### Commutative Algebra chapter IV

Chapter-IV
Projective  Modules.
Definition:4.1
A Sequence of R-modules and R-homomorphism  of the type M M1 M2→……. Mn Mn+1 is called an exact sequence if
Image  ί= Ker  ί+1,0≤ί≤n-1
Examples:  4.2
1.    The sequence o→2Z Z2→0 is an exact sequence where i is the inclusion map and þ is the projection of Z onto       Z2= Z/2Z
2.    For any two R-modules M and N, the sequence O→M M N N→O is exact where ί is the inclusion given by ί(x)=(x,0) and is the projection given by (x,y)=y,xЄM,yЄN

Definition:4.3
An exact sequence O→M M M→o of R-modules splits, if there exists an  R-homomorphism t: M"→ M such that gt=IM, the identity map on M.
Example:4.4
Example (ii) above is a split exact sequence with the splitting map t:N→M
N give by t(y)=(0,y)
Remark:4.5
o→M M M→o is exact if  is injective, g is surjective and
g induces an isomorphism of coker ( )=M/f(M) onto M″.
Proof:
Assume that the sequence is exact.
Obviously  injective and g is surjective.
Since M M″ and g is surjective  we have M/kerg M″
But kerg=Im = (M,)
Thus M/ (M) M″.
conversely suppose M/f(M)″ M.
Also M/kerg M″ kerg=f(M)=Imf.
The sequence is exact.
Theorem: 4.6
If o→M M M″→o is a split exact sequence, then M M′ M″
Proof:
Suppose  o→M M M″→o is a split exact sequence.
Let t:M″→M be a splitting so that gt= IM
This implies that t is injective.
For  i t(x)=o,x M,then
x=gt(x) ( gt=IM)
=g(0)
x=0
t is injective.
I   M, then  =tg( )+{ -tg( )}
Now g{ -tg( }=g( )-g(tg ))
=g( )-g( )
=o.
-tg( )  kerg.
=tg( )+{ -tg( )}
t(M″)+kerg.
M=t(M″)+kerg
To prove t(M) kerg={0}
Let y t (M)  kerg
T(M″) and y kerg
y=t(z),z M″and g( )=o
z=gt(z)=g(y)=0
z=o
y=t(z)=t(o)=o
y=o
t(M) kerg={o}
M=t(M) kerg
t(M)  Imf    ( kerg=Imf)
M M″ M′ since  and t  are injective maps.
Corollary:4.7
Let oM  M M″→o be an exact sequence of R-modules
which splits.Then there exists an R-homomorphism s:MM’
such that sf=IM’.
Proof:
Let oM M M″→o be an exact sequence.
Let t: M″→M be a splitting so that by above theorem,
we have M= (M) t(M)
Take 1 to be the projection of M onto (M),and let s=f-1 1
Now sf:MM  is  such that  s (x)=( -1 1) (x)
s (x)= -1( (x))=x M
s =IM

Definition4.8
Let M and N be R-modules.The set S of all R-homomorphisms of M in N has a natural R-module structure for the operations,
( +g)(x)= (x)+g(x)    x M, ,g s.
(a )(x)=a (x),     a R,x M, S.
This module is denoted by HomR(M,N).
Remark:
Let M be a fixed R-modules.Any homomorphism N of
R-modules induces a homomorphism
f*:HomR(M1N)HomR(MN) given by *( )= α,α  HomR(M,N).
Then (gf)*=g*f*,g HomR(N,N) and
I*N=Id,the identity map of HomR(M,N)
Similarly,for any fixed module N and a homomorphism g:MM, there exists an R-module homomorphism
g* : HomR (M, N) à HomR (M, N) given by g*(β) = βg, β  HomR (M, N)
Then (gh)* = h*g*, h  HomR (M, M)
and IM* = Id, the identity map of HomR (M, N)
Theorem:4.9
For any R-module M, and an exact sequence  oà N  N  N →0 ----(1)
The induced sequence 0 HomR(M, N)  HomR (M,N)  HomR (M,N″)
is exact.
Proof:
Clearly * is injective, for if
* (α) = 0, α HomR (M,N)  then f α = 0.
This implies α = 0 as  is injective.
Now g (x) = g( (x))
g (x) = 0 since Im  = kerg.
g  = 0
g* * = (g )* = 0 as g = 0
Thus Im *  kerg*
Conversely let β  ker g* so that
g*(β) = gβ=0
For any x M, gβ(x) = 0
β  kerg = Im ( the sequence (1) is exact)
Hence β(x) = (y) for some unique y M
This defines a map →N by setting α(x) = y where
β(x) = (y)
Clearly α is a homomorphism and
α(x) = y)
= β(x) x M
α = β
Hence β Im( *)
ker g* Im *
Im( *) = kerg* and hence the sequence
0→HomR (M, N)  HomR (M,N) HomR (M, N) is exact.
Remark:4.10
By a similar argument we can also show that for any
R-module N and an exact sequence of R-modules
0→M  M M → 0 the induced sequence
0 HomR (M, N) HomR (M,N) HomR (M, N) is exact.

