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

GRADUATION THESIS
Major: Algebra

Supervisor:
Dr. DO TRONG HOANG

HA NOI - 2019
HA NOI – 2019


Contents

Introduction

2

1 Fundamental concepts

4

1.1

Graded modules . . . . . . . . . . . . . . . . . . . . . . .

4

1.2

Graded complexes . . . . . . . . . . . . . . . . . . . . . .

10

2 Free resolutions

12


2.1

Graded free resolutions . . . . . . . . . . . . . . . . . . .

12

2.2

Contructions . . . . . . . . . . . . . . . . . . . . . . . . .

14

3 Betti numbers

21

3.1

Hilbert’s Syzygy Theorem . . . . . . . . . . . . . . . . .

21

3.2

Betti numbers

23

. . . . . . . . . . . . . . . . . . . . . . .


Conclusion

27

References

28

1


INTRODUCTION

The study of free resolutions is a core and beautiful area in commutative algebra. The idea to associate a free resolution to a finitely generated module was introduced in two famous papers by Hilbert in 1890,
1893. Free resolutions provide a method for describing the structure of
modules. Base on the basic knowledge about algebraic and desiring comprehensive improvement of mathematics, I would like to choose a topic
“free resolutions and Betti numbers of graded module” for my graduation
thesis.
The main goal of this thesis is to describe the structure of graded
finitely generated modules. I will focus on the algebraic invariants associated the free resolutions. Moreover, I also use computer softwares
(CoCoA) to calculate algebraic invariants and verify mathematical issues which arises in the free resolutions. The CoCoA software can be
downloaded free in the website ().
Throughout this thesis, we always denote R = k[x1 , . . . , xn ] polynomial ring with n variables x1 , . . . , xn over field k. According to my
understanding, we organize the thesis based on three books [3, 4, 5]. In
chapter 1, we will present some basic concepts for further exploration of
our topic such as graded modules, graded complexes, Hilbert functions,
Hilbert series. Chapter 2 will provide a graded free resolutions and a
contruct it for a graded finitely generated R-module M . Chapter 3 is
reserved for Betti numbers and relevant invariants.


2


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 thank
the teachers of Hanoi Pedagogical University 2 for helping me to have
the knowledge and create conditions for me to do this thesis.
Due to limitations in time and knowledge, the thesis can not avoid
errors. I hope to receive feedback from teachers and friends.

3


Chapter 1
Fundamental concepts
This chapter aims to provide all relevant definitions and some context
regarding current research topics. For a comprehensive introduction to
these concepts see [4].

1.1

Graded modules

This section will recall some definitions and notations about the graded
R-modules and Nakayama’s lemma.
Definition 1.1.1. A ring R is called graded (or more precisely, Zgraded) if there exists a family of subgroups (Ri )i∈Z of R such that
1. R = ⊕i Ri (as abelian groups), and
2. Ri Rj ⊆ Ri+j for all i, j.
Remark 1.1.2. If R = ⊕i Ri 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

n xn ,

4

where each xn ∈ Rn . Then for all


i, we have
xn xi

xi = 1.xi =
n

Moreover, we have xi = xi x0 for all i. Therefore,
xn x0

x0 = 1.x0 =
n

xn = 1.

=
n

Hence 1 = x0 ∈ R0 Now we prove that Ri is an R0 -module for all i.
Indeed, we have R0 · Rn ⊂ Rn for all i, this statement is proved.

Now, let R = ⊕i∈Z Ri be a graded ring.
Definition 1.1.3. A R-module M is called graded if it satisfies two
following conditions:
1) M = ⊕i∈Z Mi (as abelian groups), and
2) Ri Mj ⊆ 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
m belongs to Mi .
Example 1.1.4. Let K be a field, and let R = K[x1 , ..., xn ] be a polynomial ring over K. For c = (c1 , . . . , cn ) ∈ Nn , let xc = xc11 . . . xcnn . Then
R is a N-graded R-module, where
Ri =

rm xm | rm ∈ K and c1 + · · · + cn = i .

