Local Rings and Golod Homomorphisms

Transcription

1 University of South Carolina Scholar Commons Theses and Dissertations 2018 Local Rings and Golod Homomorphisms Thomas Schnibben University of South Carolina Follow this and additional works at: Part of the Mathematics Commons Recommended Citation Schnibben, T.(2018). Local Rings and Golod Homomorphisms. (Doctoral dissertation). Retrieved from This Open Access Dissertation is brought to you for free and open access by Scholar Commons. It has been accepted for inclusion in Theses and Dissertations by an authorized administrator of Scholar Commons. For more information, please contact

2 Local Rings and Golod Homomorphisms by Thomas Schnibben Bachelor of Arts Wofford College 2012 Submitted in Partial Fulfillment of the Requirements For the Degree of Doctor of Philosophy in Mathematics College of Arts and Sciences University of South Carolina 2018 Accepted by: Andrew Kustin, Major Professor Adela Varicu, Committee Member Jesse Kass, Committee Member Michael Filaseta, Committee Member Brian Habing, Committee Member Cheryl L. Addy, Vice Provost and Dean of the Graduate School

3 Abstract The Poincaré series of a local ring is the generating function of the Betti numbers for the residue field. The question of when this series represents a rational function is a classical problem in commutative algebra. Golod rings were introduced by Golod in 1962 and are one example of a class of rings that have rational Poincaré series. The idea was generalized to Golod homomorphisms by Levin in In this paper we prove two homomorphisms are Golod. The first is a class of ideals such that the natural projection to the quotient ring is a Golod homomorphism. The second deals with Golod homomorphisms between certain fiber products. To prove the second we give a construction for a resolution of a module over a fiber product. ii

4 Table of Contents Abstract ii Chapter 1 Introduction Chapter 2 Notation, Definitions and Preliminaries Algebra Retracts Fiber Products Poincaré Series and Golod Modules Large Homomorphisms Module Homomorphisms Chapter 3 An Answer to a Question of Gupta Chapter 4 Resolutions over Fiber Products Construction of Resolution The total complex of G is a minimal Resolution Chapter 5 The Poincaré Series of Module over a Fiber Product Hilbert Series Fiber Products and Poincaré Series Bibliography iii

5 Chapter 1 Introduction Let R be a local ring, m its maximal ideal, and k = R/m its residue field. Then we can define the Poincaré series of R Pk R (t) = dim k Tor R n (k, k)t n. n=0 The study of this series is a classical problem in commutative algebra. Since R is local the residue field can be resolved by free R-modules. The minimal resolution of k has the form 0 R β 0 R β 1... R βn... where β n is the nth Betti number. It follows that dim k Tor R n (k, k) = β n. So the Poincaré series of R is the generating function of the sequence of Betti numbers. in t. Of particular interest is the question of when this series equals a rational function For example, if R = k[x]/(x 2 ), then Tor R n (k, k) = 1 for all n. So a rational function. P R k (t) = n=0 t n = 1 1 t, Both Kaplansky and Serre asked whether the Poincaré series of a local ring is always a rational function. Kaplansky did not publish the question. Serre posed the question in [15, pg. 118]. There is some evidence that this may be the case as several familiar rings have rational Poincaré series. For example: 1

6 1. Let R is a regular local ring of Krull dimension n. The residue field k of R is resolved by the Koszul Complex. In this case P R k (t) = (1 + t) β Let R be a regular local ring of Krull dimension n and f 1,..., f c R a regular sequence. Then Tate shows in [16] that the complete intersection S = R/(f 1,..., f c ) has a rational Poincaré series and that P S k (t) = (1 + t)n (1 t 2 ) c. However the answer to the question of Kaplansky and Serre was shown to be no by David Anick [1] in Ancik s counterexample was the ring k[x 1,..., x 5 ] (x 2 1, x 2 2, x 2 4, x 2 5, x 1 x 2, x 4 x 5, x 1 x 3 + x 3 x 4 + x 2 x 5 ) where k is a field of characteristic not equal to two. The examples of rings with rational and irrational Poincaré series do not give much insight into which of the following should be expected: 1. Most rings have rational Poincaré series. Rings with irrational Poincaré series are rare. 2. While some special classes of rings have rational Poincaré series, most rings have irrational Poincaré series. 3. Rational and irrational Poincaré series both commonly occur. This is difficult to determine since the Poincaré series of a ring is not easily computable unless additional hypothesis about the ring are made. It was shown by Serre that if S is a regular local ring with the same residue field as R and S R is a surjective ring homomorphism then the Poincaré series of R is 2

7 term-wise bounded above by the series See [2, 3.3.2] for a proof. P S k (t) 1 t (P S R(t) 1). (1.1) A ring that meets this upper-bound is called a Golod ring, named for Evgenii Golod who first examined such rings in [4]. This upper-bound is a rational function in t since both P S k (t) and P S R(t) are polynomials. So Golod rings are another example of rings with rational Poincaré series. Such rings exhibit maximal growth of Betti numbers. The Golod property can be generalized in two ways: 1. The Poincaré series can be generalized to any R-module M. When S is a regular local ring with the same residue field as R and S R is a surjective ring homomorphism an upper-bound similar to (1.1) for P R M(t) exists. This allows us to define a Golod module. 2. The upper-bound in (1.1) still holds when the regular condition is removed from S. In Chapter 2 notation is established and key definitions are given. These include algebra retracts, fiber products, Golod homomorphisms, and large homomorphisms. Several preliminary results are also given. In Chapter 3 a question asked by Gupta in [5] is answered. Here the ideal I(s, t, u, v) = (x 1,..., x m ) s + (x 1,..., x m ) u (y 1,..., y n ) v + (y 1,..., y n ) t of k[x 1,..., x m, y 1,..., y n ] is considered. Using the results of [9] we show that the ring is Golod when 1 u < s and 1 v < t. k[x 1,..., x m, y 1,..., y n ] I(s, t, u, v) 3

8 In Chapter 4 we turn our attention to the fiber product, S R T, of S and T over R that fits the commutative diagram S R T π T T π S p T p S S R. In this chapter we produce a minimal free resolution for an S-module M over S R T using the minimal free resolutions of R and M as S-modules and R as a T -module. The definition of the resolution was inspired by the work of Moore in [13], who produced a resolution in the case that R = k, the shared residue field of S and T. To produce a resolution in the case that R is more general the hypothesis that R is an algebra retract of S and of T is added. In Chapter 5 we use the structure of the resolution given in Chapter 4 to find the following equality of the Poincaré series of an S-module M over the fiber product S R T 1 PM(t) = P S ( R(t) 1 A PM(t) S PR(t) + 1 ) S PR T (t) 1. This equality is not original to this paper. It was first shown by Dress and Krämer in [3] in the case that R = M = k. Herzog showed the equality in [6] using the assumption that the maps p S and p T are equivalent representations of R. That is, there exists maps φ S : S T and φ T : T S with the property that the diagram S φ T p S φ S T R p T commutes. Neither of these previous works produced a resolution that reflected the equality. 4

