( ρ , ρ , . . . , ρ d ) and ǫ = ( ǫ , ǫ , . . . , ǫ d ).Similarly, we have a matrix factorization Ω − MF dS ( f ) ( X ) = (Ω − , Ω − , . . . , Ω − d ), the cosyzygy of X ,and a short exact sequence of matrix factorizations induced by the commutative diagrams(2.5) 0 F k +1 F k +1 ⊕ b F k +1 b F k +1 F k F k ⊕ b F k b F k , ϕ k λ Xk +1 D ′ k η Xk +1 Ω − k λ Xk η Xk for all k ∈ Z d , where η Xk = (cid:16) − Ξ Xk b F k (cid:17) , λ Xk = F k Ξ Xk ! . The induced short exact sequence is(2.6) X I ( X ) Ω − MF dS ( f ) ( X ) , λ η here η = ( η , η , . . . , η d ) and λ = ( λ , λ , . . . , λ d ). Finally, we can represent the components ofΩ − k by Ω − k = · · · − θ ( k +1) k F k +2 · · · − θ ( k +2) k F k +3 . . . 0 0 − θ ( k +3) k ... ... . . . ... ... ...0 0 · · · F k − − θ ( k − k · · · F k − − θ ( k − k . Example 2.12.
Let ( ϕ, ψ ) ∈ MF S ( f ). Then,Ω MF S ( f ) ( ϕ, ψ ) = ( − ψ, − ϕ ) ∼ = ( ψ, ϕ )and Ω − MF S ( f ) ( ϕ, ψ ) = ( − ψ, − ϕ ) ∼ = ( ψ, ϕ ) . In particular, both Ω MF S ( f ) ( − ) and Ω − MF S ( f ) ( − ) are isomorphic to the shift functor when d = 2.In this case, ( ϕ, ψ ) is reduced if and only if Ω MF S ( f ) ( ϕ, ψ ) (respectively Ω − MF S ( f ) ( ϕ, ψ )) is reduced.However, for X ∈ MF dS ( f ) with d >
2, neither Ω MF S ( f ) ( X ) nor Ω − MF S ( f ) ( X ) will be reduced. Forinstance, if X = ( ϕ : F → F , ϕ : F → F , ϕ : F → F ) ∈ MF S ( f ) is of size n , thenΩ MF S ( f ) ( X ) = (cid:18)(cid:18) − ϕ − ϕ ϕ F (cid:19) , (cid:18) − ϕ − ϕ ϕ F (cid:19) , (cid:18) − ϕ − ϕ ϕ F (cid:19)(cid:19) and Ω − MF S ( f ) ( X ) = (cid:18)(cid:18) − ϕ ϕ F − ϕ (cid:19) , (cid:18) − ϕ ϕ F − ϕ (cid:19) , (cid:18) − ϕ ϕ F − ϕ (cid:19)(cid:19) which are of size 2 n .Another observation that can be made from these formulas is that, for each k ∈ Z , we have anisomorphism of R -modulescok (cid:18) − ϕ k +1 − ϕ k +1 ϕ k +2 F k +2 (cid:19) ∼ = cok( ϕ k +1 ϕ k +2 ) ∼ = syz R (cok ϕ k ) ⊕ R m k for some m k ≥
0. A similar statement is true for Ω − MF S ( f ) ( X ) and more generally we have thefollowing proposition. Proposition 2.13.
Let X ∈ MF dS ( f ) be of size n . Let Ω MF dS ( f ) ( X ) and Ω − MF dS ( f ) ( X ) be the matrixfactorizations constructed in 2.11. Then, for each k ∈ Z d , cok(Ω k ) ∼ = syz R (cok ϕ k ) ⊕ R m k ∼ = cok(Ω − k ) where m k = n − µ R (cok ϕ k ) . roof. Let k ∈ Z d . The diagram (2.3) induces the diagram0 0 00 b F k +1 b F k cok Ω k F k +1 ⊕ b F k +1 F k ⊕ b F k F k /f F k F k +1 F k cok ϕ k
00 0 0 Ω k D k ϕ k with exact rows and columns. The right most column displays cok Ω k as a (not necessarily reduced)syzygy of cok ϕ k over R as desired.Alternatively, we can see from the explicit formulas for Ω k and Ω − k thatcok Ω k ∼ = cok θ X ( k +1) k ∼ = cok Ω − k . Now, recall that θ ( k +1) k = ϕ k +1 ϕ k +2 · · · ϕ k − . Since ( ϕ k , θ ( k +1) k ) is a matrix factorization with2 factors, we have that cok( θ ( k +1) k ) ∼ = syz R (cok ϕ k ) ⊕ R m k for some m k ≥
0. In particular, m k = n − µ R (cok ϕ k ) by the uniqueness of minimal free resolutions over R . (cid:3) Returning to our original goal, we note that since the matrix factorizations P ( X ) ∼ = I ( X ) ∼ = L di =1 P ni are projective and injective by Lemma 2.9, the sequences (2.4) and (2.6) imply thatMF dS ( f ) has enough projectives and enough injectives. Additionally, we have the following. Proposition 2.14.
An object X ∈ MF dS ( f ) is projective if and only if it is injective.Proof. The matrix factorizations P ( X ) and I ( X ) are both projective and injective by Lemma 2.9.Let X ∈ MF dS ( f ). If X is projective, (2.4) implies that X is a summand of the injective matrixfactorization P ( X ) and therefore is injective. Conversely, if X is injective, (2.6) implies it is asummand of the projective I ( X ). (cid:3) We have therefore established our original goal of this section.
Theorem 2.15.
The category MF dS ( f ) is a Frobenius category. (cid:3) For matrix factorizations
X, X ′ ∈ MF dS ( f ), let I ( X, X ′ ) denote the set of morphisms X to X ′ that factor through an injective matrix factorization. The stable category MF dS ( f ) is formed bytaking the same objects as MF dS ( f ) and morphisms given by the quotientHom MF dS ( f ) ( X, X ′ ) = Hom MF dS ( f ) ( X, X ′ ) /I ( X, X ′ ) . A consequence of Theorem 2.15 is that MF dS ( f ) carries the structure of a triangulated category withsuspension functor given by Ω − MF dS ( f ) ( − ) [Hap88]. We call a morphism ( α , α , . . . , α d ) : X → X ′ null-homotopic if there exist S -homomorphisms s j : F j → F ′ j − , j ∈ Z d , such that α i = X k ∈ Z d θ X ′ i ( i − k ) s i − k +1 θ X ( i − k +1) i or each i ∈ Z d . We denote by HMF dS ( f ) the homotopy category of matrix factorizations which hasthe same objects as MF dS ( f ) and, for any X, X ′ ∈ MF dS ( f ), has morphismsHom HMF dS ( f ) ( X, X ′ ) = Hom MF dS ( f ) ( X, X ′ ) / ∼ , where ∼ is the equivalence relation α ∼ α ′ if and only if α − α ′ is null-homotopic. Proposition 2.16.
The stable category MF dS ( f ) and the homotopy category HMF dS ( f ) coincide.Proof. Let
X, X ′ ∈ MF dS ( f ). It suffices to show that a morphism α : X → X ′ is null-homotopicif and only if it factors through the morphism λ X : X → I ( X ). The proof relies on an explicitdescription of morphisms I ( X ) → X ′ . Indeed, if β : I ( X ) → X ′ is any morphism, then, for any k, j ∈ Z d with j = 1, we have a commutative diagram F k + j − ⊕ b F k + j − F k + j − ⊕ b F k + j − · · · F k +1 ⊕ b F k +1 F k ⊕ b F k F ′ k + j − F ′ k + j − · · · F ′ k +1 F ′ k . β k + j − D ′ k + j − β k + j − D ′ k + j − D ′ k +1 β k +1 D ′ k β k ϕ ′ k + j − ϕ ′ k + j − ϕ ′ k +1 ϕ ′ k Write the components of β i as β i = (cid:2) β ii β i ( i +1) β i ( i +2) · · · β i ( i − (cid:3) for some β i ( i + s ) : F i + s → F ′ i . Notice that the maps D ′ k , D ′ k +1 , . . . , D ′ k + j − are the identity on F k + j .Hence, β k ( k + j ) = ϕ ′ k ϕ ′ k +1 · · · ϕ ′ k + j − β ( k + j − k + j ) = θ X ′ k ( k + j − β ( k + j − k + j ) by the commutativity of the outer most rectangle of the diagram above. Therefore, β k = h θ X ′ k ( k − β ( k − k β k ( k +1) θ X ′ k ( k +1) β ( k +1)( k +2) · · · θ X ′ k ( k − β ( k − k − i . Now, if β : I ( X ) → X ′ is such that βλ X = α , then α k = β k λ Xk = − X i ∈ Z d θ X ′ k ( k − i ) β ( k − i )( k − i +1) θ X ( k − i +1) k for any k ∈ Z d . This says precisely that α is null-homotopic via the maps {− β ( j − j } j ∈ Z d .Conversely, if α is null-homotopic via maps s j : F j → F ′ j − , then it is not hard to see that themaps γ k = h θ X ′ k ( k − s k s k +1 θ X ′ k ( k +1) s k +2 · · · θ X ′ k ( k − s k − i , for k ∈ Z d , form a morphism γ = ( γ , γ , . . . , γ d ) : I ( X ) → X ′ such that α = γλ X . (cid:3) To end this section, we give explicit formula for the mapping cone of a morphism. Suppose α : X → X ′ is a morphism in MF dS ( f ). The mapping cone of α is the matrix factorization C ( α ) = (∆ , ∆ , . . . , ∆ d ) where∆ k = ϕ ′ k · · · α k · · · − θ X ( k +1) k F k +2 · · · − θ X ( k +2) k F k +3 . . . 0 − θ X ( k +3) k ... ... . . . ... ...0 0 · · · · · · F k − − θ X ( k − k : F ′ k +1 ⊕ b F k +1 → F ′ k ⊕ b F k or all k ∈ Z d . The cone of α fits into a commutative diagram X I ( X ) Ω − MF dS ( f ) ( X ) X ′ C ( α ) Ω − MF dS ( f ) ( X ) α λ X β η X q p where p k = (cid:0) b F k (cid:1) , q k = (cid:18) F ′ k (cid:19) , and β k = (cid:18) α k − Ξ Xk b F k (cid:19) for all k ∈ Z d , and p = ( p , . . . , p d ) , q =( q , . . . , q d ) , and β = ( β , . . . , β d ).3. MF dS ( f ) as a category of modules Let ( S, n , k ) be a regular local ring, f a non-zero non-unit in S , and d ≥ dS ( f ) is equivalent to the category of MCM modules over a certainnon-commutative S -algebra which is finitely generated and free as an S -module. This extends aresult of Solberg [Sol89, Proposition 1.3] for all d ≥
2. As a consequence we will conclude that, if S is complete, the Krull-Remak-Schmidt Theorem (KRS) holds in MF dS ( f ).Recall from Section 2 the projective (equivalently injective) matrix factorizations P , P , . . . , P d ,and their direct sum P = L i ∈ Z d P i . Our first step is to understand the homomorphisms from P j to P i for any i, j ∈ Z d . Definition 3.1.