This is called the standard grading on the polynomial ring K[x1 , . . . , xn ].
Notice that R0 = K and deg xi = 1 for all i.
5


Let M be Z-graded finite generated R-module. We asume M =
Rf1 + ... + Rfs . Hence there exists an onto map
Rs −→ M
ei −→ 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. The
map
H(−) : Z −→ Z
i −→ dimK (Mi )
is called the Hilbert function of M . Futhermore,

HilbM (t) =

HM (i) =
i∈Z

dimK (Mi )
i∈Z

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
HilbM (−p) (t) = tp HilbM (t).
Note that degM (−p) (x) = a ⇒ degM (x) = a + p.
Example 1.1.6. LetR = K[x, y] and I = (x3 , y 2 ). Then R/I is graded
ring in degree 0 with basis {1}, in degree 1 with basis {x, y}, in degree

6


2 with basis {x2 , xy}, in degree 3 with basis {x2 y}. 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:
HilbR/I(−7) (t) = t7 HilbR/I (t)
= t7 + 2t8 + 2t9 + t10
In order to compute the Hilbert series, in practice we can use CoCoA
software as follows:
> Use R ::= QQ[x, y];
> I:=Ideal(x3 , y 2 );

> Hilbert(R/I);
H(0) = 1
H(1) = 2
H(2) = 2
H(3) = 1
H(t) = 0 for t ≥ 4
> HilbertSeries(R/I);
(1 + 2x + 2x2 + x3 ).
Definition 1.1.7. Let N and T be graded R-modules and a homomorphism ϕ : N → T . For each m ∈ N , if deg (ϕ(m)) = i + deg(m) then we
say that ϕ has degree i.
7


Example 1.1.8.

1. Consider the homomorphism
x x2
f : R[x]/(x ) −−−−−→ R/[x].
2

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. Therefore, f is graded and has degree 1.
2. Consider the homomorphism
1 x
f : R[x](x2 ) −−−−→ R/[x].
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 and
deg(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 graded
R-module is finitely generated graded R-module, as required.
Now, we prove (1) ⇒ (2): Since M is a finitely generated graded
R-module, so
M = m1 R + ... + mk R,

8

deg(mi ) = ai .


Set
W = R(−a1 ) ⊕ ... ⊕ R(−ak ).
We consider the homomorphism
ϕ : W −→ U
ei −→ mi ,
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. Set
Ker 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. If M = JM , then M = 0.
2. If W is a graded R-submodule of M such that M = W + JM , then
M = W.

Proof. First, we prove (1). Suppose that is M = 0. We choose a finite
minimal system of homogeneous generators of M . Let m be an element
of minimal degree in that system. It follows that Mj = 0 for j < deg(m).
Since I is a proper ideal, we conclude that every homogeneous element
in IM has degree strictly greater than deg(m). This contradicts to
m ∈ M = IM .
Now, we prove (2). We have M = W + JM . Then
M/W = (W + JM )/W
= JM/W
= J(M/W )
By (1), we obtain M/W = 0. It implies that M = W .
9


1.2

Graded complexes

In order to prepare for a definition the free resolution of finitely generated modules, this section will present graded complexes and exact
sequences.
Definition 1.2.1. The sequence (F• ) of homomorphisms of R-modules
d

d

d

i
2
1

(F• ) : ... −→ Fi −→
Fi−1 −→ ... −→ F2 −→
F1 −→
F0 −→ ...

such that di−1 di = 0 (e.g. Im di ⊂ Ker di−1 ) for i ∈ Z is called a complex.
The map di is called the differential of (F• ).
f

g

Example 1.2.2. (1) 0 −→ Z −→ Z −→ Z/2Z −→ 0 is a complex
sequence, where f : n → 2n and g : m → m = m + 2Z. Indeed, we
have Im f = 2Z and Ker g = 2Z. So, Im f = Ker g.
f

g

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

g

(3) 0 −→ Z −→ Z −→ Z/4Z −→ 0 is not complex sequence, where
f : n → n and g : m → 2m + 4Z. Indeed, Im f = Z

Ker g = 2Z.