9 Chapter 2 Notation, Definitions and Preliminaries In this chapter we establish notation and definitions that will be used throughout the remainder of the paper. We also record a few results that will be used later. We use (R, m, k) for a local ring R with unique maximal ideal m and residue field k = R/m. All rings are assumed to be Noetherian and all modules are assumed to be finitely generated. If F (t) = a n t n and G(t) = b n t n are power series in t with b n a n for all n, then we say that F (t) is term-wise bounded above by F (t) and write G(t) F (t). Observation Let F (t) and G(t) be power series in t with positive coefficients and G(t) F (t). If 1/F (t) and 1/G(t) exist, then 1/F (t) 1/G(t). Proof. Let F (t) = a n t n, G(t) = b n t n, n=0 n=0 1 F (t) = n=0 c n t n, and 1 G(t) = n=0 d n t n. We will show that c n d n for all n. Note that since 1/F (t) and 1/G(t) exist a 0 and b 0 are invertible and hence non-zero. We proceed by induction on n. For n = 0 note that c 0 = 1 a 0 and d 0 = 1 b 0. 5

10 Since G(t) F (t), b 0 a 0. So c 0 d 0. Assume that c k d k for k < n. Now note that c n = 1 a 0 and d n = 1 b 0 k=1 k=1 ( a k )c n k ( b k )d n k. Since a k b k, a k b l and 1/a 0 1/b 0. By hypothesis c n k d n k for k = 1,..., n. It follows that c n d n. 2.1 Algebra Retracts Definition Let A and T be commutative rings and p : A T a ring homomorphism. The ring T is an algebra retract of A if there is a ring homomorphism i : T A with pi = id T. The ring homomorphism i is call a section of p. The homomorphism i in the above definition need not be unique. We will call any homomorphism i that satisfies the condition pi = id T a section of p. The datum of an algebra retract are: The rings T and A, a surjective homomorphism p : A T, and at least one section i : T A. We write T i A p T when T is an algebra retract of A and i is a section of p. Note that the condition that pi = id T forces i to be injective and p to be surjective. We can consider A to be an T module via the map i. So t a = i(t)a. Note that the maps p and i are T module homomorphisms: p(t a) = p(i(t)a) = p(i(t))p(a) = tp(a) = t p(a) and i(t t) = i(t )i(t) = t i(t). 6

11 Also p is an A module homomorphism: p(a a) = p(a a) = p(a )p(a) = a p(a). However i is not an A module homomorphism: i(a t) = i(p(a)t) = i(p(a))i(t) = i(p(a)) i(t). In general i(p(a))i(t) ai(t). For example if t = 1 T and a ker p non-zero. Then i(1 T ) = 1 A since i is a ring homomorphism and i(p(a)) = 0 A. Then i(p(a))i(1 T ) = 0 but ai(1 T ) = a 0 A. There is the exact sequence of T or A modules: So T = A/ ker π T as T or A modules. 0 T A ker π T 0. p 2.2 Fiber Products Definition Suppose R, S and T are rings and p S : S R and p T : T R are ring homomorphisms. The fiber product over R of S and T, denoted S R T, is the ring S R T = {(s, t) S T : p S (s) = p T (t) R}. There is a ring homomorphism π S : S R T S given by π S ((s, t)) = s and a ring homomorphism π T : S R T R given by π T ((s, t)) = t. We will call the maps π S and π T the projection homomorphisms. Note that the diagram commutes. S R T S π S π T T p S R p T (2.1) 7

12 Example Let (R, m, k) be a local ring. Define S = R[x 1,..., x m ] and T = R[y 1,..., y n ]. Let I (x 1,..., x m ) be an ideal of S and J (y 1,..., y n ) be an ideal of T. Then S I T R J = R[x 1,..., x m, y 1,..., y n ] I + J + (x 1,..., x m )(y 1,..., y n ) We record a few important details about S R T, S, T and R. Observation The kernel of π T is {(s, 0) : s ker p S } and the kernel of π S is {(0, t) : t ker p T }. Proof. Clearly if (s, t) {(s, 0) : s ker p S } then (s, t) ker π T. On the other hand suppose that (s, t) ker π T. Then π T ((s, t)) = t = 0. So (s, t) = (s, 0). Also, since (s, t) {(s, 0) : s ker p S }. So ker π T = {(s, 0) : s ker p S }. Similarly ker π S = {(0, t) : t ker p T } Observation The ideals ker π T and ker π S of A satisfy (ker π T )(ker π S ) = ker π T ker π S = 0. Proof. Since (ker π T )(ker π S ) ker π T ker π S it suffices to show that ker π T ker π S = 0. Let (s, t) ker π T ker π S. Then (s, t) ker π S. So, by (2.2.2), (s, t) = (0, t) for t ker p T. Similarly, since (s, t) ker π T, (s, t) = (s, 0) for s ker p S. It follows that (s, t) = (0, 0). Observation If p S is surjective, then so is π T. Proof. Let t T. Since p S is surjective, there exists an s S with p S (s) = p T (t). So (s, t) S R T and π T ((s, t)) = t. Similarly, if p T is surjective then so is π S. From this point forward we will assume that both p S and p T are surjective. 8

13 Observation If S, T, and R are local rings with maximal ideals m S, m T, and m R respectively, then S R T is local with maximal ideal m S R T = {(s, t) m S m T : p S (s) = p T (t) R}. Proof. First note that m S R T = πs 1 (m S ) = πt 1 (m T ). Since m S is a maximal ideal, m S R T is also. Now we will show that m S R T is the unique maximal ideal. Suppose that (s, t) S R T and (s, t) / m S R T. We will show that (s, t) is a unit. Since (s, t) / m S R T, either s / m S or t / m T. Assume with out loss of generality that s / m S. Then p S (s) / m R. Since p T (t) = p S (s), t / m T. Since both S and T are local rings s and t are invertible. It follows that (s, t) is invertible. Thus S R T is a local ring with m S R T its maximal ideal. Definition A fiber product of the form A R R is called a trivial fiber product. Note that the trivial fiber product A R R is isomorphic to A. Observation Suppose that A = S R T is a fiber product and either of the projection maps is an isomorphism. Then the fiber product is trivial. Proof. Without loss of generality we assume that π S : A S is an isomorphism. Then, by (2.2.2), 0 = ker π S = {(0, t) : t ker p T }. So ker p T = 0. Since p T is surjective by assumption, T = R. Then A = S R T = A T T a trivial fiber product. 9

14 The next theorem is useful for determining which rings can be realized as a nontrivial fiber product. Theorem Let A be a local ring. One can realize A as a fiber product if and only if there exists two non-zero ideals I and J of A with I J = 0. In this case A = A I A I+J Proof. Suppose that A is a non-trivial fiber product fiber product. So A J. A = S R T and there are ring homomorphisms π S : A S and π T : A T. Then ker π S and ker π T are ideals of A. Neither of these ideals is zero, since otherwise S R T would be a trivial fiber product by (2.2.7). By (2.2.3), ker π S ker π T = 0. Let On the other hand, suppose that I and J are two non-zero ideals with I J = 0. S = A I, T = A J, and R = A I + J. We will show that A = S R T. For a A, let ā, â, and ã denote the image of a in S, T, and R, respectively. Then we define the map θ : A S R T by θ(a) = (ā, â). This map is well-defined since the projection from A to R can be factored through the projection to S or T. If a ker θ, then ā = 0. So a I. Similarly, a J. Then a I J = 0. So θ is injective. To see that θ is surjective let (ā 1, ā 2 ) S R T. Then ã 1 = ã 2 in R. So a 1 a 2 I + J. Let a 1 a 2 = i + j for i I and j J. Let a = a 1 i = a 2 + j. Then θ(a) = (a 1 i, â 2 + j) = (ā 1, â 2 ). Thus θ is surjective and A = S R T. 10

