3. PREFACE
The book “A study on noetherian rings and modules” is intended for the PG
students in kerala university. In this book all the topic have been deal within a simple
and lucid manner. A sufficiently large number of problems have been solved by
studying this book , the student is expected to understand the concept of noetherian
rings and modules,filtered and graded modules. To do more problems involving the
noetherian rings and modules, filtered and graded modules.
Suggestion for the further improvement of this book will be highly
appreciated.
Thara Thankan
4. CHAPTER I
Definition 1.1
A group <G,*) is a set G, together with a binary operation * on G, such that the
following axioms are satisfied.
1. The binary operation * is associative
2. There is an element e in G such that e*x = x * e = x for all x+G this element e is an
identify element for * on G)
3. For each a in G, there is an element a’ in G with the property that a’ * a = a*a’=e
(The element a’ is an inverse of a with respect to *)
Definition 1.2
Subset of a group is called subgroup.
Definition 1.3
Let (S,.) be a semi group. If there is an element e in S sum that ex=x=xe ∀ , ,
that is a communicative group is called abelian group.
Definition 1.5 (Rings)
By a ring mean a nonempty set R with two binary operations + and . , called addition
and multiplication (also called product) respectively such that,
i. (R, +) is an additive abelian group.
ii. (R, .) is a multiplicative semi group.
iii. Multiplication is distributive over addition that is for all a, b, c
a.(b+c) = a.b + a.c
(a+b) . c = a.c + b.c
Definition 1.6
A commutative ring is a ring R in which multiplication is commutative, that is ab = ba for
all a, b, . If a ring is not commutative it is called non commutative.
5. Definition 1.7
Let (R,+, .) be a ring, and let S be a nonempty subset of R. Then S is called a sub ring
of (S, +, .) is it self a ring.
Definition 1.8
Let S be a subset of a ring R. Then the smallest subring of R containing S in called
the sub ring generated by S.
Definition 1.9
If there exists a positive integer n such that na = 0 for each element a of a ring R, the
smallest such positive integer is called the character of R. If no such positive integer exist R
is said to have characteristic Zero. The characteristic of R denoted char R.
Definition 1.10
A ring ( or left) ideal A in a ring R is called nilpotent. If An
= (0) for some positive
integer n.
Definition 1.11
An element a in a ring R is called nilpotent if there exists a positive integer n such
that an
= 0.
Definition 1.12
An element e in a ring R is called idempotent if e = e2
.
Definition 1.13
A ring R is called a simple ring if the only ideals of R are the Zero ideal and R itself.
Definition 1.14
A right (or left) ideal a in a ring R is called a nil ideal if each element of A is
nilpotent.
Definition 1.15
A ring R whose non – Zero elements form a group under multiplication is called a
division ring, if an addition, R is commutative, then R is called a filed.
6. Definition 1.16
A ring R is called an integral domain if xy=0, , implies X=0 or Y=0
Definition 1.17
Homomorphism
A map ∅ of group G in to a group G’ is a homomorphism if (ab)∅ = ∅. ∅ for all
a, b
Example 1.18
Let ∅: → G1
be a map defined by a∅ = 2a
, where G=R the group of reals under
addition and G=R+ the group of positive reals under multiplication.
Let a, b =R
(a+b) ∅=2a+b
=2a
.2b
=a∅. ∅
Definition 1.19
Let f be a mapping from a ring R into a ring S suchthat
i. f (a+b)=f(a)+f(b), a, b
ii. f(ab) = f (a), f(b), a, b
Then f is called a homomorphism of R info S.
Definition 1.20 (Zorn’s Lemma)
If energy chain c in a poset (s,=) has an upper bound in S, then (S,=) has a maximal
element.
Definition 1.21 (Vector Space)
Let F be a field. A vector space over f (or f-Vector Space) consists of an abelian
group V under addition together with an operation of scalar multiplication of each element
of V by each element of F on the left, such that for all a, b and ∝ the following
condition are satisfied.
1. ∝
2. a(b∝) = ( ) ∝
3. ( + ) ∝= ( ∝) + ( ∝)
4. (∝ + ) = ∝ +
7. 5. 1 ∝=∝
The elements of V are vectors and the elements of F are scalars. When only one field F is
under discussion, me drop the reference to F and refer to a vector space.