For i = j ∈ Z d , let e ij ∈ Hom S ( S, S ) d denote the d -tuple of homomorphisms suchthat the j + 1 , j + 2 , . . . , i − , i components are multiplication by f while the rest are the identityon S . For each i ∈ Z d , let e ii = 1 P i , the identity on P i .For instance, if i ∈ Z d , the i th map in e i ( i − is multiplication by f and the rest are the identityon S while e ( i − i is of the form ( f, f, . . . , f, , f, . . . , f, f ), where only the i th component is theidentity on S . Lemma 3.2.
For all i, j ∈ Z d , Hom MF dS ( f ) ( P j , P i ) = S · e ij . In particular, the morphisms from P j to P i form a free S -module of rank 1.Proof. The morphisms from P j to P i are tuples of the form ( α , . . . , α d ) for some α k ∈ S . If( α , . . . , α d ) ∈ Hom MF dS ( f ) ( P i , P i ), then it is easy to see that α = α = · · · = α d and soHom MF dS ( f ) ( P i , P i ) = S · e ii . Suppose ( α , . . . , α d ) ∈ Hom MF dS ( f ) ( P j , P i ) where i = j . Considerthe following diagram, which commutes since ( α , . . . , α d ) is a morphism of matrix factorizations: P j · · · S S · · ·
S S · · ·P i · · · S S · · ·
S S · · · α i +1 α i fα j +1 α j f We conclude that α j +1 = α j +2 = · · · = α i , α j = α j − = · · · = α i +1 , and α i = f α i +1 . Thus, eachcomponent of the morphism can be rewritten in terms of the element α i +1 ∈ S . It follows that( α , . . . , α d ) = α i +1 e ij , that is, Hom MF dS ( f ) ( P j , P i ) is generated by e ij as an S -module. Since thecomponents of e ij are given by multiplication by non-zero elements of S , a morphism s · e ij = 0 ifand only if s = 0. Hence, Hom MF dS ( f ) ( P j , P i ) is in fact a free S -module of rank 1 for all i, j ∈ Z d . (cid:3) Let Γ = End MF dS ( f ) ( P ) op where P = L di =1 P i . As S -modules, Γ ∼ = L i,j ∈ Z d Hom( P j , P i ) andtherefore, Lemma 3.2 implies that Γ is a free S -module of rank d . For each pair i, j ∈ Z d , use thesame symbol e ij to denote the image of the generator of Hom( P j , P i ) under the natural inclusion om( P j , P i ) → Γ. The set { e ij } i,j ∈ Z d forms a basis for Γ as an S -module. We record the basicrules for multiplication of the basis elements e ij . Lemma 3.3.
The basis elements { e ij } i,j ∈ Z d satisfy the following properties.(i) e ij e pq = 0 if and only if i = q (ii) e ii = e ii for all i ∈ Z d (iii) P di =1 e ii = 1 Γ (iv) e ij e ii = e ij and e jj e ij = e ij for all i, j ∈ Z d (v) e i ( i − e ji = ( f e ( i − i − if j = i − e j ( i − otherwise(vi) e ij e ( i +1) i = ( f e ( i +1)( i +1) j = i + 1 e ( i +1) j otherwise(vii) d X i =1 e i ( i − e jj = e j ( j − = e ( j − j − d X i =1 e i ( i − for all j ∈ Z d (cid:3) The element P di =1 e i ( i − ∈ Γ is of particular interest because of the following.
Lemma 3.4.
Let z = P di =1 e i ( i − and s ≥ an integer. Write s = dq + r for q ≥ and ≤ r < d .Then, z s = f q d X i =1 e i ( i − r ) . In particular, z d = f · Γ .Proof. If s = 1 there is nothing to prove. Assume the formula holds for s ≥ z s +1 .By induction, z s +1 = z · f q d X i =1 e i ( i − r ) where s = dq + r , q ≥
0, and 0 ≤ r < d . If r = d −
1, then, by Lemma 3.3, z · d X i =1 e i ( i − r ) = d X i =1 e ( i +1) i e i ( i +1) = f · Γ . Since s = dq + d −
1, we have that s + 1 = d ( q + 1). Hence, z s +1 = f q +1 d X i =1 e ii as needed. If 0 ≤ r < d −
1, then z · d X i =1 e i ( i − r ) = d X i =1 e i ( i − r − also by Lemma 3.3. In this case, z s +1 = f q d X i =1 e i ( i − ( r +1)) which completes the induction since s + 1 = dq + ( r + 1) with 0 ≤ r + 1 < d . (cid:3) emma 3.5. If i, j ∈ Z d with i = j , then e ij can be written as a product of basis elements of theform e ℓ ( ℓ − . In particular, e ij = e ( j +1) j e ( j +2)( j +1) · · · e ( i − i − e i ( i − . Proof.
Let i = j ∈ Z d . Lemma 3.3 (vi) implies that, for any ℓ = j ∈ Z d , the element e ℓ ( ℓ − can befactored out of e ℓj on the right e ℓj = e ( ℓ − j e ℓ ( ℓ − . Since i = j , we may apply this equality for ℓ = i, i − , . . . , j +2 , j +1 which gives us the factorization e ij = e ( i − j e i ( i − = e ( i − j e ( i − i − e i ( i − = · · · = e ( j +1) j e ( j +2)( j +1) · · · e ( i − i − e i ( i − . (cid:3) Let MCM(Γ) denote the full subcategory of finitely generated left Γ-modules which are free whenviewed as S -modules via the inclusion S · Γ ⊂ Γ. The rest of this section is dedicated to provingthe following.
Theorem 3.6.
The categories
MCM(Γ) and MF dS ( f ) are equivalent. In particular, if S is complete,the Krull-Remak-Schmidt Theorem holds in MF dS ( f ) . If S is complete, it is known that the Krull-Remak-Schmidt Theorem holds in the categoryMCM(Γ) (for example see [Aus86]). So, establishing the equivalence is the main concern (whichdoes not require S to be complete). We start by defining a functor H : MCM(Γ) → MF dS ( f ) usingthe element z = P di =1 e i ( i − ∈ Γ. Let M be a Γ-module in MCM(Γ). Lemma 3.3 (i)-(iii) show that e , . . . , e dd are orthogonal idempotents such that e + e + · · · + e dd = 1 Γ . Thus, M decomposes,as an S -module, into M = e M ⊕ · · · ⊕ e dd M. Since M ∈ MCM(Γ), each summand e ii M is a free S -module. Lemma 3.3 (vii) shows that left mul-tiplication by z ∈ Γ defines an S -homomorphism between free S -modules z : e ii M → e ( i − i − M for all i ∈ Z d . Proposition 3.7.
Let M ∈ MCM(Γ) . The d -tuple of S -homomorphisms ( z : e M → e M, z : e M → e M . . . , z : e M → e dd M ) forms a matrix factorization of f in MF dS ( f ) , where each map is multiplication by z = P di =1 e i ( i − ∈ Γ .Proof. In light of Lemma 3.4, the only piece that needs justification is that each of the free S -modules involved are of the same rank. To see this, let i ∈ Z d . The composition e ii M e ( i +1)( i +1) M e ii M z d − z is f times the identity on e ii M . Since e ii M and e ( i +1)( i +1) M are free over S , Lemma 1.7 impliesthat rank S ( e ii M ) = rank S ( e ( i +1)( i +1) M ). (cid:3) Following Proposition 3.7, the functor H is defined as follows: H ( M ) = ( z : e M → e M, . . . , z : e M → e dd M ) for any M ∈ MCM(Γ) and, for a homomorphism h : M → N in MCM(Γ), define H ( h ) = ( h | e M , . . . , h | e dd M ), where h | e ii M denotes the restriction of h to the S -direct summand e ii M . Since h is a Γ-homomorphism, h | e ii M maps e ii M into e ii N . Since multiplication by anelement of Γ commutes with any Γ-homomorphism, this d -tuple forms a morphism between thematrix factorizations H ( M ) → H ( N ). At this point we can prove that H is both full and faithful. Proposition 3.8.