15 2.2.9 (Fiber product of modules). Let M be an S-module, N a T -module, and P an R-module. We can consider P as an S or T -module via p S or p T, respectively. Suppose that µ : M P and ν : N P are surjective S and T -module homomorphisms, respectively. Then we define the A module M P N = {(m, n) M N : µ(m) = ν(n) P }. The A action on M P N is given by a (m, n) = (π S (a) m, π T (a) n). The above is an element of M P N since µ(π S (a) m) = p S (π S (a)) µ(m). and ν(π T (a) n) = p T (π T (a)) ν(n). Further p T (π T (a)) = p S (π S (a)), since diagram (2.1) commutes, and µ(m) = ν(n), since (m, n) M P N, so that µ(π S (a) m) = ν(π T (a) n) ( Fiber Product of Retracts). Suppose that R i T T p T R and R i S S p S R are algebra retracts. Then S R T is a T -algebra via the map µ T : T S R T given by µ T (t) = (i S (p T (t)), t). This is a well-defined map to S R T since p S i S = id R so that p S (i S (p T (t))) = p T (t). Further, T is an algebra retract of S R T and π T µ T = id T where π T : S R T T is the natural projection to T. Similarly S R T is a S-algebra and S is an algebra retract of S R T. 11

16 2.3 Poincaré Series and Golod Modules. Definition If (R, m, k) is a local ring and M is an R-module, then the Poincaré series of M is PM(t) R = dim k Tor R n (M, k)t n. n=0 Proposition Let ϕ : T R be a surjective ring homomorphism and M an R-module. Then P R M(t)P T k (t) P R k (t)p T M(t). See [2, 3.3.3] for proof. Observation Suppose that (R, m, k) is a local ring and 0 M M M 0 is a split exact sequence. Then P R M(t) = P R M (t) + P R M (t). Proof. Since the sequence is split exact M = M M. Then claim follows from the definition of the Poincaré series and the fact that Tor commutes with direct sums. If (S, m S, k) is a regular local ring and (R, m R, k) is a local ring with ϕ : S R a surjective ring homomorphisms, then Serre showed that P R k (t) P S k (t) 1 t (P S R(t) 1). (2.2) Definition The ring R is Golod if equality holds in (2.2). Golod rings have maximal growth of Betti numbers and their Poincaré series is a rational function. Golod characterized rings that meet this upper bound in [4] by introducing higher homology operations. 12

17 Definition (Trivial Massey Operation). Let A be a differentially graded algebra with H 0 (A) = k. We say that A admits a trivial Massey operation if for some k-basis b = {h λ } λ Λ of H 1 (A) there exists a function µ : i=1b i A such that µ (h λ ) = z λ Z(A) with cls(z λ ) = h λ and µ ( ) p 1 h λ1,..., h λp = µ ( ) ( ) h λ1,..., h λj µ hλj+1,..., h λp where ā = ( 1) deg a+1 a for a A. j=1 Golod showed that if the Koszul complex of a local ring (R, m R, k) admits a trivial Massey operation then the ring R is Golod. The upper bound given in (2.2) still holds when the condition that S be regular is removed. Levin generalized the Golod definition in [11]. In this paper Levin noted that the Golod property really depends on the homomorphisms between S and R, not on S and R themselves. Definition (Golod Homomorphisms). Suppose that ϕ : S R is a surjective ring homomorphism of local rings. The homomorphism ϕ is called Golod it P R k (t) = P S k (t) 1 t(p S R(t) 1). This property can be further generalized by replacing the residue field with any finitely generated R module. Definition (ϕ-golod Modules). Suppose that ϕ : S R is a surjective ring homomorphism of local rings. If M is an R-module, then it can be considered as a S-module via ϕ. M is called ϕ-golod R-module when P R M(t) = P S M(t) 1 t(p S R(t) 1). Proposition Let ϕ : T R be a surjective ring homomorphism and M an R-module. If M is ϕ-golod, then ϕ is a Golod homomorphism. 13

18 Proof. M is ϕ-golod. So by definition the equality holds. Also P R M(t) = P T M(t) 1 t(p T R (t) 1) P R M(t)P T k (t) P R k (t)p T M(t) by (2.3.2). Combining the above and the upper-bound for P R k (t) given by Serre we get PM(t) T 1 t(pr T (t) 1) P k T (t) = PM(t)P R k T (t) Pk R (t)pm(t) T PM(t) T Pk T (t) 1 t(pr T (t) 1). Since the outside terms are equal, it follows that Pk R (t)pm(t) T = PM(t) T Pk T (t) 1 t(pr T (t) 1). Dividing both sides by PM(t) T shows that ϕ is a Golod homomorphism. Let (R, m, k) be a local ring and M an R-module. Let ˆR denote the completion of R. Since ˆR is a flat R-module P R M(t) = P ˆR M (t). By the Cohen Structure Theorem there is a regular local ring T and an ideal I T with ˆR = T/I. We call the ring R Golod if π is Golod, where π is the quotient map from T to ˆR. An R-module M is called Golod if M is π-golod. Example The following are all examples of Golod homomorphisms. 1. If (R, m, k) is a regular local ring and f m 2 is a regular element, then the natural quotient map π : R R (f) is a Golod homomorphism. See [2, 5.1]. 14

19 2. If (R, m, k) is a regular local ring and I an ideal, then it is shown in [8] that the natural quotient map is a Golod homomorphisms for s 0. π : R R I s 3. The previous example can be improved when R is a polynomial ring over a field and I is a monomial ideal. In this case π is a Golod homomorphisms for s 2 [9]. 4. Let (S, m S, k), (S, m S, k), (T, m T, k), and (T, m T, k) be local rings. If ϕ : S S and ψ : T T are surjective homomorphisms, then there is a surjective homomorphism θ : S k T S k T induced by the universial property for fiber products. If ϕ and ψ are Golod homomorphisms, then θ is as well. See [13, 4.2.2] The following example illustrates an example of a non-golod homomorphism. Example Let (R, m, k) be a regular local ring of dimension n, f 1,..., f c a regular sequence in R and S = R (f 1,..., f c ). If c > 1, then the natural quotient map π : R S is not a Golod homomorphism Proof. The Poincaré series of S is P S k (t) = (1 + t)n (1 t 2 ) c. The Poincaré series for S is given in [16]. On the other hand the upper bound given by Serre is P R k (t) 1 t (P R S (t) 1) = (1 + t) n 1 t ((1 + t) c 1). Since c > 1 the denominators are different. So π is not a Golod homomorphism. 15