Definition 1.22
Let A1, A2 …………….. An be a family of right ideals in a ring R. Then the smallest
right ideal of R containing each Ai, 1=i=2 (that is intersection of all the right ideals in R
containing each Ai), is called the sum of A1, A2 …………An
Definition 1.23
A sum A = ∑ of right (or left) ideals in a ring R is called a direct sum if each
element a is uniquely expressible in the form ∑ , , 1 = = If the sum A=Ai
is a direct sum, me ………….it us
A=A1+A2+……………….+An=∑
Definition 1.24
An ideal P in a ring R is called a prime ideal if it has the following property. If A and
B are ideals in R such that ABcP, then AcP or BcP
Definition1.25 (Module)
Let A be ring. A left module over A, or left A module m is an abelian group, usually
written additively, together with an operation of A on, such that, for all a, b ∈ and
x,y∈ are have.
Let M be an A module. By a sub module N of M we mean an additive subgroup such
that ANCN. Then N is a module (with the operation induced by that of A on M)
Definition 1.26
A non empty subset N of an R module M is called an R Sub module ( or simply sub
module) of M if
i. a-b N for all a, b N
ii. ra N for all a N, r
Clearly, (o) or simply o and M are R Sub modules called trivial sub modules. In case
R is a field, N is called a subspace of M.
8. Definition 1.27
An R module M is called a cyclic module if M=(x) for some X .
Definition 1.28
Let (Ni), 1=i=k be a family of R sub modules of a module M. Then the sub module
generated by VkNii=1, that is, the smallest sub module containing the sub modules Ni =i=k
is called the sum of sub modules Ni, 1=i=k and is denoted by ∑
Definition 1.29
An R module M is called completely reducible if M = ∑ ∝∝ , where Ma are
simple R Sub modules.
Definition 1.30
Let R be a commutative ring with identity, and M a until R module. The annihilator
of M, written Ann (M) is the set
{rtR: Mr = 0 for all m }
The sub modules of generated by m1)………………mn is the set <m1, …….mr> =
{m1r1 +………+mnrn:r1…………….rn+R}
We say that M is finitely generated if there is a finite set of elements of M which
generates M. We say that M is cyclic if it is generated by just one element.
9. CHAPTER 2
NOETHERIAN RINGS AND MODULES
Definition 2.1
Let A be a ring and M a module (ie, a left A- module). we shall say that M is
Noetherian if it satisfies any one of the following three conditions.
1. Every sub module of M is finitely generated.
2. Every ascending sequence of sub modules of M.
M1 M2 M3 ………………………
such that Mi+Mi+1 is finite
3. Every non-empty set S of sub modules of M has a maximal element .
We shall now prove that the above three conditions are equivalent.
(1) (2) suppose we have an ascending sequence of sub modules of M as
above Let N be the union of all the
Mi (i =1,2 ……). Then N is finitely generated say by elements x1,
………….xr and each generator is in some Mi. Hence there exists an index
j such that
x1………………………..xrMj
Then
<x1,…………………xr>CMjCN=<x1…………….xr>,
whence equality holds and our implication is proved.
(2) (1) Let N0 be an element of S. If N0 is not maximal, it is properly contained in a
submodule N2 Inductively,if we have found Ni which is not maximal it is contained properly
in a submodule Ni+1. In this way we could construct an infinite chain, which is impossible.
(3) (1) Let N be a sub module of M. Let a0 . If N≠<a0>, then there exists an element
a1 which does not lie in <a0>. Proceeding inductively, we can find an ascending sequence
of sub modules of N, namely.
<a0><a0,a1><a0,a1,a2>…………….
10. Where the inclusion each time is proper. The set of these submodules has a maximal
element, say a submodules <a0,a1………ar>, and it is then clear that this finitely generated
submodule must be equal to N.
Preposition 2.2
Let M be a Noetherian A module. Then every sub module and every factor module of
M is Noetherian.
Proof:
Our assertion is clear for sub module. For the factor module, let N be a submodule
and f: M→ / the canonical homomorphism.
Let M 1 M 2 ……... be an ascending chain of submodules of M/N and let Mi=f-1
( M i).
Then M1 M2 M3……………. is an ascending chain of sub modules of M1 which must
have a maximal element, say Mr, so that Mi=Mr for r ≥ i. Then f (Mi) = M i and our
assertion follows.
Preposition 2.3
Let M be a module N a submodule. Assume that N and M/N are Noetherian. Then M
is Noetherian.
