Free resolutions and betti numbers of graded module

MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY N2 DEPARTMENT OF MATHEMATICS TRAN QUANG KIEN FREE RESOLUTIONS AND BETTI NUMBERS OF GRADED MODULE GRADUATION THESIS Maj

MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY N2

DEPARTMENT OF MATHEMATICS

TRAN QUANG KIEN

FREE RESOLUTIONS AND BETTI

NUMBERS OF GRADED MODULE

GRADUATION THESIS

Major: Algebra

HA NOI – 2019

MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY N2

DEPARTMENT OF MATHEMATICS

TRAN QUANG KIEN

FREE RESOLUTIONS AND BETTI

NUMBERS OF GRADED MODULE

1.1 Graded modules 41.2 Graded complexes 10

The study of free resolutions is a core and beautiful area in tative algebra The idea to associate a free resolution to a finitely gen-erated module was introduced in two famous papers by Hilbert in 1890,

commu-1893 Free resolutions provide a method for describing the structure ofmodules Base on the basic knowledge about algebraic and desiring com-prehensive improvement of mathematics, I would like to choose a topic

“free resolutions and Betti numbers of graded module” for my graduationthesis

The main goal of this thesis is to describe the structure of gradedfinitely generated modules I will focus on the algebraic invariants as-sociated the free resolutions Moreover, I also use computer softwares(CoCoA) to calculate algebraic invariants and verify mathematical is-sues which arises in the free resolutions The CoCoA software can bedownloaded free in the website (http://cocoa.dima.unige.it)

Throughout this thesis, we always denote R = k[x1, , xn] nomial ring with n variables x1, , xn over field k According to myunderstanding, we organize the thesis based on three books [3, 4, 5] Inchapter 1, we will present some basic concepts for further exploration ofour topic such as graded modules, graded complexes, Hilbert functions,Hilbert series Chapter 2 will provide a graded free resolutions and acontruct it for a graded finitely generated R-module M Chapter 3 isreserved for Betti numbers and relevant invariants

poly-This thesis was completed under the guidance of Dr Do Trong Hoang.

I would like to express my gratitude to him I would also like to thankthe teachers of Hanoi Pedagogical University 2 for helping me to havethe knowledge and create conditions for me to do this thesis

Due to limitations in time and knowledge, the thesis can not avoiderrors I hope to receive feedback from teachers and friends

Chapter 1

Fundamental concepts

This chapter aims to provide all relevant definitions and some contextregarding current research topics For a comprehensive introduction tothese concepts see [4]

Z-1 R = ⊕iRi (as abelian groups), and

2 RiRj ⊆ Ri+j for all i, j

Remark 1.1.2 If R = ⊕iRi is a graded ring, then R0 is a subring of R,

1 ∈ R0 and Ri is an R0-module for all i

Proof Since R0 · R0 = R0, R0 is closed under multiplication and thus is

a subring of R We can write P

nxn, where each xn ∈ Rn Then for all

Now, let R = ⊕i∈ZRi be a graded ring.

Definition 1.1.3 A R-module M is called graded if it satisfies twofollowing conditions:

1) M = ⊕i∈ZMi (as abelian groups), and

2) RiMj ⊆ Mi+j, for i, j ∈ Z

Then, the Mi is called the homogeneous component of M , and each m ∈

M is called a homogeneous element of degree i, denoted degM(m) = i, if

Let M be Z-graded finite generated R-module We asume M =

Rf1 + + Rfs Hence there exists an onto map

Rs −→ M

ei 7−→ fi

Hence, dimK(Mi) ≤ dimK(M ) ≤ s To measure the size of the module

M , we should first measure the sizes of its graded components

Definition 1.1.5 Let M be a Z-graded finite generated R-module Themap

is called Hilbert series of M

For p ∈ Z, denote by M (−p) the graded R-module such that M (−p)i =

Mi−p for all i We say that M (−p) is the module M shifted p degrees,and call p the shift Its Hilbert function is

Note that degM (−p)(x) = a ⇒ degM(x) = a + p

Example 1.1.6 LetR = K[x, y] and I = (x3, y2) Then R/I is gradedring in degree 0 with basis {1}, in degree 1 with basis {x, y}, in degree

