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# Betti numbers and Euler characteristics

Betti numbers and Euler characteristics
Esben Bistrup Halvorsen
January 28, 2003
This paper constitutes a course project related to the course
Homological Algebra (Math253), which I took in the spring
semester of 2002 at the University of California, Berkeley.
The presentation is based on a discussion paper by my ad-
visor at the University of Copenhagen, Hans–Bjørn Foxby,
who has guided me through the entire process.
Introduction
The purpose of this paper is to introduce and investigate the concepts of Betti
numbers and Euler characteristics, which arise in algebraic topology as invariants
describing topological spaces. In this presentation we will ignore the origin of the
concepts, however, and instead discuss them completely within the language of
homological algebra.
Betti numbers and Euler characteristics are useful tools in describing modules
and complexes. Knowing these provides enough information to calculate the ﬂat
and projective dimensions of a complex (see corollary 4.4), and to determine
whether or not a ﬁnitely generated module is free (see theorem 5.9). In this
paper we shall also obtain results that can be formulated completely without Betti
numbers and Euler characteristics, although not easily proved without them. The
most remarkable of these states that a suﬃcient condition for a ﬁnitely generated
module to have trivial annihilator is that the annihilator consists only of zero-
divisors (see theorem 5.7).
In our presentation, we will limit ourselves to consider modules over a local
and noetherian ring; on the other hand, we will allow ourselves to designate Betti
numbers and Euler characteristics not only to modules, as is traditionally the case,
but also to complexes. This complies with the dogma “It’s better to work with
complexes than modules” from Hans–Bjørn Foxby’s notes [FoxbyHHA]. Towards
the end of the discussion, however, we will return to modules and see how the
Euler characteristics behaves.
1
Rings and modules 2
1 Rings and modules
Throughout this section, Rdenotes a (nontrivial, unitary, commutative) local
ring. The unique maximal ideal of Ris denoted by mand the quotient ﬁeld
R/mis denoted by k. All modules are, unless otherwise stated, assumed to be
R-modules.
In this section we obtain a variety of results about modules over local rings.
These will be needed later to prove general facts about Betti numbers and Euler
characteristics.
1.1 Free modules
Suppose Mis a module. The quotient M/mMhas a natural structure of a k-
vector space, the scalar multiplication being deﬁned by [r] [m]M= [rm]M.
Furthermore, any homomorphism ϕ:MNof modules induces a linear map
M/mMN/mNof k-vector spaces given by [m]M7→ [ϕ(m)] N.
Deﬁnition 1.1. If Mis a module, Mdenotes the k-vector space M/mM, and
βR
0(M) denotes the dimension of M. If ϕ:MNis a homomorphism of
modules, ϕdenotes the induced k-linear map MNgiven by ϕ([m]M) =
[ϕ(m)] N.
For the next theorem, notice (see for example [Lang] proposition XVI.2.7) that
if Iis an ideal of Rand Mis a module, then M/IM
=(R/I)RM; in particular,
M
=kRM. Furthermore, it is easy to see that under this isomorphism, any
homomorphism M/I M N/IN induced by a homomorphism ϕ:MN(that
is, on the form [m]IM 7→ [ϕ(m)]IN ) corresponds to the homomorphism (R/I)Rϕ;
in particular, the k-linear map ϕ:MNcorresponds to kRϕ. In other words,
deﬁnition 1.1 simply gives new symbols to the functor kR.
Theorem 1.2. If Fis a ﬁnitely generated free module, then F
=Rnfor n=
βR
0(F).
Proof: We already know that F
=Rnfor some nN(see for instance [Lang]
corollary III.4.3), so it only remains to check that Rnis a k-vector space of
dimension n. This follows from the isomorphisms Rn
=kRRn
=(kRR)n
=
kn.
Deﬁnition 1.3. When Fis a ﬁnitely generated free module, the rank of Fis
the number rankRF=βR
0(F).
Normally one would deﬁne the rank of a ﬁnitely generated free module Fto
be the nN0such that F
=Rn, but for this to be well-deﬁned, one has to
show that Rn
=Rmimplies n=m. This is exactly what theorem 1.2 does: if
Rn
=Rm, then n=βR
0(Rn) = βR
0(Rm) = m.
Rings and modules 3
Lemma 1.4. Let 0KFG0be an exact sequence of modules,
and suppose that Fand Gare ﬁnitely generated free modules. Then Kis ﬁnitely
generated and free, and rankRF= rankRK+ rankRG.
Proof: Since Gis free and thereby projective, the sequence splits, hence there
is an isomorphism F
=KG; in particular, Kmust be ﬁnitely generated.
According to [Lang] theorem X.4.4 a ﬁnitely generated projective module over a
local ring is free, hence for the freeness of Kit suﬃces to show that Kis projective.
This, however, follows from the fact (see for example [H&S] proposition I.4.5) that
a direct sum of modules is projective if and only if each summand is projective.
We now have an isomorphism F
=KGof free and ﬁnitely generated
modules, so it follows by the remarks preceding the lemma that rankRF=
rankRK+ rankRG.
Theorem 1.5. Let Fbe a ﬁnitely generated free module and (f1, . . . , fn)a set of
generators for F. Then (f1, . . . , fn)is a basis if and only if n= rankR(F).
Proof: Let (e1, . . . , en) be the standard basis for Rn, and consider the homo-
morphism ϕ:RnFdeﬁned on the basis by ϕ(ei) = fi. We then have the
exact sequence
0ker ϕRnϕ
F0,
and from the previous lemma it follows that kerϕis a free module with
n= rankR(F) + rankR(ker ϕ).
Now, if (f1, . . . , fn) is a basis, then kerϕ= 0, and the above equation yields n=
rankR(F). Conversely, if n= rankR(F), then the equation implies rankR(ker ϕ) =
0, hence ker ϕ= 0.
A similar result is found in
Theorem 1.6. Let Mbe a ﬁnitely generated R-module and (m1, . . . , mn)a set
of generators for M. Then ([m1]M,...,[mn]M)is a basis for Mif and only if
n=βR
0(M).
Proof: Since m1, . . . , mngenerate M, [m1]M, . . . , [mn]Mgenerate Mwhich
is a k-vector space of dimension βR
0(M), and we know that a generating set in a
vector space is a basis if and only if the number of generators is the same as the
dimension.
A simple consequence of theorem 1.5 and theorem 1.6 is the following.
Corollary 1.7. Let Fbe a ﬁnitely generated free module. Then (f1, . . . , fn)is a
basis for Fif and only if ([f1]F, . . . , [fn]F)is a basis for F.
