functions of maximal Cohen-Macaulay modules over a hypersurface ring A. We ...... that if x is M-superficial then it is also Ì. M-superficial. Furthermore. I2M/JIM.
HILBERT FUNCTIONS OF COHEN-MACAULAY MODULES
A Thesis Submitted to the Faculty of Purdue University by Tony Joseph Puthenpurakal
In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy
August 2002
ii
To my parents
iii
ACKNOWLEDGMENTS I am indebted to my thesis advisor Prof. L. L. Avramov for his patience and many useful discussions. I also thank Professors J. Lipman, W. Heinzer and B. Ulrich for serving on my committee. I affectionately thank all my friends with whom I shared good times and bad times as well. Most of the research for this thesis was done in Cafe Vienna. I especially wish to thank all my Cafe Vienna friends. I am also grateful to my brother and his family who were with me all the way. This thesis is dedicated to my parents.
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TABLE OF CONTENTS Page ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1 PRELIMINARIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.1
Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.2
Basic Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
2 COHEN-MACAULAY MODULES OF MINIMAL MULTIPLICITY . . . .
19
3
. . . . . . . . . .
25
3.1
The First Hilbert Coefficient . . . . . . . . . . . . . . . . . . . . . . .
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3.2
Basic Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3.3
Second Hilbert coefficient . . . . . . . . . . . . . . . . . . . . . . . . .
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4 HILBERT FUNCTIONS OF MODULES OVER HYPERSURFACES . . .
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HILBERT COEFFICIENTS OF SYZYGY MODULES
4.1
Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4.2
Hilbert coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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LIST OF REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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VITA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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ABSTRACT Tony Joseph Puthenpurakal. Ph.D., Purdue University, August, 2002. Hilbert functions of Cohen-Macaulay modules. Major Professor: L.L. Avramov. Let (A, m) be a d-dimensional Noetherian local ring, M a finite Cohen-Macaulay A-module of dimension r and let I be an ideal of definition for M . In Chapter 2 we define the notion of minimal multiplicity of Cohen-Macaulay modules with respect to I and show that if M has minimal multiplicity with respect to I then the associated graded module GI (M ) is Cohen-Macaulay. In Chapter 3 we assume that A is CohenMacaulay, M is maximal Cohen-Macaulay and I is m-primary. We find a relation between the first Hilbert coefficient of M , A and SyzA 1 (M ). Sharper results are found when I = m. Set χ1 (M ) = e1 (M ) − e0 (M ) + µ(M ). We prove that χ1 (A)µ(M ) ≥ χ1 (M ) + χ1 (SyzA 1 (M )). ∗ ∗ When A is Gorenstein then M is the first syzygy of S A (M ) = (SyzA 1 (M )) . A
relation between the second Hilbert coefficient of M , A and S A (M ) is found when G(M ) is Cohen-Macaulay and depth G(A) ≥ d − 1. In Chapter 4 we study Hilbert functions of maximal Cohen-Macaulay modules over a hypersurface ring A. We show that if d > 0 then the Hilbert function of M with respect to m is non-decreasing. If A = Q/(f ) for some regular local ring Q, we determine a lower bound for e0 (M ) and e1 (M ). Furthermore we analyze the case when equality holds and prove that in this case G(M ) is Cohen-Macaulay. Furthermore in this case we also determine the Hilbert function of M .
1
INTRODUCTION This thesis is devoted to the study of Hilbert functions of finitely generated CohenMacaulay modules over a local ring. In this thesis all rings are commutative noetherian and all modules are assumed to be finite. Let (A, m) be a d-dimensional Noetherian local ring and M a finite A-module. Let I be an ideal of definition for M , i.e., mn M ⊆ IM for some n > 0 . We let µ(M ) denotes the minimal number of generators of M and λ(M ) its length. The Hilbert function of M with respect to I is the function H I (M, n) = λ(I n M/I n+1 M ) for all n ≥ 0 It is an important invariant of M that determines, among other things the dimension and the multiplicity of the module. When the ring A is Cohen-Macaulay then Sally ([16], [20], [20], [18], [19], [21]), Rossi and Valla ( [13], [14], [15] ), and others, show that the Hilbert function also contains information about GI (A) = L n n+1 , the associated graded ring of A with respect to I. A main theme of n≥0 I /I this thesis is to extend such results to the module case that is, to the associated L graded module GI (M ) = n≥0 I n M/I n+1 M considered as a GI (A)-module. The formal power series I HM (z) =
X
H I (M, n)z n
n≥0
is called the Hilbert series of M with respect to I. It is well known that it is a rational function of the form I (z) = HM
hIM (z) , where r = dim M (1 − z)r
and
hIM (z) = hI0 (M ) + hI1 (M )z + · · · + hIs (M )z s ∈ Z[z]
2 We call the polynomial hIM (z) the h-polynomial of M with respect to I. It follows from the definition that hI0 (M ) = λ(M/IM ) and hI1 (M ) = λ(IM/I 2 M ) − rλ(M/IM ) If f is a polynomial, then we use f (i) to denote its i-th derivative. The integers (i)
eIi (M ) = hM (1)/i! for i ≥ 0 are called the Hilbert coefficients of M . The number eI (M ) = eI0 (M ) is known as the multiplicity of M with respect to I. When I = m we drop the superscript m; for example, we write ei (M ) instead of em i (M ). When M = A there has been a lot of research regarding Hilbert coefficients and their relation to depth GI (A). Relatively little is known (see [4]) for modules in general. We describe the contents of various chapters. 1.Preliminaries In this chapter we extend to modules some of the basic techniques used to study Hilbert functions of a ring, in particular: Valabrega and Valla’s Theorem, Sally Descent, Singh’s Equality. Sally Descent is useful for arguments by induction on M . If x ∈ m then Balwant Singh’s equality relates the Hilbert function of a module M and M/xM . The proofs of these results for the ring A extends to the module case. The only theorem that needs a detailed proof in the case of modules is Valabrega and Valla’s theorem. If j is the largest integer such that a ∈ I j then a∗ denotes the image of a in I j /I j+1 ; this element is called the initial form of a. Valla and Valabrega’s theorem gives a necessary and sufficient condition for a sequence of elements x1 , . . . , xr to have the property that the sequence x∗1 , . . . , x∗r is a GI (A)-regular. We extend to modules, assuming that the initial forms of x1 , . . . , xr have degree 1. Theorem 1.2 Let (A, m) be a local ring, M an A-module and I an ideal in A. Let x1 , . . . , xr be an M -regular sequence with xi ∈ I \ I 2 for i = 1, . . . , s. Set J = (x1 , . . . , xs ). The following conditions are equivalent: i. x∗1 , . . . , x∗s is a GI (M )-regular sequence. ii. I i M ∩ JM = JI i−1 M for all i ≥ 1.
3 We prove this theorem by first treating the case when x∗1 , . . . , x∗s is a GI (A)-regular sequence. In this case the result is proved by considering the trivial extension R = An M and applying Valla and Valabrega’s theorem to the ideal L = (I, M ). To treat the general case notice that the A-algebra homomorphism A[X1 ..., Xs ] → A mapping Xi to xi for i = 1, . . . , s induces a ring homomorphism from A0 = A[X1 , . . . , Xs ](m,X1 ,...,Xs ) to A that turns M into a A0 -module. If Q = (I, X1 , . . . , Xs )A0 then GQ (M ) = GI (M ) and GQ (A) = GI (A)[X1∗ , . . . , Xs∗ ]. So the previous case is applicable and we get the result. Another basic technique used in the entire thesis is reduction modulo superficial sequence. An element x ∈ I is called superficial for M with respect to I if there exists an integer c > 0 such that (I n M :
M x)
∩ I c M = I n−1 M for all n > c.
It is well known that superficial elements exist when the residue field of A is infinite. A sequence x1 , . . . , xs is called a superficial sequence for M with respect to I if xi is superficial for M/(x1 , . . . , xi−1 ) for i = 1, . . . , s. Many important invariants behave well under reduction by superficial sequences. An application of Balwant Singh’s equality is to show that if x ∈ I is M -superficial with respect to I then ei (M/xM ) = ei (M ) for i = 0, . . . , dim M − 1. Combined with Sally Descent this yields that, if a property P of Cohen-Macaulay modules involves e0 , . . . , ei then generally the property P is true if and only if it is true for all Cohen-Macaulay modules of dimension ≤ i. 2.Cohen-Macaulay modules of minimal multiplicity In [1] Abhyankar proves the inequality e0 (A) ≥ 1+µ(A)−d for a Cohen-Macaulay local ring A. In [20] Sally describes the difference between the two sides. Namely, she proves that if x1 , . . . , xd is a maximal superficial sequence then e0 (A) = 1 + µ(M ) − d + λ(m2 /(x1 , . . . , xd )m). We extend this result to modules.
4 Theorem 2.1 Let A be a local ring, M a Cohen-Macaulay A-module of dimension r and I an ideal of definition of M . If x1 , . . . , xr is a maximal M -superficial sequence with respect to I and J = (x1 , . . . , xr ) then the following equality holds: eI0 (M ) = hI0 (M ) + hI1 (M ) + λ(I 2 M/JIM ) We call the equality above as the Abhyankar-Sally equality. We prove this theorem by first considering the case when x1 , . . . , xr is an A-regular sequence. Here the proof is similar to the ring case. To treat the general case note that none of the terms in the Abhyankar-Sally equality changes when we complete the ring with respect to m. So we may assume that the A is complete. By Cohen’s Structure theorem there exists a regular local Q such that A = Q/L. If K is the inverse image of I then GL (M ) = GI (M ). For i = 1, . . . , r pick yi ∈ K which maps to xi . As the x1 , . . . , xr is M -regular sequence so is the sequence r, hence by the New Intersection theorem it is also a Q-regular sequence. This reduces the theorem to the previous case. The previous theorem allows to make the following definition: Definition Let A be a local ring, M a Cohen-Macaulay A-module of dimension r and I an ideal of definition of M . We say M has minimal multiplicity with respect to I if eI0 (M ) = hI0 (M ) + hI1 (M ). When M = A this notion gives the usual definition of a ring of minimal multiplicity. In the process of studying rings of minimal multiplicity it is useful to also look at the first Hilbert coefficient. Northcott proved that an inequality eI1 (A) ≥ eI0 (A) − λ(A/I) holds for Cohen-Macaulay local ring (see [6]). Fillmore extended it to Cohen-Macaulay modules [2, p. 218] i.e. eI1 (M ) ≥ eI0 (M ) − λ(M/IM ) We give a different proof of this result. We show that if equality holds then M has minimal multiplicity with respect to I, this extends a theorem due to Huneke [6] and Ooishi [11]. We study the difference. χI1 (M ) = eI1 (M ) − eI0 (M ) + λ(M/IM )
5 L Set depth GI (M ) = grade(M, G(M )) where M = m/I ⊕ ( n≥1 I n /I n+1 ) is the irrelevant maximal ideal of GI (A). Using the techniques developed in the previous chapter and the Abhyankar-Sally equality, we prove following theorem which characterizes modules of minimal multiplicity. Theorem 2.4. Let A be a local ring, M a Cohen-Macaulay A-module of dimension r and I an ideal of definition of M . If M has minimal multiplicity with respect to I then GI (M ) is a Cohen-Macaulay GI (A)-module. Furthermore, the following conditions are equivalent : (i) M has minimal multiplicity with respect to I (ii) For every maximal M -superficial sequence x1 , . . . , xr there is an equality I 2 M = (x1 , . . . , xr )IM (iii) For some maximal M -superficial sequence x1 , . . . , xr there is an equality I 2 M = (x1 , . . . , xr )IM (iv) deg hIM (z) ≤ 1. (v) χI1 (M ) = 0. 3. Hilbert coefficients of Syzygy Modules. In this chapter we study a problem on Hilbert coefficients that does not arise when one restricts to rings. We assume that A is Cohen-Macaulay and I is an mprimary ideal. We let SyzA 1 (M ) denote the first syzygy module of M in a minimal free resolution of M over A. We first show that if M is maximal Cohen-Macaulay then µ(M )eI1 (A) ≥ eI1 (M ) + eI1 (SyzA 1 (M )) Sharper results are obtained when I = m. It is easy to see that that the function n+1 n 7→ TorA ) is a polynomial function (see Proposition 3.1.1. We show that 1 (M, A/m
it has degree d − 1 if M is non-free and use it to derive the following theorem.