Proof:
With every submodule L of M we associate the pair of modules
L→(L∩ , ( + )/
We contend : If E F are two submodule of M suchthat their associated pairs are equal,
then E=F. To see this, let X . By the hypothesis that (E+N)/N=(F+N)/N there exist
elements u, v and y suchthat y + u = x+v Then
x-y = u-v ∩ = ∩
Since y , it follows the x and are contention is proved. If we have an ascending
sequence
E1 E2 ……………………
Then the associated pairs form an ascending sequence of sub modules of N and M/N
respectively, and these sequence must stop. Hence our sequence
11. E1 E2…………..also stops, by our proceeding contention preposition 2.2 and 2.3 may be
summarized by saying that in an exact sequence → 0 MMM is Noetherian iff
M and M are Noetherian.
Corollary 2.4
Let M be a module, and let N, N be submodules. If M=N+ N and if both N,
N are Noetherian, then M is Noetherian .A finite direct sum of Noetherian modules is
Notherian.
Proof:
We first observe that the direct product N * N is Noetherian since it contains N as a
submodule whose factor module is isomorphic to N’ and preposition 2.2 applies. We have a
subjective homomorphism
N x N →
Suchthat the pair (x, x) with x ∈N and x∈ N maps on x+x. By preposition2.2,it
follows that M is Noetherian.
Definition 2.5
A ring A is called Noetherian if it is Noetherian as a left module over itself. This
means that every left ideal is finitely generated.
Preposition 2.6
Let A be a Noetherian ring and let M be a finitely generated module. Then M is
Noetherian.
12. Proof:
Let x1………….xn be generators of M. There exists a homomorphism.
f : A* A*……………..*A →
of the product of A with itself n times such that
f(a1, …………….an) = a1x1+…………..+anxn
This homomorphism is subjective. By the corollary of the proceeding preposition, the
product is Noetherian, and hence M is Noetherian by preposition 2.2
Preposition 2.7
Let A be a ring which is Noetherian and let ∅: → be a Surjective ring
homomorphism. Then B is Noetherian.
Proof:
Let b1 ………… bn …………..be an ascending chain of left ideals of B and let
ai =∅-1
(bi
) Then the ai form an ascending chain of left ideals of A which must stop, say at
ar since ∅ (ai)for all bi, our proposition is proved.
Definition 2.1.1 (Associated primes)
We let A be a commutative ring. Modules and homomorphism are A- modules and
A- homomorphism unless otherwise specified.
Preposition 2.1.2
Let S be a multiplicative subset of A , and assume that S does not contain O. Then
there exists an ideal of A which is maximal in the set of ideals not intersecting S, and any
such ideal is prime.
13. Proof:
The existence of such an ideal P follows from Zorn’s lemma ( the set of ideals not
meeting S is not empty, because it contains the zero ideal, and is clearly inductively
ordered). Let P be the maximal in the set. Let a,b∈ , ∈ , but ab and bp. By
hypothesis the ideal (a, p) and (b, p) generated by a and p meet S, and there exist therefore
elements s, pppAxxccSs ,,,,,, Such that
S = ca+xp and pxbcs
Multiplying these two expressions we obtain
pabccss
with some pp whence we see that s slies in p. This contradicts the fact that P does not
intersect S, and proves that p is prime.
Lemma 2.1.3
Let X be an element of module M, and let a be its annihilator. Let P be a prime
ideal of A. Then (Ax)p ≠0 if and only if P contains a.
Proof:
The lemma is an immediate consequence of the definitions.
Let a be an element of A. Let M be a module. The homomorphism
X→ Mxax , will be called the principal homomorphism associated
with a, and will be denoted by aM . We shall say that aM is locally nilpotent if
for each x∈ there exists an integers n(x)≥1 such that ( )
= 0. The
condition implies that for energy finitely generated submodule N of M, there
exists an integer n≥1 such that an
N=0. We take for n the largest power of a
annihilating a finite set of generators of N. Therefore, If M is finitely
generated, aM is locally nilpotent if and only if it is nilpotent.
14. Preposition 2.1.4
Let M be a module ≠0. Let p be a maximal element in the set of ideals which are
annihilators of elements ∈ , ≠ 0, Then P is prime.
Proof:
Let P be the annihilator of the element ≠ 0, Then P≠A. Let a, b ∈ , ∈ ,
pa Then ≠ 0. But the ideal (b, p) annihilates ax, and contains P. Since P is maximal, it
follows that ∈ , and hence P is prime.
Corollary 2.1.5
If A is Noetherian and M is a module ≠ 0, then there exists a prime associated with
M.
Proof:
The set of ideals as in preposition 2.1.4 is not empty since M≠0, and has a maximal
element because A is Noetherian.