Definition 1.2.3. A sequence
d

d

d

i
2
1
(F• ) : ... −→ F1 −→
Fdi−1 −→ ... −→ F2 −→
F1 −→
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 exact
and 2) is not exact.
10


Definition 1.2.6. A complex (F• ) is called graded if the modules Fi are
graded and each di is a homomorphism of degree 0.
Example 1.2.7. Take A = k [x, y] and B = x5 , xy . We have the
graded complex








y



4

−x
x5 xy
0 → A(−6) −−−−→ A(−5) ⊕ A(−2) −−−−−−→ A → A/B → 0.
Indeed, we obtain A(−6) and A(−5) ⊕ A(−2) are graded. Now, suppose
that f1 , f2 are the basis elements of A(−5), A(−2). Hence deg(f1 ) = 5,
deg(f2 ) = 2. We define
f : A(−5) ⊕ A(−2) → A
f1 → x5
f2 → xy.
We have deg(x5 ) = 5 = deg(f1 ) + 0 and deg(xy) = 2 = deg(f2 ) + 0, so f
has degree 0. Similarly, denote by g1 the basis element of A(−6), then
deg(g1 ) = 6. We define
g : A(−6) → A(−5) ⊕ A(−2)
g1 → −yf1 + x4 f2 .
We have deg(−yf1 + x4 f2 ) = deg(−y) + deg(f1 ) = 6 = deg(g1 ) + 0, so g
has degree 0.

11


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 a
way to find minimal free resolution, moreover a way for using a mathematical software (CoCoA).

2.1

Graded free resolutions

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

d

d

(F• ) : · · · → Fi →i Fi−1 → · · · → F2 →2 F1 →1 F0 → M → 0
such that
(1) Fi are free R-modules,
(2) Im di = Ker di−1 ,
is called a free resolution of M .
Definition 2.1.2. A free resolution (F• ) of a finitely generated graded
module M is called graded if the modules Fi are graded and each di is a
12


graded homomorphism of degree 0. The resolution (F• ) is called minimal
if
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∈Z R(−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, y 5 ). The sequence


 y


−x2



0





4
−y 


0
x
0 → A(−4) ⊕ A(−6) −−−−−−−−−→ A(−3) ⊕ A(−2) ⊕ A(−5)
x3 xy y 5

−−−−−−−−−→ A → 0
is a graded free resolution of A/B over A. Indeed, let see the detailed
proof in the Example 2.2.1.
Example 2.1.4. The resolution in example 2.1.3 is minimal. Indeed, we
have Im d1 = x3 , xy, y 5 . Moreover, x3 , xy, y 5 are all in A, so d1 (F1 ) ⊆
(x, y)F0 .
On the other hand, Im(d2 ) = y 4 f2 + xf3 , yf1 − x2 f2 . Moreover,
y 4 ∈ (x, y), f2 ∈ F1 ⇒ y 4 f2 ∈ (x, y)F1 and x ∈ (x, y), f3 ∈ F1 ⇒ xf1 ∈
F1 , so y 4 f2 + xf3 ∈ (x, y)F1 . Similarly, we have yf1 − x2 f2 ∈ (x, y)F1 .
Hence d2 (F2 ) = y 4 f2 + xf3 , yf1 − x2 f2 ⊆ (x, y)F1 .
13


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

2.2

Contructions

We construct a graded free resolution of a graded finitely generated
R-module M .
d

d

d

(F• ) : . . . → Fi →i Fi−1 → . . . → F2 →2 F1 →1 F0 → M → 0
Let M0 = M , M =< m1 , . . . , mr >, deg(mi ) = ai . Set F0 = R(−a1 ) ⊕