20 The last example leads to the following observation about Golod homomorphisms. Observation The composition of Golod homomorphisms is not necessarily Golod. Proof. Let k be a field and R = k[x, y]. Then π 1 : R R (x 2 ) and π 2 : R (x 2 ) R (x 2, y 2 ) are Golod homomorphisms by (2.3.1). However their composition π 2 π 1 : R is not a Golod homomorphisms by (2.3.2). R (x 2, y 2 ) 2.4 Large Homomorphisms Levin introduced the idea of large homomorphisms in [12]. Definition Let (R, m R, k) and (S, m S, k) be local rings. The surjective ring homomorphism ϕ : R S is large if the induced map ϕ : Tor R (k, k) Tor S (k, k) is surjective. The following theorem gives several equivalent conditions for a surjective ring homomorphism to be large. It was proven by Levin in [12]. Theorem ([12, 1.1]). Let (R, m R, k) and (S, m S, k) be local rings and ϕ : R S a local surjective ring homomorphism. Then the following are equivalent 1. The homomorphism ϕ is large. 2. For any finitely generated S-module M, considered as an R-module via ϕ, P R M(t) = P S M(t)P R S (t). 16

21 3. The homomorphism ϕ : Tor R (S, k) Tor R (k, k) induced by the canonical projection p : S k is injective. 4. For any finitely generated S-module M, regarded as an R-module via ϕ, the induced homomorphisms Tor R (M, k) Tor S (M, k) are surjective. 5. There is an exact sequence of algebras k Tor S (k, k) Tor R (k, k) Tor R (S, k) k. Example The following are large homomorphisms 1. Let (R, m, k) be a local ring. The natural projection R k is a large homomorphism. 2. Let (R, m, k) be a local ring. If x m/m 2, then R R/(x) is large when a) x is a non zero-divisor, b) x (0 : m). 3. Let (S, m S, k) and (T, m T, k) be local rings. The projections S k T S and S k T are large homomorphisms [3]. The following was proven by Herzog in [6, Theorem 1]. Proposition Suppose that T i A p T is an algebra retract. The homomorphism p is large. This proposition will play an important role in chapter 5 where we consider the Poincaré series of modules over the fiber product S R T and R is an algebra retract of both S and T. 17

22 2.5 Module Homomorphisms Here we record a few results about module homomorphisms. Theorem Let R be a commutative Noetherian ring and M be a finitely generated R module. If f : M M is a surjective R module homomorphism, then it is an isomorphism. Proof. Since R is Noetherian and M is finitely generated, M is a Noetherian module. Then we have the ascending chain of submodules of M ker f ker f 2 ker f 3... which must stabilize. So there is some n with ker f n = ker f N for n N. Now let, x ker f. Since f is surjective and the composition of surjective maps is surjective, f n is surjective. So there exists some y with x = f n (y). Then 0 = f(x) = f n+1 (y). So y ker f n+1 = ker f n. Then x = f n (y) = 0. This shows that f is injective. Since f is surjective by assumption, f is an isomorphism. Corollary Let R be a commutative Noetherian ring and M and N be a finitely generated R modules. If f : M N and g : N M are surjective R module homomorphisms, then both are isomorphisms. Proof. The maps gf : M M is a surjection. So, by (2.5.1) gf is an isomorphism. If x ker f, then gf(x) = g(f(x)) = 0. So x ker gf = 0, since gf is an isomorphism. Then ker f = 0, showing that f is injective. So f is an isomorphism. Similarly, g is an isomorphism. 18

23 Chapter 3 An Answer to a Question of Gupta Here we let k be a field of any characteristic and R = k[x 1,..., x m, y 1,..., y n ]. In [5] the author poses the following question: Let I(s, t, u, v) = (x 1,..., x m ) s + (x 1,..., x m ) u (y 1,..., y n ) v + (y 1,..., y n ) t R. [5, Remark 5.2] For which values of (s, t, u, v) N 4 is R/I(s, t, u, v) Golod? In [5] the author showed that R/I(s, t, u, v) is Golod when t = 2, v = 1 and u < s. Theorem The ring R/I(s, t, u, v) is Golod if and only if s = 1, t = 1 or 1 u < s and 1 v < t. The proof of (3.0.1) uses results established in [9]. The following definitions will be needed. The authors in [9] define for the polynomial ring k[x 1,..., x n ] d r (f) = f(0,..., 0, x r,..., x n ) f(0,..., 0, x r+1,..., x n ) x r. Note that d r (f + g) = d r (f) + d r (g) for f, g k[x 1,..., x n ]. Also note that for a monomial u we have u d r x r, if r is the smallest integer such that x r divides u, and (u) = 0, otherwise. If I = (f 1,... f m ) k[x 1,..., x n ] is an ideal, then d(i) is the ideal generated by the elements d r (f i ) for r = 1,..., n and i = 1,..., m. The operator d r depends on the 19

24 ordering of the variables. If σ is a permutation, then define d r σ(f) = σ(d r (f(x σ 1 (1),... x σ 1 (n)))). If I = (f 1,... f m ) k[x 1,..., x n ] is an ideal, then d σ (I) is the ideal generated by the elements d r σ(f i ) for r = 1,..., n and i = 1,..., m. We make use of the following theorem: Theorem ([9, 2.3]). Let I k[x 1,..., x n ] be a proper ideal with (d σ (I)) 2 I for some permutation σ. Then k[x 1,..., x n ]/I is a Golod ring. Proof of First consider the case that 1 < s, t and s u or t v then I(s, t, u, v) = (x 1,..., x m ) s + (y 1,..., y n ) t. The ring k[x 1,..., x m, y 1,..., y n ]/I(s, t, u, v) is a retract of k[x 1, y 1 ]/(x s 1, y t 1). The later ring is not Golod (see (2.3.2)). Then I(s, t, u, v) is not Golod by [5, 4.2]. Henceforth assume that u < s and v < t. In the case that s = 1 R I(s, t, u, v) = k[y 1,...,y m] (y 1,...,y m) t > 1 t k t = 1 A field is a regular ring and the identity map is obviously Golod. Then ring is Golod by [9, 3.1]. The case that t = 1 is similar. k[y 1,..., y m ] (y 1,..., y m ) t Now consider the case that 1 u < s and 1 v < t. Order the variables by x 1,..., x m, y 1,... y n. Then the following hold: 1. d((x 1,..., x m ) s ) = (x 1,..., x m ) s 1, 20

25 2. d((y 1,..., y n ) t ) = (y 1,..., y n ) t 1, and 3. d((x 1,..., x m ) u (y 1,..., y n ) v ) = (x 1,..., x m ) u 1 (y 1,..., y n ) v. Proof of (1). If x e 1 i 1... x e k ik is a generator of (x 1,..., x m ) s with i 1 < i 2 < < i k and e 1 positive, then d j ( x e 1 i 1... x e k ik ) = x e 1 1 i 1... x e k ik, if j = i 1, and 0, otherwise. In either case d j ( x e 1 i 1... x e k ik ) (x1,..., x m ) s 1. then On the other hand if x e 1 i 1... x e k ik ( d i 1 x e 1 +1 i 1 is a generator of (x 1,..., x m ) s 1 with i 1 < < i k... x e k ik ) = x e 1 i 1... x e k ik. Proof of (2). The equality is established in a similar way as (1). Proof of (3). If x e 1 i 1... x e k ik y f 1 j 1... y f l j l is a generator of (x 1,..., x m ) u (y 1,..., y n ) v with i 1 < < i k, j 1 < < j l and e 1 positive, then d j ( x e 1 i 1... x e k ik y f 1 j 1... y f l j l ) = x e 1 1 i 1... x e k ik y f 1 j 1... y f l j l, if j = i 1, and 0, otherwise. In either case d j ( x e 1 i 1... x e k ik y f 1 j 1... y f l j l ) (x1,..., x m ) u 1 (y 1,..., y n ) v. On the other hand if x e 1 i 1... x e k ik y f 1 j 1... y f l j l is a generator of (x 1,..., x m ) u 1 (y 1,... y n ) v with i 1 < < i k and j 1 < < j l then ( d i 1 x e 1 +1 i 1... x e k ik y f 1 j 1... y f l j l ) = x e 1 i 1... x e k ik y f 1 j 1... y f l j l. 21

