Local Rings and Golod Homomorphisms
|
|
- David Adams
- 5 years ago
- Views:
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
48 Bibliography [1] D. Anick, A counterexample to a conjecture of Serre, Annals of Math. 115 (1982), [2] L. Avramov, Infinite free resolutions Six lectures on commutative algebra (Bellaterra, 1996), 1 118, Progr. Math., 166, Birkhäuser, Basel, [3] A. Dress and H. Krämer, Bettireihen von Faserprodukten lokaler Ringe, Math. Ann. 215 (1975), [4] E. S. Golod. Homologies of some local rings.: Dokl. Akad. Nauk SSSR 114 (1962), [5] A. Gupta, Ascent and descent of the Golod property along algebra retracts, J. Algebra 480 (2017) [6] J. Herzog, Algebra retracts and Poincaré series, Manuscripta Mathematica 4 (1977), [7] J. Herzog, C. Huneke, Ordinary and Symbolic Powers are Golod, Advances in Mathematics 246 (2013), [8] J. Herzog, V. Welker, S. Yassemi, Homology of Powers of Ideals: Artin-Rees Numbers of Syzygies and the Golod Property, Algebra Colloq. 23 (2016) pp [9] J. Herzog and R. A. Maleki, Koszul cycles and Golod rings, Manuscripta Mathematica (2018), 1-13 [10] J. Lescot, La série de Bass d un produit fibré d anneaux locaux, Paul Dubreil and Marie-Paule Malliavin algebra seminar, (Paris, 1982), , Lecture Notes in Math., 1029, Springer, Berlin, (1983). [11] G. Levin, Local Rings and Golod Homomorphisms, Journal of Algebra 37 (1975),
CONDITIONS FOR THE YONEDA ALGEBRA OF A LOCAL RING TO BE GENERATED IN LOW DEGREES
CONDITIONS FOR THE YONEDA ALGEBRA OF A LOCAL RING TO BE GENERATED IN LOW DEGREES JUSTIN HOFFMEIER AND LIANA M. ŞEGA Abstract. The powers m n of the maximal ideal m of a local Noetherian ring R are known
More informationFormal power series rings, inverse limits, and I-adic completions of rings
Formal power series rings, inverse limits, and I-adic completions of rings Formal semigroup rings and formal power series rings We next want to explore the notion of a (formal) power series ring in finitely
More informationA NEW PROOF OF SERRE S HOMOLOGICAL CHARACTERIZATION OF REGULAR LOCAL RINGS
A NEW PROOF OF SERRE S HOMOLOGICAL CHARACTERIZATION OF REGULAR LOCAL RINGS RAVI JAGADEESAN AND AARON LANDESMAN Abstract. We give a new proof of Serre s result that a Noetherian local ring is regular if
More informationFINITE REGULARITY AND KOSZUL ALGEBRAS
FINITE REGULARITY AND KOSZUL ALGEBRAS LUCHEZAR L. AVRAMOV AND IRENA PEEVA ABSTRACT: We determine the positively graded commutative algebras over which the residue field modulo the homogeneous maximal ideal
More informationALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.
ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then
More informationHilbert function, Betti numbers. Daniel Gromada
Hilbert function, Betti numbers 1 Daniel Gromada References 2 David Eisenbud: Commutative Algebra with a View Toward Algebraic Geometry 19, 110 David Eisenbud: The Geometry of Syzygies 1A, 1B My own notes
More informationFREE RESOLUTION OF POWERS OF MONOMIAL IDEALS AND GOLOD RINGS
FREE RESOLUTION OF POWERS OF MONOMIAL IDEALS AND GOLOD RINGS N. ALTAFI, N. NEMATI, S. A. SEYED FAKHARI, AND S. YASSEMI Abstract. Let S = K[x 1,..., x n ] be the polynomial ring over a field K. In this
More informationACYCLIC COMPLEXES OF FINITELY GENERATED FREE MODULES OVER LOCAL RINGS
ACYCLIC COMPLEXES OF FINITELY GENERATED FREE MODULES OVER LOCAL RINGS MERI T. HUGHES, DAVID A. JORGENSEN, AND LIANA M. ŞEGA Abstract We consider the question of how minimal acyclic complexes of finitely
More informationCOURSE SUMMARY FOR MATH 508, WINTER QUARTER 2017: ADVANCED COMMUTATIVE ALGEBRA
COURSE SUMMARY FOR MATH 508, WINTER QUARTER 2017: ADVANCED COMMUTATIVE ALGEBRA JAROD ALPER WEEK 1, JAN 4, 6: DIMENSION Lecture 1: Introduction to dimension. Define Krull dimension of a ring A. Discuss
More informationLecture 2. (1) Every P L A (M) has a maximal element, (2) Every ascending chain of submodules stabilizes (ACC).