Rings and modules 4
If m1, . . . , mngenerate a module M, then [m1]M,...,[mn]Mgenerate the
vector space Mwhose dimension is βR
0(M). In particular we must have n
βR
0(M), that is, any generating set for Mcontains at least βR
0(M) elements. This
observation leads us to
Deﬁnition 1.8. A set (m1, . . . , mn) of generators for a module Mis said to be
minimal if n=βR
0(M).
Minimal generating sets do exist. In fact we have
Proposition 1.9. Any set of generators for a ﬁnitely generated module Mcan
be reduced to a minimal set of generators.
Proof: A generating set for Minduces a generating set for Mwhich can be
reduced to a basis for Mwith βR
0(M) elements. Thus we can ﬁnd a subset of the
generators for Mthat induce a basis for M. This subset must contain βR
0(M)
elements, and by Nakayama’s lemma (as stated in [Lang] lemma X.4.3) it will
still be generating.
We will now prove a theorem which states that any homomorphism ϕ:FG
between ﬁnitely generated free modules splits up into two parts, one being an
isomorphism, and the other being “minimal” in the sense that the induced map ϕ
is zero. This will become very useful later when we construct a similar “splitting”
of free resolutions. First we need a lemma:
Lemma 1.10. Suppose Fand Gare ﬁnitely generated free modules and ϕ:F
Gis a homomorphism such that ϕ(F)*mG. Then there exist bases (f1, . . . , fn)
and (g1, . . . , gm)for Fand Grespectively, such that ϕ(f1) = g1and ϕ(Rf2+· · ·+
Rfn)Rg2+· · · +Rgm.
Proof: Consider the induced linear map ϕ:FGof k-vector spaces. Let
n= rankR(F) and m= rankR(G), and pick an element e1Fwith g1=
ϕ(e1)/mG. Notice that we must have e1/mFas well, such that we now have
[e1]F6= 0 and [g1]G6= 0. Now extend [e1]Fto a basis ([e1]F, . . . , [en]F)
for Fand extend [g1]Gto a basis ([g1]G, . . . , [gm]G) for G. By corollary 1.7,
(e1, . . . , en) is a basis for F, and (g1, . . . , gm) is a basis for G.
We will now transform the basis (e1, . . . , en) into a basis (f1, . . . , fn) with the
desired property. Let f1=e1, and choose for i= 2, . . . , n elements ri1, . . . , rim
Rsuch that ϕ(ei) = ri1g1+· · · +rimgm. Let fi=eiri1e1. Then (f1, . . . , fn) is
a generating set for Fsince e1=f1and ei=ri1e1+fifor i= 2, . . . , n, hence it
is a basis by theorem 1.5. Furthermore, ϕ(f1) = g1, and for i= 2, . . . , n,
ϕ(fi) = ri2g2+···+rim gm,
so ϕ(Rf2+· · · +Rfn)Rg2+· · · +Rgmas desired.
Complexes 5
Theorem 1.11. Suppose Fand Gare ﬁnitely generated free modules, and that
ϕ:FGis a homomorphism. There then exist direct sum decompositions
F=KLand G=MNinto ﬁnitely generated free submodules, such that
ϕ(L)mN, and such that ϕinduces an isomorphism ϕ|K:K
=
M.
Proof: If ϕ(F)mGwe are done. If not, we can apply lemma 1.10 to obtain
bases (f1, . . . , fn) and (g1, . . . , gm) for Fand Grespectively with ϕ(f1) = g1
and ϕ(Rf2+· · · +Rfn)Rg2+· · · +Rgm. Let F0=Rf2+· · · +Rfnand
G0=Rg2+· · · +Rgm, and notice that F=Rf1F0and G=Rg1G0and that
ϕmaps Rf1isomorphically onto Rg1. Thus, if ϕ(F0)mG0, we are done. If not,
lemma 1.10 applies to F0,G0and the induced homomorphism ϕ|F0:F0G0, and
we alter the bases (f2, . . . , fn) and (g2, . . . , gm) for F0and G0(but omit changing
the symbols) such that we get ϕ(f2) = g2and ϕ(Rf3+· · ·+Rfn)Rg3+· · ·+Rgm.
Now we let F00 =Rf3+· · · +Rfnand G00 =Rg3+· · · +Rgm, and notice that
F= (Rf1+Rf2)F0and G= (Rg1+Rg2)G00 and that ϕmaps Rf1+Rf2
isomorphically onto Rg1+Rg2. Therefore, if ϕ(F00 )mG00, we are done. If not,
we repeat the process. Eventually the process will terminate, say at the i’th step,
at which point we have obtained bases (f1, . . . , fn) and (g1, . . . , gm) for Fand G
respectively, such that ϕ(f1) = g1, . . . , ϕ(fi) = gi, and ϕ(Rfi+1 +· · · +Rfn)
m(Rgi+1+· · ·+Rgm). (Notice that it is possible that i=nsuch that ϕis simply an
isomorphism in itself.) We can now let K=Rf1+· · ·+Rfi,L=Rfi+1 +···+Rfn,
M=Rg1+· · · +Rgi, and N=Rgi+1 +· · · +Rgm, and the theorem is proved.
2 Complexes
In this section we introduce notation, terminology and constructions related to
complexes and the morphism between them. The presentation is based exclu-
sively on [FoxbyHHA] sections 0, 1, 2, 4 and 7.
2.1 Categories of complexes and the homology functor
Throughout this subsection, Rdenotes a (nontrivial, unitary) ring. All mod-
ules are, unless otherwise stated, assumed to be left R-modules.
We now move our focus from modules to the more general concept of complexes.1
Recall that a complex Xis a family (Xn)nof modules together with a family
(X
n)nof homomorphisms X
n:XnXn1, which are called the diﬀerentials
of X, such that X
nX
n+1 = 0 for all nZ:
X=· · · Xn+1
X
n+1
Xn
X
n
Xn1 · · · .
1Notation and deﬁnitions are from [FoxbyHHA] sections 1 and 2.
Complexes 6
The complex Xis said to be bounded at the left if Xn= 0 for suﬃciently large n.
In that case, when presenting Xas above, we omit the superﬂuous zero modules
and simply write
X= 0 XsXs1 · · · .
Similarly, Xis said to be bounded at the right if Xn= 0 for suﬃciently small n,
and we then write
X=· · · Xt+1 Xt0.
A complex is said to be bounded if it is bounded at the left and at the right. In
the sequel, modules will always be thought of as complexes concentrated in degree
0, that is, complexes Xwith Xn= 0 for all n6= 0.
When Xand Yare complexes, a morphism α:XYis a family α= (αn)n
of homomorphisms αn:XnYnmaking the following diagram commutative.