26 The three assertions have been established; it follows that d (I(s, t, u, v)) = (x 1,..., x m ) s 1 + (x 1,..., x m ) u 1 (y 1,..., y n ) v + (y 1,..., y n ) t 1 and (d (I(s, t, u, v))) 2 = (x 1,..., x m ) 2s 2 + (x 1,..., x m ) s+u 2 (y 1,..., y n ) v + (x 1,..., x m ) s 1 (y 1,..., y n ) t 1 + (x 1,..., x m ) 2u 2 (y 1,..., y n ) 2v + (x 1,..., x m ) u 1 (y 1,..., y n ) t+v 1 + (y 1,..., y n ) 2t 2. So d (I(s, t, u, v)) 2 I when 1 < u. Thus, by [9, 2.3], I(s, t, u, v) is Golod. If u = 1 and 1 < v, then d σ (I(s, t, u, v)) 2 I(s, t, u, v) where σ is the permutation that gives the variables the order y 1,..., y n, x 1,..., x m. Thus, by [9, 2.3], R/I(s, t, u, v) is Golod. Finally when u = v = 1, then we note that the quotient ring is isomorphic to the fiber product over k of Golod Rings. k[x 1,..., x m, y 1,..., y n ] I(s, t, 1, 1) = k[x 1,..., x m ] (x 1,..., x m ) s k k[y 1,..., y n ] (y 1,..., y n ) t. Then I(s, t, 1, 1) is Golod by [10, 4.1]. 22

27 Chapter 4 Resolutions over Fiber Products 4.1 Construction of Resolution Let (R, m R, k), (S, m S, k) and (T, m T, k) be local commutative rings with surjective ring homomorphisms p S : S R and p T : T R. Define A = S R T and let π S : A S and π T : A T be the projection homomorphisms. Suppose further that R i S S p S R and R i T T p T R are algebra retracts. Then, by (2.2.10), S µ S A π S S and T µ T A π T T are algebra retracts. The diagram A µ T π T T π S µ S p T i T (4.1) S i S p S R commutes. Let M be an S-module. We consider M to be an A-module via π S : A S. Our goal is to resolve M as an A-module. Let B be a minimal resolution of R as a T -module. Let p T be the augmentation from B to R. Also let C and D be minimal resolutions of R and M as S-modules, respectively. Let p S be the augmentation from C to R. Let B = B T A, C = C S A and D = D A S. Since T A and S A are functors, B, C and D are complexes of free A-modules. Since B is minimal, Im B d m T B d 1. It follows that Im B d m A B d 1. Similarly, Im C d m A C d 1 and Im D d m A D d 1. 23

28 Observation There are isomorphisms of A-modules: 1. Im B 1 = ker π S. 2. Im C 1 = ker π T. Proof. First note that Im B 1 = ker p T. Then Im B 1 = B 1 (B 1 T A) = B 1 (B 1) T A = ker p T T A = (ker p T )A. The last isomorphisms is as A-modules given by k a µ T (k)a. We will show that (ker p T )A = ker π S = {(0, t) t ker p T }. If k ker p T and (s, t) A, then k(s, t) = (π S (µ T (k))s, π T (µ T (k))t) = (i S (p T (k))s, kt) = (0, kt) ker π S. On the other hand, if (0, k) ker π S then k ker p T and (0, k) = k(1, 1) (ker p T )A. So Im B 1 = (ker p T )A = ker π S. Similarly Im C 1 = ker π T. We now define the double complex G. We will show in (4.2.1) that the total complex of G is a minimal resolution of M as an A-module. Let G d,1 = C d. For l even define and for l > 1 odd define G d,l = G d,l = d l+2 d =1 d l+2 d =1 B d A G d d,l 1 C d A G d d,l 1. 24

29 Consider the picture.... v 3,4 v 4,4 v 5,4 v 6,4 0 G 2,3 G 3,3 G 4,3 G 5,3... h 3,3 h 4,3 h 5,3 2,3 v 3,3 v 4,3 v 5,3 v G :0 G 1,2 G 2,2 G 3,2 G 4,2... h 2,2 h 3,2 h 4,2 (4.2) v 1,2 v 2,2 v 3,2 v 4,2 0 G 0,1 G 1,1 G 2,1 G 3,1... h 1,1 h 2,1 h 3, We describe the horizontal and vertical maps of G. First define the horizontal maps. Note that G d,1 = D d. So define h d,1 = D d. If l is even d,l h d B Bd G d d,l = 1 + ( 1)d 1 d d h,l 1 if d 2 ( 1) d 1 d d h,l 1 if d = 1. If l > 1 is odd d,l h d C Cd G d d,l = 1 + ( 1)d 1 d d h,l 1 if d 2 ( 1) d 1 d d h,l 1 if d = 1. Now we define the vertical maps. If l is even d,l v 1 B 1 if d = 1 Bd G d d,l = 0 if d > 1. If l > 1 is odd d,l v 1 C 1 if d = 1 Cd G d d,l = 0 if d > 1. 25

30 Proposition The picture (4.2) is a double complex. Proof. The rows are complexes. We proceed by induction on l. When l = 1 we show that is a complex. That is G,1 : 0 G 0,1 G 1,1 G 2,1 G 3,1.... h 1,1 h 2,1 h 3,1 h 4,1 0 D 0 D 1 D 2 D D 1 D 2 D 3 D 4 which is a complex. So G,1 is a complex. Assume that G,k is a complex for k < l. The modules in G,l depend on whether l is even or odd. We show the case that l is even. Let b g G d,l with b B d and g G d d,l 1. If d 3, then ( d 1,l h h d,l (b g) ) ( = d 1,l 1 h B d (b) g + ( 1) d b d d h,l 1(g) ) ( = d B 1 B d (b) ) g + ( 1) d 1 d B (b) h d d,l 1(g) + ( 1) d d B (b) ( h d d,l 1(g) + ( 1) 2d 1 b d d h 1,l 1 h d d,l 1(g) ) = 0. The first summand is zero because B is a complex. The last summand is zero because G,l 1 is a complex by the inductive hypothesis. The middle terms are the same except with opposite signs. The cases that d = 1, 2 are similar. The case that l is odd is similar. The columns are complexes. We show that d+1,l+1 v d,l v = 0. The module G d+1,l+1 depends on whether l is even or odd. We show the case that l is odd. Let b g G d+1,l+1 with b B d and g G d,l. Then g = d l+2 d =1 c C d a c c g c 26