6 Theorem 3.1.6 Let A be a Cohen-Macaulay local ring of dimension d ≥ 1 and let M be a maximal Cohen-Macaulay A-module. Then : A 1. µ(M )e1 (A) − e1 (M ) − e1 (SyzA 1 (M )) ≥ e0 (Syz1 (M ))
2. µ(M )χ1 (A) ≥ χ1 (M ) + χ1 (SyzA 1 (M )). In the theorem below we establish similar inequalities for higher Hilbert coefficients of maximal Cohen-Macaulay modules over Gorenstein rings. For every A-module we set M ∗ = HomA (M, A). Note that if M is maximal Cohen-Macaulay then so is M ∗ . Also, type(M ) = ExtdA (k, M ) denotes the Cohen-Macaulay type of M . Theorem 3.3.1 Let (A, m) be a Gorenstein local ring. Let M be a maximal Cohen ∗ ∗ Macaulay A-module and set S A (M ) = SyzA 1 (M ) . Assume depth G(A) ≥ d − 1 and G(M ) is Cohen-Macaulay. Then the following hold: 1. type(M )e2 (A) ≥ e2 (M ) + e2 (S A (M )). 2. type(M )ei (A) ≥ ei (M ) for each i ≥ 0. Notice that we have a an exact sequence 0 → M → G → S A (M ) → 0 with G ∼ = Atype M . Furthermore if A is a hypersurface and M has no free summands, then S A (M ) = SyzA 1 (M ) and µ(M ) = type(M ). By using Sally Descent and Balwant Singh’s equality we reduce the proof of the theorem to case of dimension one. The proof of the theorem above and that of one in the next chapter involves a study of the modules: Lt (M ) =
M
n+1 TorA , M) t (A/m
n≥0
for all t ≥ 0. If x1 , . . . , xs is a sequence of elements in m, then we give Lt (M ) a structure of a graded A[X1 , . . . , Xs ]-module. Each exact sequence of A-modules 0 → N → F → M → 0 induces a long exact sequence of graded A[X1 , . . . , Xs ]modules · · · → Lt+1 (M ) → Lt (N ) → Lt (F ) → Lt (M ) → · · · → L0 (M ) → 0
7 Theorem 3.3.1 is proved by showing that if dim M = 1 and x is M -superficial, then L1 (M ) is a Cohen-Macaulay A[X]-module. 4. Hilbert function of Modules over hypersurfaces If A is regular and M is a maximal Cohen-Macaulay A-module, then it is free. The next case is that of hypersurface rings. In this chapter we assume that A is a hypersurface ring. It is easy to see that if depth G(M ) is positive, then the Hilbert function of M is non-decreasing. The main result of this chapter is Theorem 4.1.1 Let (A, m) be a hypersurface ring of positive dimension. If M is a maximal Cohen-Macaulay A-module, then the Hilbert function of M is non-decreasing. It suffices to treat the case when A is complete. So A = Q/(f ) for some regular local ring (Q, n). Note that if M is a maximal A -module then projdimQ M = 1. Let φ
0 −→ Qµ(M ) −−M → Qµ(M ) −→ M −→ 0 be a minimal presentation of M . If x, y is a M ⊕ Q-superficial sequence then we consider L0 (Q), L0 (M ) and L1 (M ) as A[X, Y ]-modules. We show that X, Y is a L0 (Q) and L1 (M ) regular sequence and then use the exact sequence 0 → L1 (M ) → L0 (Q)n → L0 (Q)n → L0 (M ) → 0. to obtain equations for the Hilbert function of M . The next goal is to give lower bounds on both the multiplicity and first Hilbert coefficient of M when A = Q/(f ) for some regular local ring (Q, n). φ
If 0 → Qn −−M → Qn → M → 0 is a minimal presentation of M the following are invariants of M . IM = ideal generated by the entries of φM i(M ) = max{i | Iφ ⊆ ni }. Theorem 4.2.5 Let (Q, n) be a regular local ring, f ∈ ne \ ne+1 , e ≥ 2, A = Q/(f ), M a maximal Cohen-Macaulay A-module and K = SyzA 1 (M ). Then 1. e(M ) ≥ µ(M )i(M ) and e1 (M ) ≥ µ(M )
i(M ) 2
.
8 2. M is a free A-module if and only if i(M ) = e. 3. If i(M ) = e − 1 then G(M ) is Cohen-Macaulay 4. The following are equivalent: i. e(M ) = µ(M )i(M ). ) ii. e1 (M ) = µ(M ) i(M . 2 iii. G(M ) is Cohen-Macaulay and hM (z) = µ(M )(1 + z + . . . + z i(M )−1 ) If these conditions hold and M is not free, then G(K) is Cohen-Macaulay and hK (z) = µ(M )(1 + z + . . . + z e−i(M )−1 ). The theorem above is easy to show when dim A = 0. To reduce the general case to dim 1 we introduce a notion of superficial element x with respect to to a map φM where φ
0 −→ Qµ(M ) −−M → Qµ(M ) −→ M −→ 0 This is a Q ⊕ M -superficial element such that i(M/xM ) = i(M ). When dim M = 1 then 1 and 2 follow from the case of dim 0. A technical lemma which relates the Hilbert function of M and M/xM gives 3. To prove 4. we prove a lemma which is interesting in its own right. Lemma 4.2.7 Let (Q, n) be a regular local ring of dimension two, f ∈ ne \ ne+1 , e ≥ 2, A = Q/(f ). If M a maximal Cohen-Macaulay A-module then hM (z) = µ(M )(1 + z + . . . + z i(M )−1 ) +
X
hi (M )z i
and hi (M ) ≥ 0 for all i
i≥i(M )
Theorem 4.2.5 is applied to the case of Ulrich modules, that is, maximal CohenMacaulay modules that satisfy e(M ) = µ(M ). It is known [7] that Ulrich A-modules exist when A is a complete hypersurface ring. Using the previous theorem we get that if M is Ulrich, then i(M ) = 1 and so G(SyzA 1 (M )) is Cohen-Macaulay.
9
1. PRELIMINARIES In this thesis all rings are commutative Noetherian and all modules are assumed finite. In this chapter we introduce notation and develop a few techniques which will be used throughout this thesis. In section 1 we define the basic notions. In Section 2 we extend to modules some of the basic techniques used to study Hilbert functions of a ring, in particular: Valabrega and Valla’s Theorem, Sally Descent, Singh’s Equality. 1.1
Notation Let (A, m) be a local ring of dimension d with residue field k = A/m. Let M be
a finite A-module. Let I be an ideal of definition of M i.e. mn M ⊆ IM for some L n > 0. Let GI (A) = n≥0 I n /I n+1 be the associated graded ring of A with respect L n n+1 to I and let GI (M ) = M be the associated graded module of M n≥0 I M/I with respect to I, considered as a graded GI (A)-module. We let λ(N ) denote the length of an A-module N and µ(N ) number of its minimal generators. The function H I (M, n) = λ(I n M/I n+1 M ), defined for all n ≥ 0, is called the Hilbert function of M . The formal power series I HM (z) =
X
H I (M, n)z n
n≥0
is called the Hilbert series of M . It is easy to see that X n≥0
λ(M/I
n+1
I HM (z) M )z = (1 − z) n
It is well known that it is a rational function of the form I HM (z) =
hIM (z) , where r = dim M (1 − z)r
and
hIM (z) = hI0 (M ) + hI1 (M )z + · · · + hIs (M )z s ∈ Z[z]
10 We call the polynomial hIM (z) the h-polynomial of M with respect to I. It follows from the definition that hI0 (M ) = λ(M/IM ) and hI1 (M ) = λ(IM/I 2 M ) − rλ(M/IM ) L We set depth GI (M ) = grade(M, GI (M )) where M = m/I ⊕ ( n≥1 I n /I n+1 ) is the irrelevant maximal ideal of GI (A). If m is a non-zero element of M and if j is the largest integer such that m ∈ I j M , then we let m∗ denote the image of m in I j M \ I j+1 M . If L is a submodule of M , then L∗ denotes the graded sub module of GI (M ) generated by all l∗ with l ∈ L. It is well known that GI (M/L) ∼ = GI (M )/L∗ . Remark 1.1.1. Let x1 , ..., xs be a sequence in A with xi ∈ I and set J = (x1 , ..., xs ). For the ring B = A/J, the ideal K = I/J and the B-module, N = M/JM there is an equality GI (N ) = GK (N ). Since GK (B) = GI (A)/J ∗ it follows that depthGI (A) GI (N ) = depthGK (B) GK (N ) Remark 1.1.2. We let R = A n M denote the trivial extension ring of A by M . As an abelian group R is the direct sum of A and M . The multiplication is defined by (a, x)(b, y) = (ab, ay + bx)
for all a, b ∈ A and x, y ∈ M.
It is known that R is a local Noetherian ring with maximal ideal n = (m, M ) and that M is an ideal of R with M 2 = 0. Set L = (I, M ). There are equalities Li = (J i , J i−1 M ) for all i ≥ 1, and so we have GL (R) = GI (A)nGI (M )(−1). Let x1 , . . . , xs be a sequence in A. Set yi = (xi , 0) for i = 1, . . . , s. It is clear that y1 , . . . , ys form an R-regular sequence if and only if x1 , . . . , xs form both a A-regular sequence and an M -regular sequence. Definition 1.1.3. A property P is called sufficiently general and an element a ∈ I is called sufficiently general element of an ideal I if the following holds: Let I be an ideal in A, minimally generated by y1 , . . . , yl . An element a ∈ I is said to be sufficiently general with respect to P if there exists a non-empty Zariski-open subset P U of k l such that whenever a = ai yi and the image of (a1 , . . . , al ) in k l lies in U , then a satisfies P.
11 An element x ∈ I is called superficial for M with respect to I if there exists an integer c > 0 such that (I n M :
M x)
∩ I c M = I n−1 M for all n > c.
We omit the phrase ” with respect to I” if the ideal is clear from the context. It is well known that if k is infinite then a sufficiently general element of I is superficial for M (see [23, p. 303]). Remark 1.1.4. If the residue field is finite we resort to the standard trick to replace A by A0 = A[X]S , I by I 0 = IA0 and M by M 0 = M ⊗A A[X]S where S = A[X] \ mA[X]. It is easy to check that the residue field of A0 is k(X), the field of rational functions over k. Furthermore 0
H I (M 0 , n) = H I (M, n)
for all n ≥ 0
depthGI 0 (A0 ) G(M 0 ) = depthGI (A) G(M ) projdimA0 M 0 = projdimA M If A is Gorenstein then A0 is also Gorenstein. If A has a canonical module ωA then ωA0 ∼ = ωA ⊗ A0 . If A is a hypersurface then A0 is Cohen-Macaulay with embedding dimension one more than the dimension, and so A0 is a hypersurface. Therefore, for many problems we may without loss of generality assume k is infinite. Remark 1.1.5. (i) If depth M > 0 then it is easy to see that every M -superficial element is also M -regular. See [5, 2.1] for the case M = A. The proof of the general case is similar, (ii) If x is superficial and M -regular, then by using the Artin-Rees lemma for M and xM one gets (I n M :
M x)
= I n−1 M for all n 0. See [17, p. 8] for the case
M = A. The proof of the general case is similar. 1.2
Basic Techniques For M = A the theorem below follows from a result of Valabrega and Valla (see
[24, 2.7]). We use it to obtain a result for modules.
12 Theorem 1.2.1. Let (A, m) be a local ring, M an A-module and I an ideal in A. Let x1 , . . . , xs be an M -regular sequence with xi ∈ I \ I 2 for i = 1, . . . , s. Set J = (x1 , . . . , xs ). The following conditions are equivalent: i. x1 , . . . , xs is a GI (M )-regular sequence. ii. I i M ∩ JM = I i−1 JM for all i ≥ 1. Proof. Assume first x∗1 , . . . , x∗s is a GI (A)-regular sequence. It is easy to see that x1 , x2 , ..., xs is an A-regular sequence. By Valabrega and Valla’s theorem for the ring A we get that I n+1 ∩ J = I n J for all n ≥ 1. Let R = A n M . Set L = (I, M ). We have Li = (I i , I i−1 M ). We set yi = (xi , 0) for all i, 1 ≤ i ≤ s. Note that Q = (y1 , . . . , ys )R = (J, JM ) and that y1 , . . . , ys form an R-regular sequence. For all i ≥ 0 we have the following equalities Li+1 ∩ Q = (I i+1 ∩ J, I i M ∩ JM )
QLi = (J, JM )(I i , I i−1 M ) = (JI i , I i−1 JM + I i JM ) = (JI i , I i−1 JM ). It follows that Li+l ∩ Q = QLi if and only if I i ∩ JM = I i−1 JM . Using Valabrega and Valla’s theorem for the ring R we see that y1∗ , . . . , ys∗ is a G(R)-regular sequence if and only if Li+l ∩ Q = QLi . It follows that x∗1 , . . . , x∗s is a GI (M )-regular sequence if and only if I n M ∩ JM = I n−1 JM for all n ≥ 1. To treat the general case we set A0 = A[X1 , . . . , Xs ](m,X1 ,...,Xs )
and Q = (I, X1 , . . . , Xs )A0
Note that GQ (A0 ) = GI (A)[X1∗ , . . . , Xs∗ ]. So X1∗ , . . . , Xs∗ is a GQ (A0 )-regular sequence. The A-algebra homomorphism A[X1 ..., Xs ] → A mapping Xi to xi for i = 1, . . . , s induces a ring homomorphism A0 → A that turns M into a A0 -module. One easily checks that GQ (M ) = GI (M ) and X1 , . . . , Xs is an M -sequence. Set K = (X1 , . . . , Xs )A0 .