· · · //Xn+1
X
n+1 //
αn+1
²²
Xn
X
n//
αn
²²
Xn1//
αn1
²²
· · ·
· · · //Yn+1
Y
n+1 //Yn
Y
n//Yn1//· · ·
All complexes together with all morphisms of complexes form a category which
is denoted by C(R). We introduce the following full subcategories of C(R) by
specifying their objects.
C(R): complexes that are bounded at the left
C(R): complexes that are bounded at the right
C(R): complexes that are bounded
C0(R): complexes that are concentrated in degree 0 (that is, modules)
Cf(R): complexes of ﬁnitely generated modules
Cl(R): complexes of modules of ﬁnite length
CI(R): complexes of injective modules
CF(R): complexes of ﬂat modules
CP(R): complexes of projective modules
CfP(R): complexes of ﬁnitely generated projective modules
CL(R): complexes of ﬁnitely generated free modules
We will freely use any combination of the subscripts {@,A,¤,0}and superscripts
{f,l,I,F,P,fP,L}be setting C?
#(R) = C?(R)∩C#(R), such that for example CL(R)
denotes the category of complexes which are bounded to the right and consist of
ﬁnitely generated free modules.
An important functor in the category C(R) is the homology functor H : C(R)
C(R) which takes a complex Xto the complex H(X) deﬁned by H(X)n=
Hn(X) = ker X
n/im X
n+1 and H(X)
n= 0 for all nZ, and a morphism α:XY
to the morphism H(α): H(X)H(Y) given by
H(α)n([x]im X
n+1 ) = Hn(α)([x]im X
n+1 ) = [αn(x)]im Y
n+1
Complexes 7
for all nZand xker X
n. H(X) is called the homology complex of X, and
Hn(X) is called the homology module in degree n.
The functor H entails a new series of subcategories of C(R). Letting ?repre-
sent any of the superscripts {f,l,I,F,P,fP,L}, we deﬁne the subcategory C(?)(R)
by
X∈ C(?)(R)def
H(X)∈ C?(R).
Letting # represent any of the subscripts {@,A,¤,0}, we likewise deﬁne the
subcategory C(#)(R) by
X∈ C(#)(R)def
H(X)∈ C#(R).
As before, the subscripts and superscripts can be combined freely by setting
C(?)
#(R) = C(?)(R)∩ C(#)(R). The names for the objects in these subcategories
are obtained by adding the words “homology” and “homologically” to the names
for the objects in the corresponding subcategories C?
#(R), such that for instance
C(f)
( )(R) is the category of complexes which are homologically bounded to the
right and have ﬁnitely generated homology modules.
The homology functor H entails another concept. Recall that a morphism
α:XYof R-complexes is said to be an isomorphism if there exists a morphism
α1:YXsuch that α1α= 1Xand αα1= 1Y. In this case we say that X
is isomorphic to Y, and we write α:X
=
Yor simply X
=Y.
Deﬁnition 2.1. Suppose α:XYis a morphism of complexes. We say that
αis a homology isomorphism, and we write α:X'
Y, if H(α): H(X)H(Y)
is an isomorphism; in other words,
X'
αYdef
H(X)
=
H(α)H(Y).
The relation X'
Yof homology isomorphism is reﬂexive and transitive,
but not necessarily symmetric. To help this we introduce
Deﬁnition 2.2. Two complexes Xand Yare said to be equivalent, and we
write X'Y, if there exist a third complex Zand homology isomorphisms
X'
Z'
Y.
One can show2(and this also shows that 'is an equivalence relation) that
X'Ywhenever there exist nNand n+ 1 complexes V(0), . . . , V (n+ 1) such
that X=V(0), Y=V(n), and we are in one of the following situations.
2This is actually the deﬁnition (or can easily be derived from the deﬁnition) of equivalence
as it is given in (1.28) in [FoxbyHHA].
Complexes 8
X=V(0) '
V(1) '
· · · '
V(n1) '
V(n) = Y,
X=V(0) '
V(1) '
· · · '
V(n1) '
V(n) = Y,
X=V(0) '
V(1) '
· · · '
V(n1) '
V(n) = Y,
X=V(0) '
V(1) '
· · · '
V(n1) '
V(n) = Y.
Isomorphic complexes are equivalent, and equivalent complexes have isomor-
phic homologies. Thus we have the implications
X
=YX'YH(X)
=H(Y).
A ﬁnal deﬁnition having to do with the homology functor is found in
Deﬁnition 2.3. Let X∈ C (R). The supremum of Xis given by supX=
sup{nZ|Hn(X)6= 0}and the inﬁmum of Xis given by inf X= inf{nZ|
Hn(X)6= 0}.
Here we follow the usual conventions that sup =−∞ and inf =, so
sup X=−∞ and inf X=whenever Xis homologically trivial, while −∞ ≤
inf Xsup X −∞ otherwise. Notice also that we obviously have X
C( )(R)sup X < , and X∈ C( )(R)inf X > −∞.
2.2 Tensor products and dot products
Throughout this subsection, Rand Sdenote (nontrivial, unitary) rings, and
Qdenotes a (nontrivial, unitary) commutative ring over which Rand Sare
algebras. All modules are, unless otherwise stated, assumed to be left R-
modules.
Recall3that for an R-module (that is, a right R-module) Mand an R-module
N, the tensor product MRNof Mand Nis a Q-module; in particular it is
an R-module if Ris commutative. If in addition Mhas an S-module structure
such that Mis an (S, R)-bimodule, then MRNhas the structure of an (S, Q)-
bimodule. Similarly, if Nhas an additional S-module structure such that Nis
a (R, S)-bimodule, then MRNhas the structure of a (Q, S)-bimodule.
The tensor product generalizes to complexes4: For X∈ C(R) and Y∈ C(R)
the tensor product of Xand Yis the Q-complex XRYwhose n’th module is
the Q-module
(XRY)n=M
`
X`RYn`,
and whose n’th diﬀerential XRY
n: (XRY)n(XRY)n1is deﬁned on
generators x`yn`X`RYn`(XRY)n,`Z, by
XRY
n(x`yn`) = X
`(x`)yn`+ (1)`x`Y
n`(yn`),
3See [FoxbyHHA] section 0.
4See [FoxbyHHA] section 4.
Complexes 9
which is an element of X`1RYn`+X`RYn`1(XRY)n1.
