The Hilbert Function of a Maximal Cohen-Macaulay Module

0.31 MB·English

by  Tony J. Puthenpurakal

THE HILBERT FUNCTION OF A MAXIMAL
COHEN-MACAULAY MODULE
5
0
0
2 TONYJ.PUTHENPURAKAL

n
a Abstract. WestudyHilbertfunctionsofmaximalCMmodulesoverCMlocal
J rings. WeshowthatifAisahypersurfaceringwithdimensiond>0thenthe
1 Hilbert function of M with respect to mis non-decreasing. If A=Q/(f) for
3 someregularlocalringQ,wedeterminealowerboundfore0(M)ande1(M)
and analyze the case when equality holds. When A is Gorenstein a relation
] between the second Hilbert coefficient of M, A and SA(M) = (SyzA1(M∗))∗
C is found when G(M) is CM and depthG(A) ≥ d−1. We give bounds for
A the first Hilbert coefficients of the canonical module of a CM local ring and
analyzewhenequalityholds. WealsogivegoodboundsonHilbertcoefficients
h. ofM whenM ismaximalCMandG(M)isCM.
t
a
m
[
Introduction

2
v Let (A,m) be a d-dimensional Noetherian local ring and M a finite A-module.
1 Let G(A) = mn/mn+1 be the associated graded module of A and G(M) =
n≥0
5 mnM/mn+1M the associated graded module of M considered as a G(A)-
0 n≥0 L
9 Lmodule. We set depthG(M) = grade(M,G(M)) where M = n≥1mn/mn+1 is
0 the irrelevant maximal ideal of G(A). If N is an A- module then µ(N) denotes its
4 minimal number of generators and λ(N) denotes its length. ThLe Hilbert function
0
of M (with respect to m) is the function
/
h
t H(M,n)=λ(mnM/mn+1M) for all n≥0
a
m In this paper we study Hilbert functions of maximal CM (= MCM) modules. If
: AisregularthenallMCMmodulesarefree. Thenextcaseisthatofahypersurface
v
ring.
i
X
Theorem 1. Let (A,m) be a hypersurface ring of positive dimension. If M is a
r
a MCM A-module, then the Hilbert function of M is non-decreasing.
This result is a corollary of a more general result (see Theorem 3.3) which also
implies that the Hilbert function of a complete intersection of codimension 2 and
positive dimension is non-decreasing (see Corollary 3.5). Another application of
Theorem3.3yieldsthat,if(A,m)isequicharacteristiclocalringofdimensiond>0,
I is an m-primary ideal with µ(I)=d+1 and M is an MCM A-module, then the
Hilbert function of M with respect to I is non-decreasing (see Theorem 3.6).
The formal power series
H (z)= H(M,n)zn
M
n≥0
X
Date:February1,2008.
1
2 TONYJ.PUTHENPURAKAL
is called the Hilbert series of M. It is well known that it is of the form
h (z)
M
H (z)= , where r =dimM and h (z)∈Z[z].
M (1−z)r M
We call h (z) the h-polynomial of M. If f is a polynomial we use f(i) to denote
M
its i-th derivative. The integers e (M) = h(i)(1)/i! for i ≥ 0 are called the Hilbert
i M
coefficients of M. The number e(M)=e (M) is the multiplicity of M. Set
0
i
χ (M)= (−1)i−je (M)+(−1)i+1µ(M) for each i≥0.
i i−j
j=0
X
Let M be a MCM module over a hypersurface ring A = Q/(f), where (Q,n) is
a regular local. If 0→Qn −φ−M→Qn →M →0 is a minimal presentation of M then
i(M)=max{i| all entries of φ are in ni} is an invariant of M.
Theorem 2. Let (Q,n) be a regular local ring, f ∈ ne\ne+1, e ≥ 2, A = Q/(f),
M a MCM A-module and K =SyzA(M). Then
1
1. e(M)≥µ(M)i(M) and e (M)≥µ(M) i(M) .
1 2
2. M is a free A-module if and only if i(M)=e.
(cid:0) (cid:1)
3. If i(M)=e−1 then G(M) is CM.
4. The following conditions are equivalent:
i. e(M)=µ(M)i(M).
ii. e (M)=µ(M) i(M) .