The functor H : MCM(Γ) → MF dS ( f ) is full and faithful. roof. Let
M, N ∈ MCM(Γ). If H ( h ) = 0 for some Γ-homomorphism h : M → N , then h | e ii M = 0for each i ∈ Z d . But this means that h = L i ∈ Z d h | e ii M = 0 implying that H is faithful.In order to show that H is full, let ( α , . . . , α d ) : H ( M ) → H ( N ) be a morphism of matrixfactorizations. So, α i : e ii M → e ii N and we have a commutative diagram(3.1) e ii M e ( i − i − Me ii N e ( i − i − N α i z α i − z for each i ∈ Z d . Let h = L dj =1 α j : M → N be the S -homomorphism given by h ( m ) = α ( e m ) + α ( e m ) + · · · + α d ( e dd m ) for all m ∈ M . We claim that h is in fact a Γ-homomorphism andfurthermore, H ( h ) = ( α , . . . , α d ). The second claim follows from the first and the definition of H and so our aim is to show that h is a homomorphism of Γ-modules. Since h is an S -homomorphismand Γ is a finitely generated free S -module with basis { e ij } i,j ∈ Z d , we would be done if we showedthat e ij h ( m ) = h ( e ij m ) for all i, j ∈ Z d and m ∈ M . By Lemma 3.5, it suffices to show that theelements of the form e k ( k − pass through h since each e ij is a product of elements of this form.Let i ∈ Z d and m ∈ M . By Lemma 3.3(vii), multiplication by z on e ii M (respectively on e ii N )coincides with multiplication by the element e i ( i − ∈ Γ. Therefore, the diagram (3.1) implies that α i − ( e i ( i − e ii m ) = e i ( i − α i ( e ii m ) . Since e i ( i − e ii m ∈ e ( i − i − M , the term on the left hand side is precisely h ( e i ( i − m ). On theother hand, since e ii h ( m ) = α i ( e ii m ), e i ( i − h ( m ) = e i ( i − e ii h ( m )= e i ( i − α i ( e ii m ) . Together, we have that e i ( i − h ( m ) = h ( e i ( i − m ) as desired. Thus, h is a Γ-homomorphism and H is full. (cid:3) To show that H is dense, we define a functor F : MF dS ( f ) → MCM(Γ) which is given by F ( (cid:3) ) = Hom MF dS ( f ) ( P , (cid:3) ), where P = L di =1 P i . For a matrix factorization X ∈ MF dS ( f ), F ( X ) isa left Γ-module by pre-composing any morphism P → X with an element of Γ. In order to showthat the image of F does indeed land in MCM(Γ), we must show that F ( X ) is a free S -module.This requires an explicit description of the morphisms P → X .Recall that P i is the matrix factorization whose i th component is multiplication by f on S while the rest are the identity on S . For k ∈ Z d , let D k : S d → S d be given by D k ( a , . . . , a d ) =( a , . . . , a k − , f a k , a k +1 , . . . , a d ) for any ( a , . . . , a d ) ∈ S d . Then, P = ( D : S d → S d , D : S d → S d , . . . , D d : S d → S d ). Lemma 3.9.
Let X = ( ϕ : F → F , . . . , ϕ d : F → F d ) ∈ MF dS ( f ) and let ( α , α , . . . , α d ) ∈F ( X ) = Hom MF dS ( f ) ( P , X ) . For each k ∈ Z d , we may write α k = (cid:2) α k α k · · · α kd (cid:3) for some α ki ∈ Hom S ( S, F k ) . Then, for any j = 0 ∈ Z d , α k ( k + j ) = ϕ k ϕ k +1 · · · ϕ k + j − α ( k + j )( k + j ) . Proof.
Similar to the proof of Lemma 2.16, the formula follows from the following commutativediagram: S d S d · · · S d S d F k + j F k + j − · · · F k +1 F k . D k + j − α k + j α k + j − D k + j − D k +1 α k +1 D k α k ϕ k + j − ϕ k + j − ϕ k +1 ϕ k he commutativity of the outermost rectangle gives us that α k D k D k +1 · · · D k + j − = ϕ k ϕ k +1 · · · ϕ k + j − α k + j .Since j = 0, the composition D k D k +1 · · · D k + j − is the identity on the ( k + j )th component of S d .Therefore, if we compare the ( k + j )th components of the homomorphisms on either side of theabove equality, we find that α k ( k + j ) = ϕ k ϕ k +1 · · · ϕ k + j − α ( k + j )( k + j ) as desired. (cid:3) Let X = ( ϕ : F → F , · · · , ϕ d : F → F d ) ∈ MF dS ( f ) be a matrix factorization. Recall fromSection 2, the homomorphisms θ Xki : F i → F k , i, k ∈ Z d , which are given by θ Xki = ( F k i = kϕ k ϕ k +1 · · · ϕ i − ϕ i − i = k. For each ( g , . . . , g d ) ∈ L di =1 F i , we associate a d -tuple of S -homomorphisms θ X ( g , . . . , g d ) := (cid:0)(cid:2) θ Xk g · · · θ Xkd g d (cid:3)(cid:1) dk =1 . Here θ Xki g i is being identified with its image in Hom S ( S, F k ) under the natural isomorphism F k ∼ =Hom S ( S, F k ). When X is clear from context, we will omit the superscripts. Lemma 3.10.
Let X = ( ϕ : F → F , · · · , ϕ d : F → F d ) ∈ MF dS ( f ) . Then, θ ( g , . . . , g d ) ∈ Hom MF dS ( f ) ( P , X ) for any ( g , . . . , g d ) ∈ L di =1 F i . Furthermore, the map θ : L di =1 F i → Hom MF dS ( f ) ( P , X ) is an isomorphism of S -modules.Proof. First, we show that θ ( g , . . . , g d ) as defined is in fact a morphism of matrix factorizationsbetween P and X . What needs to be shown is the commutativity of the diagram S d S d F k +1 F k h θ ( k +1)1 g ··· θ ( k +1) d g d i D k h θ k g ··· θ kd g d i ϕ k for all k ∈ Z d . Notice that ϕ k θ ( k +1) k = ϕ k ϕ k +1 ϕ k +2 · · · ϕ k − = f · F k = f θ kk and ϕ k θ ( k +1) i = θ ki for all i = k . Therefore, we have that ϕ k (cid:2) θ ( k +1)1 g · · · θ ( k +1) d g d (cid:3) = (cid:2) θ k g · · · f θ kk g k · · · θ kd g d (cid:3) = (cid:2) θ k g · · · θ kk g k · · · θ kd g d (cid:3) D k which implies the commutativity of the diagram as desired.In order to show θ is an S -module isomorphism, let ( α , . . . , α d ) ∈ Hom MF dS ( f ) ( P , X ), k ∈ Z d ,and denote the components of α k = (cid:2) α k . . . α kd (cid:3) ∈ Hom S ( S d , F k ) as above. By Lemma 3.9, α k ( k + j ) = θ k ( k + j ) α ( k + j )( k + j ) for each j = 0. Hence, α k = (cid:2) θ k α · · · θ kd α dd (cid:3) . It follows that themorphism ( α , . . . , α d ) depends only on the diagonal components α , α , . . . , α dd . In particular,the tuple ( α , . . . , α dd ) ∈ L dj =1 F j is a pre-image for ( α , . . . , α d ) under the map θ . Finally, θ isinjective since k th component of (cid:2) θ k g · · · θ kk g k · · · θ kd g d (cid:3) is θ kk g k = g k . (cid:3) Corollary 3.11.
For any X ∈ MF dS ( f ) of size n , the Γ -module F ( X ) = Hom MF dS ( f ) ( P , X ) is afree S -module of rank dn . In particular, F ( X ) ∈ MCM(Γ) . (cid:3) Consider how the elements e ii ∈ Γ act on a morphism α : P → X . From the Lemma 3.10,we may write α k = (cid:2) θ k g · · · θ kd g d (cid:3) for some ( g , . . . , g d ) ∈ L dj =1 F j . For i ∈ Z d , we write e ii = ( ǫ ii , . . . , ǫ dii ) where ǫ kii ( a , . . . , a d ) = (0 , . . . , , a i , , . . . ,
0) for any ( a , . . . , a d ) ∈ S d . It followsthat α k ◦ ǫ kii = (cid:2) · · · θ ki g i · · · (cid:3) where the only nonzero entry is in the i th position. Hence, e ii · α = (cid:0)(cid:2) · · · θ ki g i · · · (cid:3)(cid:1) dk =1 ∈ e ii F ( X ) and in fact this is the form of every element of e ii F ( X ): e ii F ( X ) = n(cid:0)(cid:2) · · · θ ki g i · · · (cid:3)(cid:1) dk =1 (cid:12)(cid:12)(cid:12) g i ∈ F i o . roposition 3.12. Let X = ( ϕ : F → F , . . . , ϕ d : F → F d ) ∈ MF dS ( f ) . Then, ( θ | F , . . . , θ | F d ) : X → HF ( X ) is an isomorphism of matrix factorizations.Proof. First notice that for any g i ∈ F i , θ | F i ( g i ) = θ (0 , . . . , , g i , , . . . ,
0) = (cid:0)(cid:2) · · · θ ki g i · · · (cid:3)(cid:1) dk =1 ∈ e ii F ( X ) . The restriction of θ is still injective and the preceding paragraph explains why θ restricted to F i issurjective. Therefore, it is enough to show the commutativity of the diagram F F d · · · F F e F ( X ) e dd F ( X ) · · · e F ( X ) e F ( X ) . ϕ d θ | F ϕ d − θ | Fd ϕ ϕ θ | F θ | F z z z z Let i ∈ Z d and g i +1 ∈ F i +1 . Then, θ | F i ϕ i ( g i +1 ) = (cid:0)(cid:2) · · · θ ki ϕ i ( g i +1 ) · · · (cid:3)(cid:1) dk =1 and θ ki ϕ i ( g i +1 ) = ( f g i +1 k = i + 1 ϕ k ϕ k +1 · · · ϕ i − ϕ i ( g i +1 ) k = i + 1 . To compute the other composition, recall that z · e ( i +1)( i +1) α = e ( i +1) i α for any α ∈ F ( X ). Write e ( i +1) i = ( ǫ i +1) i , ǫ i +1) i , . . . , ǫ d ( i +1) i ) where ǫ k ( i +1) i : S d → S d for each k ∈ Z d . It suffices to computethe composition of S -homomorphisms S d → F k (cid:16)(cid:2) · · · θ k ( i +1) ( g i ) · · · (cid:3) ◦ ǫ k ( i +1) i (cid:17) ( a , . . . , a d ) , for each k ∈ Z d and ( a , . . . , a d ) ∈ S d . We have that ǫ k ( i +1) i ( a , . . . , a d ) = ( (0 , . . . , f a i , . . . , k = i + 1(0 , . . . , a i , . . . , k = i + 1where the non-zero entries are in the ( i + 1)st position. Thus, the composition above is equal to a i θ k ( i +1) ( g i ) when k = i + 1 and f a i g i when k = i + 1. Comparing this with the components of θ | F i ϕ i ( g i +1 ) we conclude that z ◦ θ | F i +1 ( g i +1 ) = θ | F i ◦ ϕ i ( g i +1 ). Hence, ( θ | F , . . . , θ | F d ) : X → HF ( X )is an isomorphism of matrix factorizations. (cid:3) As a consequence of Proposition 3.12, the functor H is also dense. This completes the proofof Theorem 3.6. It is also worth mentioning that the analogous statement for the composition F H is true, that is,
F H ( M ) ∼ = M for any M ∈ MCM(Γ). This follows from observing that theisomorphism of free S -modules, θ H ( M ) : M → F H ( M ), is also a Γ-homomorphism. As in the proofof Proposition 3.8, one can show that e i ( i − θ H ( M ) ( m ) = θ H ( M ) ( e i ( i − m ) for all m ∈ M and i ∈ Z d .4. The d -fold Branched Cover Let ( S, n , k ) be a complete regular local ring, and let f be a non-zero non-unit in S . Set R = S/ ( f )and fix an integer d ≥ Definition 4.1.
