A truncated minimal free resolution of the residue field
aa r X i v : . [ m a t h . A C ] D ec A TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD
VAN C. NGUYEN AND OANA VELICHE
Abstract.
In a paper in 1962, Golod proved that the Betti sequence of the residue field of a local ringattains an upper bound given by Serre if and only if the homology algebra of the Koszul complex of thering has trivial multiplications and trivial Massey operations. This is the origin of the notion of Golodring. Using the Koszul complex components he also constructed a minimal free resolution of the residuefield. In this article, we extend this construction up to degree five for any local ring. We describe how themultiplicative structure and the triple Massey products of the homology of the Koszul algebra are involvedin this construction. As a consequence, we provide explicit formulas for the first six terms of a sequencethat measures how far the ring is from being Golod. Introduction
Throughout the paper, ( R, m , k ) is a local noetherian ring with maximal ideal m and residue field k , ofembedding dimension n and codepth c = n − depth R . Let K R be the Koszul complex of R on a minimal setof generators of m . Serre pointed out that there is always a coefficient-wise inequality between the Poincar´eseries of the R -module k and a series of rational form involving the ranks of the homologies of K R (1.0.1) P R k ( t ) := ∞ X i =0 rank k Tor Ri ( k , k ) t i (1 + t ) n − P ci =1 rank k H i ( K R ) t i +1 . In [6], Golod proved that a ring R attains this upper bound if and only if the graded-commutative algebra A := H( K R ) has trivial multiplications and trivial Massey operations; such a ring is now called a Golod ring .In the same paper, he also constructed the minimal free resolution of the R -module k in terms of K R .For any local ring, not necessarily Golod, using the components of the Koszul complex K R as buildingblocks and the graded-commutative structure of A , we give a detailed description of the minimal free resolu-tion of the R -module k , up to degree five; see Construction 3.1 and Theorem 3.2. The higher degrees of theresolution can be extended similarly for some special cases of R , but in general it requires an understandingof higher Massey products, which remains elusive and is left for future projects.Toward this goal, in Section 2 we analyze in details the multiplicative structure of the algebra A , up todegree four, in terms of bases of A i and A i · A j , and necessary maps. Moreover, in degree four the Masseyproducts appear for the first time, as ternary Massey products h A , A , A i ; we give a description of theelements of this set in Proposition 2.10. In Section 4, we obtain several direct applications of Construction 3.1by considering the difference of series P ( t ) := (1 + t ) n − P ci =1 rank k H i ( K R ) t i +1 − P R k ( t )that measures how far the ring R is from being Golod, that is, how far the Betti numbers of R are from theirmaximum possible values. In Proposition 4.3 we compute the first six coefficients of P ( t ) in terms of themultiplicative invariants of A . In Proposition 4.4, for any local ring of embedding dimension n with rationalPoincar´e series of the form (1 + t ) n /d ( t ) , we express d ( t ) in terms of those invariants, up to degree five. InSection 5, we illustrate Construction 3.1 through an example of a ring of codepth 4 examined by Avramov in[1]. In that example he provided a nontrivial indecomposable Massey product element in h A , A , A i thatdoes not come from multiplications of the homology. Using Proposition 2.10, we prove that, up to a scalar,this is the only element with this property, modulo the products in homology. Date : December 11, 2020.2020
Mathematics Subject Classification.
Key words and phrases.
Golod rings, minimal free resolution, Tor algebra, Massey products.
V. C. NGUYEN AND O. VELICHE Multiplicative structure on the homology of the Koszul algebra
Let ( R, m , k ) be a local ring of embedding dimension n and codepth c . The differential graded algebrastructure on the Koszul complex K = K R on a minimal set of generators of the maximal ideal m induces agraded-commutative algebra structure on the homology of K : A := H( K ) = A ⊕ A ⊕ A ⊕ · · · ⊕ A c . In this section we discuss this multiplicative structure up to degree four. We will use this structure extensivelyin constructing a truncated minimal free resolution of the residue field over R in the next section. Thefollowing notation is used throughout the paper: the differential map on K is denoted by ∂ K , a homogeneouselement of degree i in the Koszul complex K is denoted by π i , a representative in K i of an element in thehomology A i is denoted by p i , and elements of R are denoted by α ’s, β ’s, and γ ’s. The subscript of ahomogeneous element indicates its index in a tuple and the superscript indicates its homological degree. Set a i := rank k ( A i ) and q ij := rank k ( A i · A j ) , for all 0 ≤ i, j ≤ c. It is clear that A = k . For each 1 ≤ i ≤ a we consider z i ∈ Ker ∂ K such that { [ z i ] } i =1 ,...,a is a basisof A . In particular, every element in Ker ∂ K can be written as a sum of elements in Im ∂ K and a linearcombination of { z i } i =1 ,...,a . Let A be a k -subspace of A such that A = ( A · A ) ⊕ A . For each 1 ≤ ℓ ≤ a − q , we consider z ℓ ∈ Ker ∂ K such that { [ z ℓ ] } ℓ =1 ,...,a − q is a basis of A . Inparticular, every element in Ker ∂ K can be written as a sum of elements in Im ∂ K , a linear combination of { z i ∧ z j } i,j =1 ,...,a , and a linear combination of { z ℓ } ℓ =1 ,...,a − q .Consider the multiplication map:(2.1.1) φ : A ⊗ A → A , defined by φ ([ x ] ⊗ [ y ]) = [ x ] ∧ [ y ] . As Im φ = A · A , we have rank k (Ker φ ) = a − q . For all 1 ≤ i ≤ a and 1 ≤ s ≤ a − q , we choose e p si ∈ Ker ∂ K such that(2.1.2) n a X i =1 [ z i ] ⊗ [ e p si ] o s =1 ,...,a − q is a basis of Ker φ .Since for each s , we have P a i =1 [ z i ] ∧ [ e p si ] = 0, so there exists e π s ∈ K such that(2.1.3) ∂ K ( e π s ) = a X i =1 z i ∧ e p si . Lemma 2.2.
Let e p si ∈ Ker ∂ K be as in (2.1.2) . Then, the vectors { (cid:0) [ e p s ] , . . . , [ e p sa ] (cid:1) } s =1 ,...,a − q in A a are linearly independent.Proof. If α s is in R such that P a − q s =1 (cid:0) [ e p s ] , . . . , [ e p sa ] (cid:1) ∧ [ α s ] = 0, then a − q X s =1 [ e p si ] ∧ [ α s ] = 0 for all 1 ≤ i ≤ a . By tensoring with [ z i ] and taking the sum over i we get a X i =1 [ z i ] ⊗ a − q X s =1 [ e p si ] ∧ [ α s ] = a − q X s =1 (cid:16) a X i =1 [ z i ] ⊗ [ e p si ] (cid:17) ∧ [ α s ] = 0 . The desired conclusion now follows from (2.1.2). (cid:3)
Proposition 2.3.
Let { [ z i ] } i =1 ,...,a and { [ z ℓ ] } ℓ =1 ,...,a − q be bases for A and A respectively. If β ij and α ℓ are in R such that a X i,j =1 [ z i ] ∧ [ z j ] ∧ [ β ij ] + a − q X ℓ =1 [ z ℓ ] ∧ [ α ℓ ] = 0 , TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD 3 then the following hold: ( a ) [ α ℓ ] = 0 for all ≤ ℓ ≤ a − q ; ( b ) There exist γ s in R such that for all ≤ i ≤ a a X j =1 [ z j ] ∧ [ β ij ] = a − q X s =1 [ e p si ] ∧ [ γ s ] , where e p si is as in (2.1.2) .Proof. (a): The first sum in the hypothesis equality is in A · A . The vectors { [ z ℓ ] } ℓ =1 ,...,a − q form a basisin A that completes a basis of A · A , thus [ α ℓ ] = 0 for all 1 ≤ ℓ ≤ a − q .(b): By (a) it follows that 0 = P a i,j =1 [ z i ] ∧ [ z j ] ∧ [ β ij ] = P a i =1 [ z i ] ∧ (cid:16) P a j =1 [ z j ] ∧ [ β ij ] (cid:17) , which impliesthat P a i =1 [ z i ] ⊗ (cid:16) P a j =1 [ z j ] ∧ [ β ij ] (cid:17) is in Ker φ . The desired conclusion now follows from (2.1.2). (cid:3)
Let A be a k -subspace of A such that A = ( A · A ) ⊕ A = ( A · A · A + A · A ) ⊕ A . For each 1 ≤ t ≤ a − q , we consider z t ∈ Ker ∂ K such that { [ z t ] } t =1 ,...,a − q is basis of A . In par-ticular, every element in Ker ∂ K can be written as a sum of elements in Im ∂ K , a linear combinationof { z i ∧ z j ∧ z k } i,j,k =1 ,...,a , a linear combination of { z i ∧ z ℓ } i =1 ,...,a ℓ =1 ,...,a − q , and a linear combination of { z t } t =1 ,...,a − q .Consider the map:(2.4.1) ψ : A ⊗ A ⊗ A → (cid:0) ( A · A ) ⊗ A (cid:1) ⊕ (cid:0) A ⊗ ( A · A ) (cid:1) , defined by ψ ([ x ] ⊗ [ y ] ⊗ [ z ]) = (cid:16) [ x ] ∧ [ y ] ⊗ [ z ] , [ x ] ⊗ [ y ] ∧ [ z ] (cid:17) for all [ x ] , [ y ] , [ z ] ∈ A and set(2.4.2) b = rank k (Coker ψ ) . Remark that b = 0 if and only if ψ is surjective. If q = 0, then A · A = 0 and hence b = 0. Lemma 2.5.