From the tensor product we derive the dot product5: Given X∈ C(R) and
Y∈ C( )(R), the dot product X·RYis deﬁned as the equivalence class represented
by GRXfor any G∈ CF(R) with G'Y; this is well-deﬁned since such
complexes Gexist (as shown in [FoxbyHHA] theorem 2.6(P) which is reproduced
in theorem 3.2 of this paper), and since any G, G0∈ CF(R) satisfying G'Y'G0
must have GRX'G0RX. If instead we are given X∈ C( )(R) and
Y∈ C(R), the dot product X·RYis deﬁned as the equivalence class represented
by YRFfor any F∈ CF(R) with F'X; this is also well-deﬁned. The
two deﬁnitions do not contradict: whenever they both apply, they yield the same
equivalence class.
2.3 Localization
Throughout this subsection, Rdenotes a (nontrivial, unitary) commutative
ring. All modules are, unless otherwise stated, assumed to be R-modules.
Let Ube a multiplicatively closed subset of the commutative ring R. Recall that
the localization U1Mof Mat Uhas the structure of an R-module as well as
of an U1R-module. Notice also that U1M
=(U1R)RM(see for example
[Eisenbud] lemma 2.4) from which the preceding statement follows. It is easy
to see that under this isomorphism, any homomorphism U1ϕ:U1MU1N
induced by a homomorphism ϕ:MNcorresponds to the homomorphism
(U1R)Rϕ, so that the functor U1() is actually the same as the functor
(U1R)R. From this it follows that U1() is a covariant additive (right
exact) functor C0(R)→ C0(U1R), so by (1.51) in [FoxbyHHA] it must induce a
functor C(R)→ C(U1R), which is likewise denoted by U1(). We can therefore
generalize the notion of localization to complexes: Given an R-complex X, the
localization of Xat Uis the U1R-complex (and thereby R-complex)
U1X=· · · U1Xn+1
U1X
n+1
U1Xn
U1X
n
U1Xn1 · · · ,
where the n’th diﬀerential U1X
nmaps xn/u U1Xnto X
n(xn)/u U1Xn1.
A complex morphism α:XYis mapped by the functor U1() to the complex
morphism U1α:U1XU1Ygiven by (U1α)n(xn/u) = αn(xn)/u for xn
Xn.
According to 2.4(2) in [FoxbyHA], U1() is exact as a functor C0(R)
C0(U1R), hence it follows from (1.51) in [FoxbyHHA] that H(U1X)
=U1(H(X))
for all X∈ C(R), and that U1() preserves homology isomorphisms and equiv-
alences.
As is the case with modules we write Xfor (R\p)1Xwhenever pis a prime
ideal.
5See [FoxbyHHA] section 7.
Resolutions 10
3 Resolutions
This section is devoted to resolutions, not only of modules, but also of complexes.
The results obtained here will be fundamental to the following sections in which
we ﬁnally encounter Betti numbers and Euler characteristics.
3.1 Injective, projective, and ﬂat resolutions
Throughout this subsection, Rdenotes a (nontrivial, unitary) ring. All mod-
ules are, unless otherwise stated, assumed to be left R-modules.
Recall that a projective resolution of a module Mis a complex
P=· · · Pn
P
n
Pn1 · · · P1
P
1
P00
of projective modules together with a homomorphism α0:P0M, such that
· · · Pn
P
n
Pn1 · · · P1
P
1
P0
α0
M0
is exact. Considering Mas a complex concentrated in 0, we can formulate this
as follows: A projective resolution of Mis a complex P∈ CP(R) with Pn= 0 for
n < 0 together with a homology isomorphism α:P'
M.
P
α
²²
· · · //Pn
P
n//
²²
Pn1//
²²
· · · //P1
²²
P
1//P0//
α0
²²
0
M· · · //0//0//· · · //0//M//0
This observation (together with similar observations about ﬂat, free, and injective
resolutions) leads to the following generalization.6
Deﬁnition 3.1. An injective resolution of a complex X∈ C( )(R) is a homology
isomorphism X'
I∈ CI(R). A ﬂat resolution of a complex X∈ C( )(R) is a
homology isomorphism X'
F∈ CF(R); a projective resolution of a complex
X∈ C( )(R) is a homology isomorphism X'
P∈ CP(R); and a free resolution
by ﬁnite modules of a complex X∈ C(f)
( )(R) is a homology isomorphism X'
L∈ CL(R).
Injective and projective (and thereby ﬂat) resolutions exist for modules. The
generalization of this is found in [FoxbyHHA] theorem 2.6 which we bring here
without proof.
6See (2.5) in [FoxbyHHA].
Resolutions 11
Theorem 3.2.
(I) Every X∈ C( )(R)has an injective resolution X'
Isuch that In= 0
for all n > sup X.
(P) Every X∈ C( )(R)has a projective (and thereby ﬂat) resolution X'
P
such that Pn= 0 for all n < inf X.
(L) If Ris noetherian, then every X∈ C(f)
( )(R)has a free resolution X'
L
by ﬁnite modules such that Ln= 0 for all n < inf X.
The concepts of injective, projective, and ﬂat dimensions of modules are gen-
eralized in the following deﬁnition7(in which we follow the usual (and logical)
conventions about inﬁmum of sets containing ±∞).
Deﬁnition 3.3. If X∈ C( )(R), then the injective dimension of Xis given by
idRX= inf{sup{nZ|In6= 0} | X'I∈ CI(R)}; and if X∈ C( )(R), then
the projective dimension of Xis given by pdRX= inf{sup{nZ|Pn6= 0} | X'
P∈ CP(R)}, and the ﬂat dimension of Xis given by fdRX= inf{sup{nZ|
Fn6= 0}| X'F∈ CF(R)}.
3.2 Minimal free resolutions
Throughout this subsection, Rdenotes a (nontrivial, unitary, commutative)
local, noetherian ring with maximal ideal mand quotient ﬁeld k=R/m. All
modules are, unless otherwise stated, assumed to be R-modules.
When dealing with Betti numbers we will often be concerned with free resolutions
by ﬁnite modules; in fact we will consider the special kind of free resolutions
deﬁned below.
Deﬁnition 3.4. A complex Lis said to be minimal if im L
nmLn1for all
nZ. A free resolution X'
L∈ CL(R) by ﬁnite modules of a complex Xis
called a minimal free resolution of Xif Lis a minimal complex.
Theorem 3.5. Suppose X'
L∈ CL(R)is a free resolution by ﬁnite modules of
a complex X. Then there exist subcomplexes Aand Bof Lwith ﬁnitely generated
modules, such that Ais exact, Bis minimal, and L=AB.
Proof: By theorem 1.11 we can for each nZﬁnd direct sum decom-
positions Ln=FnGn=MnNn, where Fn,Gn,Mn, and Nnare free
ﬁnitely generated submodules of Ln, such that L
n(Gn)mNn1, and such that
L
ninduces an isomorphism L
n|Fn:Fn
=
Mn1. The properties ensure that
7See [FoxbyHHA] section 8.