. . .⊕R(−ar ). Let fi be a basis element of R(−ai ), j = 1, r, then deg(fi ) =
ai . By Theorem 1.1.9, the following homomorphism of degree 0 is defined
well as follows:
d0 : F0 −→ M
fj −→ mj .
We have Ker(d0 ) = M1 , and so M1 is a graded finitely generated Rmodule. So M1 =< l1 , . . . , ls >, deg(li ) = bi . Set F1 = R(−b1 ) ⊕ . . . ⊕
R(−bs ) and gj is a basis element of R(−cj ), deg(gj ) = cj . Define
d1 : F1 −→ M1
gj −→ lj .
Then d1 is a surjective homomorphism of degree 0. Consider the embedding map:
d1 : F1 −→ F0
gj −→ lj .
14


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 = t1 , . . . , tp ,
deg(ti ) = dI. Set Fi+1 = R(−d1 ) ⊕ . . . ⊕ R(−dp ) and ui is a basis
element of R(−di ). Define
di+1 : Fi+1 −→ Mi+1
uj −→ tj .
Then di+1 is a surjective homomorphism of degree 0. Consider the embedding map:
di+1 : Fi+1 −→ Fi
uj −→ tj .
By construction we have Ker(di ) = Im(di+q ).
Example 2.2.1. Let A = K[x, y] and B = (x3 , xy, y 5 ). We will construct a graded free resolution of A/B over A.
Set F0 = A and let d0 : A → A/B. Then Ker(d0 ) = x3 , xy, y 5 . Set
F1 = A(−3) ⊕ A(−2) ⊕ A(−5) and f1 , f2 , f3 are the basis elements of
A(−3) ⊕ A(−2) ⊕ A(−5). We defined
d1 : A(−3) ⊕ A(−2) ⊕ A(−5) → A

f1 → x 3
f2 → xy
f3 → y 5 .
We obtain that
x3 xy y 5
G = A(−3) ⊕ A(−2) ⊕ A(−5) −−−−−−−−−→ A → A/B → 0.
15


Then we get Ker(d1 ) =

af1 + bf2 + cf3 ∈ G | ax3 + bxy + cy 5 = 0 ,

where a, b, c ∈ A. We have
3

5

ax + bxy + cy = 0 ⇔

ax3 = −y(bx + cy 4 ),
cy 5 = −x(ax2 + by)

From there, y | a and x | c. Suppose that a = y˜
a, c = x˜
c. Then
a
˜x2 + b + c˜y 4 = 0. It implies that b = −˜
ax2 + c˜y 4 .
So, we can write it into discrete sections as follows

a
˜ = a y 4 + a”
b = b y 4 + b”x2 + bx2 y 2
c˜ = c + c”x2 .
We get (a + b + c”)x2 y 4 + (a” + b”)x2 + (b + c )y 4 = 0. Thus,



a + b + c” = 0
a” + b” = 0


 b + c = 0.

Therefore,
a , a”, b , b”, ¯b, c , c” =

(0, 1, 0, −1, 0, 0, 0) , (0, 0, −1, 0, 0, 1, 0) ,
(1, 0, 0, 0, −1, 0, 0) , (0, 0, 0, 0, −1, 0, 1) .

It implies that
(a, b, c) =

σ1 = y, −x2 , 0 , σ2 = 0, −y 4 , x , σ3 = y 5 , x2 y 4 , 0 ,
σ4 = 0, −x2 y 4 , x3

=

σ1 , σ2 .


We conclude that Ker (d1 ) = yf1 − x2 f2 , −y 4 f2 + xf3 . We have
deg yf1 − x2 f2

= 4

deg −y 4 f2 + xf3

= 6.