13 For every submodule N of M and for all i ≥ 1 we also have : Qi = (I i , I i−1 K, . . . , IK i−1 , K i )A0 Qi N = I i N K iN = J iN Qi M ∩ KM = I i M ∩ JM Qi−1 KM = I i−1 JM It follows that Qi M ∩ KM = Qi−1 KM if and only if I i M ∩ JM = I i−1 JM . The sequence X1∗ , . . . , Xs∗ is GQ (A0 )-regular, so by the already settled case we get that it is a GQ (M )-regular sequence if and only if Qi M ∩ KM = Qi−l KM . Since GQ (M ) = GI (M ), we conclude that the sequence x∗1 , . . . , x∗s is GI (M )-regular if and only if I i M ∩ JM = I i−1 JM for all i ≥ 1. A sequence x1 , . . . , xs is called a superficial sequence for M with respect to I if xi is superficial for M/(x1 , . . . , xi−1 ) for i = 1, . . . , s. A sequence x1 , . . . , xs in A is called a superficial M -regular sequence with respect to I if x1 , . . . , xs is both an M -regular sequence and a M -superficial sequence with respect to I. Again we will omit the phrase ”with respect to I” if I is clear from the context. The next lemma is well known. We include a proof for completeness. Lemma 1.2.2. Let (A, m) be a local ring, M an A-module and I an ideal for definition of M . Let x1 , . . . , xs ∈ I be a superficial M -regular sequence. We then have (x1 , . . . , xs )M ∩ I i M = (x1 , . . . , xs )I i−1 M for all i 0. Proof. We argue by induction on s. The case s = 1 follows from Remark 1.1.5(ii) We assume the result holds for s = n − 1 and prove it for s = n. Set J = (x1 , . . . , , xn−1 ) and N = M/JM . Since xn is both regular and superficial on N , we have (I i N :
N xn )
= I i−1 N for all i 0.
By induction hypothesis we have JM ∩ I i M = JI i−1 M for all i 0.
14 We take i large enough so that both preceding conditions hold. Choose t in (x1 , . . . , xn )M ∩ I i M and write it as t = t1 + xn l with t1 ∈ JM . The equality t = xn l implies l ∈ I i−1 N . Then l = t2 + l1 with t2 ∈ JM and l1 ∈ I i−1 M . Therefore, we have t = t1 + xn (l1 + t2 ), and so t − xn l1 = t1 + xn t2 ∈ (x1 , . . . , xn−l )M ∩ I i M = (x1 , . . . , xn−1 )M. It follows that t ∈ (x1 , . . . , xn )I i−1 M as desired. We use the preceding lemma and Theorem 3 to prove the following result. Theorem 1.2.3. Let (A, m) be a local ring, M an A module and I an ideal of definition for M . Let x1 , . . . , xs be a superficial M -regular sequence and set J = (x1 , . . . , xs ). The following conditions are equivalent: (i) x1 , . . . , xs is a GI (M )-regular sequence. (ii) (JM )∗ = (x∗1 , . . . , x∗s )GI (M ). (iii) The canonical map GI (M )/(x∗1 , . . . , x∗s )GI (M ) −→ GI (M/JM ) is an isomorphism. Proof. By definition, for each i ≥ 0 there is an equality (JM )∗i
JM ∩ I i M + I i+1 M = I i+1 M
(∗)
(i) =⇒ (ii). Theorem 3 implies JM ∩ I i M = JI i−1 M for all i ≥ 1. Therefore, the equality (*) becomes (JM )∗i =
JI i−1 M + I i+1 M = (x∗1 , . . . , x∗s )GI (M )i I i+1 M
So we get (JM )∗ = (x∗1 , . . . , x∗s )GI (M ). (ii) =⇒ (iii). Using (*) we get JM ∩ I i M = JI i−1 M
mod I i+1 M.
(**)
15 By Lemma 1.2.2 there exists a least positive integer l such that JM ∩ I i M = JI i−1 M for all i ≥ l. If l 6= 1 then JM ∩ I l−1 M % JI l−2 M . If t ∈ JM ∩ I l−1 M \ JI l−2 M , then by (**) we get t1 ∈ JI l−2 M such that t − t1 ∈ JM ∩ I l M = JI l−1 M . So we get t ∈ JI l−2 M a contradiction. Therefore l = 1, so (i) holds. (ii) =⇒ (iii) follows from the isomorphism GI (M/JM ) ∼ = GI (M )/(JM )∗ . (iii) =⇒ (ii) because (x1 , . . . , xs )GI (M ) ⊆ ((x1 , . . . , xs )M )∗ . The next theorem is useful for arguments by induction on the dimension of M . In the ring case the statement (2) below is referred to as Sally descent; we use the same name in the module case as well. Theorem 1.2.4. Let (A, m) be a local ring, M an A-module and I an ideal of definition for M . Let x1 , . . . , xs be a superficial M -regular sequence. If we let J = (x1 , . . . , xs ), (B, n) = (A/J, m/J), N = M/JM and K = I/J then 1. depthGI (A) GI (M ) ≥ s if and only if x∗1 , . . . , x∗s is a GI (M )-sequence. 2. depthGI (A) GI (M ) ≥ s + 1 if and only if depthGK (B) GK (N ) ≥ 1. Proof. (1) If x∗1 , . . . , x∗s is a GI (M )-sequence, then clearly depth GI (M ) ≥ s. Conversely, if depth GI (M ) ≥ s, then we prove by induction on s that x∗1 , . . . , x∗s is a GI (M )-sequence. We first prove the assertion for s = 1. We can choose a homogeneous element y ∈ GI (A) such that y is GI (M )-regular. Since x1 is M -superficial and M -regular there exists an integer n0 such that (0 :
∗ GI (M ) x1 )n
= 0 for all n ≥ n0 .
If p ∈ GI (M ) is homogeneous with x∗1 p = 0 then x∗1 y s p = 0, where s is chosen so that c = s deg y + deg p ≥ n0 . Therefore we get y s p ∈ (0 :
GI (M ) x1 )c
= 0. This forces
p = 0. Assume the assertion holds for s = n − 1. So x∗1 , . . . , x∗n−1 is GI (M )-regular by the induction hypothesis. Set J1 = (x1 , . . . , xn−1 ) and L = (x∗1 , . . . , x∗n−1 ). Theorem 1.2.3 yields GI (M )/LGI (M ) = GI (M/J1 M ) and so we get depth G(M/J1 M ) ≥ 1. Since xs is (M/J1 M )-superficial, the result for s = 1 shows that x∗n is regular on GI (M/J1 M ) = GI (M )/LGI (M ). Therefore, x∗1 , . . . , x∗n is a GI (M )-regular sequence. (2) We have depthGI (A) GI (N ) = depthGK (B) GI (N ), see Remark 1.1.1. Assume first
16 depthGI (A) GI (M ) ≥ s+1. By (1) we see that x∗1 , . . . , x∗s is a GI (M )- regular sequence and from Theorem 1.2.3 we get GI (M )/(x∗1 , . . . , x∗s )GI (M ) ∼ = GI (M/(x1 , . . . , xs )M ). Therefore, we get depth GI (N ) ≥ 1. Conversely if depth GI (N ) ≥ 1, then an argument similar to that for the case M = A, [5, 2.2] shows thatx∗1 , . . . , x∗s is a GI (M )regular sequence. Theorem 1.2.3 yields GI (M/JM ) ∼ = GI (M )/(x∗1 , . . . , x∗s )GI (M ). Therefore, depth GI (M ) ≥ s + 1. The first part of the following theorem is proved by Singh [22, Th. 1] for M = A. The same proof applies to the general case. We give a different proof for completeness. We use 4 and < to denote coefficient-wise inequalities of formal power series. Theorem 1.2.5. Let (A, m) be a local ring, M an A-module and I an ideal of definition for M . If x ∈ I \ I 2 , N = M/xM and K = I/xI then 1. H I (M, n) = λ(N/K n+1 N ) − λ ((I n+1 M :
M x)/I
2. x∗ is GI (M ) -regular if and only if I n+1 M :
Mx
n
M) .
= I n M for all n ≥ 0.
I 3. HM (z) 4 HNK (z)/(1 − z) with equality if and only if x∗ is GI (M )-regular.
Proof. 1. Consider the exact sequence: 0→
(I n+1 M : M x) M x M N → n → − n+1 → n+1 → 0 n I M I M I M K N
Computing lengths we get the required result. 2. First assume x∗ is GI (M ) -regular. Let p ∈ (I n+1 M :
M x).
If p ∈ / I n M then
there exists j < n such that p ∈ I j M \ I j+1 M . So x∗ p∗ = 0. Since x∗ is regular we get p∗ = 0 which is a contradiction. Therefore I n+1 M :
Mx
= I n M for all n ≥ 0.
Note that for all n ≥ 0 we have an exact sequence 0 −→ So if I n+1 M :
Mx
(I n+1 M :
M x) ∩ I nM
I n−1 M
−→
I n−1 M x∗ I n M −→ n+1 I nM I M
= I n M for all n ≥ 0 then x∗ is GI (M )-regular.
We get 3. from 1. and 2.
17 Singh’s theorem motivates the introduction of the following entities. bIn (x, M ) = λ (I n+1 M : M x)/I n M X bIn (x, M )z n bIx,M (z) =
n≥0
Notice that bI0 (x, M ) = 0. We note two useful corollaries of the previous theorem. Corollary 1.2.6. Let (A, m) be a Noetherian ring, M an A-module of dimension r and I an ideal of definition for M . Let x ∈ I be both M -superficial and M -regular. Set B = A/(x). For the B-module N = M/xM and the ideal K = I/(x) in B the following hold : I (1) dim N = dim M − 1 and hK 0 (N ) = h0 (M ).
(2) bIx,M (z) is a polynomial. r I (3) hIM (z) = hK N − (1 − z) bx,M (z). 2 (4) hI1 (M ) = hK 1 (N ) if and only if I M ∩ xM = xIM . I (5) eK i (N ) = ei (M ) for every i = 0, . . . , r − 1. r (6) eIr (M ) = eK r (N ) − (−1)
I n n≥1 bn (x, M )z .
P
(7) x∗ is GI (M )-regular if and only if bIn (x, M ) = 0 for all n ≥ 0. ∗ (8) eIr (M ) = eK r (N ) if and only if x is GI (M )-regular.
Proof. (1) is clear. Since x is both M -superficial and M -regular, we have bIn (x, M ) = 0 for n 0, which gives (2). Theorem 1.2.5.1 gives (3). Notice that bI1 (x, M ) = 0 if and only if I 2 M ∩ xM = xIM from this (4) is obvious. Both (5) and(6) follow from (3). The assertion of (7) is clear and (8) follows from (6) and (7). Corollary 1.2.7. Let (A, m) be a local ring, M an A-module of dimension r, and I a ideal of definition for M . Let x1 , . . . , xs be a superficial M -regular sequence. For N = M/(x1 , . . . , xs )M and K = I/(x1 , . . . , xs ) the following hold
18 I K I (1) hK 0 (N ) = h0 (M ) and ei (N ) = ei (M ) for every i = 0, . . . , r − s. ∗ ∗ (2) hIM (z) = hK N (z) if and only if x1 , . . . , xs is a GI (M )-regular sequence.