Corollary 2.1.6
Assume that both A and M are Noetherian, M≠0. Then there exists a sequence of sub
modules.
M=M1 M2 …………….. Mr = 0
Such that each factor module Mi/Mi+1 is isomorphic to A/pi for some prime Pi.
Proof:
Consider the set of submodules having the property described in the corollary. It is
not empty, since there exist an associated prime P of M and if P is the annihilator of x, then
Ax≈ . Let N be a maximal element in the set. If N≠M, then by the proceeding argument
applied to M/N there exists a sub module of N of containing N such that N /N is
isomorphic to A/P for some P, and this contradicts the maximality of N.
15. Preposition 2.1.7
Let A be Noetherian, and a∈ . Let M be a module. Then aM is injective if and only
if a does not lie in any associated prime of M.
Proof:
Assume that aM is a not injective, so that ax=0 for some ∈ , ≠ 0. By corollary
2.1.5 there exist an associated prime P of Ax, and a is an element of P. Conversely, if aM is
injective, then a cannot lie in any associated prime because a does not annihilate any non-
zero element of M.
Preposition 2.1.8
Let N be a sub module of M. Energy associated prime of N is associated prime of N
is associated with M also. An associated prime of M is associated with N or with M/N.
Proof:
Let P be an associated prime of M, and say p is the annihilator of the element ≠ 0.
If 0 NAx then A is isomorphic to a submodule of M/N and hence p is associated with
M/N. Suppose 0 NAx Let = ∈ with a∈ and y≠0. Then P annihilates y. We
claim P=ann(y). Let ∈ and by =0. Then ba but aP, so b∈ . Hence P is the
annihilator of y in A, and therefore is associated with N.
Definition 2.2.1( Primary Decomposition)
A primary decomposition of an ideal I in A is an expression of I as a finite
intersection of primary ideals ,say I=
n
i
I
0
i
A primary ideals in a A is in somesense a generalization of a primary number, an ideal I in
a ring capitalize primary is I A and is xyI either Ix or yn
I for some n>0
16. Example 2.2.2
Let m be a maximal ideal of A and let q be an ideal of A such that mk
q for some
positive integer k. Then q is primary and m belongs to q.
The above conclusion is not always true if m is replaced by some prime ideal P for
instance, let R be a factorial ring with a prime element t. Let A be the subring of
polynomials f(x)← [ ] such that
f(x) =a0+a1x+………………………
with a1, divisible by t. Let P = ( tX, X2
).Then P is prime but
P2
= (t2
x2
, tx3
,x4
)
is not primary, as one sees because X2
P2
but tk
P2
for all K≥1, yet t2
x2
P2
Preposition 2.2.3
Let M be module, and Q1……….Qr submodules which are P- Primary for the same
prime P. Then Q1 ………. Qr is also P-Primary.
Proof:
Let Q=Q …………… Qr. Let ap. Let ni be such that (aM,Qi)ni
= 0 for each i=1,
………..r and let n be the maximum of n1………nr. Then an
M/Q=O
so that aM/Q is nilpotent. Conversely, Suppose aP. Let , jfor some j. Then an
xQj
for all positive integers n, and consequently aM/Q is injective.
Remark 2.2.4
Let N be a sub module of M when n is written as a finite intersection of primary sub
modules, say
N=Q1 ……….. Qr
We shall call this is a primary decomposition of N.
Preposition 2.2.5
Let N be a sub module of M and let,
17. N =Q1 …….. Qr=Q1 ……….. Qs is the same. If {P1,………Pm} is the set of isolated
primes belonging to these decompositions, then Qi=Qi for i=1, ……….m in other words the
primary modules corresponding to isolated primes are uniquely determined.
Proof:
The uniqueness of the number of terms in a reduced decomposition and the
uniqueness of the family of primes belonging to the primary components will be a
consequence of theorem 2.2.6 below.
There remains to prove the uniqueness of the primary module belonging to an
isolated prime, say P1. By definition, for each j=2, ……r there exists ajpj and aj pj. Let
a=a2, ………..ar be the product. Then apj for all j>1, but
apj. We can find an integer n≥1 such that an
M/Qi=0 for j=2,……….r.
Let,
N1=set of such that an
xN We contend that Q1=N1. Let xQ1 Then
an
xQ1 ……… Qr=N so 1 conversely, let 1, so that an
, and in
particular an
1 Since 1 we know by definition that aM/Q1 is injective. Hence ∈ 1
Theorem 2.2.5
Let M be a Noetherian Module. Let N be a submodule of M. Then N admits a
primary decomposition.