1 2
iii. G(M) is CM and h (z)=µ(M)(1+z+...+zi(M)−1).
(cid:0) M(cid:1)
If these conditions hold and M is not free, then G(K) is CM and h (z) =
K
µ(M)(1+z+...+ze−i(M)−1).
IfAiscompleteandM isaCohen-Macaulay(=CM)A-modulethenthereexists
a Gorenstein local ring R such that M is a MCM R-module. So it is significant to
see how Hilbert functions of MCM modules over Gorenstein rings behave.
When A is CM, M is a MCM A-module and N =SyzA(M) then
1
(1) µ(M)e (A)≥e (M)+e (N) & µ(M)χ (A)≥χ (M)+χ (N).
1 1 1 1 1 1
For the first inequality see [10, 17(3)]. The second follows from [10, 21(1)].
In the theorem below we establish similar inequalities for higher Hilbert co-
efficients of MCM modules over Gorenstein rings. For every A-module we set
M∗ = Hom (M,A). Note that if M is MCM then so is M∗ cf.[2, 3.3.10.d]. Also,
A
type(M)=dim Extd(k,M) denotes the Cohen-Macaulay type of M.
k A
Theorem 3. Let (A,m) be a Gorenstein local ring. Let M be a MCM A-module.
Set τ =type(M) and SA(M)= SyzA(M∗) ∗. If G(M) is CM and depthG(A)≥
1
d−1 then the following hold
(cid:0) (cid:1)
1. τe (A)≥e (M)+e (SA(M)) and τχ (A)≥χ (M)+χ (SA(M)).
2 2 2 2 2 2
2. type(M)e (A)≥e (M) and type(M)χ (A)≥χ (M) for each i≥0.
i i i i
Let A be CM with a canonical module ω . Set τ = typeA. It is well known
A
that e (ω )=e (A). Using [10, Theorem18]it followsthat e (ω )≤τe (A) with
0 A 0 1 A 1
equality if and only if A is Gorenstein. Here we give a lower bound on e (ω ).
1 A
Theorem4. Let(A,m)beaCMlocalringofdimensiond≥1andwithacanonical
module ω . Set τ =typeA. We have
A
HILBERT FUNCTION 3
(1.) τ−1e (A)≤e (ω )≤τe (A).
1 1 A 1
(2.) (a.) e (ω )=τe (A) iff A is Gorenstein.
1 A 1
(b.) e (A)=τe (ω ) iff A is Gorenstein or A has minimal multiplicity.
1 1 A
(3.) If dimA=1 and G(A) is CM then
e (A)≤τe (ω ) and χ (A)≤τχ (ω ) for each i≥1.
i i A i i A
Let A = k[[x ,...,x ]]/q be CM with q ⊆ (x)2 and k an infinite field. Let
1 n
1 ≤ r ≤ d. For any two sets of r sufficiently general k-linear combinations of
x ,...,x say y ,...,y and z ,...,z we show H(A/(y),n) = H(A/(z),n) for
1 ν 1 r 1 r
each n ≥0 (see 7.6). We use it to bound Hilbert coefficients of M if G(M) is CM
(see Theorem 7.8).
Hereisanoverviewofthecontentsofthepaper. InSection1weintroducenota-
tion and discuss a few preliminary facts that we need. The proof of the Theorems
1 and 3 involves a study of the modules; L (M)= TorA(A/mn+1,M) for all
t n≥0 t
t≥0. Ifx ,...,x is asequenceofelements inm then,insection2,wegiveL (M)
1 s t
L
a structure of a graded A[X ,...,X ]-module. We prove Theorem 1 in Section 3,
1 s
Theorem2inSection4andTheorem3inSection5. WeproveTheorem4insection
6. In Section 7 we prove Lemma 7.6 and use to prove Theorem 7.8.
1. Preliminaries
In this paper all rings are Noetherian and all modules are assumed finite i.e.,
finitely generated. Let (A,m) be a local ring of dimension d with residue field
k = A/m. Let M be an A-module. If m is a non-zero element of M and if j is
the largest integer such that m ∈ mjM, then we let m∗ denote the image of m in
mjM/mj+1M. IfLisasubmoduleofM,thenL∗ denotesthegradedsubmoduleof
G(M) generated by all l∗ with l ∈L. It is well known that G(M)/L∗ =G(M/L).