Proof. (1) The result follows from Corollary 1.2.6.1 and (4), by induction on s. (2) Set E = GI (M )/(x∗1 , . . . , x∗s )GI (M ). If x∗1 , . . . , x∗s is a GI (M )-regular sequence then the h-polynomial of GI (M ) is the same as the h-polynomial of E. By Theorem 1.2.3.3 we get GK (N ) = E and so we have hIM (z) = hK N (z). I K s Conversely if hIM (z) = hK N (z) then we have HM (z) = HN (z)/(1 − z) . We get the
result by induction on s using Theorem 1.2.5.3. When dim M = 1, a classical theorem on one dimensional Cohen- Macaulay local rings extends to modules. Proposition 1.2.8. Let M be a Cohen-Macaulay module of dimension one over a local ring A and I be an ideal of definition for M . Let x ∈ I be an M -superficial element and set ρIj (M ) = λ(I j+1 M/xI j M ) for all j ≥ 0. The following equalities then hold 1. ρI0 (M ) = eI0 (M ) − hI0 (M ). 2. H I (M, j) = eI0 (M ) − ρIj (M ) 3. If deg hIM (z) = s then ρIj (M ) = 0 for all j ≥ s and hIM (z)
=
hI0 (M )
+
s X
(ρIj−1 (M ) − ρIj (M ))z j
j=1
4. eIi (M ) =
P
j≥i−1
j i−1
ρIj (M )
Proof. By corollary 1.2.6(4) we have eI0 (M ) = eK 0 (N ) = A(M/xM ). From this (1) follows. For (2) the proof given in [26, 6.18] for M = A applies to the general case. Finally notice that (2) implies (3) and that (3) implies (4). Remark 1.2.9. The previous Proposition and Corollary 1.2.7(2) implies that if M is Cohen-Macaulay and depth GI (M ) ≥ dim M − 1 then eIi (M ) ≥ 0 for all i ≥ 0.
19
2. COHEN-MACAULAY MODULES OF MINIMAL MULTIPLICITY The first theorem in this chapter illustrates the reason why it is meaningful to talk about Cohen-Macaulay modules of minimal multiplicity. Next we give a proof of Northcott’s inequality for modules. Finally we prove a theorem which characterizes modules of minimal multiplicity. When (A, m) is a Cohen-Macaulay local ring of dimension d and x1 , . . . , xd a maximal A-superficial sequence the Abhyankar-Sally equality (see [20, 2.4]) reads: e0 (A) = 1 + µ(m) − d + λ(m2 /(x1 , . . . , xd )m) Note that h0 (A) = 1 and h1 (A) = µ(m)−d. Valla extended this identity to m-primary ideals (see [25, Lemma 1]). We extend it further to modules. Theorem 2.1. Let A be a local ring, M a Cohen-Macaulay A-module of dimension r and I an ideal of definition of M . Let x1 , . . . , xr be a maximal M -superficial sequence. Set J = (x1 , . . . , xs ). We have the following equality eI0 (M ) = hI0 (M ) + hI1 (M ) + λ(I 2 M/JIM ) Proof. First we assume that x1 , . . . , xr is an A-regular sequence. Denote the Koszul complex on x1 , . . . , xr as KA (X). Using this resolution of A/J and the fact J ⊆ I we get
λ TorA 1 (A/J, M/IM ) = rλ(M/IM ). Next we tensor the exact sequence 0 −→ IM −→ M −→ M/IM −→ 0
20 with A/J to get the exact sequence 0 −→ TorA 1 (A/J, M/IM )) −→ IM/JIM −→ M/JM −→ M/IM −→ 0 Computing lengths we get e0 (M ) = λ(M/IM ) − rλ(M/IM ) + λ(IM/JIM ).
(2.1)
Finally from the sequence 0 −→ I 2 M/JIM −→ IM/JIM −→ IM/I 2 M −→ 0 we get λ(IM/JIM ) = λ(I 2 M/JIM ) + λ(IM/I 2 M ).
(2.2)
Putting (2) and (3) above together we get the desired equality. We now prove the reb m) b is a faithfully flat A b be the m-adic completion of A. Since A sult in general. Let (A, c) ∼ b Set Ib = I A. b Note that we have G b(M algebra we can assume that A ⊆ A. = GI (M ). I c-superficial. Furthermore It is also clear that if x is M -superficial then it is also M c/JI M c. Therefore we may assume as well that A is complete. By I 2 M/JIM ∼ = I 2M Cohen’s Structure Theorem there exists a regular local ring (Q, n) such that A = Q/L for some ideal L in Q. Set K to be the inverse image of the ideal I. For i = 1, . . . , r pick yi ∈ K such that yi = xi . Since x1 , . . . , xr is an M -regular sequence, we get that y1 , . . . , yr is also an M -regular sequence. By the New Intersection theorem it follows that y1 , . . . , yr form a Q-regular sequence (see [8, 6.2.3]). Note that GI (M ) = GK (M ) and that I 2 M/JIM = K 2 M/(y1 , . . . , yr )KM .The special case considered above gives the desired equality. We call the equality above the Abhyankar-Sally equality. The previous theorem allows to make the following definition: Definition 2.2. Let A be a Noetherian local ring, M Cohen-Macaulay A-module and I an ideal of definition for M . We say M has minimal multiplicity with respect to I if eI0 (M ) = hI0 (M ) + hI1 (M ).
21 When M = A we have the usual definition of Cohen-Macaulay local ring with minimal multiplicity. In the process of studying rings of minimal multiplicity it is useful to also look at the first Hilbert coefficient. Northcott proved that an inequality eI1 (A) ≥ eI0 (A) − λ(A/I) holds for Cohen-Macaulay local ring (see [10]). Fillmore extended it to Cohen-Macaulay modules [4, p. 218] i.e. eI1 (M ) ≥ eI0 (M ) − λ(M/IM ) We include a proof for completeness. Proposition 2.3. Let (A, m) be a local ring, M a Cohen-Macaulay A-module of dimension r, and I a ideal of definition for M . Then eI1 (M ) ≥ eI0 (M ) − λ(M/IM ). Proof. Notice that λ(M/IM ) = hI0 (M ). Using Corollary 1.2.7(1) it suffices to prove the proposition for dimensions 0 and 1. If dim M = 0 then hIi (M ) = H I (M, i) ≥ 0. Since eI1 (M ) =
X
ihIi (M ) = eI0 (M ) − λ(M/IM ) +
i≥1
X
(i − 1)hIi (M )
i≥2
we get eI1 (M ) ≥ eI0 (M ) − λ(M/IM ). When dim M = 1 choose x ∈ I a superficial M -superficial element. For N = M/xM , K = I/(x) we have K I eI1 (M ) ≥ eK 1 (N ) ≥ e0 (N ) − λ(N/KN ) ≥ e0 (M ) − λ(M/IM ).
The first inequality holds from 1.2.6(6) and the second one is the dimension zero case proved above. The equality is due to 1.2.6(1) and (4). We study the difference. χI1 (M ) = eI1 (M ) − eI0 (M ) + λ(M/IM ). We now prove Theorem 2.4. Let A be a local ring, M a Cohen-Macaulay A-module of dimension r and I an ideal of definition of M . If M has minimal multiplicity with respect to I then GI (M ) is a Cohen-Macaulay GI (A)-module. Furthermore the following conditions are equivalent
22 (i) M has minimal multiplicity with respect to I (ii) For every maximal M -superficial sequence x1 , . . . , xr there is an equality I 2 M = (x1 , . . . , xr )IM (iii) For some maximal M -superficial sequence x1 , . . . , xr there is an equality I 2 M = (x1 , . . . , xr )IM (iv) deg hIM (z) ≤ 1. (v) χI1 (M ) = 0. Proof. We first show that if M has minimal multiplicity then GI (M ) is CohenMacaulay. When dim M = 0 there is nothing to prove. So assume dim M = r > 0. By Remark 1.1.4 it suffices to consider the case when the residue field of A is infinite. Let x1 , . . . , xr be a maximal M -superficial sequence. Then by the Abhyankar-Sally equality we get I 2 M = JIM . Therefore I n M = JI n M for all n ≥ 2. Therefore we get I n M ∩ JM = I n−1 JM for all n ≥ 1. Since M is Cohen-Macaulay we have that x1 , . . . , xr is also an M -regular sequence. Therefore by Theorem 1.2 we get x∗1 , . . . , x∗r is a GI (M )-regular sequence and so we get GI (M ) is Cohen- Macaulay. The Abhyankar-Sally equality gives the equivalence of (i) and (ii). Clearly (ii) implies (iii). If we assume (iii), then by Theorem 1.2 we get that GI (M ) is Cohen-Macaulay and so by Theorem 1.2.4(1) we get x∗1 , . . . , x∗r is a GI (M )-regular sequence and so by I/J
I/J
Corollary 1.2.7(2) we have hIM (z) = hM/JM (z). Clearly deg hM/JM (z) ≤ 1 and so we have (iv). Clearly (iv) implies (v). We now prove (v) =⇒ (i). If dim M = 0 then hIi (M ) = H I (M, i) ≥ 0. Since X X (i − 1)hIi (M ) ihIi (M ) = eI0 (M ) − λ(M/IM ) + eI1 (M ) = i≥2
i≥1
So we get hIi (M ) = 0 for all i ≥ 2. Therefore eI0 (M ) = hI0 (M ) + hI1 (M ) and so M has minimal multiplicity. If dim M = 1 and x is M -superficial then set ρIj (M ) = λ(I j+1 M/xI j M ). By Proposition 1.2.8 we have X X ρIj (M ) eI1 (M ) = ρIj (M ) = eI0 (M ) − hI0 (M ) + j≥0
i≥1
23 So we get ρIj (M ) = 0 for all j ≥ 1. In particular ρI1 (M ) = 0 and therefore we get I 2 M = xIM and this implies that M has minimal multiplicity. As proved earlier this implies that GI (M ) is Cohen-Macaulay. When dim M = r > 1 and x1 , . . . , xr is a maximal M -superficial sequence we set N = M/(x1 , . . . , xr−1 )M . Clearly dim N = 1. By Corollary 1.2.7(1) we have eI0 (N ) = eI0 (M ) and eI1 (N ) = eI1 (M ). Therefore by the one dimensional case we get that N has minimal multiplicity and GI (N ) is CohenMacaulay. So by Sally Descent (see Theorem 1.2.4(2)), we see that GI (M ) is CohenMacaulay. From Corollary 1.2.7(2) we have hIM (z) = hIN (z), so it follows that M has minimal multiplicity. We give a few examples of modules with minimal multiplicity. Example 2.5. If A has minimal multiplicity with respect to I then all maximal Cohen-Macaulay A-modules have minimal multiplicity with respect to I. In all the remaining examples we assume I = m. Definition 2.6. Let (A, m) be a Cohen-Macaulay local ring. A maximal CohenMacaulay A-module M is called Ulrich if e(M ) = µ(M ). Clearly Ulrich modules have minimal multiplicity. The following Cohen-Macaulay rings have Ulrich modules: (a) If dim A = 1 then it is easy to see that mn is Ulrich for all n 0. (b) If A = Q/(f1 , . . . , fs ) is a strict complete intersection (i.e. f1∗ , . . . , fs∗ form a G(Q)-regular sequence. See [7]. Next we give example’s of modules over a hypersurface ring such that M has minimal multiplicity but it is not Ulrich. Example 2.7. Set Q = k[[y1 , . . . , yd+1 ]] and n to be the maximal ideal of Q. Let a, b, c, d in n be such that f = ad − bc ∈ n3 \ n4 . Set A = Q/(f ). Set a b d −b and ψ = φ= c d −c a
24 Define M and K by the exact sequences φ
ψ
0 → Q2 − → Q2 → M → 0 and 0 → Q2 − → Q2 → K → 0 Note that M and K are maximal Cohen-Macaulay A-modules and K = SyzA 1 (M ). We prove in 4.2.12 that M or K have minimal multiplicity. We also prove that M and K are not Ulrich modules. For the next example we need the following lemma which is interesting in its own right. Lemma 2.8. Let (A, m) be a Cohen-Macaulay module with a canonical module ωA . Then A has minimal multiplicity with respect to m if and only if ωA has minimal multiplicity. Proof. Since ωA is a maximal Cohen-Macaulay A-module, it follows that if A has minimal multiplicity then ωA has minimal multiplicity. To prove the converse, it suffices to consider the case when the residue field of A is infinite (see Remark 1.1.4). If dim A = 0 then m2 ωA = 0 Since ωA is a faithful A-module we get m2 = 0. Therefore A has minimal multiplicity. When dim A = d > 0 then let x1 , . . . , xd be a superficial A ⊕ ωA sequence such that (x1 , . . . , xd ) is also a minimal reduction of m. Set (B, n) = (A/J, m/J). Note that ωB = ωA /JωA . If ωA has minimal multiplicity then by Theorem 2.4 we get m2 ωA = JmωA . So n2 ωB = 0 and as shown before this yields n2 = 0. Therefore m2 ⊆ J. Since x1 , . . . , xd is analytically independent in m (cf. [2, Remark 4.6.9]) we get m2 = Jm. So A has minimal multiplicity by Theorem 2.4. Remark 2.9. We assert that e(ωA ) = e(A). When dim A = 0 it is true because λ(ωA ) = λ(A). When dim A > 0 then we go mod a maximal superficial A ⊕ ωA sequence to get the result by the dim 0 case and Corollary 1.2.7(1). Example 2.10. Let (A, m) be a Cohen-Macaulay module such that type A = 2 and e(A) = µ(M ) − d + 2. Then K = SyzA 1 (M ) has minimal multiplicity. For proof see 3.1.8. For an example of a ring of the kind above see [26, p. 336].