The ( d -fold) branched cover of R is the hypersurface ring R ♯ = S J z K / ( f + z d ) . Throughout this section, we will also assume that k is algebraically closed and that the charac-teristic of k does not divide d . In this case, the polynomial x d − ∈ k [ x ] has d distinct roots in k and the group formed by its roots is cyclic of order d . Any generator of this group is a primitive d throot of unity . Since S is complete, it also contains primitive d th roots of 1 ∈ S [LW12, CorollaryA.31]. or the rest of this section, fix an element ω ∈ S such that ω d = 1 and ω t = 1 for all 0 < t < d .The ring R ♯ carries an automorphism σ : R ♯ → R ♯ of order d which fixes S and sends z to ωz .Denote by R ♯ [ σ ] the skew group algebra of the cyclic group of order d generated by σ acting on R ♯ .Specifically, R ♯ [ σ ] = L i ∈ Z d R ♯ · σ i as R ♯ -modules with multiplication given by the rule( s · σ i ) · ( t · σ j ) = sσ i ( t ) · σ i + j for s, t ∈ R ♯ and i, j ∈ Z d . The left modules over R ♯ [ σ ] are precisely the R ♯ -modules N whichcarry a compatible action of σ , that is, an action of σ such that σ ( rx ) = σ ( r ) σ ( x ) for all r ∈ R ♯ and x ∈ N . It follows that R ♯ itself is naturally a left R ♯ [ σ ]-module with the action of σ givenby evaluating σ ( r ) for any r ∈ R ♯ . We say that a left R ♯ [ σ ]-module is maximal Cohen-Macaulay (MCM as usual) if it is MCM when it is viewed as an R ♯ -module. Denote the category of MCM R ♯ [ σ ]-modules by MCM σ ( R ♯ ).In the case d = 2, Kn¨orrer showed that the category of MCM modules over R ♯ [ σ ] is equivalent tothe category of matrix factorizations of f with 2 factors [Kn¨o87, Proposition 2.1]. The main goalof this section is to extend the equivalence given by Kn¨orrer for all d ≥ Lemma 4.2.
Let N be an R ♯ [ σ ] -module. Then, N decomposes as an S -module into N = L i ∈ Z d N ω i where N ω i = (cid:8) x ∈ N : σ ( x ) = ω i x (cid:9) . Furthermore, if N is a MCM R ♯ [ σ ] -module, then N and each summand N ω i are finitely generatedfree S -modules.Proof. In order to justify the direct sum decomposition of N , we will make repeated use of the fact P d − i =0 ω ki = 0 for any k ∈ Z d . Let x ∈ N and observe that dx = dx + d − X i =0 ω − i ! σ ( x ) + d − X i =0 ω − i ! σ ( x ) + · · · + d − X i =0 ω − ( d − i ! σ d − ( x )= d − X i =0 σ i ( x ) + d − X i =0 ω − i σ i ( x ) + · · · + d − X i =0 ω − ( d − i σ i ( x )= d − X k =0 d − X i =0 ω − ik σ i ( x ) . Also notice that, for any k ∈ Z d , σ d − X i =0 ω − ik σ i ( x ) ! = σ ( x ) + ω − k σ ( x ) + ω − k σ ( x ) + · · · + ω − ( d − k x = ω k ( x + ω − k σ ( x ) + ω − k σ ( x ) + · · · + ω − ( d − k σ d − ( x )) . That is, P d − i =0 ω − ik σ i ( x ) ∈ N ω k . Since σ is S -linear and d is invertible in S , we have that x = d − X k =0 d − X i =0 ω − ik σ i ( x ) d ∈ N + N ω + · · · + N ω d − implying that N = P d − k =0 N ω k .Next, suppose we have a sum of elements(4.1) x + x + · · · + x d − = 0 ith x i ∈ N ω i for each i ∈ Z d , and let j ∈ Z d . Notice that if k, ℓ ∈ Z d , then ω − jk σ k ( x ℓ ) = ω − jk + kℓ x ℓ = ω ( − j + ℓ ) k x ℓ . In particular, ω − jk σ k ( x j ) = x j for all k ∈ Z d . Therefore, applying ω − jk σ k to (4.1) gives us an equation ω − jk x + ω ( − j +1) k x + · · · + x j + · · · + ω ( − j − k x d − = 0 . Summing over Z d , we find that X i = j X k ∈ Z d ω k ( − j + i ) x i + dx j = 0 . Once again, since P d − k =0 ω k ( − j + i ) = 0 for all i = j , we can conclude that x j = 0. Thus, N = L d − i =0 N ω i as desired.The second statement holds since a finitely generated R ♯ -module N is MCM over R ♯ if and onlyif it is free as an S -module. (cid:3) Definition and Proposition 4.3.
Let
R, R ♯ , R ♯ [ σ ], and ω be as above. Let µ ∈ S be any root of x d + 1 ∈ S [ x ].(i) Let N be an MCM R ♯ [ σ ]-module and N ω i be as in Lemma 4.2 for each i ∈ Z d . Define amatrix factorization A ( N ) ∈ MF dS ( f ) as follows. Multiplication by µz defines an S -linearhomomorphism N ω i → N ω i +1 for all i ∈ Z d . The composition N ω d − N N ω · · · N ω d − N ω d − µz µz µz µz µz is equal to − z d = f times the identity on N ω d − . It follows that the above homomorphismsand free S -modules form a matrix factorization of f in MF dS ( f ) which we denote as A ( N ).For a homomorphism g : N → M of MCM R ♯ [ σ ]-modules, define a morphism of matrixfactorizations A ( g ) = (cid:0) g | N ωd − , g | N ωd − , . . . , g | N (cid:1) where g | N ωi denotes the restriction of g to the S -direct summand N ω i of N . Thus, we havea functor A : MCM σ ( R ♯ ) → MF dS ( f ) . (ii) Let X = ( ϕ : F → F , . . . , ϕ d : F → F d ) ∈ MF dS ( f ). Define B ( X ) = F d ⊕ F d − ⊕ · · · ⊕ F as an S -module which has the structure of a R ♯ [ σ ]-module by defining the action of z as z · ( x d , x d − , . . . , x ) = (cid:0) µ − ϕ d ( x ) , µ − ϕ d − ( x d ) , . . . , µ − ϕ ( x ) (cid:1) and the action of σ as σ · ( x d , x d − , . . . , x ) = ( x d , ωx d − , ω x d − , . . . , ω d − x ) , for any x i ∈ F i , i ∈ Z d . For a morphism of matrix factorizations α = ( α , α , . . . , α d ) : X → X ′ , where X ′ = ( ϕ ′ : F ′ → F ′ , . . . , ϕ ′ d : F ′ → F ′ d ), define B ( α ) : B ( X ) → B ( X ′ ) by B ( α )( x d , x d − , . . . , x ) = ( α d ( x d ) , α d − ( x d − ) , . . . , α ( x ))for all ( x d , x d − . . . , x ) ∈ B ( X ). Thus, we have a functor B : MF dS ( f ) → MCM σ ( R ♯ ) . Proof.
Several pieces of the definitions need justification. First we note that, since − d throot in k , we may apply [LW12, Corollary A.31] to obtain an element µ ∈ S such that µ d = − i) Multiplication by µz defines an S -linear map N ω i → N ω i +1 for any i ∈ Z d since µ ∈ S and σ ( zx ) = σ ( z ) σ ( x ) = ω i +1 zx for all x ∈ N ω i . Notice that ( µz ) d = µ d z d = f ∈ R ♯ . Therefore, the composition N ω i +1 N ω i N ω i +1 ( µz ) d − µz equals f · N ωi for all i ∈ Z d . We know that each N ω i is a free S -module by Lemma 4.2so, by applying Lemma 1.7, we have that rank S ( N ω i ) = rank S ( N ω i +1 ) for all i ∈ Z d . Thisimplies that A ( N ) ∈ MF dS ( f ).If g : N → M is a homomorphism of R ♯ [ σ ]-modules, then g ( rx ) = rg ( x ) and σg ( x ) = g ( σ ( x )) for all x ∈ N and r ∈ R ♯ . It follows that g | N ωi ( N ω i ) ⊆ M ω i and the diagram N ω i N ω i +1 M ω i M ω i +1 g | Nωi µz g | Nωi +1 µz commutes for all i ∈ Z d . In other words, A ( g ) is a morphism of matrix factorizations A ( N ) → A ( M ).(ii) First we justify that the defined action of z and σ make B ( X ) a MCM R ♯ [ σ ]-module. Recallthe homomorphisms θ Xki : F i → F k , from Section 2, which are given by θ Xki = ( F k i = kϕ k ϕ k +1 · · · ϕ i − ϕ i − i = k. We will drop the superscript X for the rest of this proof. We claim that for any s ≥ x d , x d − , . . . , x ) ∈ B ( X ), z s · ( x d , x d − , . . . , x ) = f q µ − r ( θ d ( d + r ) ( x d + r ) , . . . , θ r ) ( x r ))where s = dq + r , q ≥
0, and 0 ≤ r < d . When s = 1 the formula is precisely the definedaction of z on B ( X ). Assume the claim is true for s = dq + r ≥ q ≥ ≤ r < d and consider multiplication by z s +1 . By induction we have that z s +1 · ( x d , . . . , x ) = z · f q µ − r (( θ d ( d + r ) ( x d + r ) , . . . , θ r ) ( x r ))= f q µ − ( r +1) ( ϕ d θ r ) ( x r ) , . . . , ϕ θ r ) ( x r )) . If r = d −
1, then ϕ k − θ k ( k + r ) = f · F k − for each k ∈ Z d and therefore z s +1 = f q +1 µ − ( r +1) ( x d , x d − , . . . , x ) . If 0 ≤ r < d −
1, then 0 ≤ r + 1 < d and therefore ϕ k − θ k ( k + r ) = θ ( k − k + r ) for each k ∈ Z d .In this case, z s +1 = f q µ − ( r +1) ( θ d (1+ r ) ( x r ) , . . . , θ r ) ( x r ))which completes the induction. It follows that multiplication by z d is given by z d · ( x d , . . . , x ) = f µ − d ( x d , . . . , x ) . By definition, µ − d = −
1. Thus, ( f + z d ) B ( X ) = 0, that is, B ( X ) is an R ♯ -module. In fact,since B ( X ) is free as an S -module, it is MCM as an R ♯ -module. n order to show that B ( X ) has the structure of an R ♯ [ σ ]-module, we must show that σ ( rx ) = σ ( r ) σ ( x ) for all r ∈ R ♯ and x ∈ B ( X ). It suffices to show that σ ( zx ) = σ ( z ) σ ( x )for all x ∈ B ( X ). This follows since σ ( z ) σ ( x ) = ωz · ( x d , ωx d − , . . . , ω d − x )= z · ( ωx d , ω x d − , . . . , x )= ( µ − ϕ d ( x ) , µ − ωϕ d − ( x d ) , . . . , µ − ω d − ϕ ( x ))= σ (cid:0) µ − ϕ d ( x ) , µ − ϕ d − ( x d ) , . . . , µ − ϕ ( x ) (cid:1) = σ ( zx )for any x = ( x d , x d − , . . . , x ) ∈ B ( X ). Hence, B ( X ) ∈ MCM σ ( R ♯ ).Finally, we must show that B ( α ) forms a homomorphism of R ♯ [ σ ]-modules. This isstraightforward to verify by recalling that α k ϕ k = ϕ ′ k α k +1 for all k ∈ Z d . (cid:3) Theorem 4.4.