An element x ∈ m is said to be superficial for M if there exists an integer c > 0
such that
(mnM: x)∩mcM =mn−1M for all n>c.
M
Superficialelements alwaysexist if k is infinite [11, p. 7]. A sequencex ,x ,...,x
1 2 r
in a local ring (A,m) is said to be a superficial sequence for M if x is superficial
1
for M and x is superficial for M/(x ,...,x )M for 2≤i≤r.
i 1 i−1
Remark 1.1. Ifthe residue fieldof Ais finite then we resortto the standardtrick
to replace A by A′ = A[X] and M by M′ = M ⊗ A′ where S = A[X]\mA[X].
S A
The residue field of A′ is k(X), the field of rational functions over k. Furthermore
H(M′,n)=H(M,n) ∀n≥0 and depth G(M′)=depth G(M).
G(A′) G(A)
Clearly projdim M′ =projdim M If A is a Gorenstein (hypersurface) ring then
A′ A
A′ is also Gorenstein (hypersurface) ring. If A has a canonical module ω then A′
A
also has a canonical module ωA′ ∼=ωA⊗A′; cf. [2, Theorem 3.3.14].
Below we collect some basic results needed in the paper. For proofs see [10].
Remark 1.2. Let(A,m)beaCohen-MacaulaylocalringwithdimA=d>0. Let
M beafiniteCMA-moduleofdimensionr. Letx ,...,x beasuperficialsequence
1 s
in M with s≤r and set J =(x ,...,x ). The local ring (B,n)=(A/J,m/J) and
1 s
B-module N =M/JM satisfy:
1. x ,...,x is a M-regular sequence in A.
1 s
2. N is a CM B-module.
4 TONYJ.PUTHENPURAKAL
3. e (N)=e (M) for i=0,1,...,r−s.
i i
4. When s=1, set x=x and b (x,M)=λ(mn+1M: x)/mnM). We have:
1 n M
a. b (x,M)=0 and b (x,M)=0 for all n≫0.
0 n
b. H(M,n)= n H(N,i)−b (x,M)
i=0 n
c. e (M)=e (N)−(−1)r b (x,M).
r rP i≥0 i
d. x∗ is G(M)-regular if and only if b (x,M)=0 for all n≥0.
n
P
5. a. depthG(M)≥s if and only if x∗,...,x∗ is a G(M) regular sequence.
1 s
b. (Sally descent) depthG(M)≥s+1 if and only if depthG(N)≥1.
6. If dimM =1 then set ρ (M)=λ(mn+1M/xmnM). We have
n
a. H(M,n)=e(M)−ρ (M).
n
b. e (M)= j ρ (M)≥0 for all i≥1.
i j≥i−1 i−1 j
7. If x ,...,x is also A-regular then SyzB(N)∼=SyzA(M)/JSyzA(M)
1 sP (cid:0) (cid:1) 1 1 1
8. depthG(M)≥s if and only if h (z)=h (z).
M N
9. M has minimal multiplicity if and only if χ (M)=0.
1
Remark 1.3. If φ:(A,m)−→(B,n) is a surjective mapof localrings andif M is
a finite B-module then mnM = nnM 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.
Also note that depth G(M)=depth G(M).
Gm(A) Gn(B)
1.4. Recallthatthefunctionn7→λ(M/mn+1M)iscalledtheHilbert-Samuel func-
tion. Let p (z) be the Hilbert-Samuel polynomial. The following number
M
(2) post(M)=min{n|p (i)=λ(M/mi+1M) for all i≥n}
M
is called the postulation number of M (with respect to m).
1.5. Iff(z)= a zk ∈Z[z]thenfori≥0sete (f)=f(i)(1)/i!= k a
k≥0 k i k≥i i k
and set
P P (cid:0) (cid:1)
i
k−1
χ (f)= (−1)i−je (f)+(−1)i+1f(0)= a .
i i−j k
i
j=0 k≥i+1(cid:18) (cid:19)
X X
Itfollowsthatifa ≥0 foralli≥0 thene (f)≥0andχ (f)≥0foralli≥0. The
i i i
following Lemma can be easily proved.