The sequence A ⊗ A ⊗ A ψ −→ (cid:0) ( A · A ) ⊗ A (cid:1) ⊕ (cid:0) A ⊗ A (cid:1) µ −→ A , where ψ ([ x ] ⊗ [ y ] ⊗ [ z ]) = (cid:16) [ x ] ∧ [ y ] ⊗ [ z ] , [ x ] ⊗ [ y ] ∧ [ z ] (cid:17) , and µ (([ u ] ⊗ [ x ] , [ y ] ⊗ [ v ])) = [ u ] ∧ [ x ] − [ y ] ∧ [ v ] , for all [ x ] , [ y ] , [ z ] ∈ A , [ u ] ∈ ( A · A ) and [ v ] ∈ A , is a complex whose homology has rank a a − a q − q + b. Proof.
We first show that µ ◦ ψ = 0. µ ( ψ ([ x ] ⊗ [ y ] ⊗ [ z ])) = µ (cid:16)(cid:16) [ x ] ∧ [ y ] ⊗ [ z ] , [ x ] ⊗ [ y ] ∧ [ z ] (cid:17)(cid:17) = [ x ] ∧ [ y ] ∧ [ z ] − [ x ] ∧ [ y ] ∧ [ z ] = 0 . Hence, the sequence in the statement of the lemma is a complex. The homology of this complex hasrank k (cid:16) Ker µ Im ψ (cid:17) = rank k (Ker µ ) − rank k (Im ψ )= rank k (cid:16)(cid:0) ( A · A ) ⊗ A (cid:1) ⊕ (cid:0) A ⊗ A (cid:1)(cid:17) − rank k (Im µ ) − rank k (Im ψ )= a q + a a − q − rank k (cid:16)(cid:0) ( A · A ) ⊗ A (cid:1) ⊕ (cid:0) A ⊗ ( A · A ) (cid:1)(cid:17) + b = a q + a a − q − a q + b = a a − a q − q + b. (cid:3) V. C. NGUYEN AND O. VELICHE
To describe a basis of the homology of the complex in Lemma 2.5, it is enough to observe that Im ψ isgenerated by elements of the form ([ z i ] ∧ [ z j ] ⊗ [ z k ] , [ z i ] ⊗ [ z j ] ∧ [ z k ]) for 1 ≤ i, j, k ≤ a and that in Ker µ/ Im ψ we have (cid:2) ([ z i ] ∧ [ z j ] ⊗ [ z k ] , (cid:3) = − (cid:2) (0 , [ z i ] ⊗ [ z j ] ∧ [ z k ]) (cid:3) , as [([ z i ] ∧ [ z j ] ⊗ [ z k ] , [ z i ] ⊗ [ z j ] ∧ [ z k ])] = 0 . Wechoose e p ui ∈ Ker ∂ K such that(2.5.1) nh , (cid:16) a X i =1 [ z i ] ⊗ [ e p ui ] (cid:17)io u =1 ,...,a a − a q − q + b, is a basis of the homology of the complex.By definition of µ , for each u we have P a i =1 [ z i ] ∧ [ e p ui ] = 0 in A . Therefore, for each u there exists e π u ∈ K such that(2.5.2) ∂ K ( e π u ) = a X i =1 z i ∧ e p ui . The following result gives a method for finding e p ui from (2.5.1). Proposition 2.6.
Consider the map φ : A ⊗ A → A , defined by φ ([ x ] ⊗ [ v ]) = [ x ] ∧ [ v ] , and set A = n a X i =1 [ z i ] ⊗ [ e p si ] ∧ [ z j ] (cid:12)(cid:12) ≤ j ≤ a and ≤ s ≤ a − q o and B = n a X i =1 [ z i ] ⊗ [ e p ui ] (cid:12)(cid:12) ≤ u ≤ a a − a q − q + b o , where e p si and e p ui are as in (2.1.2) and (2.5.1) respectively. Then, the following hold: ( a ) The set B is linearly independent and Ker φ = (Span k A ) ⊕ (Span k B ) . ( b ) Span k A ⊆ A ⊗ ( A · A ) and for b defined in (2.4.2) , b = a q − rank k (Span k A ) . ( c ) If B is a k -subspace of Ker φ such that Ker φ = (Span k A ) ⊕ B, then for every basis B ′ of B the set [(0 , B ′ )] is a basis of the homology of the complex in Lemma 2.5.Proof. (a): The linear independence of B follows from (2.5.1). Next, we show that the elements of A and B generate the kernel of φ . By definitions of e p si and e p ui , it is clear that the listed elements are in the kernelof φ . Let P a i =1 [ z i ] ⊗ [ p i ] ∈ Ker φ , for some [ p i ] ∈ A . Then [(0 , P a i =1 [ z i ] ⊗ [ p i ])] ∈ Ker µ , and thus bydefinition (2.5.1) for all u, i, j there exist δ u ∈ R and p ij ∈ Ker ∂ K such that (cid:16) , a X i =1 [ z i ] ⊗ [ p i ] (cid:17) = a a − a q − q + b X u =1 (cid:16) , a X i =1 [ z i ] ⊗ [ e p ui ] ∧ [ δ u ] (cid:17) + a X i,j =1 (cid:16) [ z i ] ∧ [ p ij ] ⊗ [ z j ] , [ z i ] ⊗ [ p ij ] ∧ [ z j ] (cid:17) . (2.6.1)Comparing the first components of both sides of (2.6.1), for each j we have:0 = a X i =1 [ z i ] ∧ [ p ij ] . By (2.1.2), there exist ε js in R such that for all 1 ≤ i, j ≤ a we write:[ p ij ] = a − q X s =1 [ e p si ] ∧ [ ε js ] . TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD 5
Comparing the second components of both sides of (2.6.1), and using the above expression for [ p ij ], for each i we have:[ p i ] = a a − a q − q + b X u =1 [ e p ui ] ∧ [ δ u ] + a X j =1 [ p ij ] ∧ [ z j ] = a a − a q − q + b X u =1 [ e p ui ] ∧ [ δ u ] + a − q X s =1 a X j =1 [ e p si ] ∧ [ z j ] ∧ [ ε js ] . Thus, P a i =1 [ z i ] ⊗ [ p i ] ∈ Ker φ is a linear combination of elements of A and B . Finally, we show thatSpan k A ∩
Span k B = { } . By definition of e p si we have P a i =1 [ z i ] ∧ [ e p si ] = 0 for all 1 ≤ s ≤ a − q . Bydefinition of ψ , for all s and j we obtain(2.6.2) ψ (cid:16) a X i =1 [ z i ] ⊗ [ e p si ] ⊗ [ z j ] (cid:17) = (cid:16) , a X i =1 [ z i ] ⊗ [ e p si ] ∧ [ z j ] (cid:17) , hence (0 , A ) ⊆ Im ψ . On the other hand by (2.5.1), [(0 , B )] is a basis of the homology of the complex inLemma 2.5. Therefore, Span k A and Span k B have no nontrivial elements in common, and part (a) holds.(b): The inclusion follows from definition of A . Observe that rank k (Ker φ ) = a a − q , and by thelinear independence of B from part (a), rank k (Span k B ) = a a − q − ( a q − b ). Therefore, by part (a),rank k (Span k A ) = a q − b. (c): By parts (a) and (b), every basis B ′ of B has a a − a q − q + b elements. It is clear that(0 , B ′ ) ⊆ Ker µ , with µ as in Lemma 2.5 and that the set [(0 , B ′ )] has at most a a − a q − q + b elements.It is enough to show that it is a generating set for the homology of the complex in Lemma 2.5 to concludethat [(0 , B ′ )] has exactly a a − a q − q + b elements, hence it forms a basis for the homology. Since B ⊆
Ker φ , every element of B can be written as a linear combination of elements of A and B ′ . In particular,every basis element in [(0 , B )] can be written as a linear combination of elements in [(0 , A )] and [(0 , B ′ )].However, each element in [(0 , A )] is zero in the homology, as remarked in (2.6.2), thus [(0 , B ′ )] is a generatingset. Part (c) now holds. (cid:3) Proposition 2.7.
Let { [ z i ] } i =1 ,...,a , { [ z ℓ ] } ℓ =1 ,...,a − q , and { [ z t ] } t =1 ,...,a − q be bases for A , A , and A respectively. If γ ijk , β iℓ , and α t are in R such that a X i,j,k =1 [ z i ] ∧ [ z j ] ∧ [ z k ] ∧ [ γ ijk ] + a X i =1 a − q X ℓ =1 [ z i ] ∧ [ z ℓ ] ∧ [ β iℓ ] + a − q X t =1 [ z t ] ∧ [ α t ] = 0 , then the following hold: ( a ) [ α t ] = 0 for all ≤ t ≤ a − q ; ( b ) There exist δ u and ε js in R such that a X j,k =1 [ z j ] ∧ [ z k ] ∧ [ γ ijk ] + a − q X ℓ =1 [ z ℓ ] ∧ [ β iℓ ] = a X j =1 a − q X s =1 [ e p si ] ∧ [ z j ] ∧ [ ε js ] + a a − a q − q + b X u =1 [ e p ui ] ∧ [ δ u ] , where e p si and e p ui are as in (2.1.2) and (2.5.1) respectively.Proof. (a): The first two sums in the hypothesis equality are in A · A . As { [ z t ] } t =1 ,...,a − q is a basis in A that completes a basis of A · A , this implies [ α t ] = 0 for all 1 ≤ t ≤ a − q .(b): By (a) it follows that a X i =1 [ z i ] ∧ (cid:16) a X j,k =1 [ z j ] ∧ [ z k ] ∧ [ γ ijk ] + a − q X ℓ =1 [ z ℓ ] ∧ [ β iℓ ] (cid:17) = 0 , hence P a i =1 [ z i ] ⊗ (cid:0) P a j,k =1 [ z j ] ∧ [ z k ] ∧ [ γ ijk ] + P a − q ℓ =1 [ z ℓ ] ∧ [ β iℓ ] (cid:1) is in Ker φ . The desired assertion nowfollows from Proposition 2.6(a). (cid:3) Massey products occur in degrees four and higher. In degreefour one may obtain only triple Massey products of elements of degree one. The
Massey product of triplets[ x ] , [ y ] , [ z ] ∈ A satisfying [ x ] ∧ [ y ] = 0 and [ y ] ∧ [ z ] = 0 , V. C. NGUYEN AND O. VELICHE is a subset of A defined as follows: h [ x ] , [ y ] , [ z ] i = { [ π xy ∧ z + x ∧ π yz ] | ∂ K ( π xy ) = x ∧ y and ∂ K ( π yz ) = y ∧ z } , for some π xy , π yz ∈ K . Remark 2.9.