Resolutions 12
Mn= im L
n+1|Fn+1 im L
n+1 MnmNnand ker L
n= ker L
n|GnGn, so that
for all nZwe have the following chain of inclusions:
Mnim L
n+1 ker L
nGn.(1)
For every nZwe now let An=Fn+Mnand Bn=GnNn. We claim
that Anand Bnare free and ﬁnitely generated, and that Lnis the direct sum of
Anand Bn. To see this, notice ﬁrst that since FnGn= 0 and MnGn, we
must have FnMn= 0, and it follows that An=FnMnis a direct sum; in
particular, Anis free and ﬁnitely generated.
Now, suppose that xAnBn. Let x=m+fbe the unique decomposition
of xinto fFnand mMn. Then f=xmBn+MnGn, so we must
have f= 0 (since FnGn= 0). Consequently x=m, but since xNn, it
follows that x= 0 (since MnNn= 0). This shows that AnBn= 0.
Finally, suppose xLnis arbitrarily chosen. Let x=f+gbe the unique
decomposition of xinto fFnand gGn, and let g=m+nbe the unique
decomposition of ginto mMnand nNn. Then n=gmGn+Mn=Gn,
so nBnand x= (f+m) + nAn+Bn. This shows that An+Bn=Ln,
and we now have the (inner) direct sum Ln=AnBn. In particular there is an
exact sequence 0 BnLnAn0, and it follows from lemma 1.4 that Bn
is free and ﬁnitely generated.
The inclusion im L
nGn1from (1) implies that L
nrestricts to a homomor-
phism L
n|Bn:BnGn1. We already know that L
nmaps Gnto mNn1, so the
image of the restriction L
n|Bnis actually contained in Gn1mNn1Bn1,
hence L
nrestricts to a homomorphism BnBn1. Let us for each nZdenote
by B
nthe homomorphism BnBn1which is the restriction of L
nto Bn. We
have then obtained a complex with free and ﬁnitely generated modules:
B=· · · Bn+1
B
n+1
Bn
B
n
Bn1 · · · .
The inclusion Mnker L
nfrom (1) and the isomorphism L
n|Fn:Fn
'
Mn1
implies that L
n(An) = Mn1An1, hence L
nrestricts to a homomorphism
AnAn1which we denote by A
n. We have then obtained a complex with free
and ﬁnitely generated modules:
A=· · · An+1
A
n+1
An
A
n
An1 · · · .
From the above it follows that Lis the direct sum ABof the complexes A
and B.
It is clear from the construction of Athat im A
n+1 =Mn= ker A
n, so A
is exact. To see that Bis minimal, suppose xBnfor some nZ. Since
L
nmaps Gnto mNn1we can ﬁnd r1, . . . , rtmand n1, . . . , ntNn1such
that B
n(x) = Pt
i=1 rini. Now, for each ilet ni=ai+bibe the direct sum
Betti numbers 13
decomposition of niinto aiAn1and biBn1. We then have L
n(x) =
Pt
i=1 riai+Pt
i=1 ribi, and since L
n(x)Bn1it follows that Pt
i=1 riai= 0,
hence L
n(x) = Pt
i=1 ribimBn1. This shows that Bis minimal.
The above theorem together with theorem 3.2(L) now yields
Corollary 3.6. Every X∈ C(f)
( )(R)has a minimal free resolution X'
αB
CL(R)such that Bn= 0 for all n < inf X.
Proof: Since Ris assumed to be noetherian, we can by theorem 3.2(L) ﬁnd
a free resolution X'
αL∈ CL(R) such that Ln= 0 for all n < inf X. Using
the above theorem, we write Las a direct sum L=AB, where Ais exact
and Bis minimal free. We must of course have An=Bn= 0 for all n < inf X,
so in particular Bn∈ CL(R). Now, the homology isomorphism αinduces an
isomorphism H(X)
=
H(L). Since it is clear that the homology functor takes
direct sums to direct sums, H(L) = H(A)H(B) = H(B), and the restriction
of αto Bmust still induce an isomorphism H(X)
=
H(B). Thus α|Bis a
homology isomorphism, hence X'
α|B
Bis a minimal free resolution.
On the level of modules (or complexes concentrated in 0), corollary 3.6 sounds
as follows.
Corollary 3.7. Given any ﬁnitely generated module M, there exists an exact
sequence
· · · Fn
F
n
Fn1 · · · F1
F
1
F0
α0
M0,
such that the modules F0, F1, F2, . . . are ﬁnitely generated and free, and such that
F
n(Fn)mFn1for all nN.
4 Betti numbers
Throughout this section, Rdenotes a (nontrivial, unitary, commutative) local,
noetherian ring with maximal ideal mand quotient ﬁeld k=R/m. All modules
are, unless otherwise stated, assumed to R-modules.
We have ﬁnally reached the point where we are able to deﬁne the n’th Betti
number of a complex. Let X∈ C(f)
( )(R) and recall that X·Rkis the equivalence
class of complexes represented by the complex kRF, which (since kis a complex
concentrated in 0) is given by
kRF=· · · kRFn+1
kRF
n+1
kRFn
kRF
n
kRFn1 · · · ,
Betti numbers 14
for any complex F∈ CF(R) with F'X. Since equivalent complexes have
isomorphic homologies, it makes sense to consider the n’th homology module
Hn(X·Rk) as the isomorphism class of modules represented by Hn(kRF). The
ﬁeld kis a (k, R)-bimodule, so kRFis also a k-complex (since its diﬀerentials are
k-linear maps), and Hn(kRF) is a k-vector space; and if F0∈ CF(R) is another
complex with F0'X, then the R-isomorphism Hn(kRF)
=Hn(kRF0) is
also a k-isomorphism (since it is a bijection). It therefore makes sense to consider
Hn(X·Rk) as an isomorphism class of k-vector spaces.
Deﬁnition 4.1. Let X∈ C(f )
( )(R) and nZ. The n’th Betti number of Xis the
number βR
n(X) = dimk(Hn(X·Rk)).