25
3. HILBERT COEFFICIENTS OF SYZYGY MODULES In this chapter we assume that A is Cohen-Macaulay and I is an m-primary ideal. We let SyzA 1 (M ) denote the first syzygy module of M in a minimal free resolution of M over A. When M is maximal Cohen-Macaulay A-module then SyzA 1 (M ) is zero (when M is free) or a maximal Cohen-Macaulay A-module. Since multiplicity is additive on exact sequences of modules of the same dimension we have that µ(M )eI0 (A) = eI0 (M ) + eI0 (SyzA 1 (M )). n+1 ) . In section 1. we For eI1 (M ) we consider the function n 7→ λ TorA 1 (M, A/I show that this function a polynomial for n 0 of degree at most d − 1. A similar looking, but different result is proved by Kodiyalam [8, Theorem 2]. Sharper results are obtained when I = m. Results concerning the second Hilbert coefficient is in section 3. In section 2 we describe a construction which will be used in section 3 and in the next chapter. 3.1
The First Hilbert Coefficient
Proposition 3.1.1. Let A be a Cohen-Macaulay local ring of dimension d, I an mprimary ideal and let M be a maximal Cohen-Macaulay A-module. Set K = SyzA 1 (M ). Then 1.
n hIK (z) − µ(M )hIA (z) + hIM (z) n+1 λ TorA (M, A/I ) z = 1 (1 − z)d+1 n≥0
X
n+1 2. For n 0 the function n 7→ λ TorA ) is given by a polynomial tIM (z) 1 (M, A/I of the form: z d−1 tIM (z) = µ(M )eI1 (A) − eI1 (M ) − eI1 (K) + lower terms in z (d − 1)! 3. eI1 (A)µ(M ) ≥ eI1 (M ) + eI1 (K)
26 n+1 Proof. Set F = Aµ(M ) and Tn = λ TorA ) . Notice that hIF (z) = µ(M )hIA (z). 1 (M, A/I We tensor the exact sequence o −→ K −→ F −→ M −→ 0 with A/I n+1 to get the exact sequence n+1 0 −→ TorA ) −→ K/I n+1 K −→ F/I n+1 F −→ M/I n+1 M −→ 0 1 (M, A/I
Computing lengths and summing over all n we get (1). Set s(z) = hIK (z) − µ(M )hIA (z) + hIM (z). Since multiplicity is additive on short exact sequences we have that s(1) = eI0 (K) − µ(M )eI0 (A) + eI0 (M ) = 0. Write s(z) = (1 − z)l(z). This shows deg tIM (z) ≤ d − 1. We have l(1) = −s0 (1) = µ(M )eI1 (A) − eI1 (M ) − eI1 (K). Using this we get (2). Finally notice that (3) follows from (2) When I = m we have more precise results. Theorem 3.1.2. Let A be a Cohen-Macaulay local ring of dimension d ≥ 1 and let M be a maximal Cohen-Macaulay A-module. The following conditions are equivalent: (i) M is free. (ii) tM (z) = 0. (iii) deg tIM (z) < d − 1. (iv) µ(M )eI1 (A) = eI1 (M ) + eI1 (SyzA 1 (M )) The theorem above proves that if M is non-free maximal Cohen-Macaulay module then deg tIM (z) = d − 1. Theorem 3.1.6 gives information on its leading coefficient. For both theorems we need the following lemma: Lemma 3.1.3. If A is a Cohen-Macaulay local ring of dimension one and M is a maximal Cohen-Macaulay A-module, then n+1 λ TorA ) ≥ e0 SyzA 1 (M, A/m 1 (M ) for all n 0
27 n+1 Proof. Since dim A = 1 Proposition 3.1.1 shows that λ TorA ) is con1 (M, A/m stant (say equal to l) for all large n. Choose n0 large enough such that for all n ≥ n0 we have n+1 λ TorA ) =l 1 (M, A/m H(M, n) = e0 (M ) H(A, n) = e0 (A) Pick n > n0 and consider the exact sequence 0 −→
A A mn −→ n+1 −→ n −→ 0 n+1 m m m
We tensor it with M to get the exact sequence n TorA 1 (M, A/m ) −→ M ⊗A
mn M M −→ −→ −→ 0 mn+1 mn+1 M mn M
(∗)
Since mn /mn+1 ∼ = k e we have mn M ⊗A n+1 ∼ = m
M mM
e
Computing lengths using the exact sequence (*) we have l ≥ eµ(M ) − λ(M/mn+1 M ) + λ(M/mn M ) = eµ(M ) − λ(mn M/mn+1 M ) = eµ(M ) − e0 (M ) = e0 SyzA 1 (M ) This establishes the assertion of the lemma. Proof of Theorem 3.1.2. The implications (i) =⇒ (ii) and (ii) =⇒ (iii) are clear. By Proposition 3.1.1 we get (iii) =⇒ (iv). So it remains to (iv) =⇒ (i). We first assume d = 1. If µ(M )e1 (A) = e1 (M ) + e1 (SyzA 1 (M )) then by Propon sition 3.1.1 we get TorA 1 (M, A/m ) = 0 for all n 0. By Lemma 3.1.3 we get A e0 (SyzA 1 (M )) = 0. This implies Syz1 (M ) = 0. Therefore M is free.
28 When dim M > 1 we prove it first assuming k is infinite. Let x1 , . . . , xd be a superficial sequence for M ⊕ A ⊕ SyzA 1 (M ) . Set J = (x1 , . . . , xd−1 ), B = A/J, A M1 = M/JM and K1 = SyzA 1 (M )/J Syz1 (M ). Since x1 , . . . , xd−1 is regular on A and
M we have K1 = SyzB 1 (M1 ). Also note that µ(M ) = µ(M1 ). By Corollary 1.2.7(1) we have e1 (B) = e1 (A), e1 (M1 ) = e1 (M ) and A e1 (SyzB 1 (M1 )) = e1 (K1 ) = e1 (Syz1 (M )).
∼ Therefore by the dimension one case we get M1 is free B-module. Since TorA 1 (M, k) = 0 TorB 1 (M1 , k) = 0 we get that M is a free A-module. If k is finite then set A = A[X]S
and M 0 = M ⊗A A0 where S = A[X] \ mA[X]. The residue field of A’ is k(X) which is infinite. Furthermore H(M, n) = H(M 0 , n) for all n ≥ 0. In particular 0 A 0 ∼ 0 µ(M ) = µ(M 0 ). Since A0 is flat over A we get SyzA 1 (M ) = Syz1 (M ) ⊗A A . So we
have 0
A 0 µ(M 0 )e1 (A0 ) − e1 (M 0 ) − e1 (SyzA 1 (M )) = µ(M )e1 (A) − e1 (M ) − e1 (Syz1 (M )) = 0 0 0 Therefore by the previous case M 0 is free. So SyzA 1 (M ) ⊗A A = 0 and since A is
faithfully flat over A we get SyzA 1 (M ) = 0 and so M is free. Remark 3.1.4. The preceding theorem fails for arbitrary m-primary ideals. Let (A, m) be a non-regular Cohen-Macaulay local ring of positive dimension d, with an infinite residue field. Let M be a non-free maximal Cohen-Macaulay A-module. Let x1 , . . . , xd be a regular sequence for M , A and K = SyzA 1 (M ). Set J = (x1 , . . . , xd ). It is easy to see that hJA (z) = eJ0 (A), hJM (z) = eJ0 (M ), hJK (z) = eJ0 (K) From Proposition 3.1.1 we get tJM (z) = 0. Theorem 3.1.2 enables us to define the following: Definition 3.1.5. If (A, m) is Cohen-Macaulay of dimension d and M is maximal non-free Cohen-Macaulay then by 3.1.2 there is an equality X lM (z) n+1 λ TorA ) zn = for some lM (z) ∈ Z[z] with lM (1) 6= 0 1 (M, A/m d (1 − z) n≥0 (3.1)
29 and an equality (1 − z)lM (z) = hSyzA1 (M ) (z) − µ(M )hA (z) + hM (z)
(3.2)
Set eT (M ) = lM (1). We now prove Theorem 3.1.6. Let A be a Cohen-Macaulay local ring of dimension d ≥ 1 and let M be a maximal Cohen-Macaulay A-module. Then A 1. µ(M )e1 (A) − e1 (M ) − e1 (SyzA 1 (M )) ≥ e0 (Syz1 (M ))
2. µ(M )χ1 (A) ≥ χ1 (M ) + χ1 (SyzA 1 (M )). Proof. Note that by Corollary 1.2.7(1) it suffices to prove the result for dimension 1. Also clearly the result holds when M is free. Therefore assume M is not free. Using Proposition 3.1.1 we get that tM (z) = µ(M )e1 (A) − e1 (M ) − e1 (SyzA 1 (M )). Using Lemma 3.1.3 we get (1). To prove (2) set eT (M ) = µ(M )e1 (A)−e1 (M )−e1 (SyzA 1 (M )) and K = SyzA 1 (M ). Since µ(M )e0 (A) = e0 (M ) + e0 (K) we get µ(M )χ1 (A) = χ1 (M ) + e1 (K) − e0 (K) + eT (M ) = χ1 (M ) + χ1 (K) + eT (M ) − µ(K) Since eT (M ) ≥ e0 (K) (by (1)) and e0 (K) ≥ µ(K) we get the required result. Remark 3.1.7. If equality holds in (2) then SyzA 1 (M ) is Ulrich. Note that χ1 (A)µ(M ) = χ1 (M ) + χ1 (K) + eT (M ) − µ(K). So it follows that equality holds in (2) if and only if eT (M ) = µ(K). Since eT (M ) ≥ e(K) by 3.1.6(1) we get the required result. We use Theorem 3.1.6 in the following example. Example 3.1.8. Let (A, m) be a Cohen-Macaulay module such that type A = 2 and e(A) = µ(M ) − d + 2. Then K = SyzA 1 (M ) has minimal multiplicity. By Lemma 2.8 we get that ωA does not have minimal multiplicity. So by Theorem 2.4 χ1 (ωA ) ≥ 1.
30 Note that hA (z) = 1 + h1 (A)z + z 2 and type A < h1 (A). So e(A) ≥ 5. We also have an exact sequence 0 → K → A2 → M → 0. We first prove that K is not Ulrich. By Remark 1.1.4 it suffices to consider the case when the residue field of A is infinite. Let x1 , . . . , xd be a K ⊕ A ⊕ ωA -superficial sequence. Set (B, n) = (A/J, m/J), N = M/JM and L = K/JK. So we have an exact sequence of B-modules 0 → L → B 2 → N → 0. By Remark 2.9 it follows that e(K) = e(A). If K is Ulrich then L ∼ = k e(K) = k e(A) . Since 0 → L → B 2 , this yields that 4 = type A2 ≥ e(A) which is a contradiction. To prove that K has minimal multiplicity note that by Theorem 3.1.6 and Remark 3.1.7 we get that 2 = 2χ1 (A) > χ1 (ωA ) + χ1 (K). Since χ1 (ωA ) > 0 we get χ1 (K) = 0. So by Theorem 2.4 we get that K has minimal multiplicity. Since χ1 (ωA ) = 1 = χ1 (ωB ) it follows that e1 (ωA ) = e1 (ωB ). Consequently we get that G(ωA ) is Cohen-Macaulay. Remark 3.1.9. Note that when dim A = 0 we get that lM (z) has non-negative coefficients. This yields the following: When G(A) and G(M ) is Cohen-Macaulay then ei (A)µ(M ) ≥ ei (M ) for all i ≥ 0. In Section 3.3 it is proved that if A is Gorenstein then it suffices to assume that depth G(A) ≥ d − 1. 3.2
Basic Construction
n+1 Remark 3.2.1. For each n ≥ 0 and t ≥ 0 set Lt (M )n = TorA , M ). For t (A/m L t ≥ 0 let Lt (M ) = n≥0 Lt (M )n . If x1 , . . . , xs is a sequence of elements in m, then
we give Lt (M ) a structure of a graded A[X1 , . . . , Xs ]-module as follows: For i = 1, . . . , s let ξi : A/mn → A/mn+1 be the maps given by ξi (a + mn ) = xi a + mn+1 . These homomorphisms induces homomorphisms A A n n+1 TorA , M) t (ξi , M ) : Tort (A/m , M ) −→ Tort (A/m
Thus, for i = 1, . . . , s and each t we obtain homogeneous maps of degree 1: Xi : Lt (M ) −→ Lt (M ). For i, j = 1, . . . , s the equalities ξi ξj = ξj ξi yields equalities Xi Xj = Xj Xi . So Lt (M ) is a graded A[X1 , . . . , Xs ]-module for each t ≥ 0.
31 Proposition 3.2.2. Let M, F , and K be finite A-modules and let x1 , . . . , xs be a sequence of elements in m. If Lt (M ), Lt (F ) and Lt (K) are given the A[X1 , . . . , Xs ]module structure described in Remark 3.2.1 then 1. Every exact sequence of A-modules 0 → K → F → M → 0 induces a long exact sequence of graded A[X1 , . . . , Xs ]-modules · · · → Lt+1 (M ) → Lt (K) → Lt (F ) → Lt (M ) → · · · → L0 (M ) → 0. 2. For i = 1, . . . , s there is an equality mn+1 M : M xi Xi ker L0 (M )n−1 −→ L0 (M )n = mn M 3. If xi ∈ m \ m2 is such that x∗i is G(M )-regular then Xi is L0 (M )-regular. 4. If F is free A-module and xi is K-superficial for some i then Xi (a) ker L1 (M ) −→ L1 (M ) n = 0 for n 0 (b) If x∗i is G(K)-regular then Xi is L1 (M )-regular. Proof. To prove part 1, set n+1 n+1 n+1 βt,n = TorA , β) : TorA , K) −→ TorA ,F) t (A/m t (A/m t (A/m n+1 n+1 n+1 αt,n = TorA , α) : TorA , F ) −→ TorA , M) t (A/m t (A/m t (A/m
and consider the connecting homomorphisms n+1 n+1 δt+1,n : TorA , M ) −→ TorA , K) t+1 (A/m t (A/m
By a well known theorem in Homological algebra if X is a free resolution of K and Z is a free resolution of M then there exists a free resolution Y of F and an exact sequence of complexes of free A-modules 0 → X → Y → Z → 0 whose homology sequence is the given exact sequence 0 → K → F → M → 0. This yields for each n a commuting diagram of complexes with exact rows ; 0
/ X ⊗ A/mn
0
ξi
/ X ⊗ A/mn+1
/ Y ⊗ A/mn
ξi
/ Y ⊗ A/mn+1
/ Z ⊗ A/mn
/0
ξi
/ Z ⊗ A/mn+1
/0
32 In homology it induces the following commutative diagram : ···
···
δ
/ Lt+1 (M )n−1t+1,n−1/ Lt (K)n−1
Xi
/ Lt+1 (M )n
δt+1,n
βt,n−1
/ Lt (F )n−1
Xi
/ Lt (K)n
βt,n
αt,n−1
/ Lt (M )n−1
Xi
/ Lt (F )n
αt,n
/ ···
Xi
/ Lt (M )n
/ ···
This proves the desired assertion. Remark 3.2.3. We will use the exact diagram above often. So when there is a reference to this remark, I mean to refer the commuting diagram above. The second part is clear from the definition of the action Xi . Part 3. follows from 2. Note that if F is free, then L1 (F ) = 0, so 1. gives an exact sequence of A[X1 , . . . , Xs ]-modules 0 −→ L1 (M ) −→ L0 (K). Together with 2. and 3. this yields the assertions in 4,. 3.3
Second Hilbert coefficient In the theorem below we establish similar inequalities for higher Hilbert coefficients
of maximal Cohen-Macaulay modules over Gorenstein rings. For every A-module we set M ∗ = HomA (M, A). Note that if M is maximal Cohen-Macaulay then so is M ∗ cf.[2, 3.3.10.d] . Also, type(M ) = ExtdA (k, M ) denotes the Cohen-Macaulay type of M. Theorem 3.3.1. Let (A, m) be a Gorenstein local ring. Let M be a maximal Cohen ∗ ∗ Macaulay A-module and set S A (M ) = SyzA 1 (M ) . If G(M ) is Cohen-Macaulay and depth G(A) ≥ d − 1 then the following hold 1. type(M )e2 (A) ≥ e2 (M ) + e2 (S A (M )). 2. type(M )ei (A) ≥ ei (M ) for each i ≥ 0. Note that if A is a hypersurface ring and if M has no free summands, then S A (M ) = SyzA 1 (M ) and µ(M ) = type(M ). Before proving Theorem 3.3.1 we need a few lemmas.
33 Lemma 3.3.2. Let (A, m) be a Gorenstein local ring with infinite residue field and let M be a maximal Cohen-Macaulay A-module. If x is A-regular then for the ring B = A/(x) and the B-module N = M/xM we have 1. HomB (N, B) ∼ = M ∗ /xM ∗ . A A ∗ ∗ ∼ ∗ ∗ 2. SyzB 1 (M /xM ) = Syz1 (M )/x Syz1 (M ).
3. S B (N ) ∼ = S A (M )/xS A (M ). Proof. Since M is maximal Cohen-Macaulay x is also M -regular. To prove 1. we use the exact sequence x
0 −→ M − → M −→ N −→ 0 to get a long exact sequence x
0 −→ HomA (N, A) −→ HomA (M, A) − → HomA (M, A) −→ Ext1A (N, A) −→ Ext1A (M, A). Since M is maximal Cohen-Macaulay and A is Gorenstein we have Ext1A (M, A) = 0 cf.[2, 3.3.10.d]. Using the isomorphisms HomA (N, A) = 0
and
Ext1A (N, A) ∼ = HomB (N, B)
see [2, 3.1.16], we get the required result. Note that 2. holds since M ∗ is maximal Cohen-Macaulay. To prove 3. note that A ∗ Hom Syz (M ), A A 1 S A (M )/xS A (M ) = ∗ x HomA SyzA 1 (M ), A ∗ SyzA 1 (M ) ∼ ,B = HomB ∗ x SyzA 1 (M ) ∗ ∗ ∼ = HomB SyzB 1 (M /xM ), B
∼ = HomB SyzB 1 (HomB (N, B)), B = S B (N )
34 The next result helps in the dimension one case. Lemma 3.3.3. Let (A, m) be a Cohen-Macaulay local ring of dimension one, let M be a non-free maximal Cohen-Macaulay A-modules and let 0 −→ E −→ F −→ M −→ 0 be an exact sequence with F a finite free A-module. Let x be A ⊕ M ⊕ E-superficial. If L1 (M ) is given the A[X]-module structure described in Remark 3.2.1 then we have 1. L1 (M ) is a Noetherian A[X]-module of dimension one. 2. (1 − z)lM (z) = hE (z) − hF (z) + hM (z). 3. When G(E) is Cohen-Macaulay then a. L1 (M ) is Cohen-Macaulay. b. ei+1 (F ) − ei+1 (M ) − ei+1 (E) ≥ 0 for every i ≥ 0. Proof. 1. Since dim A = 1 and M is non-free, it follows from Lemma 3.1.3 that λ(L1 (M )n ) is a non-zero constant for large n. Since X : L1 (M )n → L1 (M )n+1 is injective for large n and since λ(L1 (M )n ) is constant for large n, it follows that L1 (M )n+1 = XL1 (M )n for large n, say for all n ≥ s. For each i = 0, 1, . . . , s choose S a finite set Pi of generators of L1 (M )i as an A-module. It is easy to see that si=0 Pi generates L1 (M ) over A[X]. Since λ(L1 (M )/XL1 (M )) < ∞ and λ(L1 (M )) = ∞ it follows that L1 (M ) has dimension one. µ(M ) ∼ 2. By Schanuel’s lemma, [9, p. 158] we have F ⊕ SyzA . Therefore 1 (M ) = E ⊕ A
(1 − z)lM (z) = hSyzA1 (M ) (z) − µ(M )hA (z) + hM (z) = hE (z) − hF (z) + hM (z) 3.a is clear from 1. and Proposition 3.2.2.3 3.b It follows from 1. that lM (z) is the h-polynomial of L1 (M ) considered as (i)
an A[X]-module. Set eTi (M ) = lM (l)/i!. Since L1 (M ) is Cohen-Macaulay all the
35 coefficients of lM (z) is non-negative. Therefore eTi (M ) ≥ 0 for all i ≥ 0. Using 2. we obtain eTi (M ) = ei+1 (F ) − ei+1 (M ) − ei+1 (E) and so we get the required result. Proof of Theorem 3.3.1. By Remark 1.1.4 it is sufficient to consider the case when the residue field of A is infinite. If M is free then M ∗ is free and so S A (M ) = 0. Since A is Gorenstein type A = 1 and so type M = µ(M ). Therefore the result holds when M is free. Next we consider the case when M is non-free. Note that µ(M ∗ ) = type(M ) ∗ cf.[2, 3.3.11.b] . Set F = Atype(M ) and L = SyzA 1 (M ). We have an exact sequence:
0 −→ L −→ F −→ M ∗ −→ 0. Dualizing this sequence we get an exact sequence 0 −→ M −→ F ∗ −→ L∗ −→ 0.
(3.3)
Note that L∗ = S A (M ) and F ∗ = Atype(M ) . When dim A = 1, assertion 1. follows from Lemma 3.3.3. When dim M = 2, let x be superficial for M ⊕ S A (M ) ⊕ A. Set N = M/xM and G = F/xF . By Lemma 3.3.2 we get S B (N ) = S A (M )/xS A (M ). Since G(M ) is Cohen-Macaulay and depth G(A) ≥ 1 we have that e2 (M ) = e2 (N ) and e2 (F ) = e2 (G). Furthermore it follows from Corollary 1.2.6.6 that e2 (S A (M )) ≤ e2 (S A (M )/xS A (M )) = e2 (S B (N )). Therefore we have: e2 (M ) + e2 (S A (M )) ≤ e2 (N ) + e2 (S A (M )/xS A (M )) = e2 (N ) + e2 (S B (N )) ≤ e2 (G) = e2 (F ) Note that the second inequality above follows from the dimension one case. When dim A > 2 let x1 , . . . , xd−2 be a M ⊕ S A (M ) ⊕ A-superficial sequence. Set J = (x1 , . . . , xd−2 ), N = M/JM, G = F/JF . Using Lemma 3.3.2.3 inductively
36 we get S B (N ) = S A (M )/JS A (M ). By Corollary 1.2.6.6 we get that e2 (S A (M )) = e2 (S B (N )), e2 (M ) = e2 (N ) and e2 (F ) = e2 (G). So the result follows from the dimension 2 case. To prove (ii) note that since G(M ) is Cohen-Macaulay and depth G(A) ≥ d − 1 it suffices to consider the case when dim A = 1. Using exact sequence (3.3) we get e(A) type(M ) ≥ e(M ). By Corollary 1.2.8.4 we get that ei (S A (M )) ≥ 0 for all i ≥ 1. So we get ei+1 (A) type(M ) ≥ ei+1 (M ) by Lemma 3.3.3.
37
4. HILBERT FUNCTIONS OF MODULES OVER HYPERSURFACES If A is regular and M is a maximal Cohen-Macaulay A-module, then it is free. The next case is that of hypersurface rings. In this chapter we assume that A is a hypersurface ring. In Section 1 we show that if M is a maximal A-module then the Hilbert function of M is non-decreasing. In Section 2 we assume A = Q/(f ), with Q-regular. When M is a maximal Cohen-Macaulay A-module we give lower bounds on e(M ) and e1 (M ) and study what happens when equality is achieved. 4.1
Monotonicity The main theorem of this section is
Theorem 4.1.1. Let (A, m) be a hypersurface ring of positive dimension. If M is a maximal Cohen-Macaulay A-module, then the Hilbert function of M is non-decreasing. The following remark will be used often. an z n be a formal power series with non-negative P coefficients. If the power series g(z) = n≥0 bn z n satisfies g(z) = f (z)/(1 − z), then P bn = ni=0 ai , and so the sequence {bn } is nondecreasing.
Remark 4.1.2. Let f (z) =
P
n≥0
The next proposition yields an easy criterion for monotonicity. Proposition 4.1.3. Let M be an A-module. If depth G(M ) ≥ 1 then the Hilbert function of M is non-decreasing. Proof. Using Remark 1.1.4 we may assume that k, the residue field of A is infinite. Since k is infinite and depth G(M ) ≥ 1 we may assume that there exists x ∈ m \ m2 , such that x∗ is G(M )-regular. So it follows that the Hilbert function of M is nondecreasing.