Let [ x ] , [ y ] , [ z ] ∈ A with [ x ] ∧ [ y ] = 0 and [ y ] ∧ [ z ] = 0 . Choose π xy and π yz in K such that ∂ K ( π xy ) = x ∧ y and ∂ K ( π yz ) = y ∧ z . Then, every element of h [ x ] , [ y ] , [ z ] i is of the form[( π xy + p xy ) ∧ z + x ∧ ( π yz + p yz )] = [ π xy ∧ z + x ∧ π yz ] + [ p xy ] ∧ [ z ] + [ x ] ∧ [ p yz ] , for some p xy and p yz ∈ Ker ∂ K . Therefore, h [ x ] , [ y ] , [ z ] i = [ π xy ∧ z + x ∧ π yz ] + A · [ z ] + [ x ] · A . The element π xy ∧ z + x ∧ π yz is called a representative of the triple Massey product h [ x ] , [ y ] , [ z ] i . Here, A · [ z ] + [ x ] · A is called the indeterminacy of the Massey operation, see e.g. May [11] and [1, Section 7].Let h A , A , A i denote the set of elements in A which are Massey products of triplets in A . The nextresult describes the elements of this set. Proposition 2.10.
Each element in h A , A , A i ⊆ A has a representative a − q X s =1 e π s ∧ p s such that a − q X s =1 [ e p si ∧ p s ] = 0 , for all i = 1 . . . , a , where p s ∈ Ker ∂ K , and e p si and e π s are defined in (2.1.2) and (2.1.3) respectively.Proof. Let [ x ] , [ y ] , [ z ] ∈ A such that [ x ] ∧ [ y ] = 0 and [ y ] ∧ [ z ] = 0 . We write [ y ] = a X i =1 [ z i ] ∧ [ α i ] for some α i ∈ R . Since [ x ] ∧ [ y ] = a X i =1 [ z i ] ∧ [ x ] ∧ [ − α i ] = 0, we have a X i =1 [ z i ] ⊗ [ x ] ∧ [ − α i ] ∈ Ker φ . Hence, by (2.1.2)there exist β s ∈ R such that [ x ] ∧ [ − α i ] = P a − q s =1 [ e p si ] ∧ [ β s ] for all i, so[ x ] ∧ [ y ] = a − q X s =1 (cid:16) a X i =1 [ z i ∧ e p si ] (cid:17) ∧ [ β s ] = a − q X s =1 [ ∂ K ( e π s )] ∧ [ β s ] = ∂ K (cid:16) a − q X s =1 e π s ∧ β s (cid:17) . Therefore, we may choose π xy = P a − q s =1 e π s ∧ β s . Similarly, using the equality [ y ] ∧ [ z ] = 0, we may choose π yz = a − q X s =1 e π s ∧ γ s , for some γ s ∈ R , where [ z ] ∧ [ α i ] = P a − q s =1 [ e p si ] ∧ [ γ s ] for all i. It follows that any element in h A , A , A i has a representative π xy ∧ z + x ∧ π yz = a − q X s =1 e π s ∧ ( z ∧ β s − x ∧ γ s ) . If we set p s = z ∧ β s − x ∧ γ s , then for each 1 ≤ i ≤ a we have a − q X s =1 [ e p si ∧ p s ] = a − q X s =1 [ e p si ] ∧ [ z ∧ β s − x ∧ γ s ] = (cid:16) a − q X s =1 [ e p si ∧ β s ] (cid:17) ∧ [ z ] + [ x ] ∧ (cid:16) a − q X s =1 [ e p si ∧ γ s ] (cid:17) = − [ x ] ∧ [ z ] ∧ [ α i ] + [ x ] ∧ [ z ] ∧ [ α i ] = 0 . (cid:3) Let A be a k -subspace of A such that A = ( A · A + A · A + Span k h A , A , A i ) ⊕ A and set a = rank k (cid:16) A · A + A · A + Span k h A , A , A i (cid:17) . TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD 7
For some choice of z r ∈ Ker ∂ K , let { [ z r ] } r =1 ,...,a − a be a basis of A . In particular, every element inKer ∂ K can be written as a sum of elements in Im ∂ K , a linear combination of { z i ∧ p i } i =1 ,...,a , a linearcombination of { z ℓ ∧ p ℓ } ℓ =1 ,...,a − q , a linear combination of { e π s ∧ p s } s =1 ,...,a − q , and a linear combinationof { z r } r =1 ,...,a − a , where p i ∈ Ker ∂ K , p ℓ ∈ Ker ∂ K and p s ∈ Ker ∂ K , such that p s is as in Proposition 2.10and e π s as in (2.1.3). Lemma 2.12.
All the elements z i , z ℓ , z t , z r , e p si , e π s , e p ui , and e π u , defined in this section, are in m K .Proof. From the inclusions Ker ∂ Kj ⊆ m K j for all j ≥
0, we obtain that z i , z ℓ , z t , z r , e p si , and e p ui are in m K .The assertions for the preimages e π s and e π u follows from the inclusions ( ∂ Kj ) − ( m K j − ) ⊆ m K j for j = 3and 4 respectively. (cid:3) Truncated minimal free resolution of the residue field
In this section, we construct the beginning of the minimal free resolution of the residue field k over localring ( R, m , k ), by using the Koszul complex K of R and the graded-commutative structure of the algebra A = H( K ) described in Section 2. Construction 3.1.
We consider the sequence of free R -modules: F : F ∂ F −−→ F ∂ F −−→ F ∂ F −−→ F ∂ F −−→ F ∂ F −−→ F , where F := K F := K F := K ⊕ K a F := K ⊕ K a ⊕ K a − q F := K ⊕ K a ⊕ K a − q ⊕ K a − q ⊕ K a − q F := K ⊕ K a ⊕ K a − q ⊕ K a − q ⊕ K a − a ⊕ K a − q ⊕ K a a − a q − q + b ⊕ K a a − a q . Using the elements described in Section 2, the differential maps of F are defined as follows.(3.1.1) ∂ F : K → K , is given by ∂ F := ∂ K . (3.1.2) ∂ F : K ⊕ K a → K , is given by ∂ F := (cid:0) ∂ K z ∧ (cid:1) , that is ∂ F π ( α i ) i =1 ,...,a ! := ∂ K ( π ) + a X i =1 z i ∧ α i . (3.1.3) ∂ F : K ⊕ K a ⊕ K a − q → K ⊕ K a , is given by ∂ F := ∂ K z ∧ − z ∧ ∂ K ) a ! , that is ∂ F π ( π i ) i =1 ,...,a ( α ℓ ) ℓ =1 ,...,a − q := ∂ K ( π ) + a X i =1 z i ∧ π i − a − q X ℓ =1 z ℓ ∧ α ℓ ( ∂ K ( π i )) i =1 ,...,a . (3.1.4) ∂ F : K ⊕ K a ⊕ K a − q ⊕ K a − q ⊕ K a − q → K ⊕ K a ⊕ K a − q , is given by ∂ F := ∂ K z ∧ z ∧ z ∧ − e π ∧ ∂ K ) a e p ∧ ) a ∂ K ) a − q , V. C. NGUYEN AND O. VELICHE that is ∂ F π ( π i ) i =1 ,...,a ( π ℓ ) ℓ =1 ,...,a − q ( α t ) t =1 ,...,a − q ( β s ) s =1 ,...,a − q := ∂ K ( π ) + a X i =1 z i ∧ π i + a − q X ℓ =1 z ℓ ∧ π ℓ + a − q X t =1 z t ∧ α t − a − q X s =1 e π s ∧ β s ( ∂ K ( π i ) + a − q X s =1 e p si ∧ β s ) i =1 ,...,a ( ∂ K ( π ℓ )) ℓ =1 ,...,a − q . (3.1.5) ∂ F : K ⊕ K a ⊕ K a − q ⊕ K a − q ⊕ K a − a ⊕ K a − q ⊕ K a a − a q − q + b ⊕ K a a − a q → K ⊕ K a ⊕ K a − q ⊕ K a − q ⊕ K a − q is given by ∂ F := ∂ K z ∧ − z ∧ z ∧ z ∧ − e π ∧ − e π ∧
00 ( ∂ K ) a e p ∧ ) a ( e p ∧ ) a − ( z ∧ ) a ∂ K ) a − q z ∧ ) a − q ∂ K ) a − q ∂ K ) a − q , that is ∂ F π ( π i ) i =1 ,...,a ( π ℓ ) ℓ =1 ,...,a − q ( π t ) t =1 ,...,a − q ( α r ) r =1 ,...,a − a ( π ′ s ) s =1 ,...,a − q ( β u ) u =1 ,...,a a − a q − q + b ( γ ℓi ) ℓ =1 ,...,a − q ; i =1 ,...,a := ∂ K ( π ) + a X i =1 z i ∧ π i − a − q X ℓ =1 z ℓ ∧ π ℓ + a − q X t =1 z t ∧ π t + a − a X r =1 z r ∧ α r − a − q X s =1 e π s ∧ π ′ s − a a − a q − q + b X u =1 e π u ∧ β u (cid:16) ∂ K ( π i ) + a − q X s =1 e p si ∧ π ′ s + a a − a q − q + b X u =1 e p ui ∧ β u − a − q X ℓ =1 z ℓ ∧ γ ℓi (cid:17) i =1 ,...,a (cid:16) ∂ K ( π ℓ ) + a X i =1 z i ∧ γ ℓi (cid:17) ℓ =1 ,...,a − q (cid:16) ∂ K ( π t ) (cid:17) t =1 ,...,a − q (cid:16) ∂ K ( π ′ s ) (cid:17) s =1 ,...,a − q . Theorem 3.2.