If Mis a module, we have already deﬁned βR
0(M) to be the dimension of the
k-vector space M. For the above deﬁnition to be consistent with the previous
one, we of course hope for the two deﬁnitions to agree whenever they both apply,
that is, whenever we consider ﬁnitely generated modules. This is fortunately the
case: If M∈ Cf
0(R), then by corollary 3.6 we can ﬁnd a minimal free (and thereby
ﬂat) resolution M'
F∈ CL(R) which by corollary 3.7 is on the form
· · · Fn
F
n
Fn1 · · · F1
F
1
F0
α0
M0,
where the modules F0, F1, F2, . . . are ﬁnitely generated and free, and where
F
n(Fn)mFn1for all nN. The latter of these properties ensures that
all the diﬀerentials in the complex kRFare zero (for recall that for all nZ
there is an isomorphism kRFn
=Fnunder which kRF
ncorresponds to
F
n). Thus H0(kRF) = kRF0=F0, so the 0’th Betti number of Mis
dimkF0= rankRF0.
On the other hand, we have an exact sequence 0 ker α0F0M0, so
application of the right exact functor kRyields the exact sequence kerα0
F0M0. Here, however, the map ker α0F0must be zero, since it is
induced by the inclusion of kerα0= im F
1mF0into F0. Consequently the map
F0Mis an isomorphism, and it follows that rankRF0=βR
0(M). This shows
that the 0’th Betti number of Mis exactly what we denoted by βR
0(M).
Proposition 4.2. Suppose that X∈ C(f )
( )(R)and that X'F∈ CF(R). Then
βR
n(X)βR
0(Fn)for all nZ, and if Fhas ﬁnitely generated modules, then
βR
n(X) = βR
0(Fn)for all nZif and only if Fis minimal.
Proof: The Betti numbers of Xare by deﬁnition the dimensions as k-vector
spaces of the homologies of the complex
kRF=· · · kRFn+1
kRF
n+1
kRFn
kRF
n
kRFn1 · · · .
Betti numbers 15
Now, as observed several times before, the above complex is the same as the
complex
· · · Fn+1
F
n+1
Fn
F
n
Fn1 · · · .
The n’th Betti number βR
n(X) is therefore the dimension of
ker F
n/im F
n+1,(2)
which is a quotient of subspaces of Fnwhose dimension is βR
0(Fn). This immedi-
ately gives the inequality βR
n(X)βR
0(Fn) for all nZ.
In the case where Fhas ﬁnitely generated modules, the vector spaces involved
in (2) are ﬁnite dimensional, so the only way for the dimension of that quotient to
equal βR
0(Fn) is if ker F
n=Fnand im F
n+1 = 0. This happens for all nZif and
only if F
n= 0 for all nZ, which again is the case if and only if F
n(Fn)mFn1
for all nZ, that is, if and only if Fis a minimal complex.
If X'
L∈ CL(R) is a minimal free resolution of X, then the above theorem
states that βR
n(X) = rankRLnfor all nZ. Together with corollary 3.6 we
therefore get
Corollary 4.3. Every X∈ C(f)
( )(R)has a minimal free resolution, and all min-
imal free resolutions X'
L∈ CL(R)must satisfy that rankRLn=βR
n(X)for
all nZ. In particular, βR
n(X)N0for all nZ, and βR
n(X) = 0 for all
n < inf X.
Proposition 4.2 and corollary 4.3 show that the Betti numbers of a complex
Xare the ranks of the modules in a minimal free resolution of X, and that these
ranks are minimal among corresponding ranks of all free resolutions of X. This
justiﬁes the terminology “minimal free resolution”.
Corollary 4.4. For all X∈ C(f)
( )(R),
fdRX= pdRX= sup{nZ|βR
n(X)6= 0}.
Proof: Let N= sup{nZ|βR
n(X)6= 0}. By corollary 4.3 we can ﬁnd a
complex L∈ CL(R) with X'Land rankRLn=βR
n(X) for all nZ. We then
have sup{nZ|Ln6= 0}=N, so Nbelongs to the set over which the inﬁmum
is taken in the deﬁnition of ﬂat dimension, hence fdRXN.
If F∈ CF(R) has X'F, then there exists a complex Zand homology
isomorphisms X'
Z'
F. The complex Zmust necessarily belong to
C(f)
( )(R), and since the composition L'
X'
Zis a minimal free resolution
of Zwhile Z'
Fis a ﬂat resolution of Z, proposition 4.2 and corollary 4.3
yield
βR
n(X) = rankR(Ln) = βR
n(Z)βR
0(Fn)
Euler characteristics 16
for all nZ. Thus βR
n(X)6= 0 implies Fn6= 0, so Nsup{nZ|Fn6= 0}.
Since this holds for all F∈ CF(R) with X'F,NfdRX. In conclusion,
N= fdRX.
Exactly the same arguments hold for the projective dimension, hence N=
pdRX.
For the next theorem, recall that since Ris (nontrivial, unitary, commutative)
local and noetherian, the localization Rat a prime ideal pis of course also (non-
trivial, unitary, commutative) local and noetherian. (See for example [Eisenbud]
corollary 2.3 for the part about being noetherian.)
Proposition 4.5. Suppose that X∈ C(f )
( )(R)and let pbe a prime ideal of R.
Then X∈ C(f )
( )(R), and βR
n(X)βR
n(X)for all nZ.
Proof: As mentioned in section 2.3, exactness of the localization functor on
modules implies that the localization functor on complexes preserves homology
isomorphisms and satisﬁes that H(X)
=(H(X)) . The latter of these properties
implies that X∈ C(f )
( )(R).
By corollary 4.3 we can ﬁnd a minimal free resolution X'
L∈ CL(R) such
that βR
n(X) = rankRLnfor all nZ. Since the localization functor preserves
homology isomorphisms, there is an induced homology isomorphism X'
L.
Now, as was also mentioned in section 2.3, the localization functor is additive.
In particular it preserves products, hence (Rt)
=(R)tfor all tN. Since
each Lnis free and ﬁnitely generated, this shows that (Ln) is free and ﬁnitely
generated as an R-module with rankR(Ln) = rankRLnfor all nZ.
We have now seen that X'
Lis a free resolution, and it follows from
proposition 4.2 that for all nZ,
βR
n(X)rankR(Ln) = rankRLn=βR
n(X).
5 Euler characteristics
Throughout this section, Rdenotes a (nontrivial, unitary, commutative) local,
noetherian ring with maximal ideal mand quotient ﬁeld k=R/m. All modules
are, unless otherwise stated, assumed to be R-modules.
Let X∈ C(f )
( )(R) be a complex with pdRX < . Notice that we must have
X∈ C(f )
( )(R), since Xis equivalent to a projective module whose n’th module
is zero for n > pdRX. Now, according to corollary 4.3, βR
n(X) = 0 whenever
n < inf X, and according to corollary 4.4, βR
n(X) = 0 whenever n > pdRX.
Euler characteristics 17
Consequently βR
n(X) is nonzero only for the ﬁnitely many nthat lie between
inf Xand pdRX, so the sum Pn(1)nβR
n(X) is well-deﬁned.