Definition:4.11
A submodule N of M is called a direct summand of M if M=N Nfor some submodule Nof M.
Result:4.12
Let K N M are sub-modules.
i)              If N is a direct summand of M, N/K is a direct summand of M/K.
ii)           If K is a direct summand of N and N is a direct summand of M, then K is a direct summand of M. Then k is a direct summand of M
iii)         If K is a direct summand of M, then K is a direct summand of N. if further N/K is a direct summand of N/K, then N is a direct summand of M.

Remark:4.13
Certain types of modules P have the property that for any surjective homomorphism g: M M, the induced homomorphism g*: HomR (P, M) HomR (P, M) is surjective. These are the projective modules defined as follows.
Definition:4.14
An R-modules M is called projective  if  it  is a direct summand of a free R-module.
Theorem:4.15
An R-module P is projective if and only if for any surjective homomorphism g: M M, the induced homomorphism
g*:HomR (P, M) HomR (P,M) is surjective.

Proof:
Assume first that P is free, with basis {ei}i I
Given α HomR (P, M), let α(ei) = xi
Since g is surjective choose yi M with    g(yi) = xi
Then the map β : P → M, defined by β(ei) = yi, i I can be extended to an R-linear map β by defining β(Ʃaiei) = Ʃaiyi­
[ (g*(β)(ei) = g*(yi)= g(yi)
= xi
g*(β)(α) = α(ei) ]
clearly g* (β)=α
This shows that g* is surjective.
If P is projective, there exist an R-module Q with P Q = F, free.
Let : F → P be the projective map
Given α  HomR(P,M), consider α HomR(F,M).
By above case, there exists β  HomR (F, M) with g* (β) = α
Then β1 = β|P : P→ M has the property that, g* (β1) = α
Hence g* is surjective.
Conversely assume that P has the property stated in the theorem.
Express P as a quotient of a free module F.
Hence there is a surjective map g: F→ P.
By our assumption the induced map,
g*: HomR (P,F) HomR (P,P) is surjective.
In particular, there exists β  HomR (P, F) with g* (β) = gβ=Ip
This shows that the exact sequence 0 kerg F P 0 split
By theorem4.6
P is isomorphic to a direct summand of F
Hence  P is projective

Example:4.16
(i)            Any free R –module is projective
(ii)          If R is a principal ideal domain then any projective
R-module is free, as any submodule of a free
R-module is free.
Theorem :4.17
P =   is projective iff  is projective for each α.
Proof :
If p is projective, it is a direct summamd of a free module F.
Since is a direct summamd of  P, it is also a direct summamd
Of  F and hence projective .
Conversely assume that each is  projective and consider a diagram
P
M M O

Let α =  i α , where iα : Pα àP is the inclusion .
Since  is projective , there  exists an R-linear map
:  à M such that g Ψ α =  α
Define Ψ : P à M by setting
Ψ (x) = (  ) if  x =
Then  Ψ is R-linear and g Ψ = , showing that p is projective .

Theorem :4.18
An R-module  p is projective if  and only  if every exact sequence
0 à M ′  M  P à 0 splits
Proof :
If P is projective , the identity map I p :P à P can be lifted to a map
: P à M such that g  = I p
i.e the exact  sequence 0 à M ′ M  Pà 0splits.
To prove the converse, suppose the exact sequence
0 à M ′  M  P à 0 splits.
Express P as the quotient of a free module F with kernal K so that the sequence 0à KàFàP à0 is exact
By assumption , this sequence splits so that  P is a direct summand of F and hence projective.
Corollary:
P is finitely generated projective if and only if P is a  direct   summand of a free module of finite rank.

PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION

SSC MATHS I PAPER SOLUTION

SSC MATHS II PAPER SOLUTION

SSC SCIENCE I PAPER SOLUTION

SSC SCIENCE II PAPER SOLUTION

SSC ENGLISH PAPER SOLUTION

SSC & HSC ENGLISH WRITING SKILL

HSC ACCOUNTS NOTES

HSC OCM NOTES

HSC ECONOMICS NOTES

HSC SECRETARIAL PRACTICE NOTES

# 2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

# HSCMaharashtraBoardPapers2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020

# SSCMaharashtraBoardPapers2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Sanskrit (Composite) MARCH 2020

Science (Paper 1) MARCH 2020

Science (Paper 2)

Geography Model Set 1 2020-2021

MUST REMEMBER THINGS on the day of Exam

Are you prepared? for English Grammar in Board Exam.

Paper Presentation In Board Exam

How to Score Good Marks in SSC Board Exams

Tips To Score More Than 90% Marks In 12th Board Exam

How to write English exams?

How to prepare for board exam when less time is left

How to memorise what you learn for board exam

No. 1 Simple Hack, you can try out, in preparing for Board Exam

How to Study for CBSE Class 10 Board Exams Subject Wise Tips?

JEE Main 2020 Registration Process – Exam Pattern & Important Dates

NEET UG 2020 Registration Process Exam Pattern & Important Dates

How can One Prepare for two Competitive Exams at the same time?

8 Proven Tips to Handle Anxiety before Exams!