Let ( R, m , k ) be a local ring. The sequence F constructed in 3.1 is a truncated minimal freeresolution of k over R , up to homological degree five.Proof. The minimality follows from Lemma 2.12. We show exactness at each degree by using the Koszulrelations in the complex K , and the basis elements and maps defined in Section 2. Exactness at degree one. Im ∂ F ⊆ Ker ∂ F : ∂ F ◦ ∂ F π ( α i ) i =1 ,...,a ! = ∂ K (cid:0) ∂ K ( π ) + P a i =1 z i ∧ α i (cid:1) = 0 . TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD 9 Im ∂ F ⊇ Ker ∂ F : If π ∈ Ker ∂ F , then there exist π ∈ K and α i ∈ R such that π = ∂ K ( π ) + a X i =1 z i ∧ α i , therefore , π = ∂ F π ( α i ) i =1 ,...,a ! . Exactness at degree two. Im ∂ F ⊆ Ker ∂ F : ∂ F ◦ ∂ F π ( π i ) i =1 ,...,a ( α ℓ ) ℓ =1 ,...,a − q = ∂ F ∂ K ( π ) + P a i =1 z i ∧ π i − P a − q ℓ =1 z ℓ ∧ α ℓ ( ∂ K ( π i )) i =1 ,...,a ! = ∂ K (cid:16) ∂ K ( π ) + a X i =1 z i ∧ π i − a − q X ℓ =1 z ℓ ∧ α ℓ (cid:17) + a X i =1 z i ∧ ∂ K ( π i )= − a X i =1 z i ∧ ∂ K ( π i ) + a X i =1 z i ∧ ∂ K ( π i ) = 0 . Im ∂ F ⊇ Ker ∂ F : If π ( α i ) i =1 ,...,a ! ∈ Ker ∂ F , then ∂ K ( π ) + P a i =1 z i ∧ α i = 0 in K . In particular, wehave P a i =1 [ z i ] ∧ [ α i ] = 0 in A . Since { [ z i ] } i =1 ,...,a is a basis of A , we get α i ∈ m . It follows that for each i there exists π i ∈ K such that(3.2.1) α i = ∂ K ( π i ) . Hence we have 0 = ∂ K ( π ) + a X i =1 z i ∧ ∂ K ( π i ) = ∂ K (cid:16) π − a X i =1 z i ∧ π i (cid:17) . In particular, there exist π ∈ K , p i ∈ Ker ∂ K , and β ℓ ∈ R such that π − a X i =1 z i ∧ π i = ∂ K ( π ) + a X i =1 z i ∧ p i + a − q X ℓ =1 z ℓ ∧ β ℓ , which implies(3.2.2) π = ∂ K ( π ) + a X i =1 z i ∧ ( π i + p i ) + a − q X ℓ =1 z ℓ ∧ β ℓ . Combining (3.2.1) and (3.2.2) we getKer ∂ F ∋ π ( α i ) i =1 ,...,a ! = ∂ F π ( π i + p i ) i =1 ,...,a − ( β ℓ ) ℓ =1 ,...,a − q . Exactness at degree three. Im ∂ F ⊆ Ker ∂ F : We show that both components of the element ∂ F ◦ ∂ F π ( π i ) i =1 ,...,a ( π ℓ ) ℓ =1 ,...,a − q ( α t ) t =1 ,...,a − q ( β s ) s =1 ,...,a − q are zero. The first component is: ∂ K (cid:16) ∂ K ( π ) + a X i =1 z i ∧ π i + a − q X ℓ =1 z ℓ ∧ π ℓ + a − q X t =1 z t ∧ α t − a − q X s =1 e π s ∧ β s (cid:17) + a X i =1 z i ∧ ∂ K ( π i ) + a X i =1 z i ∧ (cid:16) a − q X s =1 e p si ∧ β s (cid:17) − a − q X ℓ =1 z ℓ ∧ ∂ K ( π ℓ ) = − a X i =1 z i ∧ ∂ K ( π i ) + a − q X ℓ =1 z ℓ ∧ ∂ K ( π ℓ ) − a − q X s =1 ∂ K ( e π s ) ∧ β s + a X i =1 z i ∧ ∂ K ( π i ) + a − q X s =1 (cid:16) a X i =1 z i ∧ e p si (cid:17) ∧ β s − a − q X ℓ =1 z ℓ ∧ ∂ K ( π ℓ ) = 0 . The last equality follows from the definition of e π s in (2.1.3). As e p si ∈ Ker ∂ K for all i and s , we have ∂ K (cid:16) ∂ K ( π i ) + P a − q s =1 e p si ∧ β s (cid:17) = 0 , for each 1 ≤ i ≤ a , so the second component is zero.Im ∂ F ⊇ Ker ∂ F : If π ( π i ) i =1 ,...,a ( α ℓ ) ℓ =1 ,...,a − q ∈ Ker ∂ F , then ∂ K ( π ) + a X i =1 z i ∧ π i − a − q X ℓ =1 z ℓ ∧ α ℓ = 0 in K , and(3.2.3) ∂ K ( π i ) = 0 for all 1 ≤ i ≤ a . (3.2.4)It follows that for each i , there exist π i ∈ K and β ij ∈ R such that(3.2.5) π i = ∂ K ( π i ) + a X j =1 z j ∧ β ij . Thus (3.2.3) becomes:(3.2.6) ∂ K (cid:16) π − a X i =1 z i ∧ π i (cid:17) + a X i =1 z i ∧ (cid:16) a X j =1 z j ∧ β ij (cid:17) − a − q X ℓ =1 z ℓ ∧ α ℓ = 0 . In A we obtain P a i,j =1 [ z i ] ∧ [ z j ] ∧ [ β ij ] − P a − q ℓ =1 [ z ℓ ] ∧ [ α ℓ ] = 0 . By Proposition 2.3, we get [ α ℓ ] = 0 forall ℓ and P a j =1 [ z j ] ∧ [ β ij ] = P a − q s =1 [ e p si ] ∧ [ β ′ s ] for some β ′ s in R . In particular, there exist π ′ ℓ ∈ K and π ′ i ∈ K such that α ℓ = ∂ K ( π ′ ℓ ) , and(3.2.7) a X j =1 z j ∧ β ij = ∂ K ( π ′ i ) + a − q X s =1 e p si ∧ β ′ s . The equation (3.2.5) becomes:(3.2.8) π i = ∂ K ( π i + π ′ i ) + a − q X s =1 e p si ∧ β ′ s , and equation (3.2.6) now becomes:0 = ∂ K (cid:16) π − a X i =1 z i ∧ π i (cid:17) + a X i =1 z i ∧ ∂ K ( π ′ i ) + a X i =1 a − q X s =1 z i ∧ e p si ∧ β ′ s − a − q X ℓ =1 z ℓ ∧ ∂ K ( π ′ ℓ )= ∂ K (cid:16) π − a X i =1 z i ∧ ( π i + π ′ i ) − a − q X ℓ =1 z ℓ ∧ π ′ ℓ + a − q X s =1 e π s ∧ β ′ s (cid:17) , where the second equality uses the definition of e π s from (2.1.3). Hence, there exist π ∈ K , p i ∈ Ker ∂ K , p ℓ ∈ Ker ∂ K , and α ′ t ∈ R such that π − a X i =1 z i ∧ ( π i + π ′ i ) − a − q X ℓ =1 z ℓ ∧ π ′ ℓ + a − q X s =1 e π s ∧ β ′ s TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD 11 = ∂ K ( π ) + a X i =1 z i ∧ p i + a − q X ℓ =1 z ℓ ∧ p ℓ + a − q X t =1 z t ∧ α ′ t . Thus,(3.2.9) π = ∂ K ( π ) + a X i =1 z i ∧ ( π i + π ′ i + p i ) + a − q X ℓ =1 z ℓ ∧ ( π ′ ℓ + p ℓ ) + a − q X t =1 z t ∧ α ′ t − a − q X s =1 e π s ∧ β ′ s The equations (3.2.7), (3.2.8), and (3.2.9), now yieldKer ∂ F ∋ π ( π i ) i =1 ,...,a ( α ℓ ) ℓ =1 ,...,a − q = ∂ F π ( π i + π ′ i + p i ) i =1 ,...,a ( π ′ ℓ + p ℓ ) ℓ =1 ,...,a − q ( α ′ t ) t =1 , ··· ,a − q ( β ′ s ) s =1 ,...,a − q . Exactness at degree four. Im ∂ F ⊆ Ker ∂ F : We show that all three components of the element ∂ F ◦ ∂ F π ( π i ) i =1 ,...,a ( π ℓ ) ℓ =1 ,...,a − q ( π t ) t =1 ,...,a − q ( α r ) r =1 ,...,a − a ( π ′ s ) s =1 ,...,a − q ( β u ) u =1 ,...,a a − a q − q + b ( γ ℓi ) ℓ =1 ....,a − q ; i =1 ,...,a are zero. The first component is: ∂ K (cid:16) ∂ K ( π ) + a X i =1 z i ∧ π i − a − q X ℓ =1 z ℓ ∧ π ℓ + a − q X t =1 z t ∧ π t + a − a X r =1 z r ∧ α r − a − q X s =1 e π s ∧ π ′ s − a a − a q − q + b X u =1 e π u ∧ β u (cid:17) + a X i =1 z i ∧ (cid:16) ∂ K ( π i ) + a − q X s =1 e p si ∧ π ′ s + a a − a q − q + b X u =1 e p ui ∧ β u − a − q X ℓ =1 z ℓ ∧ γ ℓi (cid:17) + a − q X ℓ =1 z ℓ ∧ (cid:16) ∂ K ( π ℓ ) + a X i =1 z i ∧ γ iℓ (cid:17) + a − q X t =1 z t ∧ ∂ K ( π t ) − a − q X s =1 e π s ∧ ∂ K ( π ′ s )= − a X i =1 z i ∧ ∂ K ( π i ) − a − q X ℓ =1 z ℓ ∧ ∂ K ( π ℓ ) − a − q X t =1 z t ∧ ∂ K ( π t ) − a − q X s =1 ∂ K ( e π s ) ∧ π ′ s + a − q X s =1 e π s ∧ ∂ K ( π ′ s ) − a a − a q − q + b X u =1 ∂ K ( e π u ) ∧ β u + a X i =1 z i ∧ ∂ K ( π i ) + a − q X s =1 (cid:16) a X i =1 z i ∧ e p si (cid:17) ∧ π ′ s + a a − a q − q + b X u =1 (cid:16) a X i =1 z i ∧ e p ui (cid:17) ∧ β u − a − q X ℓ =1 a X i =1 z i ∧ z ℓ ∧ γ ℓi + a − q X ℓ =1 z ℓ ∧ ∂ K ( π ℓ )+ a − q X ℓ =1 a X i =1 z ℓ ∧ z i ∧ γ ℓi + a − q X t =1 z t ∧ ∂ K ( π t ) − a − q X s =1 e π s ∧ ∂ K ( π ′ s ) = 0 , by definitions of e π s , e p si , e π u , and e p ui . For each 1 ≤ i ≤ a , as e p si is in Ker ∂ K and e p ui is in Ker ∂ K , we have ∂ K (cid:16) ∂ K ( π i ) + a − q X s =1 e p si ∧ π ′ s + a a − a q − q + b X u =1 e p ui ∧ β u − a − q X ℓ =1 z ℓ ∧ γ ℓi (cid:17) + a − q X s =1 e p si ∧ ∂ K ( π ′ s )= − a − q X s =1 e p si ∧ ∂ K ( π ′ s ) + a − q X s =1 e p si ∧ ∂ K ( π ′ s ) = 0 . Therefore, the second component is zero. For each 1 ≤ ℓ ≤ a − q , ∂ K (cid:0) ∂ K ( π ℓ ) + P a i =1 z i ∧ γ ℓi (cid:1) = 0 . Thus,the third component is zero.Im ∂ F ⊇ Ker ∂ F : If π ( π i ) i =1 ,...,a ( π ℓ ) ℓ =1 ,...,a − q ( α t ) t =1 ,...,a − q ( β s ) s =1 ,...,a − q ∈ Ker ∂ F , then ∂ K ( π ) + a X i =1 z i ∧ π i + a − q X ℓ =1 z ℓ ∧ π ℓ + a − q X t =1 z t ∧ α t − a − q X s =1 e π s ∧ β s = 0 , (3.2.10) ∂ K ( π i ) + a − q X s =1 e p si ∧ β s = 0 for all 1 ≤ i ≤ a , and(3.2.11) ∂ K ( π ℓ ) = 0 for all 1 ≤ ℓ ≤ a − q . (3.2.12)The equality (3.2.12) implies that there exist π ′ ℓ ∈ K and γ iℓ ∈ R such that(3.2.13) π ℓ = ∂ K ( π ′ ℓ ) + a X i =1 z i ∧ γ iℓ . In A a the equality (3.2.11) becomes P a − q s =1 ([ e p s , . . . , [ e p sa ]) ∧ [ β s ] = 0 . Applying Lemma 2.2 we obtain[ β s ] = 0 for all s , thus(3.2.14) β s = ∂ K ( π ′ s ) for some π ′ s ∈ K . The equality (3.2.11) now becomes ∂ K ( π i ) + a − q X s =1 e p si ∧ ∂ K ( π ′ s ) = ∂ K (cid:16) π i − a − q X s =1 e p si ∧ π ′ s (cid:17) = 0 . Therefore, there exist π i in K , δ ijk and γ ′ iℓ in R such that for all i we have:(3.2.15) π i = ∂ K ( π i ) + a X j,k =1 z j ∧ z k ∧ δ ijk + a − q X ℓ =1 z ℓ ∧ γ ′ iℓ + a − q X s =1 e p si ∧ π ′ s . Putting together (3.2.13), (3.2.14), and (3.2.15), into the equality (3.2.10) we get:0 = ∂ K ( π ) + a X i =1 z i ∧ (cid:16) a − q X s =1 e p si ∧ π ′ s + ∂ K ( π i ) + a X j,k =1 z j ∧ z k ∧ δ ijk + a − q X ℓ =1 z ℓ ∧ γ ′ iℓ (cid:17) + a − q X ℓ =1 z ℓ ∧ (cid:16) ∂ K ( π ′ ℓ ) + X i =1 z i ∧ γ iℓ (cid:17) + a − q X t =1 z t ∧ α t − a − q X s =1 e π s ∧ ∂ K ( π ′ s )= ∂ K (cid:16) π + a − q X s =1 e π s ∧ π ′ s − a X i =1 z i ∧ π i + a − q X ℓ =1 z ℓ ∧ π ′ ℓ (cid:17) TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD 13 + (cid:16) a X i,j,k =1 z i ∧ z j ∧ z k ∧ δ ijk + a X i =1 a − q X ℓ =1 z i ∧ z ℓ ∧ ( γ ′ iℓ + γ iℓ ) + a − q X t =1 z t ∧ α t (cid:17) . In A this reduces to: a X i,j,k =1 [ z i ] ∧ [ z j ] ∧ [ z k ] ∧ [ δ ijk ] + a X i =1 a − q X ℓ =1 [ z i ] ∧ [ z ℓ ] ∧ [ γ ′ iℓ + γ iℓ ] + a − q X t =1 [ z t ] ∧ [ α t ] = 0 . By Proposition 2.7, for each t there exists π ′′ t ∈ K , and for each i there exist elements δ ′ u , ε js in R , and π ′ i ∈ K such that α t = ∂ K ( π ′′ t ) , and(3.2.16) a X j,k =1 z j ∧ z k ∧ δ ijk + a − q X ℓ =1 z ℓ ∧ ( γ iℓ + γ ′ iℓ ) = ∂ K ( π ′ i ) + a a − a q − q + b X u =1 e p ui ∧ δ ′ u + a X j =1 a − q X s =1 e p si ∧ z j ∧ ε js . Thus, (3.2.10) further becomes:0 = ∂ K (cid:16) π + a − q X s =1 e π s ∧ π ′ s − a X i =1 z i ∧ π i + a − q X ℓ =1 z ℓ ∧ π ′ ℓ (cid:17) + a X i =1 z i ∧ (cid:16) ∂ K ( π ′ i ) + a a − a q − q + b X u =1 e p ui ∧ δ ′ u + a X j =1 a − q X s =1 e p si ∧ z j ∧ ε js (cid:17) + a − q X t =1 z t ∧ ∂ K ( π ′′ t )= ∂ K (cid:16) π + a − q X s =1 e π s ∧ π ′ s − a X i =1 z i ∧ π i + a − q X ℓ =1 z ℓ ∧ π ′ ℓ − a X i =1 z i ∧ π ′ i + a a − a q − q + b X u =1 e π u ∧ δ ′ u + a X j =1 a − q X s =1 e π s ∧ z j ∧ ε js − a − q X t =1 z t ∧ π ′′ t (cid:17) . The second equality above follows from definitions of e π s and e π u in (2.1.3) and (2.5.2). Therefore, thereexist π ∈ K , δ r ∈ R , p i ∈ Ker ∂ K , p ℓ ∈ Ker ∂ K and p s ∈ Ker ∂ K with p s as in Proposition 2.10, such that π − a X i =1 z i ∧ ( π i + π ′ i ) + a − q X ℓ =1 z ℓ ∧ π ′ ℓ − a − q X t =1 z t ∧ π ′′ t + a − q X s =1 e π s ∧ (cid:16) π ′ s + a X j =1 z j ∧ ε js (cid:17) + a a − a q − q + b X u =1 e π u ∧ δ ′ u = ∂ K ( π ) + a X i =1 z i ∧ p i + a − q X ℓ =1 z ℓ ∧ p ℓ + a − q X s =1 e π s ∧ p s + a − a X r =1 z r ∧ δ r . This implies:(3.2.17) π = ∂ K ( π ) + a X i =1 z i ∧ ( π i + π ′ i + p i ) − a − q X ℓ =1 z ℓ ∧ ( π ′ ℓ − p ℓ ) + a − q X t =1 z t ∧ π ′′ t + a − a X r =1 z r ∧ δ r − a − q X s =1 e π s ∧ (cid:16) π ′ s + a X j =1 z j ∧ ε js − p s (cid:17) − a a − a q − q + b X u =1 e π u ∧ δ ′ u . By (3.2.16), the expression (3.2.15) becomes: π i = ∂ K ( π i ) + a X j,k =1 z j ∧ z k ∧ δ ijk + a − q X ℓ =1 z ℓ ∧ γ ′ iℓ + a − q X s =1 e p si ∧ π ′ s (3.2.18) = ∂ K ( π i + π ′ i ) + a − q X s =1 e p si ∧ (cid:16) π ′ s + a X j =1 z j ∧ ε js (cid:17) + a a − a q − q + b X u =1 e p ui ∧ δ ′ u − a − q X ℓ =1 z ℓ ∧ γ iℓ . We conclude that by using (3.2.13), (3.2.14), (3.2.16), (3.2.17), (3.2.18), and Proposition 2.10, the chosenkernel element is in the image of ∂ F :Ker ∂ F ∋ π ( π i ) i =1 ,...,a ( π ℓ ) ℓ =1 ,...,a − q ( α t ) t =1 ,...,a − q ( β s ) s =1 ,...,a − q = ∂ F π ( π i + π ′ i + p i ) i =1 ,...,a ( π ′ ℓ − p ℓ ) ℓ =1 ,...,a − q ( π ′′ t ) t =1 ,...,a − q ( δ r ) r =1 ,...,a − a (cid:16) π ′ s + P a j =1 z j ∧ ε js − p s (cid:17) s =1 ,...,a − q ( δ ′ u ) u =1 ,...,a a − a q − q + b ( γ ℓi ) ℓ =1 ,...,a − q ; i =1 ,...,a . (cid:3) Remark 3.3.