31 for a c A and g c G d d,l 1. If d > 1, then v d+1,l+1(b g) = 0 by definition. Consider d = 1. Then ( d,l v v d 1,l 1 (b g) ) ( = d,l v B 1 (b)g ) d l+2 = 1 B (b) d,l v a c c g c d =1 c C d d l+2 = 1 B (b) d =1 = B 1 (b) c C d c C 1 a c C 1 (c)g c (ker π T )(ker π S ) = 0. v d,l(a c c g c ) The reduction on the fourth line occurs because of the definition of v d,l. The containment on the next to last line is due (4.1.2). The case that l is even is similar. The squares commute. We will show that the square G d,l+1 G d+1,l+1 d+1,l+1 h v d,l+1 v d+1,l+1 G d 1,l h d,l G d,l commutes. The module G d+1,l+1 depends on whether l is even or odd. We show the case that l is odd. Let b g G d+1,l+1 with b B d and g G d+1 d,l. If d 2, then d,l( h d+1,l+1(b v d g)) = d,l(0) h = 0. On the other hand ( d,l+1 v h d+1,l+1 (b g) ) ( = d,l+1 v B d (b) g + ( 1) d b d+1 d h,l(g) ). Note that d B (b) g B d 1 G d+1 d,l and b d+1 d h,l(g) B d G d d,l. If d 3, then the vertical map sends both to zero since d, d 1 2. If d = 2, then v d,l+1 ( B d (b) g + ( 1) d b h d+1 d,l(g) ) = B 1 ( B 2 (b) ) g = 0 27

32 since ( B) 2 = 0. If d = 1, then d,l( h d+1,l+1(b v d g)) = d,l( h 1 B (b)g) = 1 B (b) d,l(g). h On the other hand ( d,l+1 v h d+1,l+1 (b g) ) ( = d,l+1 v b h d,l (g) ) = 1 B (b) d,l(g). h So the squares commute. The case that l is even is similar. 4.2 The total complex of G is a minimal Resolution Theorem Given the hypothesis of (4.1.1), the total complex of the double complex G is a minimal resolution of M as an A-module. Before proving (4.2.1) we consider the complex 0 ker 1 B B 2 B 3 (ker π T )B 1 2 B (ker π T )B 2 3 B (ker π T )B (4.3) First note that B 1 ((ker π T )B 1 ) = ker π T B 1 (B 1 ) = (ker π T ) (ker π S ) = 0. So (ker π T )B 1 ker B 1. Since B is a complex, Im B 2 ker B 1. So (4.3) is induced on quotients by the complex 0 ker B 1 B 2 B B 2 B 3 Lemma The complex (4.3) is exact. Proof. First note that we have (ker π T )B 1 ker B 1 B 1. Then we have the inclusion of modules Note that ker B 1 (ker π T )B 1 B 1 (ker π T )B 1. B 1 (ker π T )B 1 B 1 ker =. 1 B ker 1 B (ker π T )B 1 28

33 So we have a short exact sequence. 0 B 1 ker B 1 π B 1 ker 1 B (ker π T )B 1 (ker π T )B 1 0. i Now consider the short exact sequence of complexes E : 0 ker 1 B B 2 B 3 (ker π T )B 1 2 B (ker π T )B 2 3 B (ker π T )B 3... i id id F : 0 B 1 B 2 B 3 (ker π T )B 1 B (ker π T )B B π (ker π T )B 3... H : 0 B 1 ker B This leads us to the long exact sequence on homology:... H n 1 (E) H n (H) H n (F) H n (E) H n+1 (H).... Clearly B 1 n = 0 ker 1 H n (H) = B 0 n 0 Note that Then the following diagram commutes: B n (ker π T )B n = Bn A T = B n. B n 1 (ker π T )B n 1 = B n B n (ker π T )B n = B n 1 B n. B n 29

34 So F is isomorphic to 0 B 1 B 2 B B 3 B Then B 1 n = 0 ker 1 H n (F) = B 0 n 0. If n 0, then H n (F) H n (E) H n+1 (H) is exact and the left and right modules are zero. So H n (E) = 0 for n 0. For n = 0, the sequence 0 H 0 (H) H 0 (F) H 0 (E) 0 π is exact, where π is the map induced on homology by π. Note that π is surjective. Since B n is a free T -module for n = 0, 1, the functor B n T is exact. Applying this functor to the the exact sequence we get 0 T A ker π T 0. π T i 0 B n B n B n T ker π T 0 χ Bn B n T i where χ Bn (b a) = π T (a)b. If k ker B 1, then χ B1 (k) ker B 1 since B 0 B 1 χ B0 B 1 B 0 B 1 commutes. Let χ ker B 1 be the restriction of χ B1 to ker B 1. Then B 1 0 ker 1 B ker 1 B ker χ ker B 1 0 χ ker B 1 χ B1 i 1 i 1 a 0 B 1 B 1 B 1 T ker π T 0 χ B1 B n T i 30

35 commutes and the rows are exact. The maps i 1 and i 1 are the natural inclusion maps. So they are injective. Then, by the snake lemma, we have: 0 H 0 (F) H 0 (H) coker a 0 ϕ is exact. In particular ϕ is surjective. So we have surjections π : H 0 (F) H 0 (H) and ϕ : H 0 (H) H 0 (F). By (2.5.2) both maps are isomorphisms. Then H 0 (E) = ker π = 0. Thus E is exact. Now we proceed with the proof of (4.2.1) Proof of (4.2.1). To determine the homology of TotG we use the spectral sequence associated to G filtered by columns. Call this spectral sequence E. The zero page of E is G with only the vertical maps. Then Ed,l 1 = Hl v (G d, ). So we compute the vertical homology at each spot. For l = 1 we compute the homology of G d+1,2 v d+1,2 G d,1 Clearly ker v d,1 = G d,1. We will show that Im v d+1,2 = (ker π S )G d,1. Let (0, j)g 0. v d,1 (ker π T )G d,1. Note that Im B 1 = ker π S. Then there exists b B 1 with B 1 (b) = (0, j). Then b g G d+1,2 and v d+1,2(b g) = B 1 (b)g = (0, j)g. On the other hand if b g G d+1,2 with b B d and g G d+1 d,1 then either v d+1,2(b g) = 0 (ker π S )G d,1 31

36 or v d+1,2(b g) = B 1 (b)g (ker π S )G d,1. Then H v 1 (G d, ) = If l is even, then we compute the homology of G d,1 (ker π S )G d,1 = Gd,1 A S. G d+1,l+1 v d+1,l+1 G d,l v d,l G d 1,l 1. First note that by definition. d l+2 d =2 B d A G d d,l 1 ker v d,l If b g ker B 1 A G d 1,l 1, then v d,l(b g) = B 1 (b)g = 0. So ker B 1 A G d 1,l 1 d l+2 d =2 B d A G d d,l 1 ker v d,l Since G d 1,l 1 is a free A module, it has a basis G d 1,l 1. Then {b g : b B 1, g G d 1,l 1 } is a basis for B 1 A G d 1,l 1. Let a b,g b g + b B 1 g G d 1,l 1 with γ d B d A G d d,l 1 Then 0 = d,l v = g G d 1,l 1 = g G d 1,l 1 g G d 1,l 1 B 1 d l+2 d =2 a b,g b g + b B 1 a b,g 1 B (b)g + 0 b B 1 a b,g b g. b B 1 γ d ker v d,l d l+2 d =2 γ d 32