Deﬁnition 5.1. Let X∈ C(f )
( )(R) be a complex with pdRX < . The Euler
characteristics of Xis the number χR(X) = Pn(1)nβR
n(X).
By the above remarks, χR(X) = PpdRX
n=inf X(1)nβR
n(X); in particular if Mis
a module, χR(M) = PpdRM
n=0 (1)nβR
n(M).
Theorem 5.2. Let X∈ C(f)
( )(R)be a complex with pdRX < , and suppose
that X'
L∈ CL(R)is a bounded free resolution by ﬁnite modules. Then
χR(X) = PnrankRLn.
Proof: By theorem 3.5 we can ﬁnd subcomplexes Aand Bof Lwith free and
ﬁnitely generated modules, such that Ais exact, Bis minimal, and L=AB. As
we saw in the proof of corollar 3.6, the homology isomorphism X'
Lrestricts
to a homology isomorphism X'
B, hence, by corollary 4.3, βR
n(X) = rankRBn
for all nZ.
Since Lis bounded, Aand Bmust also be bounded. In particular there exist
s, t Zsuch that Ais an exact sequence on the form
0At
A
t
At1 · · · As+1
A
s+1
As0.
For all nZwe have an exact sequence 0 ker A
nAn
A
n
im A
n
0, so if im A
nis free and ﬁnitely generated, then lemma 1.4 gives that im A
n+1 =
ker A
nis free and ﬁnitely generated, and that rankRAn= rankR(im A
n+1) +
rankR(im A
n). But im A
s+1 is indeed free and ﬁnitely generated since it equals
As, hence it follows by induction that all the images are free and ﬁnitely generated,
and we see that we have a telescoping sum
X
n
(1)nrankRAn=X
n
(1)n(rankRim A
n+1 + rankRim A
n) = 0.
Thus
χR(X) = X
n
(1)nrankRBn
=X
n
(1)n(rankRAn+ rankRBn)
=X
n
(1)nrankRLn
as desired.
We saw in proposition 4.5 that βR
n(X)βR
n(X) for all X∈ C(f )
( )(R) and
all prime ideals pof R. When considering Euler characteristics instead of Betti
numbers, we can do much better:
Euler characteristics 18
Proposition 5.3. Suppose that X∈ C(f )
( )(R)is a complex with pdRX < , and
let pbe a prime ideal of R. Then pdRX < and χR(X) = χR(X).
Proof: We know from proposition 4.5 that X∈ C(f )
( )(R) and βR
n(X)βR
n(X)
for all nZ, so it follows from corollary 4.4 that pdRXpdRX < .
Now, by corollary 4.3 we can ﬁnd a minimal free resolution X'
L
CL(R) such that βR
n(X) = rankRLnfor all nZ. Since Xhas ﬁnite projective
dimension, it follows from corollary 4.4 that Lis bounded, that is, L∈ CL(R). As
we saw in proposition 4.5, the free resolution X'
Linduces a free resolution
X'
Lin which each (Ln) is a free R-module with rankR(Ln) = rankRLn.
As the localization Lof course also is bounded, it now follows from theorem 5.2
that
χR(X) = X
n
(1)nrankRLn=X
n
(1)nrankR(Ln) = χR(X).
We conclude this paper by investigating properties of the Euler characteristics
of modules. First a lemma:
Lemma 5.4. If pis an associated prime of R, then depth R= 0, and every
ﬁnitely generated R-module Pwith pdRP < is free with rankRP=χR(P).
Proof: Suppose that p= AnnR(r0) is a prime for some nonzero element r0
R. In the localization R, the element r0/1 is nonzero since tr06= 0 for all
tR\p. For any element r/u p, we have that (r/u)(r0/1) = rr0/u = 0
since rp= AnnR(r0), so all elements of pare zero-divisors in R. However,
all elements outside the maximal ideal pare invertible and therefore not zero-
divisors, so pis exactly the ideal Zd Rof zero-divisors of R. Another way of
saying this is that depth R= 0. According to [FoxbyHA] lemma 17.1(8) any
ﬁnitely generated module with ﬁnite projective dimension over a ring with depth
0 is free. Consequently Pis free, hence Pis its own minimal free resolution, so
βR
n(P) = 0 for all n6= 0, and it follows that rankRP=βR
0(P) = χR(P).
The lemma is particularly useful together with proposition 5.3: By [Lang]
corollary X.2.7 we can ﬁnd an associated prime pof R. Given a ﬁnitely generated
module Mwith ﬁnite projective dimension, Mwill be ﬁnitely generated with
ﬁnite projective dimension, and also free with rankRM=χR(M) = χR(M).
Thus ﬁnding the Euler characteristics of a module comes down to ﬁnding the
rank of a free module—a task which apparently is less complicated.
For the next theorem, notice that it is clear from the deﬁnition (and the
fact that the tensor product and the homology functor are additive) that Betti
numbers are additive, that is, that βR
n(XY) = βR
n(X)+βR
n(Y) for all complexes
X, Y ∈ C(f )
( )(R) and all nZ. This implies that whenever Xand Yare complexes
Euler characteristics 19
such that the Euler characteristics is deﬁned, then the Euler characteristics is
deﬁned for XY, and χR(XY) = χR(X) + χR(Y).
Theorem 5.5. Suppose 0KMN0is an exact sequence of ﬁnitely
generated modules, each having projective dimension <. Then χR(M) =
χR(K) + χR(N).
Proof: According to [Lang] corollary X.2.7 we can ﬁnd an associated prime
pof R, and by the preceding lemma, all ﬁnitely generated modules with ﬁnite
projective dimension over Rmust be free. Thus, by exactness of the localization
functor, we get an induced exact sequence 0 KMN0 of ﬁnitely
generated free R-modules. We therefore get a splitting M=KN, and
it follows from proposition 5.3 and the additivity of Euler characteristics that
χR(M) = χR(M) = χR(K) + χR(N) = χR(K) + χR(N).
Notice how the above theorem resembles lemma 1.4. The theorem actually
also holds for complexes, but proving this requires a more constructional and
cumbersome approach than the one given above, so we leave the theorem as it is,
sacriﬁcing generality for simplicity.
Proposition 5.6. If M∈ Cf
0(R)is a ﬁnitely generated module with pdRM < ,
then 0χR(M)βR
0(M).
Proof: Using (as usual) [Lang] corollary X.2.7 we can ﬁnd an associated prime
pof R, so by lemma 5.4, Mis a free R-module with rankRM=χR(M).