2 with basis {x2, xy}, in degree 3 with basis {x2y} The Hilbert series

of R/I is:

HilbR/I(t) = 1 + 2t + 2t2 + t3

By the above example, Hilbert series of R/I(−7) is:

> HilbertSeries(R/I);

(1 + 2x + 2x2 + x3)

Definition 1.1.7 Let N and T be graded R-modules and a phism ϕ : N → T For each m ∈ N , if deg (ϕ(m)) = i + deg(m) then wesay that ϕ has degree i

homomor-Example 1.1.8 1 Consider the homomorphism

Suppose that 1, x is the basis of R[x]/(x2) Hence deg(1) = 0, deg(x) =

1 We have f (1) = x ⇒ deg(f (1)) = 1, f (x) = x2 ⇒ deg(f (x)) = 2.Thus, deg(f (1)) = deg(1) + 1 and deg(f (x)) = deg(x) + 1 There-fore, f is graded and has degree 1

2 Consider the homomorphism

Hence deg(1) = 0, deg(x) = 1 We have f (1) = 1 ⇒ deg(f (1)) = 0,

f (x) = x ⇒ deg(f (x)) = 1 Thus, deg(f (1)) = deg(1) + 0 anddeg(f (x)) = deg(x) + 0 Therefore, f is graded and has degree 0

Theorem 1.1.9 The following properties are equivalent

1 M is a finitely generated graded R-module

2 M ∼= W/T , where W is a finite direct sum of shifted free R-module,

T is a graded submodule of W , and the isomorphism has degree 0

Proof (2) ⇒ (1): We know that the quotient of finitely generated gradedR-module is finitely generated graded R-module, as required

Now, we prove (1) ⇒ (2): Since M is a finitely generated gradedR-module, so

M = m1R + + mkR, deg(mi) = ai

where ei is a basis element of R(−ai) Then ϕ is graded and has degree

0 This implies that U ∼= W/ Ker ϕ We know that Ker ϕ is graded SetKer ϕ = T , then U ∼= W/T

Lemma 1.1.10 (Nakayama’s Lemma) Let J be a proper graded ideal

in R Let M be a finitely generated graded R-module

1.2 Graded complexes

In order to prepare for a definition the free resolution of finitely erated modules, this section will present graded complexes and exactsequences

gen-Definition 1.2.1 The sequence (F•) of homomorphisms of R-modules

(2) 0 −→ Z −→ Zf −→ Z/2Z −→ 0 is complex sequence, where f : n 7→g4n and g : m 7→ m+2Z Indeed, we have Im f = 4Z and Ker g = 2Z

So, Im f ⊂ Ker g

(3) 0 −→ Z −→ Zf −→ Z/4Z −→ 0 is not complex sequence, whereg

f : n 7→ n and g : m 7→ 2m + 4Z Indeed, Im f = Z * Ker g = 2Z.Definition 1.2.3 A sequence

(F•) : −→ F1 −→ Fdi di−1 −→ −→ F2 d2

−→ F1 d1

−→ F0 −→

is called exact sequence if Im di = Ker di−1

Remark 1.2.4 Every exact sequence is complex

Example 1.2.5 Following the example 1.2.2, the sequence 1) is exactand 2) is not exact

Definition 1.2.6 A complex (F•) is called graded if the modules Fi aregraded and each di is a homomorphism of degree 0.

Example 1.2.7 Take A = k [x, y] and B = x5, xy
We have thegraded complex

g : A(−6) → A(−5) ⊕ A(−2)

g1 7→ −yf1 + x4f2

We have deg(−yf1 + x4f2) = deg(−y) + deg(f1) = 6 = deg(g1) + 0, so ghas degree 0

Chapter 2

Free resolutions

In order to formally definite the graded Betti numbers, it is necessary

to take a look at graded free resolutions of modules We also provide away to find minimal free resolution, moreover a way for using a mathe-matical software (CoCoA)