Lemma 1.6. If g(z),p(z),q(z) and r(z) are polynomials with integer coefficients
that satisfy the equation (1−z)g(z)=p(z)−q(z)+r(z) then
(i) e (q)=e (p)+e (r).
0 0 0
(ii) e (q)=e (p)+e (r)+e (g) for i≥1.
i i i i−1
(iii) χ (q)=χ (p)+χ (r)+g(0).
0 0 0
(iv) χ (q)=χ (p)+χ (r)+χ (g) for i≥1.
i i i i−1
(v) If all the coefficients of g are non-negative then for i≥0 we have
e (q)≥e (p)+e (r) and χ (q)≥χ (p)+χ (r). (cid:3)
i i i i i i
2. Basic Construction
Remark 2.1. For each n ≥ 0 and t ≥ 0 set L (M) = TorA(A/mn+1,M). For
t n t
t≥0letL (M)= L (M) . Ifx ,...,x isasequenceofelementsinm,then
t n≥0 t n 1 s
we give L (M) a structure of a graded A[X ,...,X ]-module as follows:
t 1 s
L
For i = 1,...,s let ξ : A/mn → A/mn+1 be the maps given by ξ (a+mn) =
i i
x a+mn+1. These homomorphisms induces homomorphisms
i
TorA(ξ ,M):TorA(A/mn,M)−→TorA(A/mn+1,M)
t i t t
HILBERT FUNCTION 5
Thus, for i=1,...,s and each t we obtain homogeneous maps of degree 1:
X :L (M)−→L (M).
i t t
For i,j = 1,...,s the equalities ξ ξ = ξ ξ yields equalities X X = X X . So
i j j i i j j i
L (M) is a graded A[X ,...,X ]-module for each t≥0.
t 1 s
Proposition 2.2. Let M,F, and K be finite A-modules and let x ,...,x be a
1 s
sequenceofelementsinm. IfL (M),L (F)andL (K)aregiventheA[X ,...,X ]-
t t t 1 s
module structure described in Remark 2.1 then
1. Every exact sequence of A-modules 0→K →F →M →0 induces a long ex-
act sequence of graded A[X ,...,X ]-modules
1 s
···→L (M)→L (K)→L (F)→L (M)→···→L (M)→0.
t+1 t t t 0
2. For i=1,...,s there is an equality
mn+1M: x
ker L (M) −X−→i L (M) = M i
0 n−1 0 n mnM
3. If x ∈m\m2 is(cid:0)such that x∗ is G(M)-re(cid:1)gular then X is L (M)-regular.
i i i 0
4. If F is free A-module and x is K-superficial for some i then
i
(a) ker L (M)−X−→i L (M) =0 for n≫0
1 1 n
(b) If x∗ is G(K)-regular then X is L (M)-regular.
(cid:0)i (cid:1) i 1
Proof. To prove part 1, set S =A[X ,...,X ],
1 s
β =TorA(A/mn+1,β):TorA(A/mn+1,K)−→TorA(A/mn+1,F)
t,n t t t
α =TorA(A/mn+1,α):TorA(A/mn+1,F)−→TorA(A/mn+1,M)
t,n t t t
and consider the connecting homomorphisms
δ :TorA (A/mn+1,M)−→TorA(A/mn+1,K).
t+1,n t+1 t
By a well knowntheoremin Homologicalalgebraif X is a free resolutionofK 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 //Y⊗A/mn //Z⊗A/mn //0
ξi ξi ξi
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
0 //X⊗A/mn+1 // Y⊗A/mn+1 // Z⊗A/mn+1 //0
In homology it induces the following commutative diagram :
··· //L (M) δt+1,n−1// L (K) βt,n−1//L (F) αt,n−1 // ···
t+1 n−1 t n−1 t n−1
Xi Xi Xi
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
··· //L (M) δt+1,n //L (K) βt,n // L (F) αt,n // ···
t+1 n t n t n
This proves the desired assertion.