If all multiplications on the algebra A are trivial, that is, q ij = 0 for all i, j ≥
1, and allMassey operations are zero, Golod’s construction in [6] gives a minimal free resolution of the residue field k of a local ring R in terms of the Koszul complex K of R . Without these assumptions, our complex F inConstruction 3.1 generalizes Golod’s resolution up to degree five.4. Applications to the Construction
Let P ∞ i =0 b i t i denote the series on the right hand side of the inequality (1.0.1) and set β i = rank k Tor Ri ( k , k )for i ≥ k over the ring R . The sequence {P i } i ≥ defined by:(4.0.1) P i := b i − β i , for all i ≥ , gives the coefficients of the series P ( t ) defined in the Introduction. The ring R is Golod if and only if P i = 0for all i ≥
0. Our goal in this section is to give a description of the sequence {P i } ≤ i ≤ in terms of theinvariants of the multiplicative structure of the algebra A = H( K R ). First, using Theorem 3.2, we describethe expected Betti numbers β i in terms of those invariants, up to degree five. Corollary 4.1.
Let R be a local ring of embedding dimension n . Let a i , q ij , a and b be as in Section 2.Then the following equalities hold: β = 1 , β = n, β = (cid:18) n (cid:19) + a ,β = (cid:18) n (cid:19) + na + a − q , β = (cid:18) n (cid:19) + (cid:18) n (cid:19) a + na + a + a − ( n + 1) q − q ,β = (cid:18) n (cid:19) + (cid:18) n (cid:19) a + (cid:18) n (cid:19) a + na + a + na + 2 a a − (cid:16)(cid:18) n + 12 (cid:19) + 2 a (cid:17) q − ( n + 1) q + b − a. Proof.
This is a direct consequence of Theorem 3.2. The formulas for β , . . . , β are clear, and we simplifythe following expressions for β and β to obtain the ones in the statement: β = (cid:18) n (cid:19) + (cid:18) n (cid:19) a + a − q + n ( a − q ) + a − q β = (cid:18) n (cid:19) + (cid:18) n (cid:19) a + n ( a − q ) + a a − a q − q + b TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD 15 + (cid:18) n (cid:19) ( a − q ) + a a − a q + n ( a − q ) + a − a. (cid:3) Remark 4.2.
Expressions for β i with 0 ≤ i ≤ ≤ i ≤
4, but in degree 5 we adopt the invariant b = rank k (Coker ψ )in (2.4.2) which allows us to construct F in the minimal resolution F in Theorem 3.2. For definition ofdeviations, see e.g. [3, Remark 7.1.1] by Avramov, and for the formulas of deviations up to degree five interm of the multiplicative invariants, see [1, Corollary 6.2], [3, page 62], and [7, Propositions 3.3.4 and 4.4.3]by Gulliksen and Levin. Proposition 4.3.
Let R be a local ring of embedding dimension n and codepth c . Let a i , q ij , a , and b be asin Section 2. The sequence P from Definition (4.0.1) satisfies the following equalities: P = P = P = 0 , P = q , P = ( n + 1) q + q , and P = (cid:16)(cid:18) n + 12 (cid:19) + 2 a (cid:17) q + ( n + 1) q − b + a. In particular, if q = q = 0 then P i = 0 for all ≤ i ≤ and P = a . If in addition q = q = 0 ,then P is the rank of the space spanned by the ternary-Massey products h A , A , A i in A .Proof. Recall that the right hand side of the inequality (1.0.1) is P ∞ i =1 b i t i = (1+ t ) n − P ci =1 a i t i +1 . Comparing thecoefficients on both sides of (1.0.1) we have the following recursive formulas for b i : b = 1 , b = n, b i = i − X j =1 a j b i − j − + (cid:18) ni (cid:19) , for all i ≥ . In particular, we obtain: b = (cid:18) n (cid:19) + a , b = (cid:18) n (cid:19) + (cid:18) n (cid:19) a + na + a + a ,b = (cid:18) n (cid:19) + na + a , b = (cid:18) n (cid:19) + (cid:18) n (cid:19) a + (cid:18) n (cid:19) a + na + a + na + 2 a a . Now, the expression for P i = b i − β i for 0 ≤ i ≤ q ij = 0 for all 1 ≤ i + j ≤
4, and Massey products are also trivial, we have a = b = 0.In Construction 3.1 we obtain the maximum possible values of the Betti numbers, so the differences betweenthe actual Betti numbers and these maximum values are exactly the P i ’s in the proposition. (cid:3) Proposition 4.4.
Let ( R, m , k ) be a local ring of embedding dimension n and codepth c . Let a i , q ij , a, and b be as in Section 2. If the Poincar´e series of R is rational of the form P R k ( t ) = (1 + t ) n /d ( t ) , then d ( t ) = (cid:16) − c X i =1 a i t i +1 (cid:17) + q t + ( q + q ) t + ( q − b + a ) t + f ( t ) t , for some f ( t ) ∈ Z [ t ] .Proof. Let P i be as in Definition (4.0.1). Set P ( t ) = ∞ X i =0 P i t i , α ( t ) = 1 − c X i =1 a i t i +1 and γ ( t ) = d ( t ) − α ( t ) . Then P ( t ) = (1 + t ) n α ( t ) − (1 + t ) n d ( t ) = (1 + t ) n · γ ( t ) α ( t ) · ( α ( t ) + γ ( t )) ⇐⇒ γ ( t ) α ( t ) + γ ( t ) = P ( t ) · α ( t )(1 + t ) n ⇐⇒ γ ( t ) α ( t ) = P ( t ) · α ( t )(1 + t ) n − P ( t ) · α ( t ) ⇐⇒ γ ( t ) = P ( t ) · ( α ( t )) (1 + t ) n − P ( t ) · α ( t ) . We compare the coefficients of t i for all 0 ≤ i ≤ γ ( t ) · (cid:16) (1 + t ) n − P ( t ) · α ( t ) (cid:17) = P ( t ) · ( α ( t )) . By Proposition 4.3, P = P = P = 0, and thus the left and right hand sides of the above equation become:LHS = ( γ + γ t + γ t + γ t + γ t + γ t + · · · ) · (cid:18) nt + (cid:18) n (cid:19) t + (cid:18)(cid:18) n (cid:19) − P (cid:19) t + (cid:18)(cid:18) n (cid:19) − P (cid:19) t + (cid:18)(cid:18) n (cid:19) + P a − P (cid:19) t + · · · (cid:19) , RHS = ( P t + P t + P t + . . . ) · (cid:0) − a t − a t + ( a − a ) t + ( a a − a ) t + · · · (cid:1) . It is clear that γ = γ = γ = 0 and comparing the coefficients of t we get: γ = P = q . Comparing thecoefficients of t and by Proposition 4.3 we get: γ + nγ = P ⇐⇒ γ = P − nγ = ( n + 1) q + q − nq = q + q . Finally, comparing the coefficients of t and by Proposition 4.3 we get: γ + nγ + (cid:18) n (cid:19) γ = P − a P ⇐⇒ γ = P − a P − nγ − (cid:18) n (cid:19) γ = (cid:16)(cid:18) n + 12 (cid:19) + 2 a (cid:17) q + ( n + 1) q − b + a − a q − n ( q + q ) − (cid:18) n (cid:19) q = q − b + a. Therefore, the expression for d ( t ) = α ( t ) + γ ( t ) in the statement holds. (cid:3) There are many classes of local rings for which the Poincar´e series is rational of the form P R k ( t ) =(1 + t ) n /d ( t ) . In light of Proposition 4.4, we write the coefficients of the polynomial d ( t ) in terms of theinvariants a i , q ij , b and a for some special cases and provide a uniform expression of the Poincar´e series inthese cases. Corollary 4.5. If ( R, m , k ) is a local ring of embedding dimension n and codepth at most 3, then P R k ( t ) = (1 + t ) n − − t − ( a − t − ( a − q ) t + q t − bt . Proof.