16


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

x3



0




−x2



−y 4









0
x
d2 : A(−4) ⊕ A(−6) −−−−−−−−−−→ A(−3) ⊕ A(−2) ⊕ A(−5),
where g1 −→ yf1 − x2 f2 , and g2 −→ y 4 f2 + xf3 . Then
Ker(d2 ) = ug1 + vg2 ∈ F2 | uyf1 + −ux2 − vy 4 f2 + vxf3 = 0 .
Hence



uy = 0


−ux2 − vy 4 = 0 ⇒



vx = 0

u=0
v = 0.

We obtain F3 = 0. Therefore,



y

0




−x2








−y 4 


0
x
0 → A(−4) ⊕ A(−6) −−−−−−−−−→ A(−3) ⊕ A(−2) ⊕ A(−5)
x3 xy y 5
−−−−−−−−−→ A → A/B → 0.
Example 2.2.2. Let A = K[x, y] and B = (x3 , xy, y 5 ). Suppose we are
given (say by computer) the non-graded free resolution



y




−x2



0






−y 4 


x3 xy y 5
0
x
3
0 → A −−−−−−−−−→ A −−−−−−−−−→ A
2

17


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 → x3 and deg x3 = 3
f2 → xy and deg (xy) = 2
f3 → y 5 and deg y 5 = 5
and 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 Amodule generated by f2 is A(−2), and the free A-module generated by
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 the
matrix of d2 is given. Since
g1 →yf1 − x2 f2
deg yf1 − x2 f2 = deg (yf1 ) = deg(y) + deg(f1 ) = 4
g2 → − y 4 f2 + xf3
deg −y 4 f2 + xf3 = deg y 4 f2 = deg(y 4 ) + deg(f2 ) = 6
and since we want d2 to be homogeneous of degree 0, we set
deg(g1 ) = 4 and deg(g2 ) = 6.

Hence the free A-module generated by g1 is A(−4) and the free A-module
generated by g2 is A(−6). Thus, A2 is identified with A(−4) ⊕ A(−6).
18


Therefore, we obtain the graded free resolution


 y


−x2




0





4
−y 


x3 xy y 5
0
x
0 → A(−4)⊕A(−6) −−−−−−−−−→ A(−3)⊕A(−2)⊕A(−5) −−−−−−−−−→ A.
In order to compute the free resolution, in practice we can use Cocoa
software
> Use R ::= QQ[x,y];
> I:=Ideal(x3 , xy, y 5 );
> Res(I);
0 − − > R(−4) ⊕ R(−6) − − > R(−2) ⊕ R(−3) ⊕ R(−5).
Theorem 2.2.3. The graded free resolution constructed above is minimal if and only if at each step we choose a minimal homogenous system
of generators of the kernel of the differential.
Proof. We use the notation introduced in Construction 2.2.1 and set
Ker(d1 ) = U . We will prove the construct resolution is minimal. On the
contrary that for some i ≥ −1, we have chosen a non-minimal homogeneous system l1 , l2 , . . . , ls of generators of Ker(di ). Assume
l1 =


rj lj
2≤j≤s

for some rj ∈ R. Since
di+1 : Fi+1 → Fi
gj → lj .
19


So l1 =

rj lj . Then
2≤j≤s

rj di+1 (gj ) ⇒ di+1 (g1 −

di+1 (g1 ) =

2≤j≤s

2≤j≤s

Hence, g1 −

2≤j≤s rj gj

rj gj ) = 0.

∈ Ker(di+1 ) = Im(di+2 ). Since the resolu-


tion is minimal, we have that Im(di+2 ⊆ (x1 , . . . , xn )Fi+1 . Hence, g1 −
2≤j≤s rj gj

∈ (x1 , . . . , xn )Fi+1 , which is a contradiction.