Proof:
We consider the set of sub modules of M which do not admit a primary
decomposition. If this set is not empty, then it has a maximal element because M is
Noetherian. Let N be this maximal element. Then N is not primary, and there exists ∈
such that aM/N is neither injective nor nilpotent.
The increasing sequence of modules
Ker aM/N ker a2
M/N Ker a3
M/N ………………………..
18. Stops, say at ar
M/N. Let ∅: / → / be the endomorphism ∅ = ar
M/N. Then ker∅2
=ker∅.
Hence 0=ker∅ ∩Im∅ in M/N, and either the kernel nor the image of ∅ 0. Taking the
inverse image in M, we see that N is the intersection of two sub modules of M, unequal to
N. We conclude from the maximality of N that each one of these sub modules admits a
primary decomposition, and there fore that N admits one also, contradiction.
Definition 2.3.1(NAKAYAMA’S LEMMA)
Let M be a finitely generated A module and I an ideal of A
contained in the Jacobson radical R (A). Then IM=0 M=0
Lemma 2.3.2
Let a be an ideal of A which is contained in every maximal ideal of A. Let E be a
finitely generated A-module. Suppose that aE=E. Then E={0}.
Proof:
Induction on the number of generators of E. Let X1,…………Xs be generators of E.
By hypothesis, there exist elements a1…………as such that
Xs=a1x1+……as xs ,
So there is an element a (namely as) in a such that (1+a) xs lies in the module generated by
the first s-1 generators. Further more 1+a is a unit in A, otherwise 1+a is contained in some
maximal ideal, and since a lies in all maximal ideals, we conclude that 1 lies in a maximal
ideal, which is not possible. Hence xs itself lies in the module generated by s-1 generators
and the proof is complete by induction.
Lemma 2.3.3
Let A be a local ring, let E be a finitely generated A-module, and F a sub module. If
E =F + mE , then E=F.
19. Proof:
Apply Lemma 2.3.2 to E/F
Lemma 2.3.4
Let A be a local ring. If x1………….xn are generators for E mod mE then they are
generators for E.
Proof:
ℎ 1 … … . .
Theorem 2.3.5
Let A be a local ring and E a finite projective A-module. Then E is free. In fact, if
x1……xn are elements of E whose residue classes x 1,………. x n are a basis of E/mE over
A/m, then x1……xn are a basis of E over A. If x1……xr are such that x 1,………. x r are
linearly independent over A/m, then they can be completed to a basis of E over A.
Proof:
By induction on r, by one of the definitions of projective modules, there is a module
F such that E F is free, say with basis (ej). Suppose first r=1. If a is such that ax1=0,
then a would be a divisor of 0 in the free module EF, which implies that a=0. Now let
a1…………ar be such that
a1x1+……………….arxr=0
There are elements bij such that
xi=∑ bijej whence ∑ ( ∑ ai bij)ej=0.
Therefore
∑ aibij=0 for all j=1, ………………s.
Since xr mE it follows that brj m for some j and this brj is therefore a unit.
Dividing by brj we can then find elements C1, ………..Cr-1 such that
20. ar=c1a1+………+Cr-1ar-1
We multiply this relation by xr on the right and substitute in the original relation ∑ aixi =0
to find
0=∑ aixi=a1(x1+c1xr)+………….+a r-1(x r-1+c r-1xr).
But x1+c1xr…………….,xr-1+cr-1xr are linealy independent mod mE, and so by induction we
conclude ai=0 for i=1,…….r-1. Hence ar=0.
Let E be a module over a local ring A with maximal ideal m. We let E(m)=E/mE. If
f: E→ is a homomorphism, then f induces a homomorphism.
f(m) : E(m) → ( ).
If f is subjective, then if follows trivially that f(m) is surjective.
Preposition 2.3.6
Let f: E → be a homomorphism of modules, finite over a local ring A
Then:
(i) If f(m) is subjective, so is f
(ii) Assume f is injective. If f(m) is subjective then f is an homomorphism.
(iii) Assume that E, F are free, If f(m) is injective (resp. an isomorphism) then f is
injective (resp. an isomorphism).
Proof:
The proofs are immediate consequences of Nakayama’s Lemma. The For instance,
in the first statement, consider the exact sequence.
E→ → /Im → 0
and apply Nakayama to the term on the right.