By [2, Theorem 3.5] the Poincar´e series of the ring R is rational given by P R k ( t ) = (1 + t ) n /d ( t ) withdeg d ( t ) ≤ d ( −
1) = 0. Since codepth R ≤ a = 0 and a = 0. Thus, Proposition 4.4 gives: P R k ( t ) = (1 + t ) n − a t − ( a − q ) t − ( a − q − q ) t − ( b − q ) t − bt . Simplifying the fraction by the common factor (1 + t ), we get the desired conclusion. (cid:3) Remark 4.6.
In [4, Lemma 3.6] Avramov defined the invariant τ for non-Gorenstein ring R of codepth 3 asfollows: τ = 1 if R is of class T , and τ = 0 otherwise. By comparing [4, (3.6.2)] and the Poincar´e expressionin Corollary 4.5 we see that τ is our b = rank k (Coker ψ ) in (2.4.2), which is a multiplicative invariant of thehomology algebra A . It follows from [2, Theorem 3.5] that b = 1 for a ring of class C (3) or T , and b = 0 forrings of class C (1) , C (2) , S , B , G ( r ) , and H ( q , q ). Corollary 4.7. If ( R, m , k ) is a Gorenstein local ring, not a complete intersection, of embedding dimension n and codepth 4, then P R k ( t ) = (1 + t ) n − − t − ( a − t + ( q − t + ( − q + q − t − ( q − q − b + a − t . Proof.
Since R is Gorenstein we have a = 2 a − , a = a , and a = 1 . The statement follows from [2,Theorem 3.5] and Proposition 4.4. (cid:3)
TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD 17
Remark 4.8.
Using the Avramov’s notation from [2] for the classification of Gorenstein local rings R ofcodepth 4 given by Kustin and Miller [9] and comparing the Poincar´e expressions from [2, Theorem 3.5] andCorollary 4.7 , we obtain the following table of algebra invariants of R :Class q q q q a b C (4) 6 4 1 1 1 4 GT GS GH ( p ) p p + 1 1 1 1 0We provide more examples of rings for which we can calculate the multiplicative invariants q ij , a, b fromthe Poincar´e series of the ring. Corollary 4.9.
Let k be a field, I be an ideal of k [ x, y, z, w ] such that ( x, y, z, w ) ⊆ I ⊆ ( x, y, z, w ) , andset R = k [ x, y, z, w ] /I . If R has a = 3 and its Poincar´e series is of the form P R k ( t ) = − t )(1 − t ) , then q = a − , q = 8 , a = 3 , and b = 0 . Proof.
By hypothesis,P R k ( t ) = (1 + t ) (1 − t )(1 − t )(1 + t ) = (1 + t ) − t − t + 3 t + 8 t + 3 t . Proposition 4.4 thus gives:(4.9.1) a = 7 , a − q = 8 , a − q − q = − , and a − q + b − a = − . From the second equality we obtain q = a − q = a − a + 11. Since a = 3and a = 7 , we get a = a + 3 and thus q = 8. The last equality in (4.9.1) becomes a − b = 3. By thedefinitions of a and b we have 0 ≤ a ≤ a = 3 and b ≥
0, thus a = 3 and b = 0. (cid:3) Examples 4.10.
Rings satisfying the hypotheses of Corollary 4.9 are discussed by Yoshino in [14] andChristensen and Veliche in [5]. For examples, consider the following ideals in Q = Q [ x, y, z, w ]: I = ( yw, xw + zw + w , z + w , xz + zw + w , y + yz, xy + zw, x + zw ) I = ( zw + w , yw, z + w , yz + xw + w , xz + w , xy + y + xw + w , x + xw + w ) I = ( zw, yw, xw − w , yz, xz, xy − z , x − y ) I = ( w , yw + zw, xw, yz + z , y + zw, xy + xz, x + zw ) . The rings
Q/I i satisfy the hypotheses of Corollary 4.9 and have a = 10 + i for 1 ≤ i ≤ Example 4.11.
In an unpublished note, Roos [13], inspired by a paper of Katth¨an [8], constructed severalexamples of non-Golod rings R of codepth 4 with trivial algebra multiplications on A . We provide here twoof them. For each one, there exists a Golod homomorphism from a complete intersection ring, hence it hasrational Poincar´e series. The algebra multiplication on A was checked using the DGAlgebras package [12] of
Macaulay2 [10]. Proposition 4.4 confirms that indeed q = q = 0, and moreover it gives us the exact sizeof the space generated by the triple Massey products h A , A , A i , known to be nonzero.Consider the following ideals in Q = Q [ x, y, z, w ]: J = ( w , xy , xz + yz , x w, x y + y w, y z + z w ) J = ( w , xy , xz + yz , x w + zw , y w + xzw, y z + yz ) . The rings
Q/J i with i = 1 , QQ/J i ( t ) = 1 + 6 t + (10 + i ) t + (7 + i ) t + 2 t , P Q/J i k ( t ) = (1 + t ) − t − (10 + i ) t − (7 + i ) t − t + t . For both rings
Q/J i , q = 0 = q by Proposition 4.4, hence b = 0 and a = 1. Since q = q = 0, thespace spanned by the ternary Massey products h A , A , A i has rank one. An example illustrating the Construction
In this section we consider the codepth 4 artinian local ring R = Q [ x, y, z, w ] / ( x , y , z − xy , x z , xyz , y w, w )from [1, Section 7]. This ring is of a particular interest to us, since its algebra A = H( K R ) has nontrivialhomology multiplication and a nontrivial ternary Massey product that does not come from this multiplication.The components { F i } i =0 ,..., are given in terms of the Koszul algebra components { K i } i =0 ,..., , the ranks a i of A i , and the multiplicative invariants q ij , a and b . The differential maps of the complex F in Construction3.1 are given in terms of elements z i , z ℓ , z t , z r , e p si , e π s , e p ui , and e π u . We explicitly describe them all for thisring. The bases of A i = H i ( K ) are computed with Macaulay2 [10]. We use our results from Section 2 toobtain the other elements needed in Construction 3.1.
The ring R is graded artinian with R i = 0 for all i ≥
6. The other graded components have thefollowing basis elements: R : 1 R : x, y, z, wR : x , xy, xz, xw, y , yz, yw, z , zwR : x y, x z, x w, xyz, xyw, xz , xzw, y z, yz , yzw, z , z wR : x yz, x yw, x zw, xyzw, xz , xz w, y z , yz w, z R : x yzw, xz . The Koszul algebra of R has the form K = V ( R ) ∼ = R ⊕ R ⊕ R ⊕ R ⊕ R. Let { T , T , T , T } be thestandard ordered vector basis of the free module K = R . Set T ij = T i ∧ T j and consider the ordered basis { T , T , T , T , T , T } for the free module K = V ( R ) ∼ = R . Set T ijk = T i ∧ T j ∧ T k and consider theordered basis { T , T , T , T } for the free module K = V ( R ) ∼ = R . Set T = T ∧ T ∧ T ∧ T and consider the basis { T } for the free module K = V ( R ) ∼ = R . By Macaulay2 , we obtain bases of A i = H i ( K ) and ranks a = 7 , a = 15 , a = 14 , a = 5 . A basis of A is { [ z i ] } i =1 ,..., where: z = wT , z = x T , z = ywT , z = y T ,z = y T − z T , z = yz T , z = xz T . Using Koszul and ring relations we obtain that the space A · A has rank q = 7 and its basis is givenby the classes of z ∧ z = − x wT , z ∧ z = z wT , z ∧ z = − yz wT , z ∧ z = − xz wT ,z ∧ z = x ywT , z ∧ z = − yz wT , z ∧ z = − y z T . Therefore, a basis of A is { [ z ℓ ] } ℓ =1 ,..., , where z = ywT , z = y T , z = xz T , z = yz T ,z = x yT − xz T , z = x zT , z = yz wT , z = z T . All other products among the basis elements of A are zero in A , except for the Koszul relations ofthe nonzero products A · A above. It follows that the kernel of the multiplication map φ : A ⊗ A → A is generated by a − q = 42 elements. We record the indices of basis elements of A · A by the set S = { (1 , , (1 , , (1 , , (1 , , (2 , , (3 , , (4 , } . A basis of Ker φ , as defined in (2.1.2), is given by elements of two types:(1) [ z i ] ⊗ [ z j ] , for 1 ≤ i, j ≤ , ( i, j ) S and ( j, i ) S ;(2) [ z i ] ⊗ [ z j ] + [ z j ] ⊗ [ z i ] , for ( i, j ) ∈ S .Therefore, according to the types above, one defines e p s ( i,j ) = ( e p s ( i,j ) k ) k =1 ,..., as: TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD 19 (1) e p s ( i,j ) k = ( z j if k = i k = i, for all 1 ≤ i, j ≤ , ( i, j ) S and ( j, i ) S ;(2) e p s ( i,j ) k = z j if k = iz i if k = j k = i, j, for all ( i, j ) ∈ S . Next, we find the nonzero elements e π s ∈ K defined in (2.1.3) as follows. For the elements coming fromKoszul relations in K , we choose e π s = 0 for those elements. The only nonzero products in K come fromelements of type (1) and they are z ∧ z = x y T = − z ∧ z and z ∧ z = xz T = − z ∧ z . Therefore, we choose the following nontrivial liftings in K : e π s (2 , = xz T = − e π s (4 , and e π s (5 , = − x yzT = − e π s (7 , . We describe next the elements of A . Claim 1.
All the products in A · A · A are zero. Proof of Claim 1.