Definition 2.1.1 Given a finitely generated R-module M A sequence

is called a free resolution of M

Definition 2.1.2 A free resolution (F•) of a finitely generated gradedmodule M is called graded if the modules Fi are graded and each di is a

graded homomorphism of degree 0 The resolution (F•) is called minimalif

di+1(Fi+1) ⊆ (x1, x2, , xn) Fi

If each module Fi is a free finitely generated graded R-module, then

we can write it as

Fi = ⊕p∈ZR(−p)ci,p

Fix a homogeneous basis of each free module Fi Then the differential di

is given by a matrix Di, whose entries are homogeneous elements in R.These matrices are called differential matrices (note that they depend

on the chosen basis)

Example 2.1.3 Given A = k[x, y] and B = (x3, xy, y5) The sequence

On the other hand, Im(d2) = 4f2 + xf3, yf1 − x2f2 Moreover,

y4 ∈ (x, y), f2 ∈ F1 ⇒ y4f2 ∈ (x, y)F1 and x ∈ (x, y), f3 ∈ F1 ⇒ xf1 ∈

F1, so y4f2 + xf3 ∈ (x, y)F1 Similarly, we have yf1 − x2f2 ∈ (x, y)F1.Hence d2(F2) = 4f2 + xf3, yf1 − x2f2 ⊆ (x, y)F1

Therefore, di+1(Fi+1) ⊆ (x, y)Fi for all i ≥ 0, so the resolution in theexample 2.1.3 is minimal.

Assume by induction, that Fi and di are defined Set Mi+1 = Ker(di).

Mi+1 is a graded finitely generated R-module So M1 = ht1, , tpi,deg(ti) = dI Set Fi+1 = R(−d1) ⊕ ⊕ R(−dp) and ui is a basiselement of R(−di) Define

d0i+1 : Fi+1 −→ Mi+1

By construction we have Ker(di) = Im(di+q)

Example 2.2.1 Let A = K[x, y] and B = (x3, xy, y5) We will struct a graded free resolution of A/B over A

con-Set F0 = A and let d0 : A → A/B Then Ker(d0) = 3, xy, y5 Set

F1 = A(−3) ⊕ A(−2) ⊕ A(−5) and f1, f2, f3 are the basis elements ofA(−3) ⊕ A(−2) ⊕ A(−5) We defined

d1 : A(−3) ⊕ A(−2) ⊕ A(−5) → A

Then we get Ker(d1) = af1 + bf2 + cf3 ∈ G | ax3 + bxy + cy5 = 0 ,where a, b, c ∈ A We have

ax3 + bxy + cy5 = 0 ⇔

"

ax3 = −y(bx + cy4),

cy5 = −x(ax2 + by)From there, y | a and x | c Suppose that a = y˜a, c = x˜c Then

Set F2 = A (−4) ⊕ A (−6) and g1, g2 are the basis elements of A (−4)and A (−6) Consider

−−−−−−−−−−→ A(−3) ⊕ A(−2) ⊕ A(−5),

where g1 7−→ yf1 − x2f2, and g2 7−→ y4f2 + xf3 Then

Ker(d2) =ug1 + vg2 ∈ F2 | uyf1 + −ux2 − vy4
f2 + vxf3 = 0

of the module A/B over A We will determine the grading Denote by

f1, f2, f3 the basis of A3 with respect to which the matrix of d1 is given.Since

f1 7→ x3 and deg x3
= 3

f2 7→ xy and deg (xy) = 2

f3 7→ y5 and deg y5
= 5and since we want d1 to be homogeneous of degree 0, we set

deg(f1) = 3, deg(f2) = 2, deg(f3) = 5

Therefore, the free A-module generated by f1 is A(−3), the free module generated by f2 is A(−2), and the free A-module generated by

A-f3 is A(−5) Thus, A3 is identified with A(−3) ⊕ A(−2) ⊕ A(−5).Furthermore, denote by g1, g2 the basis of A2 with respect to which thematrix of d2 is given Since

deg(g1) = 4 and deg(g2) = 6

Hence the free A-module generated by g1 is A(−4) and the free A-modulegenerated by g2 is A(−6) Thus, A2 is identified with A(−4) ⊕ A(−6)