Remark 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.
6 TONYJ.PUTHENPURAKAL
The second part is clear from the definition of the action X . Part 3. follows
i
from 2. If F is free, then L (F) = 0, so 1. gives an exact sequence of S-modules
1
0→L (M)→L (K). Together with 2. and 3. this yields the assertions in 4. (cid:3)
1 0
Remark 2.4. If(A,m)isCMofdimensiondandM ismaximalnon-freeCMthen
by [10, Remark 23] there is an equality
l (z)
(3) λ TorA(M,A/mn+1) zn = M here l (z)∈Z[z] and l (1)6=0
1 (1−z)d M M
n≥0
X (cid:0) (cid:1)
(4) (1−z)l (z)=h (z)−µ(M)h (z)+h (z).
M SyzA(M) A M
1
We study the case when dimA=0.
Lemma 2.5. If dimA=0 and M is any finite A-module then for all i≥0
1. µ(M)e (A)≥e (M)+e (SyzA(M)) and µ(M)χ (A)≥χ (M)+χ (SyzA(M)).
i i i 1 i i i 1
2. µ(M)e (A)≥e (M) and µ(M)χ (A)≥χ (M).
i i i i
Proof. Note that when dimA=0 we get that l (z) has non-negative coefficients.
M
Using (4 ) and Lemma 1.6.v we get 1. and 2. The assertion 3. follows from 1.
and 2. since N = SyzA(M) has dimension zero and so h (z) has non-negative
1 N
coefficients. Therefore e (N) and χ (N) are non-negative for i≥0 (see (1.5) ). (cid:3)
i i
2.6. ItfollowsfromLemma2.5.3thatifG(A)andG(M)isCMthene (A)µ(M)≥
i
e (M) for all i≥0.
i
Remark 2.7. In view of the Remark 2.4 and (2.6) it is quite important to under-
standL1(M)= TorA(M,A/mn+1)whenM is MCM.Inthe nextLemma we
n≥0 1
answer the question when dimM =1.
L
Lemma 2.8. 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 L (M) is given the A[X]-module structure described in Remark 2.1 then we have
1
1. There is an m-primary ideal q such that qA[X]L (M) = 0. Furthermore
1
L (M) is a Noetherian A/q[X]-module of dimension one.
1
2. (1−z)l (z)=h (z)−h (z)+h (z).
M E F M
3. If G(E) is CM then X is L (M)-regular. Furthermore
1
e (F)≥e (M)+e (E) and χ (F)≥χ (M)+χ (E) for all i≥0.
i i i i i i
Proof. 1. Since dimA = 1 and M is non-free, it follows from Lemma 2.8 that
λ(L (M) ) is a non-zero constant for large n. Since X: L (M) → L (M)
1 n 1 n 1 n+1
is injective for large n and since λ(L (M) ) is constant for large n, it follows
1 n
that L (M) = XL (M) for large n, say for all n ≥ s. For n ≥ 0 set q =
1 n+1 1 n n
ann L (M) . Note that q is m-primary for all n. Since the map X: L (M) →
A 1 n n 1 n
L (M) isbijectiveforalln≥swehaveq =q foreachn≥s. Setq=∩s q .
1 n+1 n s n=0 n
Clearly qL (M) =0 for each n≥0. Thus L (M) is an A/q[X] module. For each
1 n 1
i=0,1,...,s choose a finite set P of generators of L (M) as an A-module. It is
i 1 i
easy to see that s P generates L (M) over A[X]. Since λ(L (M)/XL (M))<
i=0 i 1 1 1
∞ and λ(L (M))=∞ it follows that dimL (M)=1.
1 1
S
HILBERT FUNCTION 7
2.BySchanuel’slemma,[9,p.158]wehaveF⊕SyzA(M)∼=E⊕Aµ(M). Therefore
1
(1−z)l (z)=h (z)−µ(M)h (z)+h (z)=h (z)−h (z)+h (z).