In (iii), use the lifting of bases as in Theorem 2.3.5
21. CHAPTER 3
FILTERED AND GRADED MODULES
Definition 3.1
Let A be a commutative ring and E a module. By a filtration of E one means a
sequence of sub modules.
E = E0 E1 E2 ………… En ………….
strictly speaking, this should be called a descending filtration.
Example 3.2
Let a be an ideal of a ring A, and E an A-module. Let
En=an
E
Then the sequence of sub modules {En} is a filtration.
More generally, let {En} be any filtration of a module E. We say that it is an a-
filtration if aEn En+1 for all n. The preceeding example is an a filtration. We say that an
a-filtration is a-stable or stable if we have aEn = En+1 for all n sufficiently large.
Preposition 3.3
Let {En} and { En } be stable a- filtrations of E. Then there exists a positive integer
d such that
E n+d E n and E n+d En for all n≥0.
Proof:
It suffices to prove the preposition when En = an
E. Since aEn En+1 for all n, we
have an
E En. By the stability hypothesis, there exists d such that
En+d = an
Ed an
E .
Definition 3.4:
A ring A is called graded (by the natural numbers) if one can write A as a direct sum
(as abelian group)
22. A=
n=0
such that for all integers m,n≥0 me have AnAm An+m . It follows in particular that A0 is a
subring, and that each component An is an A0- module.
Definition3.5:
Let A be a graded wing. A module E is called a graded module if E can be
expressed as a direct sum (as abelian group)
E=
such that AnEm En+m. In particular En is an A0-module. Elements of En are then called
homogenous of degree n.
Example 3.6
Let K be a field, and let X0,………Xr be independent variables. The polynomial ring
A= K[X0……….Xr] is a graded algebra with K=A0. The homogeneous elements of degree n
are the polynomials generated by the monomials in X0,……….Xr of degree n, that is
X0
do
……..Xr
dr
with ∑ =
An ideal I of A is called homogeneous if it is graded, as an A module. Then the factor ring
A/I is also a graded ring.
Definition 3.7
Let A be a ring and a an ideal. We view A as a fitered ring, by the power an
. We
define the first associated graded ring to be
Sa(A) = S=
+
= 0
.
Lemma 3.8
Let A be a Noetherian ring, and E a finitely generated module, with an a -
filtration. Then Es is finite over S if and only if the filtration of E is a-stable.
23. Proof:
Let,
Fn=
+
= 0
and let,
Gn = E0 ………..En aEn a2
En a3
En …………
Then Gn is an S-Sub module of Es and is finite over S since Fn is finite over A. We have,
Gn Gn+1 and Gn = Es
Since S is Noetherian, we get:
Es is finite over S Es =GN for some N
EN+m =am
E for all m ≥0
the filtration of E is a –stable
This proves the lemma.
Definition3.9
Let A be a ring and a an ideal of A. We define the second associated graded ring
gra (A)=
+
= 0
an
/an+1
Preposition 3.10
Assume that A is Noetherian, and let a be an ideal of A. Then gra(A) is Noetherian.
If E is a finite A- module with a stable a -filtration, then gr(E) is a finite gra(A) -module.
Proof:
Let x1……xs be generators of a. Let x i be the residue class of xi in a/a2
. Then
gra (A) = (A/a)[ x 1,……… x s]
24. is Noetherian, thus proving the first assertion. For the second assertion we have for
some d,
Ed+m=am
Ed, for all m≥0
Hence gr(E) is generated by the finite direct sum.
gr(E)0 …………..gr(E)d
But each gr(E)n= En/En+1 is finitely generated over A, and annihilated by a, so is a finite
A/a- module. Hence the above finite direct sum is a finite A/a-module, so gr(E) is a finite
gra(A) –module.
Definition 3.1.1(. The Hilbert Polynomial)
A graded Noetherian ring together with a finite graded A-module E, so
A=
E=
n=0 and n=0
We have seen in preposition 3.4 that A0 is Noetherian, and that A is a finitely
generated A0-algebra. The same type of argument shows that E has a finite number of
homogeneous generators, and En is a finite A0-module for all n≥0.
Let ∅ be an Euler- poin care Z-valued function on the class of all finite A0-modules.
Theorem 3.1.2 (Hilbert-Serre)
Let S be the number of generators of A as A0-algebra. Then P(E, t) is a rational
function of type.
P(E,t) =
( )
( )
With suitable positive integers di and f (+) ∈ [ ].
Proof:
Induction on S. for S=0 the assertion is trivially true. Let S≥1. Let
= [ … … … … . ] deg. = ≥ 1. Multiplication by Xs on E gives rise to an exact
sequence.