37 Since G d 1,l 1 is a basis, So ( b B 1 a b,g b) ker B 1. Then Thus g G d 1,l 1 a b,g b g + b B 1 B 1 d l+2 d =2 a b,g b = 0. b B 1 γ d = g G d 1,l 1 ker B 1 A G d 1,l 1 ker d,l v = ( ) d l+2 ker 1 B A G d 1,l 1 a b,g b g + b B 1 d =2 d l+2 d =2 B d A G d d,l 1. d l+2 d =2 γ d B d A G d d,l 1. We will show that Im v d+1,l+1 = (ker π T )G d,l. Let (s, 0)g (ker π T )G d,l. Note that Im C 1 = ker π T. Then there exists c C 1 with C 1 (c) = (s, 0). Then c g G d+1,l+1 and v d+1,l+1(c g) = C 1 (c)g = (s, 0)g. On the other hand if c g G d+1,2 with c C d and g G d+1 d,1 then either v d+1,2(c g) = 0 (ker π T )G d,l or v d+1,2(c g) = C 1 (c)g (ker π T )G d,l. Then H v l (G d, ) = = ( ) ker B 1 A G d 1,l 1 d l+2 d =2 B d A G d d,l 1 (ker π T )G d,l ker B 1 A G d 1,l 1 ker π T (B 1 A G d 1,l 1 ) ( d l+2 d =2 B d A G d d,l 1 ) A T. By a similar argument ( ) ker C Hl v 1 A G d 1,l 1 d l+2 d (G d, ) = =2 C d A G d d,l 1 (ker π S )G d,l ker C ( = 1 A G d l+2 ) d 1,l 1 ker π S (C 1 A G d 1,l 1 ) C d A G d d,l 1 A S. d =2 33

38 for l > 1 odd. Now we will find E 2. We take homology at each spot of the complex... 0 H v 3 (G 2, ) H v 3 (G 3, ) H v 3 (G 4, )... h 3,3 h 4,3 E 1 : 0 H v 2 (G 1, ) H v 2 (G 2, ) H v 2 (G 3, )... h 2,2 h 3,2 0 H v 1 (G 0, ) H v 1 (G 1, ) H v 1 (G 2, )... h 1,1 h 2, where h, is the map in map induced on homology by h,. For l = 1 we have the complex 0 G 0,1 A S G 1,1 A S G 2,1 A S.... 1,1 h 2,1 h (4.4) The module G d,1 = D d. Then G d,1 A S = D d A S = D d as S-modules. So (4.4) is isomorphic to D as complexes of S-modules. The homology does not change when we treat (4.4) as a complex of A-modules. Thus M if d = 0 Hd h (H1 v (G, )) = H d (D ) = 0 if d > 0. For l even we consider the homology of 0 ker B 1 AG l 2,l 1 ker π T(B 1 A G l 2,l 1) h l,l ker B 1 AG l 1,l 1 ker π T(B 1 A G l 1,l 1) B 2 A G l 2,l 1 A T... (4.5)... h d,l ker B 1 AG d 1,l 1 ker π T(B 1 A G d 1,l 1) ( d l+2 d =2 B d A G d d,l 1) A T

39 Note that (4.5) is isomorphic to the tensor product of the complexes G,l 1 : 0 G l 2,l 1 G l 1,l 1... and 0 ker B 1 ker π T B 1 B 2 A T B 3 A T.... (4.6) By (4.2.2), (4.6) is exact. The homology does not change when we consider it as a complex of A-modules. Then (4.5) is exact since it is the tensor product of an exact complex and another complex. So H h d (H v l (G, )) = 0 for l even. By a similar argument Hd h (Hl v (G, )) = 0 for l > 1 odd. Thus the spectral sequence collapses on page 2. Then H n (Tot G) = ( Hp+q h H v q+1 (G, ) ) M if p = q = 0 = p+q=n 0 otherwise. Thus Tot G is a resolution of M by free A-modules. Since Im B m A B, Im C m A C, and Im D m A D, the resolution is minimal. 35

40 Chapter 5 The Poincaré Series of Module over a Fiber Product The purpose of this chapter is to establish the relationship between Poincaré series over the fiber product A = S R T and Poincaré series over S and T. As stated before this result is already known [3],[6]. Here we show that the structure of the resolution from Chapter 4 reflects this equality. We will then consider the case that S, T, R and S, T, R satisfy the conditions of (4.1.1) with surjective ring homomorphisms ϕ : S S and ψ : T T. The universal property for fiber products induces a surjective ring homomorphisms θ : S R T S R T. We will give conditions for this homomorphism to be Golod. 5.1 Hilbert Series Definition Let M be a graded k-vector space with M d = 0 for d 0 and dim k M d finite for all d. Then the Hilbert series of M is the formal Laurent series HS M (t) = d dim k M d t d. If HS M (t) is defined, then we say that M has a Hilbert series. Let T l k(m) = M kl = M k M k k... k M }{{} l times where Tk 0 (M) = k. Then the tensor algebra of M is T k (M) = T l k(m). l=0 36

41 Define and note that T l,d k (M) = d 1 +d 2 + +d l =d M d1 k M d2 k... k M dl T l k(m). T l k(m) = d=0 We record two properties of Hilbert series T l,d k (M) Proposition Let M and N be graded vector spaces with Hilbert series. Then HS M k N(t) = HS M (t)hs N (t). Proof. By the definition of the Hilbert series HS M k N(t) = d dim k (M k N) d t d. The k-vector space (M k N) d = M i k N j. i+j=d Then dim k (M k N) d = dim k M i dim k N j. i+j=d The right hand side is precisely the coefficient of t d in the product HS M (t)hs N (t). Proposition Let T k (M) be the tensor algebra of M over k. If M d = 0 for d < 0, then Proof. First we have Since HS Tk (M)(t) = 1 1 HS M (t). HS Tk (M)(t) = T l k(m). l=0 T l k(m) = M k M k k... k M }{{} l times the Hilbert series of Tk l (M) can be determined by (5.1.2). So HS T l k (M) (t) = (HS M (t)) l. 37

42 Thus HS Tk (M)(t) = HS T l (M) (t) k l=0 = (HS M (t)) l l=0 1 = 1 HS M (t). 5.2 Fiber Products and Poincaré Series Definition Let k be a field and C be a set. Then let k C be the k-vector space with basis C. Corollary Given the assumptions of we have 1 PM(t) = P S ( R(t) 1 A PM(t) S PR(t) + 1 ) S PR T (t) 1. We prove using the following lemma Lemma The k-vector space k A G d,l is isomorphic to if l = 2l + 1 and if l = 2l + 2. d 1 +d 2 =d d 1 +d 2 +d 3 =d d 1 1 k k T l,d 1 k ( k C 1 k k B 1 ) k k D d2 k B d1 k T l,d 2 k ( k C 1 k k B 1 ) k k D d3 Proof of We proceed by induction on l. For l = 1 we have G d,l = D d, which has basis D d. The k vector space k k k k k D d has basis {1 1 d : d D d }. Then the map d 1 1 d fo d D d gives the isomorphism. 38