38 Remark 4.1.4. If φ : (A, m) −→ (B, n) is a surjective map of local rings and if M is a finite B-module then mn M = nn M for all n ≥ 0. Therefore Gm (M ) = Gn (M ). The notation G(M ) will be used to denote this without any reference to the ring. We deduce Theorem 4.1.1 from the following result. Theorem 4.1.5. Let (Q, m) be a Noetherian local ring and let M be a finite Qmodule with projdimQ M ≤ 1. If depth G(Q) ≥ 2 then the Hilbert function of M is non-decreasing. Proof. It is sufficient to consider the case when the residue field of Q is infinite (see Remark 1.1.4). Since projdimQ M ≤ 1 we have a presentation of M 0 −→ Qn −→ Qm −→ M −→ 0.
(4.1)
with 0 ≤ n ≤ m. Let x, y be elements in m \ m2 such that x∗ , y ∗ is a G(Q)-regular sequence. Let L0 (Q), L0 (M ) and L1 (M ) be the Q[X, Y ]-modules described in Remark 3.2.1. The element X is L0 (Q)-regular by Proposition 3.2.2.3. Set B = Q/(x) and note that M Q L0 (Q) = = L0 (B). n+1 ) XL0 (Q) (x, m n≥0 Since G(B) = G(Q)/x∗ G(Q) we see that y ∗ is G(B)-regular. Proposition 3.2.2.3 shows that Y is L0 (B)-regular. Thus, the sequence X, Y is a L0 (Q)-regular sequence. Using the exact sequence (4.1) and Proposition 3.2.2.1, we obtain an exact sequence of graded Q[X, Y ] modules 0 −→ L1 (M ) −→ L0 (Q)n −→ L0 (Q)m −→ L0 (M ) −→ 0. Since X, Y is an L0 (Q)- regular sequence, we see that it is also L1 (M )-regular. The regularity of X, Y imply equalities
(4.2)
39
X i≥0
X i≥0
u(z) λ(L0 (Q)i )z = (1 − z)2 i
where
X u(z) = λ
where
X v(z) = λ
i≥0
v(z) λ(L1 (M )i )z = (1 − z)2 i
i≥0
L0 (Q)i (X, Y )L0 (Q)i−1
L1 (M )i (X, Y )L1 (M )i−1
zi.
zi.
Using the exact sequence (4.2) we get X
λ(L0 (M )i )z i = (m − n)
i≥0
The equality HM (z) = (1 − z)
P
i≥0
u(z) v(z) + (1 − z)2 (1 − z)2
λ(L0 (M )i )z i yields
HM (z) = (m − n)u2 (z)/(1 − z) + v2 (z)/(1 − z). Now Remark 4.1.2 shows that the Hilbert function of M is non-decreasing. We obtain Theorem 4.1.1 as a corollary to the previous theorem. Proof of Theorem 4.1.1 . We may assume that A is complete and so A ∼ = Q/(f ) for some regular local ring (Q, n) and f ∈ n2 . Then depth M = dim Q − 1 and projdimQ M = 1. Using Theorem 4.1.5 it follows that the Hilbert function of M is non-decreasing. Since the Hilbert function is increasing if depth G(M ) > 0, we construct a maximal Cohen-Macaulay module M over a hypersurface ring A such that depth G(M ) = 0. Example 4.1.6. Set Q = k[[x, y]] and n = (x, y). Define M by the exact sequence φ
→ Q2 −→ M −→ 0 0 −→ Q2 − where φ=
x −y
y 2
0
Set (A, m) = (Q/(y 3 ), n/(y 3 )). Since the determinant of φ annihilates M we get y 3 M = 0. Note that M is a maximal Cohen-Macaulay A-module. Set K = SyzA 1 (M )
40 and E = M ⊕ K. Note that G(Q) = k[x∗ , y ∗ ]. Since y 3 E = 0 , we have that if P ∈ AssG(Q) (G(E)) then P ⊇ (y ∗ ). So we get that x∗ ∈ / P if P is a relevant associated prime of G(E). Therefore x∗ is an E = M ⊕ K-superficial element. We first show that G(M ) is not Cohen-Macaulay. Assuming it is, then 1.2.4.1 shows that x∗ is G(M )-regular. However if m1 , m2 are the generators of M then xm1 = y 2 m2 ∈ n2 M and this implies m1 ∈ (n2 M :
M x)
= nM , which is a contradiction.
We now show that G(K) is Cohen-Macaulay. Note that M defines a matrix factorization (φ, ψ) where
0
ψ= y2
−y x
and φψ = ψφ = y 3 ε3 . Since all the entries of ψ are in the maximal ideal we have ψ
0 −→ G − → F −→ K −→ 0 where F = G = Q2 Set bi (K) = λ((ni+1 K :
i K x)/n K).
(4.3)
Since y 3 K = 0 we get n3 K = xn2 K. Therefore
ni+1 K = xni K for all i ≥ 2. So we get bi (K) = 0 for all i ≥ 2. Therefore to prove G(K) is Cohen-Macaulay it is sufficient to show that b1 (K) = 0. Let p ∈ (n2 K :
K x).
We want to show that p ∈ nK. Let u = (u1 , u2 ) be the
pre-image of p in F . Using the commutative diagram in Remark 3.2.3 we see that there exists v = (v1 , v2 ) ∈ G such that xu + n2 G = ψ(v) + n2 F Therefore we get −yv2 = xu1
mod n2
y 2 v1 + xv2 = xu2
mod n2
From the first equation we obtain u1 , v2 ∈ n. Substituting this in the second equation we get xu2 ∈ n2 . Since x∗ is G(Q)-regular we get u2 ∈ n, hence p ∈ nK. This shows b1 (K) = 0, and as asserted before, this proves that G(K) is Cohen-Macaulay. This example also has yields an example of an maximal Cohen-Macaulay module P over a hypersurface such that G(P ) is Cohen-Macaulay but G(SyzA 1 (P )) is not
41 Cohen-Macaulay. Simply take P = K in the previous example. Then note that G(SyzA 1 (P )) = M . The next corollary partly overlaps with a result of Elias [3]: all equicharacteristic Cohen-Macaulay rings of dimension 1 and embedding dimension 3 have nondecreasing Hilbert functions. Corollary 4.1.7. If (A, m) be a complete intersection of positive dimension and codimension 2 then the Hilbert function of A is non-decreasing. Proof. We may assume that A is complete and hence A = Q/(f, g) for a regular sequence f, g in a regular local ring (Q, q). Set (R, n) = (Q/(f ), q/(f )). Then G(R) is Cohen-Macaulay, dim A = dim R − 1 and projdimR A = 1. Therefore by Theorem 4.1.5 we get that the Hilbert function of A is non-decreasing. The following example from [26, p. 337] illustrates an example when (A, m) is a complete intersection of dimension 1 and codimension 2 such that depth G(A) = 0. Example 4.1.8. Set A = k[[t6 , t7 , t15 ]]. We have A = k[X, Y, Z]/(Y 3 −XZ, X 5 −Z 2 ). Furthermore G(A) = k[X, Y, Z]/(XZ, Y 6 , Y 3 Z, Z 2 ) and hA (z) = 1 + 2z + z 2 + z 3 + z 5 . 4.2
Hilbert coefficients In the rest of this paper εs denotes the s × s identity matrix. Let (Q, n) be a
regular local ring, f ∈ ne \ ne+1 , e ≥ 2, A = Q/(f ), M a maximal Cohen-Macaulay A-module and K = SyzA 1 (M ). By a matrix-factorization of f we mean a pair (φ, ψ) of square-matrices with elements in Q such that φψ = ψφ = f ε. If M is an A-module then projdimQ M = 1. Also a presentation of M φ
0 −→ Qn − → Qn −→ M −→ 0 yields a matrix factorization of f . See [27, p. 54] for details.
42 In the sequel (φM , ψM ) will denote a matrix factorization of f such that φ
0 −→ Qn −−M → Qn −→ M −→ 0 is a minimal presentation of M . Note that ψ
0 −→ Qn −−M → Qn −→ SyzA 1 (M ) −→ 0 is a not-necessarily minimal presentation of SyzA 1 (M ). If φ : Qn −→ Qm is a linear map then we set Iφ = ideal generated by the entries of φ iφ = max{i | Iφ ⊆ ni }. If M has the minimal presentations: φ0
φ
0 → Qn − → Qn → M → 0 and 0 → Qn − → Qn → M → 0, then it is well known that Iφ = Iφ0 , iφ = iφ0 and det(φ) = u det(φ0 ) with u a unit. We set i(M ) = iφ and det(M ) = (det(φ)). For g ∈ Q, g 6= 0, set vQ (g) = max{i | g ∈ ne }. For convenience set vQ (0) = ∞. Note that e(Q/(g)) = vQ (g) for any g 6= 0. We first consider the case when dim A = 0. Remark 4.2.1. Let (Q, n) be a DVR, vQ (f ) = e, A = Q/(f ) and M a finite Amodule. If n = (y) then f = uy e , where u is a unit. Therefore µ(M )
M∼ =
M
Q/(y ai ) with 1 ≤ a1 ≤ . . . ≤ aµ(M ) ≤ e.
i=1
This yields a minimal presentation of M : ψ
0 → Qn − → Qn → M → 0 where ψij = δij y ai . This yields 1. i(M ) = a1 . 2. hM (z) = µ(M )(1 + z + . . . + z i(M )−1 )+ higher powers of z
43 3. h0 (M ) ≥ h1 (M ) ≥ . . . ≥ hs (M ). 4. M is free if and only if i(M ) = e. 5. If K = SyzA 1 (M ) then µ(M )
K∼ =
M
Q/(y e−ai )
i=1
6. e(M ) ≥ µ(M )i(M ) and e1 (M ) ≥ µ(M )
i(M ) 2
.
7. vQ (det φ) ≥ i(M )µ(M ) with equality if and only if e(M ) = i(M )µ(M ). We note an immediate corollary to assertion 3. in the previous remark. Corollary 4.2.2. Let (A, m) be a hypersurface ring. Let M be a maximal CohenMacaulay A-module such that G(M ) is Cohen-Macaulay. If hM (z) = h0 (M ) + h1 (M )z + · · · + hs (M )z s is the h-polynomial of M then h0 (M ) ≥ h1 (M ) ≥ . . . ≥ hs (M ). Proof. Set d = dim A. We may assume that A is complete with infinite residue field, hence A = Q/(f ) for some regular local ring (Q, n). Consider M as a Q-module. Let x1 , . . . , xd be a Q⊕M -superficial sequence. Set J = (x1 , . . . , xd ), (R, q) = (Q/J, n/J), B = A/J and N = M/JM . Note that R is a DVR. Since G(M ) is Cohen-Macaulay we also have hM (z) = hN (z) and so the result follows from Remark 4.2.1.3. To use the other assertions in Remark 4.2.1 we need the following definitions. Definition 4.2.3. Let (Q, n) be a regular local ring, f ∈ ne \ ne+1 , e ≥ 2, A = Q/(f ) φ
and M a maximal Cohen-Macaulay A-module. Let 0 → Qn − → Qn → M → 0 be a minimal presentation of M . We say that x ∈ n is φ-superficial if 1. x is (Q ⊕ M ⊕ A)-superficial. 2. If φ = (φij ) then vQ (φij ) = vQ/xQ (φij ) 3. vQ (det(φ)) = vQ/xQ (det(φ)).
44 Since e(Q/(g)) = vQ (g) for any g 6= 0 it follows that if x is Q ⊕ M ⊕ A ⊕ L Q/(φ ) ⊕ Q/ det(φ) -superficial then it is φ-superficial. So φ-superficial eleij ij ments exist if the residue field of Q is infinite. If x is φ-superficial, then clearly i(M ) = i(M/xM ). Also note that Q/xQ is regular and we have an exact sequence n n φ⊗Q Q/xQ Q M Q −−−−−−→ −→ 0 −→ −→ 0 xQ xQ xM This enables the following definition: Definition 4.2.4. Let (Q, n) be a regular local ring, f ∈ ne \ ne+1 , e ≥ 2, A = Q/(f ) φ
and M a maximal Cohen-Macaulay A-module Let 0 → Qn − → Qn → M → 0 be a minimal presentation of M . We say that x1 , . . . , xr is a φ-superficial sequence if xi is (φ ⊗Q Q/(x1 , . . . , xi−1 ))-superficial for i = 1, . . . , r. Theorem 4.2.5. Let (Q, n) be a regular local ring, f ∈ ne \ ne+1 , e ≥ 2, A = Q/(f ), M a maximal Cohen-Macaulay A-module and K = SyzA 1 (M ). Then 1. e(M ) ≥ µ(M )i(M ) and e1 (M ) ≥ µ(M )
i(M ) 2
.