→ → → + → + → 0
25. Let
nKK and nLL
Then K, L are finite A-module (being sub modules and factor modules of E), and are
annihilated by Xs, so are in fact graded [ ………….. ] modules. By definition of an Euler
poin care function we get
∅( ) − ∅( ) + ∅( ) − ∅( ) = 0 . Multiplying by tn+ds
and summing
over n, we get
(1 − ) ( , ) = ( , ) − ( , ) + ( ), Where g(t) is a polynomial in Z(t).
The theorem follows by induction.
Remark 3.1.3
In theorem 3.1.2, if = [ … … . . ] then di = degxi as shown in the proof. The
next result shows what happens when all the degree are equal to 1.
Theorem 3.1.4
Assume that A is generated as an A0 – algebra by homogeneous elements of degree1.
Let d be the order of the pole of P (E,t) at t=1. Then for all sufficiently large n, ∅( ) is a
polynomial in n of degree d-1 (for this statement, the Zero polynomial is assumed to have
degree-1)
Proof:
By theorem 3.1.2, ∅( ) is the coefficient of tn
in the rational function
( , ) = ( )/(1 − )
Cancelling powers of 1-t, we write ( , ) =
( )
( )
, and ℎ( ) ≠ 0 with ℎ( ) ∈ ( ). Let
ℎ( ) = ∑ .
We have the binomial expansion
(1 − ) = (
+ − 1
− 1
)
For convenience we let
−1
= 0 for ≥ 0 and
−1
=1 for n = -1 we then get
∅( ) = ∑ (
+ − − 1
− 1
) for all n≥m
26. The sum on the right hand side is a polynomial in n with leading coefficient.
( )
( − 1)1
= 0
This proves the theorem.
The polynomial of theorem 3.1.4 is called the Hilbert polynomial of the graded
module E with respect to ∅.
We now put together a number of results of this chapter, and give an application of
theorem 3.1.4 to certain filtered modules.
Let A be a Noetherian local ring with maximal ideal m. Let q be an m-primary ideal.
Then A/q is also Noetherian and local. Since some power of M is contained in q, it follows
that A/q has only one associated prime, viewed as module over itself, namely m/q itself
similarly, If M is a finite A/q-module then M has only one associated prime and the only
simple A/q- module is in fact an A/m module which is one-dimensional. Again since some
power of m is contained in q, it follows that A/q has finite length, and M also has finite
length. We now use the length function as an Euler poin care function in applying Theorem
3.1.4
Theorem 3.1.5
Let A be a Noetherian local ring with maximal ideal m. Let q be an m-primary ideal,
and let E be a finely generated A module, with a stable q-filtration. Then
i. E/En has finite length for n≥0
ii. For all sufficiently large n, this length is a polynomial g(n) of degree ≤ S, where S
is the least number of generators of q.
iii. The degree and leading coefficient of g(n) depend only on E and q, but not on the
chosen filtration.
Proof:
Let,
= ( ) = /
27. Then gr(E)= / is graded G module and = / . By preposition 3.11, G is
Noetherian and gr(E) is a finite G module. By the remarks proceeding the theorem, E/En has
finite length, and if ∅ denotes the length, then
∅ = ∑ ∅ ( )
If … … … generate q, then the image –
1
………
−
in q/q2 generate gas A/q algebra and
each – has degree 1. By theorem 3.1.4 we see that
∅ = ℎ )
is a polynomial in n of degree ≤ − 1 for sufficiently large n, since
∅ − ∅ = ℎ( ),
is follows by lemma 3.1.6 below that ∅ is a polynomial g(n) of degree ≤ for all
integer n. The last statement concerning the independent of the degree of g and its leading
coefficient from the chosen filtration follows immediately from preposition 3.3 and will be
left to the reader. This conclude the proof.
From the theorem, we see that there is a polynomial XE,q such that
XE,q(n) = length (E/qn
E)
for all sufficiently large n. If E=A, then XA,q is usually called the characteristic polynomial
of q. In particular, we see that
XA,q (n) = length (A/qn
)
for all sufficiently large n.