The functors A : MCM σ ( R ♯ ) → MF dS ( f ) and B : MF dS ( f ) → MCM σ ( R ♯ ) arenaturally inverse and establish an equivalence of the categories MCM σ ( R ♯ ) ≈ MF dS ( f ) .Proof. Let X = ( ϕ : F → F , . . . , ϕ d : F → F d ) ∈ MF dS ( f ). Then, B ( X ) = F d ⊕ F d − ⊕ · · · ⊕ F with the action of σ on B ( X ) given by σ ( x d , . . . , x ) = ( x d , ωx d − , . . . , ω d − x ) for each x i ∈ F i .For each i ∈ Z d , the S -module F i is embedded into B ( X ) via the natural inclusion map which wewill denote as q i : F i → B ( X )Notice that the action of σ on B ( X ) implies that B ( X ) ω d − i = { (0 , . . . , , x i , , . . . ,
0) : x i ∈ F i } = q i ( F i ) . Therefore, the matrix factorization AB ( X ) is given by B ( X ) ω d − B ( X ) · · · B ( X ) ω d − B ( X ) ω d − µz µz µz µz which is isomorphic to X via the morphism F F d · · · F F B ( X ) ω d − B ( X ) · · · B ( X ) ω d − B ( X ) ω d − . q ϕ d q d ϕ d − ϕ q ϕ q µz µz µz µz Explicitly, the diagram commutes since if k ∈ Z d and x ∈ F k +1 , then µzq k +1 ( x ) = µq k ( µ − ϕ k ( x )) = q k ϕ k ( x ) . To show AB is naturally isomorphic to the identity, suppose we have a morphism α = ( α , . . . , α d ) : X → X ′ where X ′ = ( ϕ ′ : F ′ → F ′ , . . . , ϕ ′ d : F ′ → F ′ d ) ∈ MF dS ( f ). The matrix factorizations X ′ is isomorphic to AB ( X ′ ) via the morphism ( q ′ , q ′ , . . . , q ′ d ) where q ′ i : F ′ i → B ( X ′ ) is the naturalinclusion. Recall that the homomorphism B ( α ) is given by B ( α )( x d , x d − , . . . , x ) = ( α d ( x d ) , α d − ( x d − ) , . . . , α ( x )) . Applying the functor A forms a morphism of matrix factorizations by restricting B ( α ) to thesubmodules B ( X ) ω d − i . The images of these restrictions land in the submodules B ( X ′ ) ω d − i . Inother words, the k th component of the morphism AB ( α ) is given by the composition B ( X ) ω d − k F k F ′ k B ( X ′ ) ω d − k p k α k q ′ k here p k is the natural projection onto F k . Therefore, AB ( α ) ◦ ( q , q , . . . , q d ) = ( q ′ α p , q ′ α p , . . . , q ′ d α d p d ) ◦ ( q , q , . . . , q d )= ( q ′ α , q ′ α , . . . , q ′ d α d )and this implies the commutativity of the diagram X X ′ AB ( X ) AB ( X ′ ) . α ( q ,...,q d ) ( q ′ ,...,q ′ d ) AB ( α ) Next, let N be an MCM R ♯ [ σ ]-module. As an S -module, BA ( N ) = N ⊕ N ω ⊕ · · · ⊕ N ω d − and in fact the natural S -isomorphism Ψ N : BA ( N ) → N given by ( n , n , . . . , n d − ) P i ∈ Z d n i is also an R ♯ [ σ ]-homomorphism. To see this, let ( n , n , . . . , n d − ) ∈ BA ( N ). Then, Ψ N is a R ♯ -homomorphism sinceΨ N ( z · ( n , n , . . . , n d − )) = Ψ N ( µ − µzn d − , µ − µzn , . . . , µ − µzn d − )= Ψ N ( zn d − , zn , . . . , zn d − )= z ( n + n + · · · n d − )= z Ψ N ( n , n , . . . , n d − )and a R ♯ [ σ ]-homomorphism sinceΨ N ( σ ( n , n , . . . , n d − )) = Ψ N ( n , ωn , . . . , ω d − n d − )= n + ωn + · · · + ω d − n d − = σ ( n ) + σ ( n ) + · · · + σ ( n d − )= σ ( n + n + · · · n d − )= σ (Ψ N ( n , n , . . . , n d − )) . (cid:3) In Section 3 we showed that the category of MCM modules over the endomorphism ring of theprojective object P = P ⊕ P ⊕ · · · ⊕ P d is equivalent to the category of matrix factorizations of f with d ≥ σ ( R ♯ ) and MCM(End MF dS ( f ) ( P ) op ). In fact, the two rings are isomorphic which wewill see below. Of course, both statements require that the characteristic of k does not divide d .Recall, also from Section 3, that Γ = End MF dS ( f ) ( P ) op is a free S -module with basis given by theelements { e ij } i,j ∈ Z d . The main rules for multiplication in Γ are given in Lemmas 3.3, 3.4, and 3.5. Proposition 4.5.
The rings R ♯ [ σ ] and Γ = End MF dS ( f ) ( P ) op are isomorphic.Proof. The set { z i σ j } i,j ∈ Z d forms a basis for R ♯ [ σ ] over S . As in 4.3, let µ ∈ S be any root of x d + 1 ∈ S [ x ]. Define a map ψ : R ♯ [ σ ] → Γ by ψ ( z ) = µ P i ∈ Z d e i ( i − and ψ ( σ ) = P i ∈ Z d ω − i e ii .Extend ψ multiplicatively, that is, define ψ ( z i σ j ) = ψ ( z ) i ψ ( σ ) j for all i, j ∈ Z d . Since { z i σ j } i,j ∈ Z d is an S -basis, ψ extends uniquely to a well defined S -linear homomorphism. Using the fact that σz = σ ( z ) σ = ωzσ ∈ R ♯ [ σ ], it is not hard to see that ψ is also a homomorphism of rings.As S -modules, both R ♯ [ σ ] and Γ are free of rank d . Therefore, to conclude that ψ is anisomorphism, it suffices to check surjectivity. First, we show that the element e kk is in the image f ψ for each k ∈ Z d . Indeed, if j ∈ Z d , then ψ ( σ j ) = X i ∈ Z d ω − ji e ii . Thus, for any k ∈ Z d , ψ d X j ∈ Z d ω jk σ j = 1 d X j ∈ Z d ω jk ψ ( σ ) j = 1 d X j ∈ Z d X i ∈ Z d ω j ( k − i ) e ii = 1 d X i = k X j ∈ Z d ω j ( k − i ) e ii + 1 d X j ∈ Z d e kk = e kk . Hence, the elements e , e , . . . , e dd , and P i ∈ Z d e i ( i − are in the image of ψ . It follows that e k ( k − ∈ Im ψ for all k ∈ Z d since e kk P i ∈ Z d e i ( i − = e k ( k − by Lemma 3.3 (iv). Finally, Lemma3.5 allows us to conclude that e ij ∈ Im ψ for all i, j ∈ Z d , implying that ψ is surjective as desired. (cid:3) The syzygy of a matrix factorization
Let ( S, n , k ) be a complete regular local ring, f ∈ S a non-zero non-unit, and set R = S/ ( f ). Fora matrix factorization X = ( ϕ : F → F , . . . , ϕ d : F → F d ), we study its syzygy, Ω MF dS ( f ) ( X ) =(Ω , Ω , . . . , Ω d ), defined in Section 2. Note that, by Section 3, the additional assumption of S being complete allows us to utilize the Krull-Remak-Schmidt Theorem in MF dS ( f ). As in previoussections, let P = L i ∈ Z d P i and Γ = End MF dS ( f ) ( P ) op .The exact structure on MF dS ( f ) ensures that Ω MF dS ( f ) ( X ) is stably equivalent to any matrixfactorization K such that there exists a short exact sequence K P X where P is projective. This follows from the appropriate version of Schanuel’s Lemma in MF dS ( f ). Lemma 5.1. (Schanuel’s Lemma) Let X ∈ MF dS ( f ) and suppose K P X q p and K ′ P ′ X q ′ p ′ are short exact sequences of matrix factorizations with P and P ′ projective. Then, P ⊕ K ′ ∼ = K ⊕ P ′ . (cid:3) We omit the proof as it follows from [B¨u10, Proposition 2.12]. The next Lemma follows directlyfrom Lemma 5.1, KRS, and the sequences (2.4) and (2.6).
Lemma 5.2.
Let
X, X ′ ∈ MF dS ( f ) . Then,(1) Ω MF dS ( f ) ( X ⊕ X ′ ) ∼ = Ω MF dS ( f ) ( X ) ⊕ Ω MF dS ( f ) ( X ′ ) (2) Ω − MF dS ( f ) ( X ⊕ X ′ ) ∼ = Ω − MF dS ( f ) ( X ) ⊕ Ω − MF dS ( f ) ( X ′ ) (3) Ω MF dS ( f ) ( X ) (respectively Ω − MF dS ( f ) ( X ) ) is projective if and only if X is projective. (cid:3) As a consequence, both Ω MF dS ( f ) ( − ) and Ω − MF dS ( f ) ( − ) define additive functors from the stablecategory MF dS ( f ) to itself. Proposition 5.3.