By proposition 5.3 we therefore conclude that χR(M) = χR(M) = βR
0(M). It
now follows from proposition 4.5 that 0 χR(M)βR
0(M).
It is somewhat surprising that the alternating sum deﬁning the Euler charac-
teristics turns out to be nonnegative for modules. A natural question to ask next
is when the Euler characteristics of a module is zero. The question is answered
in
Theorem 5.7. If M∈ Cf
0(R)is a ﬁnitely generated module with pdRM < ,
then the following conditions are equivalent:
(i) χR(M)>0
(ii) SuppRM= Spec R
(iii) AnnRM= 0
(iv) AnnRMZd R
Proof: If M= 0 then all conditions are false, and the theorem holds. We can
therefore assume that M6= 0.
Euler characteristics 20
“(i) (ii)”. If pis a prime ideal not in SuppRM, then M= 0, and it follows
from proposition 5.3 that χR(M) = χR(M) = χR(0) = 0, which contradicts
condition (i).
“(ii) (iii)”. Suppose that AnnRM6= 0. There exists according to [Lang]
corollary X.2.7 a prime ideal associated with AnnRM, that is, we can ﬁnd a
nonzero element r0AnnRMsuch that AnnR(r0) = pis a prime ideal. Consider
now the element r0/1R, which must be nonzero since tr06= 0 for all tR\p.
For all m/u Mwe have (r0/1)(m/u) = 0 since r0m= 0, so r0/1AnnRM,
hence AnnRM6= 0. However, since pis an associated prime (of AnnRMR),
it follows from lemma 5.4 that Mis free, and the only free module with nontrivial
annihilator is the zeromodule, so M= 0, that is, p/SuppRMwhich contradicts
condition (ii).
“(iii) (iv)”. Obvious.
“(iv) (i)”. According to [Eisenbud] theorem 3.1, there are only ﬁnitely
many primes associated with R, and their union is exactly the set of zero-divisors
of R. From the assumption it therefore follows that AnnRMis contained in a
ﬁnite union of associated primes of R, hence by the prime avoidance theorem (see
for example [Eisenbud] lemma 3.3) AnnRMis contained in one of the associated
primes p. It follows (see for example [Eisenbud] corollary 2.7) that M6= 0.
On the other hand, since pis an associated prime, we have by lemma 5.4 that
Mis a free R-module with rankRM=χR(M) = χR(M). Consequently
χR(M)>0.
Notice that conditions (ii), (iii) and (iv) are formulated completely without
reference to Betti numbers or Euler characteristics; the fact that they are equiva-
lent makes sense (but would be hard to prove) in a world without Betti numbers
and Euler characteristics. A consequence of the theorem is given in the corollary
below, which can also be formulated without the mention of Betti numbers or
Euler characteristics, simply by replacing “χR(K)>0” by one of its equivalent
conditions.
Corollary 5.8. Suppose that Kis a ﬁnitely generated submodule of a free and
ﬁnitely generated module F, and that pdRK < . Then either K= 0 or
χR(K)>0.
Proof: Suppose that the latter condition fails to hold. Then by the above
theorem, AnnRK*Zd R, so we can ﬁnd rRsuch that rannihilates Kbut
is not a zero-divisor of R. Let (f1, . . . , fn) be a basis for F, and let kbe any
element of K. We can then ﬁnd r1, . . . , rnRsuch that k=r1f1+· · · +rnfn,
and we must therefore have rr1f1+· · · +rrnfn=rk = 0. Since the fi’s formed
a basis, it follows that rr1=· · · =rrn= 0, but rwas not a zero-divisor, so we
conclude that r1=· · · =rn= 0, that is, k= 0. Consequently K= 0.
Theorem 5.7 answered the question of when the Euler characteristics χR(M)
of a module Mreaches its lower bound of 0. The last theorem of this paper
References 21
answers the question of when the Euler characteristics reaches its upper bound
of βR
0(M).
Theorem 5.9. If M∈ Cf
0(R)is a ﬁnitely generated module with pdRM < ,
then χR(M) = βR
0(M)if and only if Mis free.
Proof: If Mis free, then Mis its own minimal free resolution, and it follows
that βR
n(M) = 0 for all n6= 0, hence χR(M) = βR
0(M).
Suppose conversely that χR(M) = βR
0(M). By corollary 3.7 we can ﬁnd an
exact sequence
· · · Fn
F
n
Fn1 · · · F1
F
1
F0
α0
M0,
such that the modules F0, F1, F2, . . . are ﬁnitely generated and free, and such
that F
n(Fn)mFn1for all nN. In particular, the module ker α0= im F
1
is ﬁnitely generated and contained in mF0. Furthermore, the exact sequence
· · · FnFn1 · · · F1ker α00 is a minimal free resolution of ker α0,
so it follows from corollary 4.3 and corollary 4.4 that pdR(ker α0)pdRM < .
We now have an exact sequence
0ker α0F0
α0
M0,(3)
and it follows from theorem 5.5 that χR(F0) = χR(ker α0) + χR(M). But F0is
free, so χR(F0) = βR
0(F0) = rankRF0by the previous, and χR(M) = βR
0(M) by
assumption, so the equation now sounds
rankRF0=χR(ker α0) + βR
0(M) (4)
Application of the right exact functor kRto the exact sequence (3) yields
the exact sequence ker α0F0M0 of k-vector spaces. Here, the ho-
momorphism ker α0F0is induced by the inclusion of kerα0mF0into F0
and therefore zero. Thus there is an isomorphism F0
=M, and it follows that
rankRF0=βR
0(M). Equation (4) therefore yields χR(ker α0) = 0, and since ker α0
is a ﬁnitely generated submodule of the free module F0and has pdR(ker α0)<,
it follows from corollary 5.8 that kerα0= 0. Thus F0
=M, so Mis free.
References
[Eisenbud] David Eisenbud: Commutative Algebra with a View Toward Alge-
braic Geometry, Springer 1995.
[FoxbyB&E] Hans–Bjørn Foxby: Bettital & Eulerkarakteristik, discussion paper
(in Danish) providing the basis for this course project, University
of Copenhagen 2002.
References 22
[FoxbyHA] Hans–Bjørn Foxby: Homologisk Algebra, lecture notes (in Danish)
for the graduate course Mat4ha, University of Copenhagen 1999.
[FoxbyHHA] Hans–Bjørn Foxby: Hyperhomological Algebra & Commutative
Rings, unpublished notes, University of Copenhagen 1998.
[H&S] P.J. Hilton & U. Stammbach: A Course in Homological Algebra,
Second Edition, Springer 1997.
[Lang] Serge Lang: Algebra, Third Edition, Addison–Wesley 1999.