We shall now study a particularly important special case having to do with
polynomial ideals. Let K be a field, and let
= … … … … … .
be the polynomial ring N+1 variable. Then A is graded, the elements of degree n being the
homogenous polynomials of degree n. We let a be a homogenous ideal of A, and for an
integer ≥ 0 we define
28. ∅( ) =
∅( , ) =
( , ) = = −
= ∅( ) − ∅( , )
As earlier in this section, An denotes the K space of homogeneous elements of degree n in
A, and similarly for an.Then we have
∅( ) =
+
we shall consider the binomial polynomial (1) =
( )………….( )
= + lower terms
If f is a function, we define the difference function ∆ by
∆ ( ) = ( ) − ( )
Then one verifies directly that
(2) ∆ = (
− 1
)
Lemma 3.1.6
Let ∈ ( ) be a polynomial of degree d with rational coefficients. (a) If ( )
∈
for all sufficiently large integers n, then there exist integers … … … …
Such that
( ) = +
− 1
+ ⋯ … … … +
In particular, ( ) ∈ for all integers n. (b) If : → is any function, and if there
exists a polynomial ( ) ∈ ( ) such that ( ) and ∆ ( ) = ( ) for all n sufficiently
large, then there exists a polynomial P as in (a) such that ( ) = ( ) for all n sufficiently
large.
Proof:
We prove (a) by induction. If the degree of P is 0, then the assertion is obvious.
Suppose deg ≥ 1. By (1) there exist rational numbers , … … . . such that P(T) has the
expression given in (a). But ∆ has degree strictly smaller than deg P using (2) and
29. induction, we conclude that , … … . . must be integers. Finally Cd is an integer
because ( ) ∈ for n sufficiently large. This proves (a).
As for (b) using (a), we can write
( ) =
− 1
+ ⋯ … … … … +
with integers C0……….Cd-1 . Let P1 be the “ integral” of Q, that is
( ) = + ⋯ … + (
1
), So ∆ .Then ∆( − )( ) = 0 for all n
sufficiently large. Hence (f-P1) (n) is equal to a constant Cd for all n sufficiently large, so we
let P=P1+Cd to conclude the proof.
Preposition 3.1.7
Let a, b be homogenous ideals in A. Then
∅( , + ) = ∅( , ) + ∅( , ) − ∅( , )
( , + ) = ( , ) + ( , ) − ( , )
Proof:
The first is immediate, and the second follows from the definition of X.
Theorem 3.1.8
Let F be a homogeneous polynomial of degree d. Assume that F is not a divisor of Zero
mod a, that : If ∈ , ∈ , then ∈ , Then
, + ( ) = ( , )
− ( , )
Proof:
First observe that trivially
∅ , ( ) = ∅( − )
because the degree of a product is the sum of the degree. Next, using the hypothesis that F is
not divisor of 0 mod a, we conclude immediately.
∅ , ( ) = ∅( − , )
Finally, by preposition 3.1.7 (the formula for X), we obtain
, + ( ) = ( , ) + , ( ) − ( , ( ))
= ( , ) + ∅( ) − ∅( , ( ))
30. −∅( ) + ∅( , ( ))
= ( , ) − ∅( − ) + ∅( − , )
= ( , ) − ( − , )
thus proving the theorem
We denote by m the maximal ideal m = (X0,………XN) in A. We can m the
irrelevant prime ideal. An ideal is called irrelevant if some positive power of m is contained
in the ideal. In particular, a primary ideal q is irrelevant if and only if m belongs to q. Note
that by the Hilbert mustellensat Z, the addition that some power of m is contained in a is
equivalent with the condition that the only Zero of a (in some algebraically closed field
containing K) is the trivial Zero.
Preposition 3.1.9
Let a be a homogeneous ideal
(a) If a is irrelevant, then X(n,a)=0 for n sufficiently large.
(b) In general there is an expression a=q1n………..nqs as a reduced primary decomposition
such that all q1 are homogeneous.
(c) If an irrelevant primary ideal occurs in the decomposition, let b be the intersection of all
other primary ideals.
Then,
X(n,a) = X (n,b)
for all n sufficiently large
Proof:
For (a), by assumption we have An=an for n sufficiently large, so the assertion (a) is
obvious. We leave (b) as an exercise As to (c), say qs is irrelevant, and let b=q1n………..nqs-
1 . By preposition 3.1.7 we have,
( , + ) = ( , ) + ( , ) − ( , ) But b + qs is irrelevant, So (c) follows
from (a), thus concluding the proof.
31. BIBLIOGRAPY
1. P.B. BHATTACHARYA, S.K.JAIN : BASIC ABSTRACT ALGEBRA
2. N. JACOBSON : BASIC ALGEBRA
3. BALAKRISHNAN : TEXT BOOK OF MODERN
ALGEBRA