Let X ∈ MF dS ( f ) be of size n and P = L di =1 P i . Then, we have isomorphisms i) Ω MF dS ( f ) (Ω − MF dS ( f ) ( X )) ∼ = X ⊕ P ( d − n ,(ii) Ω − MF dS ( f ) (Ω MF dS ( f ) ( X )) ∼ = X ⊕ P ( d − n , and(iii) Ω MF dS ( f ) ( X ) ∼ = Ω − MF dS ( f ) ( X ) .Proof. Since Ω − MF dS ( f ) ( X ) is of size ( d − n , there is a short exact sequenceΩ MF dS ( f ) (Ω − MF dS ( f ) ( X )) P ( d − n Ω − MF dS ( f ) ( X ) . By applying Schanuel’s lemma to this sequence and the sequence (2.6), we find thatΩ MF dS ( f ) (Ω − MF dS ( f ) ( X )) ⊕ P n ∼ = X ⊕ P ( d − n . We may cancel one copy of P n from both sides by KRS to obtain the first statement. Dually, thesecond statement follows from the injective version of Schanuel’s Lemma.In order to prove (iii), we construct an explicit isomorphism. For each k ∈ Z d define an S -homomorphism α k : b F k → b F k by α k = F k +1 θ X ( k +1)( k +2) θ X ( k +1)( k +3) · · · θ X ( k +1)( k − F k +2 θ X ( k +2)( k +3) · · · θ X ( k +2)( k − F k +3 . . . ...... ... ... . . . θ X ( k − k − · · · F k − . It is not hard to see that each α k is an isomorphism and that the diagram b F k +1 b F k b F k +1 b F kα k +1 Ω k α k Ω − k commutes for all k ∈ Z d . Hence, we have an isomorphism of matrix factorizations ( α , . . . , α d ) :Ω MF dS ( f ) ( X ) → Ω − MF dS ( f ) ( X ). (cid:3) Remark 5.4.
In the case d = 2, no projective summands occur in the first two isomorphisms. Thisagrees with what we saw in Example 2.12 which said that the syzygy and cosyzygy operations areisomorphic to the shift functor:Ω MF S ( f ) ( ϕ, ψ ) ∼ = ( ψ, ϕ ) ∼ = Ω − MF S ( f ) ( ϕ, ψ )for any ( ϕ, ψ ) ∈ MF S ( f ). From this we can see that all three statements of Proposition 5.3 areimmediate when d = 2. In particular, the isomorphism in Proposition 5.3 (iii) is actually theidentity. In contrast, the isomorphism constructed in Proposition 5.3 (iii) when d = 3 is F ⊕ F F ⊕ F F ⊕ F F ⊕ F F ⊕ F F ⊕ F F ⊕ F F ⊕ F F ϕ F ! − ϕ − ϕ ϕ F ! F ϕ F ! − ϕ − ϕ ϕ F ! F ϕ F ! − ϕ − ϕ ϕ F ! F ϕ F ! − ϕ ϕ F − ϕ ! − ϕ ϕ F − ϕ ! − ϕ ϕ F − ϕ ! . he next Proposition uses Proposition 5.3 to show that each object in MF dS ( f ) has a projectiveresolution which is periodic of period at most 2. A projective resolution in this context is withrespect to the exact structure on MF dS ( f ) (see [B¨u10, 12.1]). Proposition 5.5.
Let X ∈ MF dS ( f ) . Then, X has a projective resolution which is periodic withperiod at most 2: · · · I ( X ) P ( X ) I ( X ) P ( X ) X. q p q p Proof.
Set Ω( X ) = Ω MF dS ( f ) ( X ) and Ω − ( X ) = Ω − MF dS ( f ) ( X ). Let α : Ω( X ) → Ω − ( X ) be theisomorphism constructed in Proposition 5.3. Then, we have two diagrams(5.1) I ( X ) P ( X ) I ( X )Ω( X ) X α − η X ρ X ǫ X λ X and(5.2) P ( X ) I ( X ) P ( X ) X Ω( X ) ρ X α − η X λ X ǫ X . The middle sequence in (5.2) is short exact since we have a commutative diagram
X I ( X ) Ω( X ) X I ( X ) Ω − ( X ) λ X α − η X αλ X η X with vertical isomorphisms. The desired resolution follows by splicing together (5.1) and (5.2), thatis, by setting p = ǫ X α − η X and q = λ X ρ X . (cid:3) Corollary 5.6.
Let P = L i ∈ Z d P i and Γ = End MF dS ( f ) ( P ) op . Then, every finitely generated left Γ -module has a projective resolution which is eventually periodic of period at most 2.Proof. Let N be a finitely generated Γ-module and set r = dim S . Let M = syz r Γ ( N ) be an arbitrary r th syzygy of N over Γ, and let0 M P r − P r − · · · P P N r − N for some finitely generated projective Γ-modules P i , i = 0 , , . . . , r −
1. Recall that finitely generated projective Γ-modules are in MCM(Γ), that is,they are finitely generated free S -modules. Thus, the Depth Lemma implies that depth S ( M ) = r .Since MCM S -modules are free, we have that M ∈ MCM(Γ) as well. Now, by Section 3, thereexists X ∈ MF dS ( f ) of size n such that F ( X ) = Hom MF dS ( f ) ( P , X ) ∼ = M . Since P is projective inMF dS ( f ), the functor F is exact. In particular, applying F to the periodic resolution constructedin Proposition 5.5 yields an exact sequence of MCM Γ-modules · · · F ( P ( X )) F ( I ( X )) F ( P ( X )) M . F ( p ) F ( q ) F ( p ) Actually, this is a free resolution of M over Γ since F ( P ( X )) ∼ = F ( L i ∈ Z d P ni ) ∼ = Γ n and similarly F ( I ( X )) ∼ = Γ n . Thus, splicing together this periodic free resolution of M and the projectiveresolution of N , we get an eventually periodic resolution of N . (cid:3) roposition 5.6 and Lemma 5.7 below give a homological description of Γ which resembles thatof a (commutative) hypersurface ring. Recall that a Noetherian ring Λ is said to be Iwanaga-Gorenstein if injdim Λ Λ and injdim Λ op Λ are both finite.
Lemma 5.7.
Let r be the Krull dimension of S . The ring Γ = End MF dS ( f ) ( P ) op is Iwanaga-Gorenstein of dimension r .Proof. First we show that any short exact sequence(5.3) 0 Γ
M M ′ q p with M, M ′ ∈ MCM(Γ) splits. To see this, first note that H (Γ) = HF ( P ) ∼ = P is injective inMF dS ( f ). Therefore, the short exact sequence of matrix factorizations H (Γ) H ( M ) H ( M ′ ) H ( q ) H ( p ) is split. Since H is full and faithful, there exists t : M → Γ such that H ( t ) H ( q ) = 1 H (Γ) and tq = 1 Γ which implies that (5.3) is split.To finish the proof, we apply results from [Aus86] which apply to both Γ and Γ op . By [Aus86,Lemma 1.1] we have that Γ ∼ = Hom S ( Q, S ) for some projective Γ op -module Q . The functorHom S ( (cid:3) , S ) : MCM(Γ) → MCM(Γ op ) defines a duality and therefore Q ∼ = Hom S (Hom S ( Q, S ) , S ) ∼ = Hom S ( Γ Γ , S ) . Thus, Hom S ( Γ Γ , S ) is Γ op -projective which happens if and only if Hom S ( Γ op Γ , S ) is Γ-projectiveaccording to [Aus86, Lemma 5.1]. In this case, [Aus86, 5.2] says that injdim Γ Γ = injdim S S whichis equal to r since S is Gorenstein. Interchanging the roles of Γ and Γ op we find that injdim Γ op Γ = r as well. (cid:3) Remark 5.8. If k is algebraically closed and its characteristic does not divide d , both Proposition5.6 and Lemma 5.7 can be restated for the skew group algebra over the d -fold branched cover R ♯ [ σ ]defined in Section 4.Over the hypersurface ring R = S/ ( f ), the reduced syzygy of an indecomposable non-free MCM R -module is again indecomposable. Proposition 5.14 gives an analogous result for matrix factor-izations. First, we describe the relationship between projective summands in MF dS ( f ) and freesummands in MCM( R ). Recall that a matrix factorization X ∈ MF dS ( f ) is stable if cok ϕ k is astable MCM R -module for all k ∈ Z d . Proposition 5.9.