Now, suppose that at each step we choose a minimal homogeneous
system of generators of the kernel of the differential. We want to show
that the obtained resolution is minimal. Assume the contrary. There
exists 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 −

2≤j≤s rj gj

∈ Ker(di+1 ) for

some rj ∈ R. Hence
di+1 (g1 ) =

rj di+1 (gj ).
2≤j≤s

Hence, l1 =

rj lj . This contradicts to the fact that we have chosen
2≤j≤s


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

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Chapter 3
Betti numbers
In this chapter, I will give the definition of the Betti numbers. Hilbert’s
Syzygy Theorem will also be presented. Two invariants, regularity and
projective dimension can be associated with the resolutions that measure
”shape” and ”size” if it is interpreted geometrically.

3.1

Hilbert’s Syzygy Theorem

We know that the free resolution of finitely generated module can
be finite or infinite. However, Hilbert’s Syzygy Theorem 3.1.4, which
says that the minimal free resolution of every finitely generated graded
module over a polynomial ring is finite.
Definition 3.1.1. The length of a free resolution (F• ) of a finitely generated M is
max {i | rank Fi = 0} .
We say that the free resolution (F• ) is finite if its length is finite. Otherwise, (F• ) is infinite. The projective dimension of M is
pdR (M ) = max {i| rank (Fi ) = 0} .
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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, y 5 ). We have the
free resolution is



y

0




−x2








−y 4 


0
x
0 → A(−4) ⊕ A(−6) −−−−−−−−−→
x3 xy y 5
A(−3) ⊕ A(−2) ⊕ A(−5) −−−−−−−−−→ A → A/B → 0.
So, pdR (A/B) = 3.
Example 3.1.3. Let M = R = k[x]/(x2 ). The following sequence
·x


·x

·x

·x

· · · −→ M −→ M −→ M −→ · · ·
is minimal free resolution.
Following the above example, length of the minimal free resolution
can be infinity. The idea to associate a resolution to a finitely generated
R-module M was introduced in Hilbert’s famous papers [1, 2].
Theorem 3.1.4 (Hilbert’s Syzygy Theorem). The minimal graded free
resolution of a finitely generated graded R-module is finite and its length
is at most n.
Proof. We know that the Koszul complex K• (x1 , . . . , xn ) is a graded
free resolution of R/m = k of length n, where m = (x1 , . . . , xn ). Thus
pdR (k) = n. Now we prove pdR (M ) ≤ n. Assume that
d

d

i
1
(F• ) · · · −→ Fi −→
Fi−1 → · · · → F1 −→
F0 → M → 0

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is the minimal graded free resolution of M . Tensoring (F• ) with R/m = k
yields the complex
d ⊗1

d ⊗1

i
1
· · · → Fi ⊗ k −→
Fi−1 ⊗ k → · · · → F1 ⊗ k −→
F0 ⊗ k → M ⊗ k → 0.

Using Im(di ) ⊆ mFi−1 for all i ≥ 1, one obtains that all the maps di ⊗ 1
are zero. Hence
Ker(di+1 ⊗ 1)
= Fi ⊗ R/m
Im(di ⊗ 1)
for i ≥ 1. In particular, Fi ⊗ R/m ∼
= Fi /mFi = 0 for i ≥ n + 1. By
Nakayama’s lemma 1.1.10, one obtains Fi = 0 for all i ≥ n + 1.

3.2

Betti numbers

It is quite difficult to obtain a description of the differential in a graded
free resolution. In such cases, we can use the Betti numbers to obtain
some information about the numerical invariants of the resolution.
Definition 3.2.1. The i-th Betti number of M over R is
bR

i (M ) = rank(Fi ).
Remark 3.2.2. The Betti numbers do not depend on the choice of the
minimal graded fre resolution of M .
Proof. By Theorem 2.2.3, we know that to construct a minimal graded
free resolution of a graded finitely generated R-module, at each step of
construction in section 2.2, we choose a minimal homogenous system of
generators of the kernel of the differential. Hence, there exists a unique
minimal graded free resolution of a graded finitely generated R-module.
Therefore, The Betti numbers do not depend on the choice of the minimal
graded free resolution of M .
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