M SyzA(M) A M E F M
1
3. It is clear from 1. and Proposition 2.2.3 that X is L (M)-regular. It also
1
follows from 1. that l (z) is the h-polynomial of L (M) considered as an A[X]-
M 1
module. Set eT(M)=l(i)(l)/i!. Since X is L (M)-regular and dimL (M)=1 we
i M 1 1
have that l (z) is the Hilbert series of L (M)/XL (M). Thus all the coefficients
M 1 1
ofl (z)isnon-negative. Using2. andLemma1.6.vwegetthedesiredinequalities.
M
(cid:3)
3. Monotonicity
The following remark will be used often.
Remark 3.1. Let f(z)= a zn be a formal power series with non-negative
n≥0 n
coefficients. If the power series g(z) = b zn satisfies g(z) = f(z)/(1−z),
P n≥0 n
then b = n a , and so the sequence {b } is nondecreasing.
n i=0 i Pn
The nexPt proposition yields an easy criterion for monotonicity.
Proposition 3.2. Let M be an A-module. Set k =A/m. If depthG(M)≥1 then
the Hilbert function of M is non-decreasing.
Proof. Using Remark 1.1 we may assume that k is infinite. Thus there exists
x ∈ m\m2, such that x∗ is G(M)-regular. It follows that the Hilbert function of
M is non-decreasing. (cid:3)
We deduce Theorem 1 from the following result.
Theorem 3.3. Let (Q,m) be a local ring with depthG(Q) ≥ 2 and let M be a
Q-module. If projdim M ≤1 then the Hilbert function of M is non-decreasing.
Q
Proof. Itissufficienttoconsiderthe casewhenthe residuefieldofQis infinite (see
Remark 1.1). Since projdim M ≤1 we have a presentation of M
Q
(5) 0−→Qn −→Qm −→M −→0 with 0≤n≤m.
Let x,y be elements in m\m2 such that x∗,y∗ is a G(Q)-regular sequence. Let
L (Q), L (M) and L (M) be the Q[X,Y]-modules described in Remark 2.1.
0 0 1
By Proposition2.2.3 we get X is L (Q)-regular. Set B =Q/(x) and notice
0
L (Q) Q
0
= =L (B).
XL (Q) (x,mn+1) 0
0
n≥0
M
Since G(B) = G(Q)/x∗G(Q) we see that y∗ is G(B)-regular. Proposition 2.2.3
shows that Y is L (B)-regular. Thus X,Y is a L (Q)-regular sequence.
0 0
Usingtheexactsequence(5)andProposition2.2.1,weobtainanexactsequence
of graded Q[X,Y] modules
(6) 0−→L (M)−→L (Q)n −→φ L (Q)m −→L (M)−→0.
1 0 0 0
Set K = imageφ. Since X is L (Q) regular we see that it is both K and L (M)-
0 1
regular. So the exact sequence 0 → L (M) → L (Q)n → K → 0 yields the exact
1 0
sequence
L (M) L (Q)n K
1 0
0−→ −→ −→ −→0
XL (M) XL (Q)n XK
1 0
8 TONYJ.PUTHENPURAKAL
Since Y is L (Q)n/XL (Q)n regular it follows that Y is L (M)/XL (M)-regular.
0 0 1 1
Thus X,Y is an L (M)- regular sequence.
1
The regularity of X,Y implies equalities
u(z) L (Q)
λ(L (Q) )zi = where u(z)= λ 0 i zi.
0 i (1−z)2 (X,Y)L (Q)
i≥0 i≥0 (cid:18) 0 i−1(cid:19)
X X
v(z) L (M)
λ(L (M) )zi = where v(z)= λ 1 i zi.
1 i (1−z)2 (X,Y)L (M)
i≥0 i≥0 (cid:18) 1 i−1(cid:19)
X X
Using the exact sequence (6) we get
u(z) v(z)
λ(L (M) )zi =(m−n) +
0 i (1−z)2 (1−z)2
i≥0
X
The equality H (z)=(1−z) λ(L (M) )zi yields
M i≥0 0 i
P
H (z)=(m−n)u(z)/(1−z)+v(z)/(1−z).
M
Now Remark 3.1 shows that the Hilbert function of M is non-decreasing. (cid:3)
We obtain Theorem 1 as a corollary to the previous theorem.