Therefore, we obtain the graded free resolution

mini-of generators mini-of the kernel mini-of the differential

Proof We use the notation introduced in Construction 2.2.1 and setKer(d1) = U We will prove the construct resolution is minimal On thecontrary that for some i ≥ −1, we have chosen a non-minimal homoge-neous system l1, l2, , ls of generators of Ker(di) Assume

resolu-tion is minimal, we have that Im(di+2 ⊆ (x1, , xn)Fi+1 Hence, g1 −P

Now, suppose that at each step we choose a minimal homogeneoussystem of generators of the kernel of the differential We want to showthat the obtained resolution is minimal Assume the contrary Thereexists an i ≥ −1 such that Im(di+2) * (x1, , xn)Fi+1 Therefore,Ker(di+1) = Im(di+2) contains a homogeneous element that is not in(x1, , xn)Fi+1 We can assume that g1 −P

rjlj This contradicts to the fact that we have chosen

l1, , ls to be a minimal homogeneous system of generators of Ker(di)

Chapter 3

Betti numbers

In this chapter, I will give the definition of the Betti numbers Hilbert’sSyzygy Theorem will also be presented Two invariants, regularity andprojective dimension can be associated with the resolutions that measure

”shape” and ”size” if it is interpreted geometrically

We know that the free resolution of finitely generated module can

be finite or infinite However, Hilbert’s Syzygy Theorem 3.1.4, whichsays that the minimal free resolution of every finitely generated gradedmodule over a polynomial ring is finite

Definition 3.1.1 The length of a free resolution (F•) of a finitely erated M is

Thus, pdR(M ) is the length of the minimal free resolution (F•) of M

Example 3.1.2 Let A = K[x, y] and B = (x3, xy, y5) We have thefree resolution is

is minimal free resolution

Following the above example, length of the minimal free resolutioncan be infinity The idea to associate a resolution to a finitely generatedR-module M was introduced in Hilbert’s famous papers [1, 2]

Theorem 3.1.4 (Hilbert’s Syzygy Theorem) The minimal graded freeresolution of a finitely generated graded R-module is finite and its length

is the minimal graded free resolution of M Tensoring (F•) with R/m = kyields the complex

It is quite difficult to obtain a description of the differential in a gradedfree resolution In such cases, we can use the Betti numbers to obtainsome information about the numerical invariants of the resolution

Definition 3.2.1 The i-th Betti number of M over R is

We have that pdR(M ) is the length of the shortest graded free lution of M So we have the following proposition:

reso-Proposition 3.2.3 The length of the minimal free resolution (F•) of

M is

pdR(M ) = maxi | bR

i (M ) 6= 0 Definition 3.2.4 The Poincar´e series of M over R is

We obtain pd(A/B) = 2, and moreover, b0(A/B) = 1, b1(A/B) = 3,

b2(A/B) = 2, and bi(A/B) = 0 for all i ≥ 3 Hence, the Poincarepolynomial of A/B is

PA/B(t) = 1 + 3t + 2t2

In order to find Betti numbers of R/I, we can use CoCoA software

as the following example:

Example 3.2.6 Let R = k[x, y, z] and I = x2, xy, xz, y2

> Use R == ZZ/32003[x, y, z];

We define the graded Betti numbers of M by

bRi,p(M ) = number of summands in Fi of the form R(−p)

Definition 3.2.8 The regularity is defined by

reg(M ) = max{j − i bRi,j(M ) 6= 0},

and the graded Poincar´e series of M over R is

PMR(t, z) = X

i≥0,p∈Z

bRi,p(M )tizp

We typically write the Betti numbers in a matrix called the Betti table

of M The entry in the i-th column and the p-th row is bi,i+p(M )

0 1 · · · i · · · pd(M )0

1

In this thesis, we have systematically presented the following results:

(1) I have recalled the definition and basic properties of some importantconcepts in homological algebra such as complexes, free resolutions

(2) A graded free resolution of a graded finitely generated R-module isconstructed

(3) Hilbert’s Syzygy Theorem have also been presented and proved

(4) The definition of Betti number and some information about the merical invariants associated to the resolution

nu-(5) In order to compute the above invariants, CoCoA software could beeasily used for studying It has already led to theoretical results