Let X = ( ϕ , . . . , ϕ d ) ∈ MF dS ( f ) and set M i = cok ϕ i for all i ∈ Z d . Then, X has a projective summand isomorphic to P i if and only if M i has a free R -summand.Proof. The statement holds when d = 2 (for instance see [Yos90, 7.5]). So, assume d ≥
3. Onedirection is immediate: If X ∼ = X ′ ⊕ P i for some X ′ = ( ϕ ′ , . . . , ϕ ′ d ) ∈ MF dS ( f ) and i ∈ Z d , then M i ∼ = cok ϕ ′ i ⊕ R .A matrix factorization Y is a summand of X if and only if T j ( Y ) is a summand of T j ( X ) forany j ∈ Z d . Therefore, for the converse, we may assume i = 1. That is, assume M ∼ = M ⊕ R forsome MCM R -module M . By Proposition 1.6 (i), there exists ( ϕ : G → F, ψ : F → G ) ∈ MF S ( f ),with ϕ minimal, such that cok ϕ ∼ = M . Then,0 G ⊕ S F ⊕ S M ϕ f ! s a minimal free resolution of M over S . Thus, there exists isomorphisms α and β and a commu-tative diagram 0 F F M G ⊕ S ⊕ S m F ⊕ S ⊕ S m M β ϕ α ϕ f
00 0 1 Sm for some m ≥
0. In other words, we have an isomorphism of matrix factorizations in MF S ( f ):( ϕ , ϕ ϕ · · · ϕ d ) ∼ = ϕ f
00 0 1 S m , ψ f · S m ! . The isomorphisms α and β also give us an isomorphism of matrix factorizations in MF dS ( f ): X ∼ = ( αϕ , ϕ , . . . , ϕ d − , ϕ d α − ) ∼ = ( αϕ β − , βϕ , . . . ϕ d − , ϕ d α − ) . Let p : F ⊕ S ⊕ S m → S and p : G ⊕ S ⊕ S m → S be projection onto the middle components of F ⊕ S ⊕ S m and G ⊕ S ⊕ S m respectively. Consider the diagram F ⊕ S ⊕ S m F d · · · F F G ⊕ S ⊕ S m F ⊕ S ⊕ S m S S · · ·
S S S S. p ϕ d α − p βϕ ϕ ··· ϕ d − ϕ d − ϕ p βϕ ϕ ϕ p βϕ βϕ p αϕ β − p f The two right most squares commute, the first since αϕ β − = ϕ f
00 0 1 S m and the second byconstruction. Similarly, for k = 3 , , . . . , d −
1, the square F k +1 F k S S p βϕ ϕ ··· ϕ k − ϕ k ϕ k p βϕ ϕ ··· ϕ k − also commutes. Since p αϕ β − = f p , we have that f p = p f = p αϕ β − βϕ ϕ · · · ϕ d − ϕ d α − = f p βϕ ϕ · · · ϕ d − ϕ d α − . We may cancel f on the left to conclude that the left most square commutes as well. Thus, wehave a morphism X ∼ = ( αϕ β − , βϕ , . . . , ϕ d − , ϕ d α − ) → P . We claim that this morphism is an admissible epimorphism. By Lemma 2.3, it suffices to show thateach of the vertical maps depicted above are surjective. By the commutativity of the diagram, p = ( p βϕ ϕ · · · ϕ k )( ϕ k +1 ϕ k +2 · · · ϕ d − ϕ d α − )for each k = 2 , , . . . , d −
1. Since p is surjective, this implies p βϕ ϕ · · · ϕ k is surjective for each k = 2 , , . . . , d − P is projective, the admissible epimorphism X ։ P impliesthat X has a direct summand isomorphic to P . (cid:3) e note that KRS was not needed for the proof of Proposition 5.9. The result therefore holdsfor an arbitrary regular local ring S . The following corollary also holds without the completenessof S . Corollary 5.10.
Let S be a regular local ring, f a non-zero non-unit in S , and d ≥ .(i) The objects P , P , . . . , P d ∈ MF dS ( f ) are the only indecomposable projectives (equivalentlyinjectives) up to isomorphism.(ii) A matrix factorization X ∈ MF dS ( f ) is stable if and only if it has no non-zero projectivedirect summands.Proof. Let Q = ( Q , Q , . . . , Q d ) ∈ MF dS ( f ) be an indecomposable projective of size n . Then, wehave a short exact sequence of the form (2.4), and more specifically, Q is a direct summand of P ( Q ) ∼ = L i ∈ Z d P ni . It follows that, for any k ∈ Z d , cok Q k is either 0 or has a direct summandisomorphic to R . Since Q is non-zero matrix factorization, there exists j ∈ Z d such that cok Q j = 0.Thus, cok Q j has a direct summand isomorphic to R . By Proposition 5.9, this implies that Q hasa direct summand isomorphic to P j . Since Q is indecomposable, we have that Q ∼ = P j . Thestatement about indecomposable injectives follows immediately because of Lemma 2.14.The second statement follows by combining (i) and Proposition 5.9. (cid:3) Proposition 5.9 and Corollary 5.10 give us a clearer picture of the structure of matrix factoriza-tions and the MCM R -modules they encode. Proposition 5.11.
Let X = ( ϕ , . . . , ϕ d ) ∈ MF dS ( f ) . Then, (5.4) X ∼ = ˜ X ⊕ P s ⊕ P s ⊕ · · · ⊕ P s d d for some stable matrix factorization ˜ X = ( ˜ ϕ , . . . , ˜ ϕ d ) and integers s k ≥ , k ∈ Z d . The integers s k are uniquely determined and the direct summand ˜ X is unique up to isomorphism. The correspondingMCM R -modules are cok ϕ k ∼ = cok ˜ ϕ k ⊕ R s k where cok ˜ ϕ k is a stable MCM R -module for all k ∈ Z d .Proof. Lemma 5.10 says that the only indecomposable projective matrix factorizations are P , P , . . . , P d .Since S is complete, KRS implies that we have a decomposition X ∼ = ˜ X ⊕ P s ⊕ P s ⊕ · · · ⊕ P s d d for some unique integers s k ≥ X which has no projective summands. Finally, usingProposition 5.9, each copy of P k contributes a free R -summand to cok ϕ k . (cid:3) Corollary 5.12.
Let X = ( ϕ , . . . , ϕ d ) ∈ MF dS ( f ) of size n . Then, (5.5) Ω MF dS ( f ) ( X ) ∼ = ˜Ω ⊕ P m ⊕ · · · ⊕ P m d d where m k = n − µ R (cok ϕ k ) and ˜Ω ∈ MF dS ( f ) is stable. Furthermore, ˜Ω is of size P dk =1 µ R (cok ϕ k ) − n .Proof. The isomorphism (5.5) follows by combining Propositions 5.11 and 2.13. Let ℓ ≥ MF dS ( f ) ( X ) is of size ( d − n , we have that( d − n = ℓ + d X k =1 m k = ℓ + dn − d X k =1 µ R (cok ϕ k ) . Thus, ℓ = P dk =1 µ R (cok ϕ k ) − n . (cid:3) We continue with two more important properties of the stable part of the syzygy of a matrixfactorization. emma 5.13. Let X ∈ MF dS ( f ) and suppose K P X is a short exact sequence with P projective. Then, ˜Ω is isomorphic to a direct summand of K where Ω MF dS ( f ) ( X ) ∼ = ˜Ω ⊕ Q is a decomposition of the form (5.5) .Proof. This follows directly from Lemma 5.1 and Proposition 5.11. (cid:3)
Proposition 5.14.
Let X ∈ MF dS ( f ) be a indecomposable non-projective matrix factorization andlet Ω MF dS ( f ) ( X ) ∼ = ˜Ω ⊕ P be a decomposition of the form (5.5) . Then, ˜Ω is indecomposable.Proof. First, note that, ˜Ω = 0. Indeed, if Ω MF dS ( f ) ( X ) was projective, then Proposition 5.2 wouldimply that X is projective as well, which is not the case. So, assume ˜Ω = Y ⊕ Y for some non-zero Y , Y ∈ MF dS ( f ). Since ˜Ω is stable, the direct summands Y and Y are also stable. Then,Ω − MF dS ( f ) (Ω MF dS ( f ) ( X )) ∼ = Ω − MF dS ( f ) ( ˜Ω) ⊕ Ω − MF dS ( f ) ( P ) ∼ = Ω − MF dS ( f ) ( Y ) ⊕ Ω − MF dS ( f ) ( Y ) ⊕ Ω − MF dS ( f ) ( P ) . For i = 1 ,
2, decompose Ω − MF dS ( f ) ( Y i ) ∼ = U i ⊕ P i , for some stable U i and projective P i . Now, applyingProposition 5.3, we have that X ⊕ P ( d − n ∼ = U ⊕ U ⊕ P ⊕ P ⊕ Ω − MF dS ( f ) ( P )where n is the size of X . Since both sides of this isomorphism are decomposed into the form (5.4),we have that U ⊕ U ∼ = X . But X is indecomposable, so one of U or U must be zero. Re-indexingif necessary, we may assume U = 0. This implies that Ω − MF dS ( f ) ( Y ) is projective and therefore Y is projective. However, this is a contradiction since Y is a non-zero stable matrix factorization.Hence, ˜Ω is indecomposable. (cid:3) So far, we have refrained from assuming that X is a reduced matrix factorization. On the otherhand, if we do assume that X ∈ MF dS ( f ) is reduced, we obtain more concise versions of 2.13, 5.11,5.12, and 5.14. Corollary 5.15.
Let X ∈ MF dS ( f ) be reduced. Then, the following hold.(i) cok Ω k ∼ = syz R (cok ϕ k ) for each k ∈ Z d .(ii) Both X and Ω MF dS ( f ) ( X ) are stable.(iii) If X is indecomposable, then Ω MF dS ( f ) ( X ) is indecomposable.Proof. By Proposition 1.6(ii), there is a one-to-one correspondence between reduced matrix factor-izations in MF S ( f ) and stable MCM R -modules (Lemma 1.6). If X = ( ϕ , ϕ , . . . , ϕ d ) ∈ MF dS ( f ) isreduced, then ( ϕ k , ϕ k +1 ϕ k +2 · · · ϕ k − ) is a reduced matrix factorization in MF S ( f ) for each k ∈ Z d .Hence, cok ϕ k is a stable MCM R -module for each k ∈ Z d and cok( ϕ k +1 ϕ k +2 · · · ϕ k − ) is its reducedfirst syzygy. Since, by Proposition 2.13, cok Ω k ∼ = cok( ϕ k +1 ϕ k +2 · · · ϕ k − ), the first statement fol-lows. The second statement follows from Proposition 5.9 and the third follows from the second andProposition 5.14. (cid:3) Examples
Let S be a regular local ring with maximal ideal n and let f be a non-zero non-unit in S . Animportant distinction between the categories MF S ( f ) and MF dS ( f ), for d >
2, is the existence ofstable non-reduced matrix factorizations. We illustrate this with our first example. xample 6.1. Let d > f ∈ S can be written as f = f f · · · f d for some f i ∈ n .Then, ( f , f , . . . , f d ) is an indecomposable reduced matrix factorization of size 1 in MF dS ( f ). Since d >
2, the syzygy of ( f , f , . . . , f d ) in MF dS ( f ) is not reduced. However, Corollary 5.15 impliesthat Ω( f , f , . . . , f d ) is indecomposable and stable.It is tempting to think that the presence of unit entries must make Ω( f , . . . , f d ) redundant insome way. Actually, we can not replace Ω( f , . . . , f d ) with a reduced matrix factorization in thefollowing sense.Set Ω( f , . . . , f d ) = (Ω , Ω , . . . , Ω d ) and let b f k = f · · · f k − f k +1 · · · f d for each k ∈ Z d . Then, byProposition 2.13, we have that cok Ω k ∼ = R/ ( b f k ). Suppose X = ( ϕ , . . . , ϕ d ) is a reduced matrixfactorization of f such that cok ϕ k ∼ = R/ ( b f k ) for all k ∈ Z d . Since ϕ k is minimal, the short exactsequence 0 F k +1 F k R/ ( b f k ) 0 ϕ k is a minimal free resolution of R/ ( b f k ) over S . By tensoring with R , it follows that rank S F k = µ R ( R/ ( b f k )) = 1. That is, X is a matrix factorization of size 1. Thus, the homomorphism ϕ k isgiven by multiplication by b f k up to a unit, say ϕ k = v k b f k for v k ∈ S invertible. Since X is a matrixfactorization of f , we have that f = ϕ ϕ · · · ϕ d = v b f b f · · · b f d = vf d − where v = v v · · · v d . Since d >
2, this is a contradiction. We conclude that no reduced matrixfactorization has the same cokernels as Ω( f , . . . , f d ).With Example 6.1 in mind, we continue to focus on stable matrix factorizations. However, weintroduce an additional assumption (which implies stability) in order to avoid particularly trivialfactorizations. Definition 6.2.