43 Assume that the claim is true for l < l. We will show that it holds for l. If l = 2l + 2, then G d,l A k = = d l+2 d 1 =1 d l+2 d 1 =1 d l+2 = = d 1 =1 d l+2 d 1 =1 d l+2 = = If l = 2l + 1, then G d,l A k = = B d1 A G d d1,l 1 A k (B d1 A k) k (G d d1,l 1 A k) (B d1 A k) k k k T l,d2 k ( k C 1 k k B 1 ) k k D d3 d 2 +d 3 =d d 1 d 2 +d 3 =d d 1 (B d1 A k) k d 1 =1 d 2 +d 3 =d d 1 d 1 +d 2 +d 3 =d d 1 1 d l+2 d 0 =1 d l+2 d 0 =1 d l+2 = = d 0 =1 d l+2 d 0 =1 d l+2 = = d 0 =1 ( k k T l,d 2 k ( k C 1 k k B 1 ) k k D d3 ) k B d1 k T l,d 2 k ( k C 1 k k B 1 ) k k D d3 k B d1 k T l,d 2 k ( k C 1 k k B 1 ) k k D d3. C d0 A G d d0,l 1 A k (C d0 A k) k (G d d0,l 1 A k) (C d0 A k) k d 1 +d 2 +d 3 =d d 0 d 1 +d 2 +d 3 =d d 0 d 0 +d 1 +d 2 +d 3 =d d 0 1 k C d0 k d 1 +d 2 +d 3 =d d 0 k B d1 k T l 1,d 2 k ( k C 1 k k B 1 ) k k D d3 ( k B d1 k T l 1,d 2 k ( k C 1 k k B 1 ) k k D d3 ) k C d0 k k B d1 k T l 1,d 2 k ( k B 1 k k C 1 ) k k D d3 k k T l,d 0 +d 1 +d 2 k ( k B 1 k k C 1 ) k k D d3. Proof of Let B = k B, C = k C and D = k D. Then P T R (t) = HS B (t), P S R(t) = HS C (t), P S M(t) = HS D (t) 39

44 and P A M(t) = HS G A k(t). Now G A k = d,l G d,l A k = k B k T k ( k B 1 k k C 1 ) k k D. The second isomorphism follows from (5.2.3). Then by (5.1.2) and (5.1.3). That is HS G A k(t) = P A M(t) = Reciprocate both sides to finish the proof. HS B (t)hs D (t) 1 (HS B (t) 1) (HS C (t) 1) P T R (t)p S M(t) 1 (P T R (t) 1)(P S R(t) 1). (5.1) Let S, T, R and S, T, R both satisfy the hypothesis of (4.1.1). Let ϕ : S S and ψ : T T be surjective ring homomorphisms. Then we have the diagram of rings A := S R T θ T A = S R T T ψ (5.2) ϕ S S R. Here θ is the map induced from the universal property of fiber products. Lemma Suppose that ϕ : S S is a surjective ring homomorphism, p S : S R and p S : S R are large homomorphisms. Then the following are equivalent 1. The homomorphism ϕ is Golod. 2. When treated as an S -module via p S, R is ϕ-golod. 3. Every finitely generated S -module M is ϕ-golod. 40

45 Proof Suppose that ϕ if a Golod homomorphisms. Then P S k (t) = Pk S(t) 1 t (PS S (t) 1). Since p S and p S are large homomorphisms we have P S k (t) = P R k (t)p S R(t) and P S k (t) = P R k (t)p S R (t) by [12, 1.1]. Then P R k (t)p S R (t) = Pk R(t)P R(t) S 1 t (PS S (t) 1). Divide by P R k (t) to show that R is a ϕ-golod S -module Suppose that R is a ϕ-golod S -module and let M be a finitely generated S -module. Since p S and p S are large homomorphisms we have P S M(t) = P R M(t)P S R(t) and P S M (t) = P R M(t)P S R (t) by [12, 1.1]. Since R is ϕ-golod we have P S R (t) = PR(t) S 1 t (PS S (t) 1). Multiply both sides by P R M(t) to get that M is ϕ-golod Let M = k to see that ϕ is a Golod homomorphism. Theorem In (5.2), θ is a Golod homomorphism if and only if ϕ and ψ are Golod homomorphisms. Proof. First note that the map p S, p T, p S and p T are large homomorphisms by (2.4.2). Then by (5.2.5) it suffices to show that R is a θ-golod A -module if and only if R is a ϕ-golod S -module and a ψ-golod T -module. Now we have the short exact sequence of R-modules 0 R S T A 0. p S p T i 41

46 Also we have an R-module homomorphisms χ : R S T given by χ(r) = (i S (r), 0) which satisfies (p S p T ) χ = id R. So the sequence is split exact. Then, by (2.3.3) P R A (t) = P R S (t) + P R T (t) P R R (t). Since R is an algebra retract of A we have the following P A A (t) = P R A (t)p A R (t), P A S (t) = P R S (t)p A R (t), P A T (t) = P R T (t)p A R (t), P A R (t) = P R R (t)p A R (t). Multiply both sides of P R A (t) = P R S (t) + P R T (t) P R R (t) by P A R (t) and using the above equalities of Poincaré series to get P A A (t) = P A S (t) + P A T (t) P A R (t). The ring homomorphisms π S : A S and π T : A T are large homomorphisms by (2.4.2), since S and T are algebra retracts of A. Combining this and the above we get P A A (t) = P A S (t) + P A T (t) P A R = P S S (t)p A S (t) + P T T (t)p A T (t) P S R(t)P A S (t) = PS S (t) PR T (t) 1 (PR T (t) 1)(PR(t) S 1) + P T T (t) PR(t) S 1 (PR T (t) 1)(PR(t) S 1) PR(t) S PR T (t) 1 (PR T (t) 1)(PR(t) S 1) = P S S (t)p R T (t) + PT T (t)p R(t) S PR(t)P S R T (t). PR T (t) + PR(t) S PR T (t)pr(t) S The third equality follows from (5.1). Hence (PR T (t)+pr(t) P S R T (t)pr(t))(p S A A (t) 1) = P R(t)(P S T T (t) 1)+P R T (t)(ps S (t) 1). (5.3) 42

47 Now we have the following sequence of (in)equalities: 1 t(pa A (t) 1) PR A (t) ( 1 t(p A A (t) 1) ) (PR T (t) + PR(t) S PR T (t)pr(t)) S = P S R(t)P T R (t) = P R(t) S + PR T (t) PR T (t)pr(t) S t ( PR(t)(P S T T (t) 1) + P R T (t)(ps S (t) 1)) PR(t)P S R T (t) = P R(t)(1 S t(pt T (t) 1)) + P R T (t)(1 t(ps S (t) 1)) P R T (t)pr(t) S PR(t)P S R T (t) = 1 t(p T T (t) 1) + 1 t(p S S (t) 1) 1 PR T (t) PR(t) S 1 P T R (t) + 1 PR S (t) 1 = 1 PR A (t). The first and last equalities follow from (5.2.2). The second equality comes from (5.3). The term-wise inequality follows from Serre s upperbound and (2.0.1). The term-wise inequality is an equality if and only if R is ϕ-golod and ψ-golod. This theorem can be compared to [13, 4.2.2]. Here we have replaced k with R at the cost of adding the hypothesis of (4.1.1). 43

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