2. M is a free A-module if and only if i(M ) = e. 3. If i(M ) = e − 1 then G(M ) is Cohen-Macaulay. 4. The following conditions are equivalent: i. e(M ) = µ(M )i(M ). ) ii. e1 (M ) = µ(M ) i(M . 2 iii. G(M ) is Cohen-Macaulay and hM (z) = µ(M )(1 + z + . . . + z i(M )−1 ) If these conditions hold and M is not free, then G(K) is Cohen-Macaulay and hK (z) = µ(M )(1 + z + . . . + z e−i(M )−1 ).
45 We need a few preliminaries before we prove this theorem. Notation Let M be an A- module.If x is A ⊕ M superficial (or more generally it is superficial with respect to to an injective map θ : Qn → Qn ) then set (B, n) = (A/(x), m/(x)) and N = M/xM . Lemma 4.2.6. Let (A, m) be a Cohen-Macaulay local ring of dimension d > 0 with infinite residue field. Let M be a Cohen-Macaulay A-module of dimension 1 with a preφ
sentation G − → F −→ M −→ 0 such that all entries in φ are in ml . If depth G(A) ≥ 1 and x is a A ⊕ M -superficial element then then 1. (mi+1 M :
i M x)/m M
= 0 for i = 0, . . . , l − 1.
2. Furthermore if ml M ⊆ xM then depth G(M ) ≥ 1. Proof. Set bi (M ) = λ((mi+1 M :
i M x)/m M ).
Since all the entries of φ are in ml we
have that φ0,j−1 = φ ⊗ A/mj = 0 for j = 1, . . . , l. 1. Note that b0 (K) = 0 for any A-module K. Fix i with 1 ≤ i ≤ l − 1. Let p ∈ (mi+1 M :
M x).
Let u be the pre-image of p in F . Using the commutative diagram
3.2.3 and since φ0,i = 0 and φ0,i+1 = 0 we obtain that xu ∈ mi+1 F . Since x is A-superficial and depth G(A) ≥ 1 we get by Theorem 1.2.4.1 that x∗ is also G(A)regular. Therefore u ∈ mi G and so p ∈ mi M , and so bi (M ) = 0. P 2. Since ml M ⊆ xM we get that nl N = 0 and so l−1 i=0 H(N, i) = e(N ) = e(M ). Using Theorem 1.2.5.2 we get that H(M, l − 1) =
l−1 X
H(N, i) − bl−1 (M ) = e(M )
i=0
Using Proposition 1.2.8.2 we obtain ml M = xml−1 M . So we have that mi+1 M = xmi M for all i ≥ l − 1. So we obtain that bi (M ) = 0 for al i ≥ l − 1. This combined with 1. yields that x∗ is G(M )-regular. An interesting consequence of the lemma above is the following lemma which gives information about the Hilbert function of a maximal Cohen-Macaulay module over a hypersurface ring of dimension 1.
46 Lemma 4.2.7. Let (Q, n) be a regular local ring of dimension two, f ∈ ne \ ne+1 , e ≥ 2, A = Q/(f ). If M a maximal Cohen-Macaulay A-module. then hM (z) = µ(M )(1 + z + . . . + z i(M )−1 ) +
X
hi (M )z i
and hi (M ) ≥ 0 for all i
i≥i(M )
Proof. The Hilbert function of M is non-decreasing by Corollary 4.1.1. Since dim M = 1 we have that the Hilbert series of M is equal to hM (z)/(1 − z). This implies that all the coefficients of hM (z) are non-negative. Set bi (M ) = λ((mi+1 M :
i M x)/m M ).
Since φ⊗A
An −−→ An −→ M −→ 0 is exact and all the entries of φ are in i(M ) we get by Lemma 4.2.6.1 that bi (M ) = 0 for i = 0, . . . , i(M ) − 1. This and Remark 4.2.1 yields that hM (z) = µ(M )(1 + z + . . . + z i(M )−1 ) +
X
hi (M )
i≥i(M )
Next we get an upper bound on l such that ml M ⊆ xM holds. Remark 4.2.8. If dim A = 1 and x is A-superficial then note that since the ring B ha length e0 (A) we get that me0 (A) ⊆ (x). Therefore if dim A = 1, M a maximal A-module and x is A ⊕ M -superficial then me0 (A) M ⊆ (x)M The next lemma deals with the case when M is a syzygy of a maximal CohenMacaulay A-module. Lemma 4.2.9. Let (A, m) be a Cohen-Macaulay A-module of dimension 1and let L be a maximal non free Cohen-Macaulay A-module and set M = SyzA 1 (L). If x is (A ⊕ M ⊕ L)-superficial then me0 (A)−1 M ⊆ (x)M . Proof. Note that by Remark 4.2.8 we have me0 (A) ⊆ (x). We also have an exact sequence : 0 → M → F → L → 0 where F is a free A-module. Set G = F/xF and W = L/xL. Going mod x we get 0 → N → G → W → 0. Note that N ⊆ nG. Therefore ne0 (A)−1 N ⊆ ne0 (A) G = 0. Therefore it follows that me0 (A)−1 M ⊆ xM .
47 Proof of Theorem 4.2.5. Clearly we may assume that the residue field of Q is infinite. φ
Let 0 → Qn − → Qn → M → 0 be a minimal presentation of M over Q. If dim A ≥ 2, then choose x1 , . . . , xd to be a maximal φ-superficial sequence. Set J = (x1 , . . . , xd−1 ). Since all the invariants considered in the theorem remain same modulo J it suffices to assume dim A ≤ 1. When dim A = 0 then all the all the results follow easily by Remark 4.2.1. Therefore assume that dim A = 1. Let x be φ-superficial. Set R = Q/xQ, N = M/xM , f = the image of f in R and B = A/xA = R/(f ). Note that a. i(M ) = i(N ). b. e(M ) = e(N ) and e1 (M ) ≥ e1 (N ) ( by Corollary 1.2.6.5 and 6) c. R is a DVR with maximal ideal say q = (y). d. vR (f ) = vQ (f ) = e. So 1. follows from Remark 4.2.1 and (c) above. 2. If i(M ) = e then note that since i(M ) = i(N ) we get by Remark 4.2.1.4 that N is a free B module. So M is a free A-module. Clearly if M is free then i(M ) = e. 3. Let M = F ⊕ L where F is a free A-module and L has no free summands. Note that i(L) = i(M ). Since G(A) is Cohen-Macaulay it suffices to show G(L) is A Cohen-Macaulay. Since L = SyzA 1 Syz1 (L) it follows from Lemma 4.2.9 that if x is a A ⊕ L ⊕ N -superficial element, then me0 (A)−1 L ⊆ xL. Since An → An → L → 0 is exact and depth G(A) = 1 we get by Lemma 4.2.6.2 that G(L) is Cohen-Macaulay. 4. By Proposition 4.2.7 we get that hM (z) = µ(M )(1 + z + . . . + z i(M )−1 ) +
X
hi (M )z i
and hi (M ) ≥ 0 for all i.
i≥i(M )
So it follows that (i) and (ii) are equivalent. The assertion (iii) =⇒ (ii) is clear. (i) =⇒ (iii).. Note that µ(N ) = µ(M ) and hN (z) = µ(M )(1 + z + . . . + z i(M )−1 ) +
X i≥i(M )
hi (N )z i .
48 Also all the coefficients are non-negative. Therefore it follows that e(M ) = e(N ) = i(M )µ(M ) holds if and only if hM (z) = hN (z) = µ(M )(1 + z + . . . + z i(M )−1 ) Since hM (z) = hN (z) we also get that G(A) is Cohen-Macaulay (see Corollary1.2.7.1). Note that since M is not free i(M ) < e. We first assert that M has no free summands. Otherwise M = F ⊕ W where F is free. This yields hM (z) = hF (z) + hW (z). Since all the coefficients of hF (z) and hW (z) are non-negative we get that coefficient of z e−1 is non-zero. This contradicts (c). Therefore if (φ, ψ) is a matrixfactorization of M then we have a minimal presentation of K ψ
0 −→ Qn − → Qn −→ K −→ 0. Let x be both φ and ψ-superficial. Set N = M/xM . Since (a) holds then note that µ(M ) N∼ . Then = R/(y i(M ) ) e−i(M ) µ(M ) ∼ SyzB ) . 1 (N ) = R/(y ∼ Since SyzB 1 (N ) = K/xK we get that i(K) = e − µ(M ) and e(K) = e(K/xK) = µ(K/xK)i(K) = µ(K)i(K). Therefore by the equivalence of (i) and (iii) we get the required result. Remark 4.2.10. Theorem 4.2.5 can be applied to the case of Ulrich modules, that is, maximal Cohen-Macaulay modules that satisfy e(M ) = µ(M ). It is known [7] that Ulrich A-modules exist when A is a complete hypersurface ring. Using the previous theorem we get that if M is Ulrich, then i(M ) = 1 and so G(SyzA 1 (M )) is Cohen-Macaulay. Furthermore hSyzA1 (M ) = µ(M )(1 + z + . . . + z e−2 ). Therefore every complete hypersurface ring has a maximal Cohen-Macaulay A- module such that G(M ) and G(SyzA 1 (M )) is Cohen-Macaulay.
49 An easy way to test the hypothesis of the previous theorem in the equicharacteristic case is the following: Proposition 4.2.11. Let Q = k[[y1 , . . . , yd+1 ]]. Let M be a Q-module with a minimal φ
presentation 0 → Qn − → Qn → M → 0. Set φ=
X
φi
where φi are forms of degree i
i≥i(M )
Then det φi(M ) 6= 0 if and only if hM (z) = µ(M )(1 + z + . . . + z i(M )−1 ). Proof. Note that det φi(M ) 6= 0 if and only if vQ (det φ) = i(M )µ(M ). Let f = det φ. Note that M is a maximal A = Q/(f )-module. Let x1 , . . . , xd be a maximal φ-superficial sequence. Set J = (x1 , . . . , xd ), R = Q/J, f = image of f in R, N = M/JM and φ = φ ⊗ Q/J. Note that i(N ) = i(M ) and vR (det φ) = vQ (det φ). If det φi(M ) 6= 0 then vQ (det φ) = i(M )µ(M ). So vR (det φ) = i(M )µ(M ). Therefore by Remark 4.2.1.7 we get that e(N ) = µ(N )i(N ). This yields that e(M ) = i(M )µ(M ) and so by Theorem 4.2.5.4 we get the required assertion. Conversely if hM (z) = µ(M )(1 + z + . . . + z i(M )−1 ) then by Theorem 4.2.5 G(M ) is Cohen-Macaulay. So hN (z) = hM (z). So we get e(N ) = µ(N )i(N ). Therefore by Remark 4.2.1.7 we get vR (det φ) = i(N )µ(N ) = i(M )µ(M ). So vQ (det φ) = i(M )µ(M ). We now give an application of the proposition proved above. Example 4.2.12. Set Q = k[[y1 , . . . , yd+1 ]] and n to be the maximal ideal of Q. Let a, b, c, d be in n be such that f = ad − bc 6= 0. Set A = Q/(f ). Set a b d −b and ψ = φ= c d −c a Define M and K by the exact sequences φ
ψ
0 → Q2 − → Q2 → M → 0 and 0 → Q2 − → Q2 → K → 0 Note that M and K are A-modules and K = SyzA 1 (M ).
50 1. If f ∈ m2 then M is an Ulrich A-module. 2. If f ∈ m3 \m4 then M or K has minimal multiplicity. However both are not Ulrich. To prove 1. note that i(M ) = 1 and det(φ1 ) 6= 0. So by Proposition 4.2.11 we get that hM (z) = µ(M ). So M is Ulrich. 2. We first show that M and K are not Ulrich. Note that since f ∈ n3 \ n4 we still have i(M ) = 1. Since f ∈ n3 we also get det(φ1 ) = 0. So by Proposition 4.2.11 we get that hM (z) 6= µ(M ). So M is not Ulrich. Similar argument yields that K is not Ulrich. Note that hA (z) = 1 + z + z 2 . To prove that M or K has minimal multiplicity note that by Theorem 3.1.6 and Remark 3.1.7 we get that 2 = 2χ1 (A) > χ1 (M ) + χ1 (K). So we get χ1 (M ) = 0 or χ1 (K) = 0. So by Theorem 2.4 we get that M or K has minimal multiplicity.
51
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53
VITA Mr. Tony Joseph Puthenpurakal was born in Ulhasnagar, India on August 12, 1972. He received his Masters degree in mathematics from the Indian Institute of Technology, Bombay in the year 1995. He joined the graduate program in mathematics at Purdue University in August, 1995.