First, all the products in A · A · A not involving [ z ] have the coefficients of [ T ijk ] ofdegree six, and R = 0. Thus, all such products are zero. Second, the only pairs ( i, j ) with 1 < i < j ≤ z i ] ∧ [ z j ] is nonzero in A · A are in the set { (2 , , (3 , , (4 , } , but [ z ∧ z ] = [ z ∧ z ] = 0.Hence, all products involving [ z ] at least once are zero. The claim now follows. (cid:3) Therefore, A · A = A · A , it has rank q = 10, and its basis is given by the classes of z ∧ z = xz wT , z ∧ z = yz wT , z ∧ z = x ywT − xz wT ,z ∧ z = x zwT , z ∧ z = x ywT , z ∧ z = x y T ,z ∧ z = − x yzwT , z ∧ z = yz wT , z ∧ z = y z T ,z ∧ z = − xz T = z ∧ z . A basis of A is { [ z t ] } t =1 ,..., , where z = yz wT , z = y z T , z = yz wT , z = z T . Using Proposition 2 .
6, we describe the elements e p ui , for 1 ≤ u ≤ a a − a q − q + b = 46 + b and1 ≤ i ≤
7, as defined in (2.5.1).
Claim 2.
Let A , B be as in Proposition 2.6 and let b be as in Section 2. ThenSpan k A = A ⊗ ( A · A ) and b = 0 . Proof of Claim 2.
It is clear that Span k A ⊆ A ⊗ ( A · A ). For 1 ≤ i, j, h ≤
7, any nonzero element[ z i ] ⊗ [ z j ] ∧ [ z h ] in A ⊗ ( A · A ) has ( j, h ) ∈ S or ( h, j ) ∈ S. We show that each such element is in Span k A . In the case { ( i, j ) , ( i, h ) } 6⊆ S we have[ z i ] ⊗ [ z j ] ∧ [ z h ] = X k =1 [ z k ] ⊗ [ e p s ( i,j ) k ] ∧ [ z h ] , as in case (1) of 5.5.By Koszul relation, the case ( i, h ) ∈ S reduces to the case ( i, j ) ∈ S in which we have[ z i ] ⊗ [ z j ] ∧ [ z h ] = [ z i ] ⊗ [ z j ] ∧ [ z h ] + [ z j ] ⊗ [ z i ] ∧ [ z h ] = X k =1 [ z k ] ⊗ [ e p s ( i,j ) k ] ∧ [ z h ] , as in case (2) of 5.5. The first equality above follows from the proof of Claim 1. This concludes Span k A = A ⊗ ( A · A ), so rank k (Span k A ) = a q . By Proposition 2.6(b) we get b = 0. (cid:3) Remark that Claim 2 implies Claim 1, since Span k A ⊆
Ker φ .By definition of A and respectively Claim 2, the following equalities hold: A ⊗ A = (cid:0) A ⊗ ( A · A ) (cid:1) ⊕ ( A ⊗ A ) = (Span k A ) ⊕ ( A ⊗ A ) . By Proposition 2.6, Ker φ = (Span k A ) ⊕ B , where B = (Ker φ ) ∩ ( A ⊗ A ). In order to find a basis for B , we record the indices of basis elements of A · A by the set U = { (1 , , (1 , , (1 , , (1 , , (2 , , (2 , , (3 , , (5 , , (5 , , (5 , , (4 , } . A basis of B is given by elements of two types:(1 ′ ) [ z i ] ⊗ [ z ℓ ] for all 1 ≤ i ≤ ≤ ℓ ≤ i, ℓ ) U ;(2 ′ ) [ z ] ⊗ [ z ] − [ z ] ⊗ [ z ].Therefore, according to the types above, one defines e p u ( i,ℓ ) = ( e p u ( i,ℓ ) k ) k =1 ,..., as:(1 ′ ) e p u ( i,ℓ ) k = ( z ℓ if k = i k = i, for all 1 ≤ i ≤ ≤ ℓ ≤ i, ℓ ) U ;(2 ′ ) e p u (4 , k = z if k = 4 − z if k = 50 if k = 4 , . It is easy to check that in K we have the equalities:(1 ′ ) z i ∧ z ℓ = 0 for all 1 ≤ i ≤ ≤ ℓ ≤ i, ℓ ) U ;(2 ′ ) z ∧ z − z ∧ z = 0.Therefore, we may choose e π u = 0 for all 1 ≤ u ≤ . A similar argument as in the proof of Claim 1 gives A · A = 0, and hence q = 0. Moreover, A · A has rank q = 2 and its basis is given by the classes of z ∧ z = − x yzwT and z ∧ z = − xz T . As A · A = 0, any element in h A , A , A i ⊆ A has a representative in K given by(5.10.1) X s =1 e π s ∧ p s such that X s =1 [ e p sk ∧ p s ] = 0 , for all 1 ≤ k ≤ , where p s ∈ Ker ∂ K is as in Proposition 2.10, and e p si and e π s are defined in (2.1.2) and (2.1.3) respectively.As described above, the only nontrivial lifting elements e π s ∈ K are e π s (2 , = − e π s (4 , and e π s (5 , = − e π s (7 , .Since all of them contain T , only the component of p s that contains z = wT contributes to a nonzeroMassey product. Thus, there exist α, β ∈ R such that X s =1 [ e π s ∧ p s ] = [ e π s (2 , ∧ p s (2 , ] + [ e π s (5 , ∧ p s (5 , ]= [ e π s (2 , ∧ z ] ∧ [ α ] + [ e π s (5 , ∧ z ] ∧ [ β ]= [ xz wT ] ∧ [ α ] + [ x yzwT ] ∧ [ β ]= h [ z ] , [ z ] , [ z ] i ∧ [ α ] − [ z ∧ z ] ∧ [ β ] . The last equality follows from the following computation: h [ z ] , [ z ] , [ z ] i = { [( ∂ K ) − ( z ∧ z ) ∧ z + z ∧ ( ∂ K ) − ( z ∧ z )] } = { [ e π s (2 , ∧ z + z ∧ e π s (4 , ] } = { [ xz wT ] } . By abusing notation, we write h [ z ] , [ z ] , [ z ] i = [ xz wT ], which is not in A · A + A · A , as showed in[1, Section 7]. It is clear now that the rank of the k vector space A · A + A · A + Span k h A · A · A i TRUNCATED MINIMAL FREE RESOLUTION OF THE RESIDUE FIELD 21 is a = 3 and its basis is given by[ z ] ∧ [ z ] , [ z ] ∧ [ z ] , and h [ z ] , [ z ] , [ z ] i . Thus, a basis of A is { [ z r ] } r =1 , where z = yz wT and z = y z T . We conclude that for the ring R discussed in this section, and by Proposition 4.3 we have: a = 7 , a = 15 , a = 14 , a = 5 ,q = 7 , q = 10 , q = 0 , q = 2 ,a = 3 , b = 0 , P = 7 , P = 45 , P = 221 . Acknowledgement
The authors thank Hailong Dao for inspiring them to work on this project and for very fruitful discussions.They also thank Frank Moore for helping them with the
DGAlgebras package [12] to check the computations.The first author was supported by the Naval Academy Research Council in Summer 2020.
References [1] L. L. Avramov,
On the Hopf algebra of a local ring , Izv. Akad. Nauk SSSR Ser. Mat. (1974), 253–277.[2] L. L. Avramov, Homological asymptotics of modules over local rings , Commutative algebra (Berkeley, CA, 1987), Math.Sci. Res. Inst. Publ., Springer, New York, vol. 15, (1989), 33–62.[3] L. L. Avramov,
Infinite free resolutions , Six Lectures on Commutative Algebra (Bellaterra, 1996), Progr. Math., (1998),Birkh¨auser, Basel, 1–118.[4] L. L. Avramov,
A cohomological study of local rings of embedding codepth
3, J. Pure Appl. Algebra, (2012), no. 11,2489–2506.[5] L. W. Christensen; O. Veliche,
Acyclicity over local rings with radical cube zero , Illinois J. Math. (2007), no. 4, 1439–1454.[6] E. S. Golod, On the homology of some local rings , Dokl. Akad. Nauk SSSR (1962), 479–482 (in Russian); Englishtranslation: Soviet Math. Dokl. (1962), 745–748.[7] T. H. Gulliksen; G. Levin, Homology of local rings , Queen’s Papers in Pure Appl. Math. (1969), x+192 pp.[8] L. Katth¨an, A non-Golod ring with a trivial product on its Koszul homology , J. Algebra, (2017), 244–262.[9] A. R. Kustin; M. Miller,
Classification of the Tor-Algebras of Codimension Four Gorenstein Local Rings , Math. Z., (1985), 341–355.[10] D. R. Grayson; M. Stillman,
Macaulay2, a software system for research in algebraic geometry , Available at .[11] J. P. May,
Matric Massey products , J. Algebra, (4) (1969), 533–568.[12] F. Moore, DGAlgebras: Data type for DG algebras. Version 1.0.1 , A
Macaulay2 package available at https://github.com/Macaulay2/M2/tree/master/M2/Macaulay2/packages .[13] J.-E. Roos,
On some unexpected rings that are close to Golod rings , (2016), unpublished notes.[14] Y. Yoshino,
Modules of G-dimension zero over local rings with the cube of maximal ideal being zero , Commutative algebra,singularities and computer algebra (Sinaia, 2002), 255–273, NATO Sci. Ser. II Math. Phys. Chem., , Kluwer Acad.Publ., Dordrecht, 2003.
Department of Mathematics, United States Naval Academy, Annapolis, MD 21402, U.S.A.
Email address : [email protected] URL : https://sites.google.com/view/vcnguyen Department of Mathematics, Northeastern University, Boston, MA 02115, U.S.A.
Email address : [email protected] URL ::