Let S be a regular local ring, f a non-zero non-unit in S , and d ≥
2. A non-zeromatrix factorization X = ( ϕ , ϕ , . . . , ϕ d ) ∈ MF dS ( f ) is pseudoprojective if cok ϕ k = 0 for some k ∈ Z d .We aim to avoid matrix factorizations with pseudoprojective summands. Notice that any inde-composable projective (equivalently injective) in MF dS ( f ) is pseudoprojective (See 5.10).An indecomposable matrix factorization in MF S ( f ) is pseudoprojective if and only if it is projec-tive; the only possibilities are (1 , f ) and ( f, X ∈ MF S ( f ) is reduced if and onlyif it is stable if and only if it has no pseudoprojective summands (compare with 1.6 (ii)). However,for d >
2, these properties are not equivalent. The following lemma and examples give the preciserelationship.
Lemma 6.3.
Let X ∈ MF dS ( f ) with d ≥ .(i) If X is reduced, then X has no pseudoprojective summands.(ii) If X has no pseudoprojective summands, then X is stable.Proof. In order to prove (i), suppose X ∼ = X ′ ⊕ X ′′ for some X ′ = ( ϕ ′ , . . . , ϕ ′ d ) and X ′′ =( ϕ ′′ , . . . , ϕ ′′ d ) with X ′ pseudoprojective. Then, cok ϕ ′ k = 0 for some k ∈ Z d . Equivalently, ϕ ′ k isan isomorphism and therefore, ϕ k ∼ = ϕ ′ k ⊕ ϕ ′′ k is not minimal.Corollary 5.10(ii) showed that X is stable if and only if it has no non-zero projective summands.Since indecomposable projective matrix factorizations are pseudoprojective, (ii) follows. (cid:3) Examples 6.4 and 6.5 below show that the converses of (ii) and (i) respectively fail in general for d > Example 6.4.
Let ( ϕ : G → F, ψ : F → G ) ∈ MF S ( f ) be a reduced matrix factorization.(i) ( ϕ, ψ, F ) ∈ MF S ( f ) is stable but is itself pseudoprojective. ii) (cid:18)(cid:18) ϕ F (cid:19) , (cid:18) ψ F (cid:19) , (cid:18) F ϕ (cid:19) , (cid:18) F ψ (cid:19)(cid:19) ∈ MF S ( f ) is stable but it is a direct sum ofpseudoprojectives. Example 6.5.
Let S = k J x, y K , with k an uncountable algebraically closed field of characteristicdifferent from 2. Let f = x y ∈ S and, as usual, set R = S/ ( f ). Consider the following matrixfactorization of f with 3 factors: X = ( ϕ , ϕ , ϕ ) = (cid:18)(cid:18) x y − x (cid:19) , (cid:18) yx − x (cid:19) , (cid:18) x y (cid:19)(cid:19) . Notice that X is not reduced. We claim that X is indecomposable. To see this, first note thatthe R -module cok (cid:18) x y − x (cid:19) is an indecomposable MCM over the D ∞ hypersurface singularity R = k J x, y K / ( x y ) [BGS87, Proposition 4.2]. If X ∼ = Y ⊕ Y ′ for some non-zero matrix factorizations Y = ( u, v, w ) , Y ′ = ( u ′ , v ′ , w ′ ) ∈ MF S ( f ), both necessarily of size 1, thencok (cid:18) x y − x (cid:19) ∼ = cok( u ) ⊕ cok( u ′ ) . This implies that one of cok( u ) or cok( u ′ ) must be zero. Rearranging the summands if necessary,we may assume cok( u ′ ) = 0. Equivalently, we have that u ′ is multiplication by a unit in S . Thiswould imply a matrix equivalence (cid:18) x y − x (cid:19) ∼ (cid:18) u u ′ (cid:19) which is not possible since the matrix on the right has a unit entry. Therefore, no such decomposi-tion is possible. Since X is indecomposable and the matrices ϕ , ϕ , and ϕ are non-invertible, itfollows that X has no pseudoprojective summands.Set µ i = µ R (cok ϕ i ), i ∈ Z . Then, µ = 2 , µ = 2 , and µ = 1. Corollary 5.12 says thatwe should expect a direct sum decomposition of Ω MF S ( f ) ( X ) into a stable matrix factorization ofsize 3 and a projective isomorphic to P = (1 , , x y ). Indeed, using invertible row and columnoperations, one can construct an isomorphismΩ MF S ( f ) ( X ) = − y − xy − y − x x xy , − − x − y − x − y − x
01 0 0 00 1 0 0 , − x − y − x y x x − x ∼ = x xy − y − xy − y , − y xy − x − y , x y x − x . Proposition 5.14 implies that the stable part of Ω MF S ( f ) ( X ),˜Ω = x xy − y − xy − y , − y xy − x − y , x − x , is also an indecomposable matrix factorization of f = x y . Clearly, none of the matrices that makeup ˜Ω are invertible and therefore ˜Ω has no pseudoprojective summands.One benefit of having no pseudoprojective summands is the following. Lemma 6.6.
Assume X = ( ϕ , ϕ , . . . , ϕ d ) ∈ MF dS ( f ) has no pseudoprojective summands. If cok ϕ j is an indecomposable MCM R -module for some j ∈ Z d , then X is an indecomposable matrixfactorization. roof. Suppose X ∼ = X ′ ⊕ X ′′ for non-zero X ′ = ( ϕ ′ , . . . , ϕ ′ d ) , X ′′ = ( ϕ ′′ , . . . , ϕ ′′ d ) ∈ MF dS ( f ).Then, cok ϕ j ∼ = cok ϕ ′ j ⊕ cok ϕ ′′ j . Since cok ϕ j is indecomposable, it follows that one of ϕ ′ j or ϕ ′′ j is an isomorphism. That is, one of X ′ or X ′′ is pseudoprojective, a contradiction. Hence, X isindecomposable. (cid:3) Any partial converse to Lemma 6.6 would be very interesting but would likely need more condi-tions on X . The only result we have in this direction is a special case of Proposition 5.9 which saysthat the cokernels of an indecomposable non-projective matrix factorization have no free summands.In the case that f is an irreducible in S , we have another interesting consequence. Proposition 6.7.
Assume f ∈ S is irreducible. Then, any X ∈ MF dS ( f ) which is not pseudopro-jective must be of size at least d .Proof. Let X = ( ϕ : F → F , . . . , ϕ d : F → F d ) and set n to be the size of X . Since ϕ ϕ · · · ϕ d = f · F , we have that (det ϕ )(det ϕ ) · · · (det ϕ d ) = f n . Since f is irreducible, for each k ∈ Z d , there exists units u k ∈ S and integers r k ≥ ϕ k = u k f r k . Moreover, we have that n = P k ∈ Z d r k .Let k ∈ Z d . Since ( ϕ k , ϕ k +1 ϕ k +2 · · · ϕ k − ) ∈ MF S ( f ), [Eis80, Proposition 5.6] implies thatrank R (cok ϕ k ) = r k , where R = S/ ( f ) as usual. The non-negative integer r k = 0 if and only if ϕ k is an isomorphism. However, since X is not pseudoprojective, we have that r k >
0. Thus, n = P k ∈ Z d r k ≥ d . (cid:3) In the case d = 2, this proposition is simply a restatement of the definition of an irreducibleelement; if ( u, v ) ∈ MF S ( f ) is of size 1, then one of u or v must be a unit in S . Example 6.8.
Let k be an algebraically closed field of characteristic 0 and set S = k J x, y K . Let ω ∈ k be a primitive 3rd root of unity. Consider the simple curve singularity R = S/ ( x + y ) oftype E . Following the notation of [Yos90, Chapter 9], the indecomposable MCM R -module B isgiven by the matrix factorization( β, α ) = y xx − y x − y , y x xy xy − y x x − xy − y ∈ MF S ( x + y )in the sense that B = cok β . The matrix α factors further which gives rise to a matrix factorizationof x + y with 3 factors:( β, β , β ) := y xx − y x − y , − y ωxωx − y ωx y , − y ω xω x y ω x − y As we can see, the entries of β, β , and β lie in the maximal ideal of S , that is, ( β, β , β ) ∈ MF S ( x + y ) is reduced. Combining Lemma 6.3, Lemma 6.6, and the fact that B = cok β is anindecomposable MCM R -module, we conclude that ( β, β , β ) is indecomposable. We also notethat, since det β = x + y , the matrix α is the adjoint of β . Hence, α = β β is a factorization ofthe adjoint.The matrix factorization ( β, β , β ) is essentially the one constructed in [BES17, Proposition 2.1].Similar examples are also found using this construction. For instance,( E ) Let f = x + xy ∈ S . y x − x xy − x y , xy ωx − ωx y − ωx y , y ω x − ω x y − ω x xy ∈ MF S ( x + xy ) E ) Let f = x + y ∈ S . y − x y − xx y , y − ωx y − ωxωx y , y − ω x y − ω xω x y ∈ MF S ( x + y )and y − x y − xx y , y − ωx y − ωxωx y , y − ω x y − ω xω x y ∈ MF S ( x + y ) . Acknowledgements
The author would like to thank his advisor, Graham J. Leuschke, for his dedication to thisproject and for his guidance and encouragement throughout its preparation.
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