Proof of Theorem 1 . We may assume that A is complete and so A ∼= Q/(f) for
some regular local ring (Q,n) and f ∈ n2. Then depthM = dimQ − 1 and
projdim M = 1. Using Theorem 3.3 it follows that the Hilbert function of M
Q
is non-decreasing. (cid:3)
SincetheHilbertfunctionisincreasingifdepthG(M)>0,weconstructaMCM
module M over a hypersurface ring A such that depthG(M)=0.
Example 3.4. Set Q=k[[x,y]] and n=(x,y). Define M by the exact sequence
0−→Q2 −→φ Q2 −→M −→0 where
x y
φ=
−y2 0
(cid:18) (cid:19)
Set (A,m) = (Q/(y3),n/(y3)). Note y3 = det(φ) annihilates M. So M is a MCM
A-module. Set K = SyzA(M). Note that G(Q) = k[x∗,y∗]. Since y3M = 0 , we
1
have that if P ∈ Ass (G(M)) then P ⊇ (y∗). So we get that x∗ ∈/ P if P is
G(Q)
a relevant associated prime of G(M). Therefore x∗ is an M-superficial element.
We show depthG(M) = 0. Otherwise by 1.2.5.a we get that x∗ is G(M)-regular.
However if m ,m are the generators of M then xm = y2m ∈ n2M and this
1 2 1 2
implies m ∈(n2M: x)=nM, which is a contradiction.
1 M
ThenextcorollarypartlyoverlapswitharesultofElias[5]: allequicharacteristic
CM rings of dimension 1 and embedding dimension 3 have non-decreasing Hilbert
functions. See [13, p. 337] for an example of a complete intersection ring (A,m) of
dimension 1 and codimension 2 such that depthG(A)=0.
Corollary 3.5. If (A,m) be a complete intersection of positive dimension and
codimension 2 then the Hilbert function of A is non-decreasing.
HILBERT FUNCTION 9
Proof. We may assume that A is complete and hence A = Q/(f,g) for a regular
sequencef,g inaregularlocalring(Q,q). Set(R,n)=(Q/(f),q/(f)). ThenG(R)
isCohen-Macaulay,dimA=dimR−1andprojdim A=1. ThereforebyTheorem
R
3.3 we get that the Hilbert function of A is non-decreasing. (cid:3)
Another application of Theorem 3.3 yields the following:
Theorem3.6. Let(A,m)beaNoetherianequicharacteristiclocalringofdimension
d > 0 and let M be a MCM A-module. Let I be an m-primary ideal in A with
µ(I)=d+1. Then the Hilbert function of M with respect to I is non-decreasing.
Proof. Without any loss of generality we may assume that A is complete. Let
I =(x ,...,x ). SinceAiscompleteandequicharacteristicitcontainsasubfield
1 d+1
k ∼=A/m. SetR=k[[T ,...,T ]]andletnbeitsuniquemaximalideal. Consider
1 d+1
the local homomorphism φ : R → A defined by φ(T ) = x . Then A becomes an
i i
R-moduleviaφ. SinceA/nA=A/I hasfinitelengthwegetM isafiniteR-module.
It can be easily checked that M is a CM R-module of dimension d.
SinceRisregular,projdim M isfinite. Soprojdim M =depthR−depthM =
R R
1. Therefore by Theorem3.3 it follows that Hn(M,j) is non-decreasing. Note that
njM = IjM for each j ≥ 1 and so Hn(M,j) = λ(InM/In+1M) for each j ≥ 0.
This establishes the assertion of the theorem. (cid:3)
4. Hilbert coefficients
4.1. Inthissectionε denotesthes×sidentitymatrix. Let(Q,n)bearegularlocal
s
ring, f ∈ne\ne+1, e≥2, A=Q/(f), M a MCM A-module and K =SyzA(M).
1
By a matrix-factorization of f we mean a pair (φ,ψ) of square-matrices with
elements in Q such that
φψ =ψφ=fε.
If M is an A-module then projdim M =1. Also a presentation of M
Q
0−→Qn −→φ Qn −→M −→0
yields a matrix factorization of f. See [4, p. 53] for details.