Lecture 2 1. Noetherian and Artinian rings and modules Let A be a commutative ring with identity, A M a module, and φ : M N an A-linear map. Then ker φ = {m M : φ(m) = 0} is a submodule of M and im φ is
More informationMATH 326: RINGS AND MODULES STEFAN GILLE
MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called
More informationOn the vanishing of Tor of the absolute integral closure
On the vanishing of Tor of the absolute integral closure Hans Schoutens Department of Mathematics NYC College of Technology City University of New York NY, NY 11201 (USA) Abstract Let R be an excellent
More informationCOHEN-MACAULAY RINGS SELECTED EXERCISES. 1. Problem 1.1.9
COHEN-MACAULAY RINGS SELECTED EXERCISES KELLER VANDEBOGERT 1. Problem 1.1.9 Proceed by induction, and suppose x R is a U and N-regular element for the base case. Suppose now that xm = 0 for some m M. We
More informationHomological Dimension
Homological Dimension David E V Rose April 17, 29 1 Introduction In this note, we explore the notion of homological dimension After introducing the basic concepts, our two main goals are to give a proof
More informationCohen-Macaulay Dimension for Coherent Rings
University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Dissertations, Theses, and Student Research Papers in Mathematics Mathematics, Department of 5-2016 Cohen-Macaulay Dimension
More informationFILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS.
FILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS. Let A be a ring, for simplicity assumed commutative. A filtering, or filtration, of an A module M means a descending sequence of submodules M = M 0
More informationQuadraticity and Koszulity for Graded Twisted Tensor Products
Quadraticity and Koszulity for Graded Twisted Tensor Products Peter Goetz Humboldt State University Arcata, CA 95521 September 9, 2017 Outline Outline I. Preliminaries II. Quadraticity III. Koszulity
More informationHomological Methods in Commutative Algebra
Homological Methods in Commutative Algebra Olivier Haution Ludwig-Maximilians-Universität München Sommersemester 2017 1 Contents Chapter 1. Associated primes 3 1. Support of a module 3 2. Associated primes
More informationCourse 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra
Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................
More informationGeneralized Alexander duality and applications. Osaka Journal of Mathematics. 38(2) P.469-P.485
Title Generalized Alexander duality and applications Author(s) Romer, Tim Citation Osaka Journal of Mathematics. 38(2) P.469-P.485 Issue Date 2001-06 Text Version publisher URL https://doi.org/10.18910/4757
More informationOn the Existence of Non-Free Totally Reflexive Modules
University of South Carolina Scholar Commons Theses and Dissertations 5-2017 On the Existence of Non-Free Totally Reflexive Modules J. Cameron Atkins University of South Carolina Follow this and additional
More informationMath 210B. Artin Rees and completions
Math 210B. Artin Rees and completions 1. Definitions and an example Let A be a ring, I an ideal, and M an A-module. In class we defined the I-adic completion of M to be M = lim M/I n M. We will soon show
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime
More informationarxiv: v2 [math.ac] 29 Sep 2013
FREE RESOLUTION OF POWERS OF MONOMIAL IDEALS AND GOLOD RINGS arxiv:1309.6351v2 [math.ac] 29 Sep 2013 N. ALTAFI, N. NEMATI, S. A. SEYED FAKHARI, AND S. YASSEMI Abstract. Let S = K[x 1,...,x n ] be the polynomial
More informationA Primer on Homological Algebra
A Primer on Homological Algebra Henry Y Chan July 12, 213 1 Modules For people who have taken the algebra sequence, you can pretty much skip the first section Before telling you what a module is, you probably
More informationJournal of Algebra 226, (2000) doi: /jabr , available online at on. Artin Level Modules.
Journal of Algebra 226, 361 374 (2000) doi:10.1006/jabr.1999.8185, available online at http://www.idealibrary.com on Artin Level Modules Mats Boij Department of Mathematics, KTH, S 100 44 Stockholm, Sweden
More informationMATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA
MATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA These are notes for our first unit on the algebraic side of homological algebra. While this is the last topic (Chap XX) in the book, it makes sense to
More informationMATH 8253 ALGEBRAIC GEOMETRY WEEK 12
MATH 8253 ALGEBRAIC GEOMETRY WEEK 2 CİHAN BAHRAN 3.2.. Let Y be a Noetherian scheme. Show that any Y -scheme X of finite type is Noetherian. Moreover, if Y is of finite dimension, then so is X. Write f
More informationGorenstein Injective Modules
Georgia Southern University Digital Commons@Georgia Southern Electronic Theses & Dissertations Graduate Studies, Jack N. Averitt College of 2011 Gorenstein Injective Modules Emily McLean Georgia Southern
More informationCOURSE SUMMARY FOR MATH 504, FALL QUARTER : MODERN ALGEBRA
COURSE SUMMARY FOR MATH 504, FALL QUARTER 2017-8: MODERN ALGEBRA JAROD ALPER Week 1, Sept 27, 29: Introduction to Groups Lecture 1: Introduction to groups. Defined a group and discussed basic properties
More informationDESCENT OF THE CANONICAL MODULE IN RINGS WITH THE APPROXIMATION PROPERTY
PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 124, Number 6, June 1996 DESCENT OF THE CANONICAL MODULE IN RINGS WITH THE APPROXIMATION PROPERTY CHRISTEL ROTTHAUS (Communicated by Wolmer V. Vasconcelos)
More informationCOHOMOLOGY OVER FIBER PRODUCTS OF LOCAL RINGS
COHOMOLOGY OVER FIBER PRODUCTS OF LOCAL RINGS W. FRANK MOORE Abstract. Let S and T be local rings with common residue field k, let R be the fiber product S k T, and let M be an S-module. The Poincaré series
More informationAzumaya Algebras. Dennis Presotto. November 4, Introduction: Central Simple Algebras
Azumaya Algebras Dennis Presotto November 4, 2015 1 Introduction: Central Simple Algebras Azumaya algebras are introduced as generalized or global versions of central simple algebras. So the first part
More informationMath 429/581 (Advanced) Group Theory. Summary of Definitions, Examples, and Theorems by Stefan Gille
Math 429/581 (Advanced) Group Theory Summary of Definitions, Examples, and Theorems by Stefan Gille 1 2 0. Group Operations 0.1. Definition. Let G be a group and X a set. A (left) operation of G on X is
More informationHomework 2 - Math 603 Fall 05 Solutions
Homework 2 - Math 603 Fall 05 Solutions 1. (a): In the notation of Atiyah-Macdonald, Prop. 5.17, we have B n j=1 Av j. Since A is Noetherian, this implies that B is f.g. as an A-module. (b): By Noether
More informationCohomology and Base Change
Cohomology and Base Change Let A and B be abelian categories and T : A B and additive functor. We say T is half-exact if whenever 0 M M M 0 is an exact sequence of A-modules, the sequence T (M ) T (M)
More informationA family of local rings with rational Poincaré Series
arxiv:0802.0654v1 [math.ac] 5 Feb 2008 A family of local rings with rational Poincaré Series Juan Elias Giuseppe Valla October 31, 2018 Abstract In this note we compute the Poincare Series of almost stretched
More informationInjective Modules and Matlis Duality
Appendix A Injective Modules and Matlis Duality Notes on 24 Hours of Local Cohomology William D. Taylor We take R to be a commutative ring, and will discuss the theory of injective R-modules. The following
More informationREPRESENTATION THEORY WEEK 9
REPRESENTATION THEORY WEEK 9 1. Jordan-Hölder theorem and indecomposable modules Let M be a module satisfying ascending and descending chain conditions (ACC and DCC). In other words every increasing sequence
More informationTROPICAL SCHEME THEORY
TROPICAL SCHEME THEORY 5. Commutative algebra over idempotent semirings II Quotients of semirings When we work with rings, a quotient object is specified by an ideal. When dealing with semirings (and lattices),
More informationRing Theory Problems. A σ
Ring Theory Problems 1. Given the commutative diagram α A σ B β A σ B show that α: ker σ ker σ and that β : coker σ coker σ. Here coker σ = B/σ(A). 2. Let K be a field, let V be an infinite dimensional
More informationNotes on p-divisible Groups
Notes on p-divisible Groups March 24, 2006 This is a note for the talk in STAGE in MIT. The content is basically following the paper [T]. 1 Preliminaries and Notations Notation 1.1. Let R be a complete
More informationIwasawa algebras and duality
Iwasawa algebras and duality Romyar Sharifi University of Arizona March 6, 2013 Idea of the main result Goal of Talk (joint with Meng Fai Lim) Provide an analogue of Poitou-Tate duality which 1 takes place
More information10. Noether Normalization and Hilbert s Nullstellensatz
10. Noether Normalization and Hilbert s Nullstellensatz 91 10. Noether Normalization and Hilbert s Nullstellensatz In the last chapter we have gained much understanding for integral and finite ring extensions.
More informationALGEBRA HW 4. M 0 is an exact sequence of R-modules, then M is Noetherian if and only if M and M are.
ALGEBRA HW 4 CLAY SHONKWILER (a): Show that if 0 M f M g M 0 is an exact sequence of R-modules, then M is Noetherian if and only if M and M are. Proof. ( ) Suppose M is Noetherian. Then M injects into
More informationHILBERT FUNCTIONS. 1. Introduction
HILBERT FUCTIOS JORDA SCHETTLER 1. Introduction A Hilbert function (so far as we will discuss) is a map from the nonnegative integers to themselves which records the lengths of composition series of each
More informationNOTES ON BASIC HOMOLOGICAL ALGEBRA 0 L M N 0
NOTES ON BASIC HOMOLOGICAL ALGEBRA ANDREW BAKER 1. Chain complexes and their homology Let R be a ring and Mod R the category of right R-modules; a very similar discussion can be had for the category of
More informationGORENSTEIN DIMENSIONS OF UNBOUNDED COMPLEXES AND CHANGE OF BASE (WITH AN APPENDIX BY DRISS BENNIS)
GORENSTEIN DIMENSIONS OF UNBOUNDED COMPLEXES AND CHANGE OF BASE (WITH AN APPENDIX BY DRISS BENNIS) LARS WINTHER CHRISTENSEN, FATIH KÖKSAL, AND LI LIANG Abstract. For a commutative ring R and a faithfully
More informationFROBENIUS AND HOMOLOGICAL DIMENSIONS OF COMPLEXES
FROBENIUS AND HOMOLOGICAL DIMENSIONS OF COMPLEXES TARAN FUNK AND THOMAS MARLEY Abstract. It is proved that a module M over a Noetherian local ring R of prime characteristic and positive dimension has finite
More informationINJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA
INJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA These notes are intended to give the reader an idea what injective modules are, where they show up, and, to
More informationNotes for Boot Camp II
Notes for Boot Camp II Mengyuan Zhang Last updated on September 7, 2016 1 The following are notes for Boot Camp II of the Commutative Algebra Student Seminar. No originality is claimed anywhere. The main
More informationGROWTH IN THE MINIMAL INJECTIVE RESOLUTION OF A LOCAL RING
GROWTH IN THE MINIMAL INJECTIVE RESOLUTION OF A LOCAL RING LARS WINTHER CHRISTENSEN, JANET STRIULI, AND OANA VELICHE Abstract. Let R be a commutative noetherian local ring with residue field k and assume
More informationON THE REPRESENTABILITY OF Hilb n k[x] (x) Roy Mikael Skjelnes
ON THE REPRESENTABILITY OF Hilb n k[x] (x) Roy Mikael Skjelnes Abstract. Let k[x] (x) be the polynomial ring k[x] localized in the maximal ideal (x) k[x]. We study the Hilbert functor parameterizing ideals
More informationLIVIA HUMMEL AND THOMAS MARLEY
THE AUSLANDER-BRIDGER FORMULA AND THE GORENSTEIN PROPERTY FOR COHERENT RINGS LIVIA HUMMEL AND THOMAS MARLEY Abstract. The concept of Gorenstein dimension, defined by Auslander and Bridger for finitely
More informationDUALITY FOR KOSZUL HOMOLOGY OVER GORENSTEIN RINGS
DUALITY FO KOSZUL HOMOLOGY OVE GOENSTEIN INGS CLAUDIA MILLE, HAMIDEZA AHMATI, AND JANET STIULI Abstract. We study Koszul homology over Gorenstein rings. If an ideal is strongly Cohen-Macaulay, the Koszul
More informationBuchsbaumness in Rees Modules Associated to Ideals of Minimal Multiplicity in the Equi-I-Invariant Case
Journal of Algebra 25, 23 255 (2002) doi:0.006/jabr.2002.94 Buchsbaumness in Rees Modules Associated to Ideals of Minimal Multiplicity in the Equi-I-Invariant Case Kikumichi Yamagishi Faculty of Econoinformatics,
More informationLINKAGE CLASSES OF GRADE 3 PERFECT IDEALS
LINKAGE CLASSES OF GRADE 3 PERFECT IDEALS LARS WINTHER CHRISTENSEN, OANA VELICHE, AND JERZY WEYMAN Abstract. While every grade 2 perfect ideal in a regular local ring is linked to a complete intersection
More information4.4 Noetherian Rings
4.4 Noetherian Rings Recall that a ring A is Noetherian if it satisfies the following three equivalent conditions: (1) Every nonempty set of ideals of A has a maximal element (the maximal condition); (2)
More informationLectures on Grothendieck Duality. II: Derived Hom -Tensor adjointness. Local duality.
Lectures on Grothendieck Duality II: Derived Hom -Tensor adjointness. Local duality. Joseph Lipman February 16, 2009 Contents 1 Left-derived functors. Tensor and Tor. 1 2 Hom-Tensor adjunction. 3 3 Abstract
More informationTCC Homological Algebra: Assignment #3 (Solutions)
TCC Homological Algebra: Assignment #3 (Solutions) David Loeffler, d.a.loeffler@warwick.ac.uk 30th November 2016 This is the third of 4 problem sheets. Solutions should be submitted to me (via any appropriate
More information(1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d
The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers
More informationAlgebraic Geometry Spring 2009
MIT OpenCourseWare http://ocw.mit.edu 18.726 Algebraic Geometry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.726: Algebraic Geometry
More informationLectures on Grothendieck Duality II: Derived Hom -Tensor adjointness. Local duality.
Lectures on Grothendieck Duality II: Derived Hom -Tensor adjointness. Local duality. Joseph Lipman Purdue University Department of Mathematics lipman@math.purdue.edu February 16, 2009 Joseph Lipman (Purdue
More information0.1 Universal Coefficient Theorem for Homology
0.1 Universal Coefficient Theorem for Homology 0.1.1 Tensor Products Let A, B be abelian groups. Define the abelian group A B = a b a A, b B / (0.1.1) where is generated by the relations (a + a ) b = a
More informationMATH 221 NOTES BRENT HO. Date: January 3, 2009.
MATH 22 NOTES BRENT HO Date: January 3, 2009. 0 Table of Contents. Localizations......................................................................... 2 2. Zariski Topology......................................................................
More informationCovering Subsets of the Integers and a Result on Digits of Fibonacci Numbers
University of South Carolina Scholar Commons Theses and Dissertations 2017 Covering Subsets of the Integers and a Result on Digits of Fibonacci Numbers Wilson Andrew Harvey University of South Carolina
More informationDedekind Domains. Mathematics 601
Dedekind Domains Mathematics 601 In this note we prove several facts about Dedekind domains that we will use in the course of proving the Riemann-Roch theorem. The main theorem shows that if K/F is a finite
More information4. Noether normalisation
4. Noether normalisation We shall say that a ring R is an affine ring (or affine k-algebra) if R is isomorphic to a polynomial ring over a field k with finitely many indeterminates modulo an ideal, i.e.,
More informationIntegral Extensions. Chapter Integral Elements Definitions and Comments Lemma
Chapter 2 Integral Extensions 2.1 Integral Elements 2.1.1 Definitions and Comments Let R be a subring of the ring S, and let α S. We say that α is integral over R if α isarootofamonic polynomial with coefficients
More informationEXTERIOR AND SYMMETRIC POWERS OF MODULES FOR CYCLIC 2-GROUPS
EXTERIOR AND SYMMETRIC POWERS OF MODULES FOR CYCLIC 2-GROUPS FRANK IMSTEDT AND PETER SYMONDS Abstract. We prove a recursive formula for the exterior and symmetric powers of modules for a cyclic 2-group.
More informationERRATA for An Introduction to Homological Algebra 2nd Ed. June 3, 2011
1 ERRATA for An Introduction to Homological Algebra 2nd Ed. June 3, 2011 Here are all the errata that I know (aside from misspellings). If you have found any errors not listed below, please send them to
More informationMath 762 Spring h Y (Z 1 ) (1) h X (Z 2 ) h X (Z 1 ) Φ Z 1. h Y (Z 2 )
Math 762 Spring 2016 Homework 3 Drew Armstrong Problem 1. Yoneda s Lemma. We have seen that the bifunctor Hom C (, ) : C C Set is analogous to a bilinear form on a K-vector space, : V V K. Recall that
More informationOn some properties of elementary derivations in dimension six
Journal of Pure and Applied Algebra 56 (200) 69 79 www.elsevier.com/locate/jpaa On some properties of elementary derivations in dimension six Joseph Khoury Department of Mathematics, University of Ottawa,
More informationp,q H (X), H (Y ) ), where the index p has the same meaning as the
There are two Eilenberg-Moore spectral sequences that we shall consider, one for homology and the other for cohomology. In contrast with the situation for the Serre spectral sequence, for the Eilenberg-Moore
More informationMatrix Factorizations. Complete Intersections
Matrix Factorizations for Complete Intersections Irena Peeva (Cornell University) Linear Algebra: Vector Spaces finite dimensional over a Field e.g. C tool: basis Algebra: Modules finitely generated over
More informationALGEBRA HW 3 CLAY SHONKWILER
ALGEBRA HW 3 CLAY SHONKWILER (a): Show that R[x] is a flat R-module. 1 Proof. Consider the set A = {1, x, x 2,...}. Then certainly A generates R[x] as an R-module. Suppose there is some finite linear combination
More informationREMARKS ON REFLEXIVE MODULES, COVERS, AND ENVELOPES
REMARKS ON REFLEXIVE MODULES, COVERS, AND ENVELOPES RICHARD BELSHOFF Abstract. We present results on reflexive modules over Gorenstein rings which generalize results of Serre and Samuel on reflexive modules
More information8. Prime Factorization and Primary Decompositions
70 Andreas Gathmann 8. Prime Factorization and Primary Decompositions 13 When it comes to actual computations, Euclidean domains (or more generally principal ideal domains) are probably the nicest rings
More informationThe Depth Formula for Modules with Reducible Complexity
The Depth Formula for Modules with Reducible Complexity Petter Andreas Bergh David A Jorgensen Technical Report 2010-10 http://wwwutaedu/math/preprint/ THE DEPTH FORMULA FOR MODULES WITH REDUCIBLE COMPLEXITY
More information2. Intersection Multiplicities
2. Intersection Multiplicities 11 2. Intersection Multiplicities Let us start our study of curves by introducing the concept of intersection multiplicity, which will be central throughout these notes.
More informationPacific Journal of Mathematics
Pacific Journal of Mathematics GROUP ACTIONS ON POLYNOMIAL AND POWER SERIES RINGS Peter Symonds Volume 195 No. 1 September 2000 PACIFIC JOURNAL OF MATHEMATICS Vol. 195, No. 1, 2000 GROUP ACTIONS ON POLYNOMIAL
More informationDivision Algebras and Parallelizable Spheres III
Division Algebras and Parallelizable Spheres III Seminar on Vectorbundles in Algebraic Topology ETH Zürich Ramon Braunwarth May 8, 2018 These are the notes to the talk given on April 23rd 2018 in the Vector
More informationHomotopy-theory techniques in commutative algebra
Homotopy-theory techniques in commutative algebra Department of Mathematical Sciences Kent State University 09 January 2007 Departmental Colloquium Joint with Lars W. Christensen arxiv: math.ac/0612301
More informationTWO IDEAS FROM INTERSECTION THEORY
TWO IDEAS FROM INTERSECTION THEORY ALEX PETROV This is an expository paper based on Serre s Local Algebra (denoted throughout by [Ser]). The goal is to describe simple cases of two powerful ideas in intersection
More informationGorenstein homological dimensions
Journal of Pure and Applied Algebra 189 (24) 167 193 www.elsevier.com/locate/jpaa Gorenstein homological dimensions Henrik Holm Matematisk Afdeling, Universitetsparken 5, Copenhagen DK-21, Denmark Received
More informationarxiv:math/ v1 [math.ac] 3 Apr 2006
arxiv:math/0604046v1 [math.ac] 3 Apr 2006 ABSOLUTE INTEGRAL CLOSURE IN OSITIVE CHARACTERISTIC CRAIG HUNEKE AND GENNADY LYUBEZNIK Abstract. Let R be a local Noetherian domain of positive characteristic.
More informationRelative Left Derived Functors of Tensor Product Functors. Junfu Wang and Zhaoyong Huang
Relative Left Derived Functors of Tensor Product Functors Junfu Wang and Zhaoyong Huang Department of Mathematics, Nanjing University, Nanjing 210093, Jiangsu Province, China Abstract We introduce and
More informationTHE GHOST DIMENSION OF A RING
PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 00, Number 0, Pages 000 000 S 0002-9939(XX)0000-0 THE GHOST DIMENSION OF A RING MARK HOVEY AND KEIR LOCKRIDGE (Communicated by Birge Huisgen-Zimmerman)
More informationHARTSHORNE EXERCISES
HARTSHORNE EXERCISES J. WARNER Hartshorne, Exercise I.5.6. Blowing Up Curve Singularities (a) Let Y be the cusp x 3 = y 2 + x 4 + y 4 or the node xy = x 6 + y 6. Show that the curve Ỹ obtained by blowing
More informationMULTIPLICITIES OF MONOMIAL IDEALS
MULTIPLICITIES OF MONOMIAL IDEALS JÜRGEN HERZOG AND HEMA SRINIVASAN Introduction Let S = K[x 1 x n ] be a polynomial ring over a field K with standard grading, I S a graded ideal. The multiplicity of S/I
More informationABSTRACT. Department of Mathematics. interesting results. A graph on n vertices is represented by a polynomial in n
ABSTRACT Title of Thesis: GRÖBNER BASES WITH APPLICATIONS IN GRAPH THEORY Degree candidate: Angela M. Hennessy Degree and year: Master of Arts, 2006 Thesis directed by: Professor Lawrence C. Washington
More informationSchool of Mathematics and Statistics. MT5836 Galois Theory. Handout 0: Course Information
MRQ 2017 School of Mathematics and Statistics MT5836 Galois Theory Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT3505 (or MT4517) Rings & Fields Lectures: Tutorials: Mon
More informationMATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 53
MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 53 10. Completion The real numbers are the completion of the rational numbers with respect to the usual absolute value norm. This means that any Cauchy sequence
More informationALGEBRA II: RINGS AND MODULES OVER LITTLE RINGS.
ALGEBRA II: RINGS AND MODULES OVER LITTLE RINGS. KEVIN MCGERTY. 1. RINGS The central characters of this course are algebraic objects known as rings. A ring is any mathematical structure where you can add
More information11. Finitely-generated modules
11. Finitely-generated modules 11.1 Free modules 11.2 Finitely-generated modules over domains 11.3 PIDs are UFDs 11.4 Structure theorem, again 11.5 Recovering the earlier structure theorem 11.6 Submodules
More informationRees Algebras of Modules
Rees Algebras of Modules ARON SIMIS Departamento de Matemática Universidade Federal de Pernambuco 50740-540 Recife, PE, Brazil e-mail: aron@dmat.ufpe.br BERND ULRICH Department of Mathematics Michigan
More informationALGEBRA EXERCISES, PhD EXAMINATION LEVEL
ALGEBRA EXERCISES, PhD EXAMINATION LEVEL 1. Suppose that G is a finite group. (a) Prove that if G is nilpotent, and H is any proper subgroup, then H is a proper subgroup of its normalizer. (b) Use (a)
More informationBUILDING MODULES FROM THE SINGULAR LOCUS
BUILDING MODULES FROM THE SINGULAR LOCUS JESSE BURKE, LARS WINTHER CHRISTENSEN, AND RYO TAKAHASHI Abstract. A finitely generated module over a commutative noetherian ring of finite Krull dimension can
More informationSEQUENCES FOR COMPLEXES II
SEQUENCES FOR COMPLEXES II LARS WINTHER CHRISTENSEN 1. Introduction and Notation This short paper elaborates on an example given in [4] to illustrate an application of sequences for complexes: Let R be
More informationON THE ISOMORPHISM CONJECTURE FOR GROUPS ACTING ON TREES
ON THE ISOMORPHISM CONJECTURE FOR GROUPS ACTING ON TREES S.K. ROUSHON Abstract. We study the Fibered Isomorphism conjecture of Farrell and Jones for groups acting on trees. We show that under certain conditions
More information