In the sequel (φ ,ψ ) will denote a matrix factorization of f such that
M M
0−→Qn −φ−M→Qn −→M −→0
is a minimal presentation of M. Note that
0−→Qn −ψ−M→Qn −→SyzA(M)−→0
1
is a not-necessarily minimal presentation of SyzA(M).
1
If φ:Qn −→Qm is a linear map then we set
i =max{i| all entries of φ are in ni}.
φ
If M has minimal presentations: 0→Qn −→φ Qn →M →0 and
0 → Qn −φ→′ Qn → M → 0, then it is well known that i = i and det(φ) =
φ φ′
udet(φ′) with u a unit. We set i(M) = i and det(M) = (det(φ)). For g ∈ Q,
φ
g 6= 0, set v (g) = max{i | g ∈ ne}. For convenience set v (0) = ∞. Note that
Q Q
e(Q/(g))=v (g) for any g 6=0. We first consider the case when dimA=0.
Q
10 TONYJ.PUTHENPURAKAL
Remark 4.2. Let (Q,n) be a DVR, v (f) = e, A = Q/(f) and M a finite A-
Q
module. If n=(y) then f =uye, where u is a unit. Therefore as an Q-module
µ(M)
M ∼= Q/(yai) with 1≤a ≤...≤a ≤e.
1 µ(M)
i=1
M
This yields a minimal presentation of M:
0→Qn −→ψ Qn →M →0 where ψ =δ yai.
ij ij
This yields
(1) i(M)=a .
1
(2) h (z)=µ(M)(1+z+...+zi(M)−1)+ higher powers of z
M
(3) h (M)≥h (M)≥...≥h (M).
0 1 s
(4) M is free if and only if i(M)=e.
(5) As an Q-module K ∼= µ(M)Q/(ye−ai).
i=1
(6) e(M)≥µ(M)i(M) and e (M)≥µ(M) i(M) .
L1 2
(7) v (detφ)≥i(M)µ(M) with equality iff e(M)=i(M)µ(M).
Q
(cid:0) (cid:1)
We note an immediate corollary to assertion 3. in the previous remark.
Corollary4.3. Let(A,m)beahypersurfaceringofdimensiond. LetM beaMCM
A-module such that G(M) is CM. If h (z)=h (M)+h (M)z+···+h (M)zs is
M 0 1 s
the h-polynomial of M then h (M)≥h (M)≥...≥h (M).
0 1 s
Proof. We may assume that A is complete with infinite residue field, hence A =
Q/(f)forsomeregularlocalring(Q,n). ConsiderM asaQ-module. Letx ,...,x
1 d
be a Q⊕M-superficial sequence. Set J = (x ,...,x ), (R,q) = (Q/J,n/J), B =
1 d
A/J and N = M/JM. Note that R is a DVR. Since G(M) is CM we also have
h (z)=h (z) and so the result follows from Remark 4.2(3). (cid:3)
M N
Tousethe otherassertionsinRemark4.2weneedthefollowingdefinitions. The
notion of superficial sequence is extremely useful in the study of Hilbert functions.
We need to generalize it to deal also with a presentation of a module.
Definition 4.4. Let(Q,n)be aregularlocalring,f ∈ne\ne+1,e≥2,A=Q/(f)
andM aMCMA-module. Let0→Qn −→φ Qn →M →0beaminimalpresentation
of M. We say that x∈n is φ-superficial if
(1) x is (Q⊕M ⊕A)-superficial.
(2) If φ=(φ ) then v (φ )=v (φ )
ij Q ij Q/xQ ij
(3) v (det(φ))=v (det(φ)).
Q Q/xQ
Since e(Q/(g)) = v (g) for any g 6= 0 it follows that if x is Q ⊕ M ⊕ A ⊕
Q
Q/(φ ) ⊕ Q/det(φ) -superficial then it is φ-superficial. So φ-superficial
ij ij
(cid:16)elements exist(cid:17)if the residue field of Q is infinite.
L
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
0−→ Q −φ−⊗−Q−Q−−/x−Q→ Q −→ M −→0
xQ xQ xM
(cid:18) (cid:19) (cid:18) (cid:19)
This enables the following definition: