Module Theory Postgraduate Course by Assistant Professor Dr. Akeel Ramadan Mehdi Semester: 2 Academic Year: 2013-2014 Syllabus: 1- Definition and examples of modules and Submodules 2- Finitely generated modules, Internal direct sums, direct summands, Quotient modules, Homomorphisms of modules and Isomorphism theorems 3- Exact and split sequences of modules 4- Direct sum and product of modules, Homomorphisms of direct products and sums 5- Free modules and Finitely presented modules 6- Simple and Semisimple modules 7- Essential submodules, Maximal submodules, The Jacobson radical of modules and The Socle of modules 8- Injective Modules, Baer’s Criterion 9- Injective Hulls 10- Bilinear mapping, Tensor product of modules 11- Properties of Tensor product 12- Flat modules 13- Character module References [1] F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, Springer- Verlag, 1992. [2] P. E. Bland, Rings and Their Modules, Walter de Gruyter GmbH and Co. KG, Berlin/New York, 2011. [3] P. A. Grillet, Abstract Algebra, Springer Science and Business Media, 2007. [4] Thomas W. Hungerford, Algebra, Springer-Verlag, New York, 1974. [5] F. Kasch, Modules and Rings, Academic Press, London, New York, 1982. [6] T. Y. Lam, Lectures on Modules and Rings, Springer-Verlag, New York, 1999. [7] T. S. Plyth, Module theory an approach to linear algebra, 1977. 1
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Module TheoryPostgraduate Course
by
Assistant Professor Dr. Akeel Ramadan Mehdi
Semester: 2Academic Year: 2013-2014
Syllabus:1- Definition and examples of modules and Submodules2- Finitely generated modules, Internal direct sums, direct summands, Quotient
modules, Homomorphisms of modules and Isomorphism theorems3- Exact and split sequences of modules4- Direct sum and product of modules, Homomorphisms of direct products and
sums5- Free modules and Finitely presented modules6- Simple and Semisimple modules7- Essential submodules, Maximal submodules, The Jacobson radical of modules
and The Socle of modules8- Injective Modules, Baer’s Criterion9- Injective Hulls10- Bilinear mapping, Tensor product of modules11- Properties of Tensor product12- Flat modules13- Character module
References[1] F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, Springer-
Verlag, 1992.[2] P. E. Bland, Rings and Their Modules, Walter de Gruyter GmbH and Co. KG,
Berlin/New York, 2011.[3] P. A. Grillet, Abstract Algebra, Springer Science and Business Media, 2007.[4] Thomas W. Hungerford, Algebra, Springer-Verlag, New York, 1974.[5] F. Kasch, Modules and Rings, Academic Press, London, New York, 1982.[6] T. Y. Lam, Lectures on Modules and Rings, Springer-Verlag, New York, 1999.[7] T. S. Plyth, Module theory an approach to linear algebra, 1977.
1
1 Definition and examples of modules and Submodules
Definition and examples of modules
Definition 1.1. Let R be a ring. A left R-module or a left module over R is a set M togetherwith
(1) a binary operation + on M under which M is an abelian group, and(2) a mapping • : R ×M →M (is called a module multiplication) denoted by r •m ,
for all r ∈R and for all m ∈M which satisfies(a) (r + s ) •m = r •m + s •m , for all r, s ∈R , m ∈M ,(b) (r s ) •m = r • (s •m ), for all r, s ∈R , m ∈M , and(c) r • (m +n ) = r •m + r •n, for all r ∈R , m , n ∈M .If the ring R has an identity element 1R and(d) 1 •m =m , for all m ∈M , then M is said to be a unitary left R-module.
Remark 1.2. The descriptor "left" in the above definition indicates that the ring elementsappear on the left. A right R-modules can be defined analogously as follows.
Definition 1.3. Let R be a ring. A right R-module or a right module over R is a set Mtogether with
(1) a binary operation + on M under which M is an abelian group, and(2) a mapping • : M ×R →M (is called a module multiplication) denoted by m • r ,
for all r ∈R and for all m ∈M which satisfies(a) m • (r + s ) =m • r +m • s , for all r, s ∈R , m ∈M ,(b) m • (r s ) = (m • r ) • s , for all r, s ∈R , m ∈M , and(c) (m +n ) • r =m • r +n • r , for all r ∈R , m , n ∈M .If the ring R has an identity element 1R and(d) m •1=m , for all m ∈M , then M is said to be a unitary left R-module.
The notation R M (resp. M R ) denotes to left (resp. right) R-module M .
Exercise: If the ring R is commutative, then a module M is left R-module if andonly if it is a right R-module.
Lemma 1.4. Let R be a ring with 1, let M be a left R-module and let r, s ∈ R, m , n ∈M .Then:
(1) r 0M = 0M ;(2) 0R m = 0M ;(3) (−1)m =−m ;(4) −(r m ) = (−r )m = r (−m );(5) (r − s )m = r m − s m ;(6) r (m −n ) = r m − r n.
Proof. (1) r 0M = r (0M +0M ) = r 0M + r 0M ⇒ r 0M +(−(r 0M )) = r 0M + r 0M +(−(r 0M ))⇒ 0M = r 0M .(2) 0R m = (0R + 0R )m = 0R m + 0R m . Thus 0R m + (−(0R m )) = 0R m + 0R m +
(−(0R m )). Hence 0M = 0R m .(3) 0M = 0R m (by (2)) = (1+(−1))m =m +(−1)m . Thus (−1)m =−m .
2
(4) Exercise.(5) Exercise.(6) Exercise.
Example 1.5. Modules over a field F and vector spaces over F are the same.
Proof. By the definitions of modules and vector spaces over a field F.
Example 1.6. Every left ideal (I ,+, .) of a ring (R ,+, .) is a left R-module.
Proof. Let (I ,+, .) be a left ideal of a ring (R ,+, .). Thus (I ,+) is an abelian group (why?).Define • : R × I → I by •(r, a ) = r.a , for all r ∈ R and for all a ∈ I . Since (R ,+, .) is a
ring, (r +s )•a = (r +s ).a = r.a +s .a = r •a +s •a , (r.s )•a = (r.s ).a = r.(s .a ) = r • (s •a )and r • (a +b ) = r.(a +b ) = r.a + r.b = r •a + r •b for all r, s ∈R , a ,b ∈ I .
Example 1.7. Every right ideal (I ,+, .) of a ring (R ,+, .) is a right R-module.
Proof. Exercise.
Example 1.8. Every ring (R ,+, .) is a left and right R-module.
Proof. Since (R ,+, .) is an ideal of (R ,+, .) it follows that (R ,+, .) is a left and right idealof (R ,+, .). By Example 1.6 and Example 1.7, R is a left and right R-module.
Example 1.9. If M is a unitary left R-module and S is a subring of R with 1S = 1R , then Mis a unitary left S-module as well. For instance the field R is an R-module, a Q-moduleand a Z-module.
Proof. Since M is a unitary left R-module, (M ,+) is an abelian group and there is amodule multiplication • : R ×M → M with 1R •m = m . Define ∗ : S ×M → M bys ∗m = s •m , for all s ∈S and m ∈M .It is clear that ∗ is a module multiplication (Why?). Thus M is a left S-module. Since1S = 1R it follows that 1S ∗m = 1R •m =m and hence M is a unitary left S-module.
Example 1.10. Every abelian group is Z-module.
Proof. Let M be any abelian group, let a ∈M and let n ∈ Z. Define the module multi-plication na :Z×M →M as follows:
If n > 0, then na = a + a + · · ·+ a (n times); if n = 0, then na = 0; if n < 0, thenna =−((−n )a ) =−a −a − · · ·−a (| n | times).
Let n , m ∈Z and let a ∈M , thus(n+m )a = a+a+· · ·+a (n+m times) = (a +a + · · ·+a )
Let n ∈ Z and let a ,b ∈ M , thus n (a + b ) = (a + b ) + (a + b ) + · · ·+ (a + b ) (ntimes) = a +a + · · ·+a︸ ︷︷ ︸
n times
+b +b + · · ·+b︸ ︷︷ ︸
n times
= na +nb .
Thus the multiplication na : Z×M →M is a module multiplication and hence Mis a Z-module. Since 1a = a it follows that M is a unitary Z-module
3
Example 1.11. Let R be a ring with 1 and let n ∈ Z+. Define Rn = {(a 1, a 2, · · · , a n ) | a i ∈R , for all i }. Make Rn into a left R-module by componentwise addition and multiplica-tion by elements of R as follows:(a 1, a 2, · · · , a n ) + (b1,b2, · · · ,bn ) = (a 1 +b1, a 2 +b2, · · · , a n +bn ) and r (a 1, · · · , a n ) =
(r a 1, · · · , r a n ), for all r ∈R and (a 1, a 2, · · · , a n ), (b1,b2, · · · ,bn )∈Rn .
Proof. Exercise
Example 1.12. If M n (R) is the set of n × n matrices over a ring R, then M n (R) is anadditive abelian group under matrix addition. If (a i j ) ∈ M n (R) and r ∈ R, then theoperation r (a i j ) = (r a i j ) makes M n (R) into a left R-module. M n (R) is also a right R-module under the operation (a i j )r = (a i j r ).
Proof. Exercise
Example 1.13. Let I be a left ideal of a ring R. Then R/I is a left R-module.
Proof. Exercise.
Example 1.14. Let M be an abelian group and End(M ) its endomorphism ring. Then Mis a left unitary End(M )-module.
Proof. Exercise.
4
SubmodulesDefinition 1.15. Let R be a ring and let M be a left R-module. A left submodule of Mis a subgroup N of M such that r • n ∈ N , for all r ∈ R , and for all n ∈ N , where • isthe module multiplication defined on M . We will use N ,→M to denote that N is a leftsubmodule of M .
Proposition 1.16. Let R be a ring, let M be a left R-module and let N be a nonemptysubset of M . Then the following statements are equivalent:(1) N is a left submodule of M ;(2) For all a ,b ∈N and for all r ∈R we have that:
(i) a −b ∈N and(ii) r •a ∈N , where • is the module multiplication defined on M ;
(3) r •a − s •b ∈N , for all a ,b ∈N and for all r, s ∈R.
Proof. Exercise
Proposition 1.17. Let R be a ring, let M be a left R-module and let N be a subset ofM . Then N is a left submodule of M if and only if N is a left R-module with the sameaddition and module multiplication on M .
Proof. (⇒) Suppose that N is a left submodule of M . Thus (by Definition 1.15) we havethat N is a subgroup of (M ,+) such that r •n ∈ N , for all r ∈ R , n ∈ N , where • is themodule multiplication defined on M . Since (M ,+) is an abelian group, (N ,+) is anabelian group.
Define ∗ : R ×N →N by r ∗n = r •n , for all r ∈ R , n ∈N . It is easy to check that∗ is a module multiplication (H.W.). Thus N is a left R-module with the same additionand module multiplication on M .(⇐) Suppose that N is a left R-module with the same addition and module multi-
plication on M . Thus (N ,+) is an abelian group and the multiplication ∗ : R ×N → Ndefined by r ∗n = r •n , for all r ∈ R , n ∈N is a module multiplication. Since N is asubset of M it follows that (N ,+) is a subgroup of (M ,+).
Let r ∈ R and let n ∈ N , thus r •n = r ∗n ∈ N . By Definition 1.15, N is a left sub-module of M .
Example 1.18. Every left R-module M contains at least two submodules (trivial sub-modules): M ,→M and 0 ,→M .
Example 1.19. Let R be a ring with 1R . Then the left submodules of R as a left R-moduleare exactly the left ideals of a ring (R ,+, .).
Proof. Let (I ,+, .) be a left ideal of a ring (R ,+, .). By Example 1.6, I is a left R-modulewith the same addition and module multiplication on R R . Since I is a subset of R itfollows from Proposition 1.17 that I is a left submodule of a left R-module R R . Henceevery left ideal of a ring (R ,+, .) is a submodules of R as a left R-module.
Suppose that N is a left submodules of R as a left R-module. By Proposition 1.16we have that a −b ∈ N and r •a ∈ N for all a ,b ∈ N and for all r ∈ R , where • is themodule multiplication defined on R R . Since r •a = r.a for all a ,b ∈N and for all r ∈R ,thus N is a left ideal of a ring R . Therefore, the left submodules of R as a left R-moduleare exactly the left ideals of a ring (R ,+, .).
5
Exercise: Find all left submodules of the following left R-modules:(1) ZZ; (2) F F , where F is a field; (3) RR; (4) Zp as a left Zp -module, where p
is a prime number; (5) Z10 as a left Z10-module.
Example 1.20. Let F be a field and let M be a left F -module. Then the left submodulesof a left F -module M are exactly the subspaces of an F -vector space M .
Proof. Since F is a field it follows from the definitions of submodules and subspacesthat there is no difference between subspaces and submodules of a left F -module M .
Example 1.21. Let M be a left Z-module. Then the left submodules of a left Z-moduleM are exactly the subgroups of an abelian group M .
Proof. Exercise
Proposition 1.22. Let M be a left R-module and let N1 and N2 be two left submodulesof M .
Define N1 ∩N2 = {x ∈M | x ∈N1 andx ∈N2} andN1+N2 = {y ∈M | y = a +b with a ∈N1 andb ∈N2}. Then N1 ∩N2 and N1+N2 are
left submodules of M .
Proof. (a) We will prove that N1 ∩N2 is a submodules of M . Since N1 and N2 are sub-groups of an abelian group M it follows that N1 and N2 are contain 0 and hence N1∩N2
is a nonempty subset of M .Let a ,b ∈ N1 ∩N2 and let r ∈ R . Thus a ,b ∈ N1 and a ,b ∈ N2. Since N1 ,→M and
N2 ,→M it follows from Proposition 1.16 that a −b , r a ∈N1 and a −b , r a ∈N2 and thisimplies that a −b , r a ∈N1 ∩N2. By Proposition 1.16, N1 ∩N2 is a left submodule of M .
(b) We will prove that N1+N2 is a submodule of M . Since N1 and N2 are subgroupsof an abelian group M it follows that N1 and N2 are contain 0 and hence 0 = 0+ 0 ∈N1+N2. Thus N1+N2 is a nonempty subset of M .
Let x , y ∈N1+N2 and let r ∈R . Thus x = a 1+b1 and y = a 2+b2, where a 1, a 2 ∈N1
and b1,b2 ∈ N2. Thus x − y = a 1 + b1 − a 2 − b2 = a 1 − a 2 + b1 − b2 and r x = r (a 1 +b1) = r a 1 + r b1. Since N1 ,→ M and N2 ,→ M it follows from Proposition 1.16 thata 1 − a 2, r a 1 ∈ N1 and b1 − b2, r b1 ∈ N2 and this implies that x − y , r x ∈ N1 +N2. ByProposition 1.16, N1+N2 is a submodule of M .
Proposition 1.23. Let M be a left R-module and let {Ni }i∈I be a family of left submod-ules of M .
Define⋂
i∈INi = {x ∈M | x ∈Ni for all i ∈ I }. Then
⋂
i∈INi is a left submodule of M .
Proof. Exercise.
Theorem 1.24. (Modular Law) If M is a left R-module and if A, B ,C are left submodulesof M with B ,→C , then(A + B )∩C = (A ∩C )+ (B ∩C ) = (A ∩C )+ B.
Proof. (1) We will prove that (A + B ) ∩C = (A ∩C ) + B . Let x ∈ (A + B ) ∩C , thus x ∈A + B and x ∈ C and hence x = a +b , where a ∈ A and b ∈ B . Since B ,→ C it followsthat x ,b ∈ C and hence x −b ∈ C . Since a ∈ A and a = x −b it follows that a ∈ A ∩C
6
and hence a +b ∈ (A ∩C )+ B . Thus x ∈ (A ∩C )+ B and this implies that (A + B )∩C ⊆(A ∩C )+ B .
Since A ∩C ⊆ C it follows that (A ∩C ) + B ⊆ C + B . Since A ∩C ⊆ A it follows that(A ∩C )+ B ⊆ A + B and hence (A ∩C )+ B ⊆ (A + B )∩ (C + B ) = (A + B )∩C . Therefore,(A + B )∩C = (A ∩C )+ B .
(2) We will prove that (A ∩C ) + (B ∩C ) = (A ∩C ) + B . Since B ,→ C it follows thatB ∩C = B and hence (A ∩C )+ (B ∩C ) = (A ∩C )+ B .
From (1) and (2) we get that (A + B )∩C = (A ∩C )+ (B ∩C ) = (A ∩C )+ B .
Remarks 1.25. (1) In Theorem 1.24, if we remove the condition B ,→C , then we alreadyhave (A ∩C )+ (B ∩C ) ,→ (A + B )∩C .
Proof. Since A ,→ A + B it follows that A ∩C ,→ (A + B )∩C . Also, since B ,→ A + B itfollows that B ∩C ,→ (A + B )∩C and hence (A ∩C )+ (B ∩C ) ,→ (A + B )∩C .
(2) The reverse inclusion in (1) above does not necessarily hold, for example:(Exercise).
Exercise: Let M be a left R-module and let A, B ,C are left submodules of M such thatA ⊆ B , A +C = B +C , A ∩C = B ∩C . By using Modular Law, prove that A = B .
Exercise: Is the union of any two submodules of a left R-module M is a submodule ofM ?
7
2 Finitely generated modules, Internal direct sums, di-rect summands, Quotient modules, Homomorphismsof modules and Isomorphism theorems
Finitely generated modules
Definition 2.1. If X is a subset of a left R-module M then < X > will denote the inter-section of all the submodules of M that contain X . This is called the submodule of Mgenerated by X , while the elements of X are called generators of <X >.
Lemma 2.2. Let X be a subset of a left R-module M . Then < X > is the smallest leftsubmodule of M that contains a subset X .
Proof. By Proposition 1.23, < X > is a submodule of M . Let N be a submodule of Mcontains X . By definition of < X > we have that < X >=
⋂
A ,→Mwith X⊆A
A. Since N ,→ M and
X ⊆ N it follows that < X >⊆ N . Thus < X > is the smallest submodule of M thatcontains a subset X .
Lemma 2.3. Let X be a nonempty subset of a left R-module M and let
N = {n∑
i=1ri x i | ri ∈R ,x i ∈X , n ∈Z+}. Then N ,→M .
Proof. Exercise
Proposition 2.4. Let X be a subset of a left R-module M . Then
<X >=
{n∑
i=1ri x i | ri ∈R ,x i ∈X , n ∈Z+}
0
, if X 6=φ
, if X =φ
Proof. Suppose that X =φ. Since {0} ,→M and X =φ ⊆ {0}, thus⋂
A ,→Mwithφ⊆A
A = 0.
Since <φ >=⋂
A ,→Mwithφ⊆A
A⇒<X >=<φ >= 0.
Suppose that X 6=φ and let N = {n∑
i=1ri x i | ri ∈R ,x i ∈X , n ∈Z+}.
We will prove that <X >=N . Let a ∈N ⇒ a =m∑
i=1s i yi with s i ∈R , yi ∈X , m ∈Z+.
Let A be any submodule of M contains X . Since yi ∈ X and s i ∈ R ⇒ s i yi ∈ A,∀i =
1, 2, ..., m ⇒ a =m∑
i=1s i yi ∈ A ⇒ N ⊆ A, for all submodule A of M contains X ⇒ N ⊆
⋂
A ,→Mwith X⊆A
A.
Since <X >=⋂
A ,→Mwith X⊆A
A⇒N ⊆<X > .
8
By Lemma 2.2, < X > is the smallest left submodule of M contains X . Since N is aleft submodule of M (by Lemma 2.3) and X ⊆N ((H.W) Why?) it follows that<X >⊆N .
Thus <X >=N = {n∑
i=1ri x i | ri ∈R ,x i ∈X , n ∈Z+}.
Definition 2.5. (1) A left R-module M is said to be finitely generated if it generated by afinite subset X , that is M =<X >.
(2) A left R-module M is said to be cyclic if it generated by a subset X = {a } containsone element only, that is M =< {a }>.
Remarks 2.6. (1) If M is a finitely generated left R-module generated by a subset X ={a 1, a 2, ..., a n} for some n ∈ Z+, then we will write M =< a 1, a 2, ..., a n >. Also, from
Proposition 2.4 we get that M =< a 1, a 2, ..., a n >= {n∑
i=1ri a i | ri ∈R}.
(2) If M is a cyclic left R-module generated by a subset X = {a }, then we will writeM =< a >. Also, from Proposition 2.4 we get that M =< a >= {r a | r ∈R}=Ra .
Examples 2.7. (1) Every left R-module M has a generated set M .(2) Every ring R with identity 1 is a cyclic left R-module, since R R =< 1>.(3) Let M =Z24 as Z24-module.Then < 6, 12>=Z24∩< 2>∩< 3>∩< 6>= {0, 6, 12, 18}=< 6>.Also, Z24 =< 1>=< 5>=< 7> as Z24-module.
Proposition 2.8. Let {Ni }i∈I be a family of submodules of a left R-module M . Then
<⋃
i∈INi >=
{∑
i∈I′a i | a i ∈Ni , I ′ ⊆ I with I ′ is finite}
0
, if I 6=φ
, if I =φ
That is, in the case I 6=φ, <⋃
i∈INi > is the set of all finite sums
∑
a i with a i ∈Ni .
Proof. If I =φ, then⋃
i∈INi =φ. By Proposition 2.4, <
⋃
i∈INi >= 0.
If I 6= φ, then⋃
i∈INi 6= φ. By Proposition 2.4, <
⋃
i∈INi >= {
n∑
j=1rj x j | rj ∈ R ,x j ∈
⋃
i∈INi , n ∈Z+}. Let x ∈<
⋃
i∈INi >, thus x =
n∑
j=1rj x j , for some rj ∈R , x j ∈
⋃
i∈INi and n ∈Z+.
If we bring together all summands rj x j which lie in a fixed Ni to form a sum x′
i and if
we treat with the remaining summands similarly then it follows that x =n∑
j=1rj x j =∑
i∈I′x′
i .
Thus <⋃
i∈INi >⊆ {∑
i∈I′a i | a i ∈Ni , I ′ ⊆ I with I ′ is finite}.
Conversely, let x ∈ {∑
i∈I′a i | a i ∈ Ni , I ′ ⊆ I with I ′ is finite}, thus x =
∑
j∈I′b j with b j ∈
N j , I ′ ⊆ I with I ′ is finite . We can write I ′ = {1, 2, ..., n}, for some n ∈Z+. Thus x =n∑
j=1b j
with b j ∈⋃
i∈INi ⇒ x ∈<⋃
i∈INi > and hence {
∑
i∈I′a i | a i ∈ Ni , I ′ ⊆ I with I ′ is finite} ⊆<
⋃
i∈INi >. Thus <⋃
i∈INi >= {∑
i∈I′a i | a i ∈Ni , I ′ ⊆ I with I ′ is finite}.
9
Definition 2.9. (Sum of submodules) Let {Ni }i∈I be a family of submodules of a left R-module M . We will use the notation
∑
i∈INi to denote the sum of the submodules {Ni }i∈I
and defined by∑
i∈INi =<⋃
i∈INi >.
Remark 2.10. Let {N1, N2, ..., Nn} be a family of submodules of a left R-modules M . Thenn∑
i=1Ni = {
n∑
i=1a i | a i ∈Ni }.
Internal direct sums and direct summandsDefinition 2.11. (Internal direct sum) Let M be a left R-module and let {Ni }i∈I be afamily of left submodules of M . We say that M is the internal direct sum of the submod-ules {Ni }i∈I and denoted by M =
⊕
i∈INi if the following two conditions are hold:
(1) M =∑
i∈INi and
(2) N j ∩∑
i∈Ii 6=j
Ni = 0, for all j ∈ I .
In case I is a finite set (for example I = {1, 2, ..., n}), then M =N1⊕N2⊕ ...⊕Nn .
Example 2.12. Let M =R3 as a left R-module and letN1 = {(x , 0, 0) | x ∈R}, N2 = {(0, y , 0) | y ∈R} and N3 = {(0, 0, z ) | z ∈R}.Prove that:(1) N1, N2 and N3 are left submodules of M ;(2) M is the internal direct sum of the submodules N1, N2 and N3.
Proof. (1) Exercise.(2) Since N1, N2 and N3 are left submodules of M ⇒ N1+N2+N3 ⊆M . Let a ∈M ,
thus a = (x , y , z )with x , y , z ∈R ⇒ a = (x , 0, 0)+ (0, y , 0)+ (0, 0, z )∈N1+N2+N3⇒M ⊆N1+N2+N3. Thus M =N1+N2+N3.
Also, N1 +N2 = {(x , y , 0) | x , y ∈ R}, N1 +N3 = {(x , 0, z ) | x , z ∈ R}and N2 +N3 ={(0, y , z ) | y , z ∈R}. Thus N1 ∩ (N2+N3) = 0, N2 ∩ (N1+N3) = 0, and N3 ∩ (N1+N2) = 0.
Hence M =N1⊕N2⊕N3.
Example 2.13. Let M be a left R-module. Then M is the internal direct sum of the trivialsubmodules 0, M .
Example 2.14. Let M = Z6 as a left Z6-module and let N1 =< 2 >= {0, 2, 4} and N2 =<3>= {0, 3}. Then M is the internal direct sum of the submodules N1 and N2.
Proof. Exercise.
Proposition 2.15. Let M be a left R-module and let {Ni }i∈I be a family of left submod-ules of M . Then the following two statements are equivalent:
(1) N j ∩∑
i∈Ii 6=j
Ni = 0, for all j ∈ I .
(2) For every x ∈ M the representation x =∑
i∈I′b i with b i ∈ Ni , I ′ ⊆ I , I ′ finite, is
unique in the following sense:If x =∑
i∈I′b i =∑
i∈I′c i with b i , c i ∈Ni , then b i = c i for all i ∈ I ′ .
10
Proof. (1)⇒ (2). Let x ∈M such that x =∑
i∈I′b i =∑
i∈I′c i with b i , c i ∈Ni I ′ ⊆ I , I ′ finite ⇒
∑
i∈I′b i −∑
i∈I′c i = 0.
Thus for all j ∈ I ′ we have that b j +∑
i∈I′
i 6=j
b i − (c j +∑
i∈I′
i 6=j
c i ) = 0 ⇒ b j − c j =∑
i∈I′
i 6=j
(c i −b i ).
Since∑
i∈I′
i 6=j
(c i−b i )∈∑
i∈I′
i 6=j
Ni ⇒ b j −c j ∈∑
i∈I′
i 6=j
Ni . Since b j −c j ∈N j ⇒ b j −c j ∈N j ∩∑
i∈I′
i 6=j
Ni .
Since N j ∩∑
i∈I′
i 6=j
Ni ⊆N j ∩∑
i∈Ii 6=j
Ni and since N j ∩∑
i∈Ii 6=j
Ni = 0 ⇒ N j ∩∑
i∈I′
i 6=j
Ni = 0 ⇒b j −c j = 0 ⇒
b j = c j , for all j ∈ I ′ .(2)⇒ (1). Let b ∈N j ∩
∑
i∈Ii 6=j
Ni , thus b = b j ∈N j andb ∈∑
i∈Ii 6=j
Ni ⇒ there is a finite subset
I ′ ⊆ I with j /∈ I ′ and b =b j =∑
i∈I′b i , b i ∈ Bi .
If we add to the left-hand side the summands 0 ∈ Bi , i ∈ I ′ and to the right-handside the summand 0∈ B j , then the same finite index set I ′ ∪{j } appears on both sidesand from uniqueness in (2) it follows that b =b j = 0. Thus N j ∩
∑
i∈Ii 6=j
Ni = 0.
Definition 2.16. (Direct summand) A submodule N of a left R-module M is said to bea direct summand of M if there is a submodule K of M such that M =N ⊕K .
In other word, there is a submodule K of M such that M =N +K and N ∩K = 0.
Example 2.17. Let M =Z6 as a left Z -module. Find all direct summands of M .
Proof. M , 0, N1 =< 2>= {0, 2, 4} and N2 =< 3>= {0, 3} are all direct summands of M .
Example 2.18. Let F be a field. Find all direct summands of F F .
Proof. M and 0 are all direct summands of F F .
Example 2.19. Find all direct summands of the following Z -module:(1) M =Z30.(2) M =Z25.
Proof. Exercise.
Example 2.20. Let M =Z as a leftZ-module. Prove that< 0> and ZZ are the only directsummands of M =Z Z.
Proof. Assume that N is a direct summand of M with N 6= 0 and N 6= ZZ. Thus N =<n > with n ∈ Z and n 6= 0, n 6= 1, n 6= −1. Since N is a direct summand of M , there isa submodule K =<m > of M for some m ∈ Z such that ZZ =< n > ⊕ <m >⇒< n >∩ < m >= 0. Since nm ∈< n > ∩ < m >⇒ nm = 0. Since n 6= 0 and Z is an integraldomain⇒m = 0. Since ZZ =< n > + <m >⇒ ZZ =< n >. Since either ZZ =< 1 > or
ZZ=<−1>⇒ either n = 1 or n =−1 and this is a contradiction. Thus< 0> and ZZ arethe only direct summands of M = ZZ.
11
Quotient modules and Homomorphisms of modules
Proposition 2.21. Let R be a ring, let M be a left R-module and let N be a left submoduleof M . The (additive, abelian) quotient group M/N can be made into a left R-module bydefining a module multiplication • : R ×M/N →M/N by r • (x +N ) = (r x )+N , for allr ∈R, x +N ∈M/N .
Proof. Exercise.
Definition 2.22. The left R-module M/N is defined in Proposition 2.21 is called quotient(or factor) module.
Definition 2.23. Let N and M be left R-modules.(1) A function f : N →M is said to be a left R-module homomorphism (or just left
R-homomorphism) iffor all a ,b ∈N and r ∈R, then f (a +b ) = f (a )+ f (b ) and f (r a ) = r f (a ).
(2) A left R-module homomorphism is a monomorphism if it is injective and is anepimorphism if it is surjective. A left R-module homomorphism is an isomorphism ifit is both injective and surjective. The modules N and M are said to be isomorphic,denoted N ∼=M , if there is some left R-module isomorphism ϕ : N →M .
(3) If f : N →M is a left R-module homomorphism, let ker( f ) = {n ∈ N | f (n ) = 0}(the kernel of f ) and let im( f ) = f (N ) = {m ∈M |m = f (n ) for some n ∈N } (the image off ). The R-module M/im( f ) is called the cokernel of f . The cokernel of f will be denotedby coker( f ).
(4) Define HomR (N , M ) to be the set of all left R-module homomorphisms from Ninto M and when M =N , HomR (N , M ) will be written as EndR (M ) (the set of all endo-morphisms of M ).
Proposition 2.24. If f : M →N is a left R-homomorphism, then f (0M ) = 0N and f (−x ) =− f (x ) for each x ∈M .
Proof. f (0M ) = f (0M +0M ) = f (0M )+ f (0M )⇒ 0N = f (0M ).(Exercise) f (−x ) =− f (x ) for each x ∈M .
Proposition 2.25. Let f : M →N be a left R-homomorphism.(1) If B is a left submodule of N , then f −1(B ) is a left submodule of M .(2) If A is a left submodule of M , then f (A) is a left submodule of N .
Proof. Exercise.
Corollary 2.26. Let f : M → N be a left R-homomorphism. Then ker( f ) is a left sub-module of M and im( f ) is a left submodule of N .
Proof. Since ker( f ) = {x ∈ M | f (x ) = 0} = f −1(0) and since 0 ,→ N it follows fromProposition 2.25(1) that ker( f ) is a left submodule of M . Also, since im( f ) = {y ∈ N |y = f (x ) for some x ∈M }= f (M ) and since M ,→M it follows Proposition 2.25(2) thatim( f ) is a left submodule of N .
12
Example 2.27. [Hungerford, p. 170] For any modules the zero map 0 : A → B given by0(a ) = 0B for all a ∈ A is a left R-homomorphism.
Proof. Exercise.
Example 2.28. Let M be a left R-module and let N be a left submodule of M . The inclu-sion mapping i N : N →M defined by i N (x ) = x , for all x ∈N is a left R-monomorphism.
Proof. Let x , y ∈ N and let r ∈ R . Thus i N (x + y ) = x + y = i N (x ) + i N (y ) and i N (r x ) =r x = r i N (x ). Hence inclusion mapping i N is a left R-homomorphism. Since i N is aninjective mapping⇒ i N is a left R-monomorphism.
Example 2.29. Let M be a left R-module. The identity mapping IM : M →M defined byIM (x ) = x for all x ∈M is a left R-isomorphism.
Proof. Exercise.
Example 2.30. Let M be a left R-module and let N be a left submodule of M . Thenthe natural mapping π : M → M/N defined by π(x ) = x +N for all x ∈ M is a left R-epimorphism with kernel N .
Proof. Let x , y ∈ M and let r ∈ R , thus π(x + y ) = (x + y ) +N = (x +N ) + (y +N ) =π(x ) +π(y ) and π(r x ) = r x +N = r (x +N ) = rπ(x ). Thus π : M → M/N is a leftR-homomorphism. Since π is a surjective mapping ⇒π is an R-epimorphism.
Example 2.31. Let (G1,∗) and (G2,◦) be abelian groups and let f : G1→G2 be a function.Then f is group homomorphism if and only if it is left Z-homomorphism.
Proof. (⇒) Suppose that f : G1 → G2 is group homomorphism, thus f (a ∗b ) = f (a ) ◦f (b ), for all a ,b ∈G1. Let n ∈Z, a ∈G1, thus f (na ) = f (a ∗a ∗ · · · ∗a
︸ ︷︷ ︸
)n−times
= f (a ) ◦ f (a ) ◦ · · · ◦ f (a )︸ ︷︷ ︸
n−times
=
n f (a ). Thus f is left Z-homomorphism.(⇐) Suppose that f : G1 → G2 is Z-homomorphism, thus for all x , y ∈ G1 we have
that f (x ∗ y ) = f (x ) ◦ f (y ). Hence f is group homomorphism
Example 2.32. Let V1 and V2 be vector spaces over a field F and let f : V1 → V2 be afunction. Then f is a left F -homomorphism if and only if it is linear transformationover F .
Proof. Exercise.
Examples 2.33. Let f : Z → Z defined by f (x ) = 2x , for all x ∈ Z. Then f is a left Z -module homomorphism but it is not ring homomorphism, since f (1) = 2 6= 1.
Exercise: Give an example of a ring homomorphism but not a left R-module homo-morphism, for some ring R .
Example 2.34. Let R be a ring. Define f : R→M 2×2(R) by f (x ) =
�
x 00 x
�
for all x ∈R.
Then f is a left R-monomorphism.
Proof. Exercise.
13
Proposition 2.35. Let N be a left R-submodule of a left R-module M . Then A is a leftR-submodule of a left R-module M/N if and only if there is a unique left R-submoduleB of M such that N ⊆ B and A = B/N .
Proof. Exercise.
Proposition 2.36. Letα : N →M be a left R-homomorphism. Thenα is R-monomorphismif and only if ker(α) = {0N }.
Proof. Exercise.
Proposition 2.37. Let f : N → M and g : M → K be left R-homomorphisms. Theng ◦ f : N → K is a left R-homomorphism.
Proof. Let a ,b ∈ N and let r ∈ R , thus (g ◦ f )(a +b ) = g ( f (a +b )) = g ( f (a ) + f (b )) =g ( f (a ))+ g ( f (b )) = (g ◦ f )(a )+ (g ◦ f )(b ).
Also, (g ◦ f )(r a ) = g ( f (r a )) = g (r f (a )) = r g ( f (a )) = r ((g ◦ f )(a )). Thus g ◦ f : N → Kis a left R-homomorphism.
Isomorphism theorems
Theorem 2.38. (First Isomorphism Theorem for Modules) If f : M → N is a left R-homomorphism, then M/ker( f )∼= im( f ).
Proof. Define ϕ : M/ker( f )→ im( f ) by ϕ(x +ker( f )) = f (x ), for all x ∈M .If x +ker( f ) = y +ker( f ), then x −y ∈ ker( f ) ⇒ f (x −y ) = 0N ⇒ f (x )− f (y ) = 0N ⇒
f (x ) = f (y ) ⇒ ϕ(x +ker( f )) =ϕ(y +ker( f )). Thus ϕ is well-defined.Let x+ker( f ), y+ker( f )∈M/ker( f ) and let r ∈R , thusϕ((x+ker( f ))+(y+ker( f ))) =
ϕ((x + y )+ker( f )) = f (x + y ) = f (x )+ f (y ) =ϕ(x +ker( f ))+ϕ(y +ker( f )).Also, ϕ(r (x +ker( f ))) =ϕ(r x +ker( f )) = f (r x ) = r f (x ) = rϕ(x +ker( f )). Hence ϕ is
a well-defined left R-homomorphism.Let y ∈ im( f ) ⇒ ∃x ∈M 3 f (x ) = y . Since x+ker( f )∈M/ker( f ) andϕ(x+ker( f )) =
f (x ) = y , thus ϕ is a surjective mapping and hence ϕ is an R-epimorphism.Let x + ker( f ) ∈ ker(ϕ), thus ϕ(x + ker( f )) = 0N ⇒ f (x ) = 0N ⇒ x ∈ ker( f ) ⇒
x + ker( f ) = ker( f ) = 0(M/ker( f )) ⇒ ker(ϕ) = 0(M/ker( f )). By Proposition 2.36, ϕ is an R-monomorphism. Therefore, ϕ is an R-isomorphism and hence M/ker( f )∼= im( f ).
Corollary 2.39. If f : M →N is a left R-epimorphism, then M/ker( f )∼=N .
Definition 2.40. Let M be a left R-module and let m ∈ M . The annihilator of m isdenoted by annR (m ) and defined as follows: annR (m ) = {r ∈R | r m = 0}.
Lemma 2.41. Let M be a left R-module and let m ∈M . Then annR (m ) is a left ideal of aring R.
Proof. Exercise.
Proposition 2.42. A left R-module M is cyclic if and only if M ∼= R/annR (m ) for somem ∈M .
14
Proof. (⇒) Suppose that a left R-module M is cyclic, thus there is m ∈M such thatM =< m >. Define f : R →M by f (r ) = r m for every r ∈ R . It is clear that f is a leftR-epimorphism (H.W.). By Corollary 2.39, R/ker( f )∼=M .Since ker( f ) = {r ∈R | f (r ) = 0}= {r ∈R | r m = 0}= annR (m ), thusM ∼=R/annR (m ).(⇐) Suppose that M ∼= R/annR (m ) for some m ∈ M . Since R/annR (m ) is a cyclic
left R-module generated by 1+annR (m ), thus M is a cyclic left R-module.
Theorem 2.43. (Second Isomorphism Theorem for Modules) If M 1 and M 2 are left sub-modules of a left R-module M such that M 1 ⊆M 2, then M 2/M 1 is a left submodule ofM/M 1 and (M/M 1)/(M 2/M 1)∼=M/M 2.
Proof. Exercise.
Example 2.44. Since < 4>,→< 2>,→ ZZ thus Theorem 2.43 implies thatZ/ < 2>∼= (Z/ < 4>)/(< 2> /< 4>).
Theorem 2.45. (Third Isomorphism Theorem for Modules) If M 1 and M 2 are left sub-modules of a left R-module M , then M 1/(M 1 ∩M 2)∼= (M 1+M 2)/M 2.
Proof. Define ϕ : M 1→ (M 1+M 2)/M 2 by ϕ(x ) = x +M 2, for all x ∈M 1. We can provethat ϕ is a left R-epimorphism (H.W.). By Corollary 2.39, M 1/ker(ϕ)∼= (M 1+M 2)/M 2.
Definition 3.2. A sequence of left R-modules and left R-homomorphism of the form
S : · · · →M n−1f n−1→ M n
f n→M n+1→ ·· · , n ∈Z,is said to be an exact sequence if it is exact at M n between a pair of R-homomorphisms
for each n ∈Z.
Proposition 3.3. Let A, B and C be left R-modules. Then
(1) The sequence 0→ Af→ B is exact (at A) if and only if f is injective.
(2) The sequence Bg→C → 0 is exact (at C ) if and only if g is surjective.
Proof. (1) (⇒) Suppose that the sequence 0 → Af→ B is exact (at A), thus im(0) =
ker( f ) ⇒ ker( f ) = 0A ⇒ f is injective, by Proposition 2.36.(⇐) Suppose that f is injective, thus ker( f ) = 0A (by Proposition 2.36). Since im(0) =
0A = ker( f ) ⇒ the sequence 0→ Af→ B is exact (at A).
(2) Exercise.
Corollary 3.4. The sequence 0→ Af→ B
g→C → 0 of left R-modules is exact if and only if
f is injective, g is surjective, and im( f ) = ker(g ).
Proof. (⇒) Suppose that the sequence 0→ Af→ B
g→ C → 0 of left R-modules is exact,
thus the sequences 0→ Af→ B , A
f→ B
g→ C and B
g→ C → 0 are exact. By proposi-
tion 3.3, f is injective, g is surjective and im( f ) = ker(g ).(⇐) Suppose that f is injective, g is surjective, and im( f ) = ker(g ). By proposi-
tion 3.3, the sequences 0→ Af→ B , A
f→ B
g→ C and B
g→ C → 0 are exact. Thus the
sequence 0→ Af→ B
g→C → 0 of left R-modules is exact.
Definition 3.5. The exact sequence of the form 0 → Af→ B
g→ C → 0 is called a short
exact sequence.
Example 3.6. Let N be a submodule of a left R-module M . Then the sequence
0→Ni→M
π→M/N → 0 is a short exact sequence, where i is the inclusion mappingand π is the natural mapping.
Proof. Since im(i ) = N and ker(π) = N (by Example 2.30), thus im(i ) = ker(π). Since i
is injective and π is surjective it follows from Corollary 3.4 that the sequence 0→Ni→
Mπ→M/N → 0 is a short exact sequence.
Example 3.7. Let f : M →N be a left R-homomorphism. Then
(1) the sequence 0→ ker( f )i→M
π→M/ker( f )→ 0 is a short exact sequence;
(2) the sequence 0→ f (M )i→N
π→N / f (M )→ 0 is a short exact sequence;
(3) the sequence 0→ ker( f )i→M
f→N
π→N / f (M )→ 0 is an exact sequence;
(4) the sequence 0 → ker( f )i→ M
f ′→ f (M ) → 0 is a short exact sequence, where
f ′ : M → f (M ) is defined by f ′(x ) = f (x ), for all x ∈M .
16
Proof. (1) By Example 3.6.(2) Since f (M ) is a left R-submodule of N (by Corollary 2.26) it follows from Exam-
ple 3.6 that the sequence 0→ f (M )i→N
π→N / f (M )→ 0 is a short exact sequence.(3) Since i is injective and π is surjective, thus by Proposition 3.3 we get that 0→
ker( f )i→ M and N
π→ N / f (M )→ 0 are exact sequences. Since im(i ) = ker( f ) ⇒ the
sequence ker( f )i→ M
f→ N is exact. Since im( f ) = f (M ) = ker(π) ⇒ the sequence
Mf→N
π→N / f (M ) is exact. Hence the sequence 0→ ker( f )i→M
f→N
π→N / f (M )→ 0is an exact sequence.
(4) Exercise.
Proposition 3.8. Let α : M →N and β : N → K be left R-homomorphisms. Then(1) ker(βα) =α−1(ker(β ));(2) im(βα) =β (im(α)).
Proof. (1) Let x ∈ ker(βα) ⇒ (βα)(x ) = 0K ⇒ α(x ) ∈ ker(β ) ⇒ x ∈ α−1(ker(β )) ⇒ker(βα)⊆α−1(ker(β )).
(2) Let x ∈ im(βα) ⇒ ∃a ∈M 3 (βα)(a ) = x ⇒ β (α(a )) = x . Since α(a ) ∈ im(α) ⇒x ∈β (im(α)) ⇒ im(βα)⊆β (im(α)).
Conversely, let y ∈ β (im(α)) ⇒ ∃x ∈ im(α) 3 y = β (x ) ⇒ ∃a ∈M 3 α(a ) = x ⇒ y =β (α(a )) = (βα)(a )∈ im(βα) ⇒ β (im(α))⊆ im(βα). Thus im(βα) =β (im(α)).
Corollary 3.9. Let α : M →N and β : N → K be left R-homomorphisms.(1) If β is an R-monomorphism, then ker(βα) = ker(α).(2) If α is an R-epimorphism, then im(βα) = im(β ).
(2) Suppose that α is an R-epimorphism, thus im(α) = N . By Proposition 3.8(2),im(βα) =β (im(α)) =β (N ) = im(β ).
Proposition 3.10. Let S1 : 0→M 1α→M
f→ N → 0 and S2 : 0→ N
g→M 2
β→M 3 → 0 be
short exact sequences of left R-modules. Then the sequence S3 : 0→M 1α→M
g f→M 2
β→
M 3→ 0 is exact.
Proof. Exercise.
Definition 3.11. The diagram
A Bf //
C Dβ //
A
C
α
��
B
D
g
��
of left R-modules and left R-homomorphisms is said to be commutative if g f =βα.
17
Theorem 3.12. (The four lemma) Suppose that the diagram of left R-modules and leftR-homomorphisms
A Bf // B C
g // C Dh //
A ′ B ′f ′ // B ′ C ′
g ′ // C ′ D ′h ′ //
A
A ′
α1
��
B
B ′
α2
��
C
C ′
α3
��
D
D ′
α4
��
is commutative and has exact rows. Then we have(1) if α1 is an epimorphism and α2, α4 are monomorphisms, then α3 is a monomor-
phism;(2) if α1 and α3 are epimorphisms and α4 is a monomorphism, then α2 is an epimor-
phism.
Proof. Exercise. See [Plyth, Theorem 3.9, p. 32]).
Theorem 3.13. (The five lemma) Suppose that the diagram of left R-modules and leftR-homomorphisms
A Bf // B C
g // C Dh // D El //
A ′ B ′f ′ // B ′ C ′
g ′ // C ′ D ′h ′ // D ′ E ′l ′ //
A
A ′
α1
��
B
B ′
α2
��
C
C ′
α3
��
D
D ′
α4
��
E
E ′
α5
��
is commutative and has exact rows. Then we have(1) ifα1 is an epimorphism andα2 andα4 are monomorphisms, thenα3 is a monomor-
phism;(2) if α5 is a monomorphism and α2 and α4 are epimorphisms, then α3 is an epimor-
phism;(3) if α1, α2, α4 and α5 are isomorphisms , then α3 is an isomorphism.
Proof. (1) Suppose that α1 is an epimorphism and α2 and α4 are monomorphisms. Byapplying Theorem 3.12(1) (The four lemma) to the left-hand three squares we see thatα3 is a monomorphism.
(2) Suppose that α5 is a monomorphism and α2 and α4 are epimorphisms. By ap-plying Theorem 3.12(2) (The four lemma) to the right-hand three squares we see thatα3 is an epimorphism.
(3) Suppose thatα1, α2, α4 andα5 are isomorphisms. By (1) above,α3 is a monomor-phism. Also, by (2) above we get that α3 is an epimorphism. Hence α3 is an isomor-phism.
Corollary 3.14. (The short five lemma) Suppose that the diagram of left R-modules andleft R-homomorphisms
0 B// B Cf // C D
g // D 0//
0 B ′// B ′ C ′f ′ // C ′ D ′
g ′ // D ′ 0//
B
B ′
α
��
C
C ′
β
��
D
D ′
γ
��
is commutative and has exact rows. Then we have(1) if α and γ are monomorphisms, then β is a monomorphism;(2) if α and γ are epimorphisms, then β is an epimorphism;(3) if α and γ are isomorphisms , then β is an isomorphism.
18
Proof. Take A = A ′ = E = E ′ = 0 in Theorem 3.13 (The five lemma).
Proposition 3.15. Let the diagram
A Bf // B C
g //
A ′ B ′f ′ // B ′ C ′
g ′ //
A
A ′
α
��
B
B ′
β
��
C
C ′
γ
��
be commutative and let α, β and γ be isomorphisms. Then the sequence Af→ B
g→ C is
exact if and only if the sequence A ′f ′→ B ′
g ′→C ′ is exact.
Proof. (⇒) Suppose that the sequence Af→ B
g→ C is exact, thus im( f ) = ker(g ). Since
α is an epimorphism it follows from Corollary 3.9(2) that im( f ′) = im( f ′α).Since im( f ′α) = im(β f ) (because the above diagram is commutative)=β (im( f )) (by Proposition 3.8(2))=β (ker(g )) (by hypothesis)= ker(g β−1) (by Proposition 3.8(1))= ker(γ−1 g ′) (because the above diagram is commutative)= (g ′)−1(ker(γ−1)) (by Proposition 3.8(1))= (g ′)−1(0) (because γ−1 is a monomorphism)
= ker(g ′), thus im( f ′) = ker(g ′) and hence the sequence A ′f ′→ B ′
g ′→C ′ is exact.
(⇐) Exercise.
Definition 3.16. A left R-monomorphism α : M →N is said to be split monomorphismif im(α) is a direct summand in N .
Definition 3.17. A left R-epimorphism β : N → K is said to be split epimorphism ifker(β ) is a direct summand in N .
Examples 3.18. (1) Let i :< 2 >→ ZZ6 be the inclusion Z -homomorphism. Then i is asplit monomorphism.
(2) Let π :Z Z6→ ZZ6/ < 2> be the natural Z -epimorphism. Then π is a split epimor-phism.
Lemma 3.19. Let α : M →N be a left R-homomorphism. Then we have:(1) if A is a left R-submodule of M , then α−1(α(A)) = A +ker(α).(2) if B is a left R-submodule of N , then α(α−1(B )) = B ∩ im(α).
Proof. Exercise.
Lemma 3.20. Let the diagram
A Bα // B
C
β
������
����
�A
C
λ
��
be commutative, (in other word λ=βα). Then(1) im(α)+ker(β ) =β−1(im(λ));(2) im(α)∩ker(β ) =α(ker(λ)).
be commutative, (in other word λ=βα). Then(1) if λ is an epimorphism, then im(α)+ker(β ) = B;(2) if λ is a monomorphism, then im(α)∩ker(β ) = 0B ;(3) if λ is an isomorphism, then im(α)⊕ker(β ) = B.
Proof. (1) Suppose that λ is an epimorphism, thus im(λ) = C . By Lemma 3.20(1),im(α)+ker(β ) =β−1(im(λ)) =β−1(C ) = B .
(2) Suppose thatλ is a monomorphism, thus ker(λ) = 0A . By Lemma 3.20(2), im(α)∩ker(β ) =α(ker(λ)) =α(0A) = 0B .
(3) By (1) and (2) above.
Proposition 3.22. Letα : M →N be a left R-homomorphism. Thenα is a split monomor-phism if and only if there exists a homomorphism β : N →M with βα= IM .
Proof. (⇒) Suppose that α : M → N is a split monomorphism, thus im(α) is a directsummand of N ⇒ there is a submodule B of N such that N = im(α)⊕ B . Let π : N →im(α) be the projection of N onto im(α) defined by π(α(a ) +b ) = α(a ), for all α(a ) ∈im(α) and b ∈ B . It is clear that π is an epimorphism (H.W)?. Define α0 : M →α(M ) byα0(a ) =α(a ), for all a ∈M . Thus α0 is an isomorphism (H.W.)?.
Put β = α−10 π : N →M . Since α−1
0 and π are left R-homomorphisms, thus β is a leftR-homomorphism.
For all a ∈M we have that (βα)(a ) =β (α(a )) =β (α0(a )) =α−10 (π(α0(a ))) =α−1
0 (α0(a )) =a = IM (a ). Thus βα= IM and hence there is a left R-homomorphism β : N →M suchthat βα= IM .(⇐) Suppose that there is a left R-homomorphism β : N →M such that βα = IM .
Thus α is a monomorphism (H.W.)?. Since IM is an isomorphism it follows from Corol-lary 3.21(3) that im(α)⊕ker(β ) =N and hence im(α) is a direct summand of N . Thus αis a split monomorphism.
Proposition 3.23. Let α : M →N be a left R-homomorphism. Then α is a split epimor-phism if and only if there exists a homomorphism β : N →M with αβ = IN .
Proof. (⇒) Suppose that α : M →N is a split epimorphism, thus ker(α) is a direct sum-mand of M ⇒ there is a submodule A of M such that M = ker(α)⊕A. Let i : A→M bethe inclusion mapping. Define α0 : A →N by α0(a ) = α(a ), for all a ∈ A. It is clear thatα0 is a left R-homomorphism (H.W.)?.
Let a ,b ∈ A such thatα0(a ) =α0(b ) ⇒ α(a ) =α(b ) ⇒ α(a−b ) = 0N ⇒ a−b ∈ ker(α).Since a −b ∈ A ⇒ a −b ∈ A ∩ker(α). Since A ∩ker(α) = 0A ⇒ a −b = 0A ⇒ a =b ⇒ α0
is a left R-monomorphism.
20
Let b ∈ N . Since α : M → N is an epimorphism, thus there is a ∈ M such thatα(a ) = b . Since M = ker(α)⊕A ⇒ a = a 1+b1 with a 1 ∈ A, b1 ∈ ker(α). Thus b = α(a ) =α(a 1+b1) = α(a 1) +α(b1) = α(a 1) + 0N = α(a 1) = α0(a 1). Hence α0 is an epimorphismand thus α0 : A→N is an isomorphism.
Put β = i α−10 : N →M . Since i and α−1
0 are left R-homomorphisms, thus β is a leftR-homomorphism.
For all x ∈N we have that (αβ )(x ) =α(β (x )) =α(i α−10 (x )) =α(α
−10 (x )) =α0(α−1
0 (x )) =x = IN (x ). Thus αβ = IN and hence there is a left R-homomorphism β : N →M suchthat αβ = IN .(⇐) Suppose that there is a left R-homomorphism β : N →M such that αβ = IN .
Thus α is an epimorphism (H.W.)?. Since IN is an isomorphism it follows from Corol-lary 3.21(3) that im(β )⊕ ker(α) =M and hence ker(α) is a direct summand of M . Thusα : M →N is a split epimorphism.
Definition 3.24. An exact sequence Af→ B
g→ C is said to be split on whether im( f ) =
ker(g ) is a direct summand in B.
Definition 3.25. Let n ≥ 2. An exact sequence A1f 1→ A2
f 2→ A3→ ·· ·→ Anf n→ An+1 is said to
be split on whether im( f i ) = ker( f i+1) is a direct summand in A i+1 for all i = 1, 2, ..., n−1.
Corollary 3.26. Let 0→ Af→ B
g→ C → 0 be a short exact sequence of left R-modules.
Then the following statements are equivalent:
(1) the short exact sequence 0→ Af→ B
g→C → 0 is split;
(2) the exact sequence Af→ B
g→C is split;
(3) there is a left R-homomorphism α : B→ A such that α f = IA ;(4)there is a left R-homomorphism β : C → B such that g β = IC ;
Proof. By Proposition 3.22 and Proposition 3.23. (H.W)?
21
4 Direct sum and product of modules, Homomorphismsof direct products and sums
Definition 4.1. Let {M i }i∈I be a family of left R-modules. The direct product of {M i }i∈I
is the cartesian product:∏
i∈IM i = {(a i )i∈I | a i ∈M i for all i ∈ I }
in which (a i )i∈I = (b i )i∈I if and only if a i =b i for all i ∈ I .
Proposition 4.2. Let {M i }i∈I be a family of left R-modules.(1) Define addition on
∏
i∈IM i as follows:
(a i )i∈I + (b i )i∈I = (a i +b i )i∈I , for all (a i )i∈I , (b i )i∈I ∈∏
i∈IM i . Then (∏
i∈IM i ,+) is an
abelian group.(2)∏
i∈IM i is a left R-module.
Proof. (1)Exercise.(2) Define • : R ×
∏
i∈IM i →∏
i∈IM i by r • (a i )i∈I = (r a i )i∈I , for all r ∈ R and for
all (a i )i∈I ∈∏
i∈IM i . Then • is a module multiplication, since for all r, s ∈ R and for all
(a i )i∈I , (b i )i∈I ∈∏
i∈IM i we have that r • ((a i )i∈I + (b i )i∈I ) = r • ((a i + b i )i∈I ) = (r (a i +
b i ))i∈I = (r a i + r b i )i∈I = (r a i )i∈I +(r b i )i∈I = r • (a i )i∈I + r • (b i )i∈I and(r + s ) • (a i )i∈I = ((r + s )a i )i∈I = (r a i + s a i )i∈I = (r a i )i∈I + (s a i )i∈I = r • (a i )i∈I + s •
(a i )i∈I .Also, (r s )• (a i )i∈I = ((r s )a i )i∈I = (r (s a i ))i∈I = r • (s a i )i∈I = r • (s • (a i )i∈I ). Thus • is a
module multiplication and hence∏
i∈IM i is a left R-module.
Definition 4.3. Let {M i }i∈I be a family of left R-modules. The external direct sum of{M i }i∈I is denoted by
∐
i∈IM i and defined as follows:
∐
i∈IM i = {(x i )i∈I ∈
∏
i∈IM i | x i = 0M i for all i but finite many i ∈ I }.
Proposition 4.4. Let {M i }i∈I be a family of left R-modules. Then∐
i∈IM i is a left submod-
ule of a left R-module∏
i∈IM i .
Proof. Since 0M i ∈M i for all i ∈ I , thus 0= (0M i )i∈I ∈∐
i∈IM i and hence
∐
i∈IM i 6=φ. It is
clear that∐
i∈IM i ⊆∏
i∈IM i . Let (x i )i∈I , (yi )i∈I ∈
∐
i∈IM i and let r ∈ R , thus x i = 0M i for all i
but finite many i ∈ I and yi = 0M i for all i but finite many i ∈ I and (x i )i∈I − (yi )i∈I =(x i − yi )i∈I = (z i )i∈I with z i = 0 for all i but finite many i ∈ I and hence (z i )i∈I ∈
∐
i∈IM i .
Thus (x i )i∈I − (yi )i∈I ∈∐
i∈IM i .
Also, r • (x i )i∈I = (r x i )i∈I = (w i )i∈I with w i = 0M i for all i but finite many i ∈ I . Thus(w i )i∈I ∈∐
i∈IM i and hence r • (x i )i∈I ∈
∐
i∈IM i . Therefore,
∐
i∈IM i is a left submodule of a
left R-module∏
i∈IM i .
Corollary 4.5. If the index set I is finite, then∏
i∈IM i =∐
i∈IM i .
22
Proof. Let I = {1, 2, ..., n}, for some n ∈Z+. We will prove thatn∏
i=1M i =
n∐
i=1M i .
By Proposition 4.4,n∐
i=1M i ⊆
n∏
i=1M i . Let x ∈
n∏
i=1M i = M 1 ×M 2 × ...×M n , thus x =
(x1,x2, ...,xn ) with x i ∈M i for all i ∈ I . Since x i = 0 for all i but finite many i ∈ I , thus
x ∈n∐
i=1M i and hence∏
i∈IM i ⊆∐
i∈IM i . Thus∏
i∈IM i =∐
i∈IM i .
Notation 4.6. Let I be a non-empty set and let M be a left R-module. Then letM I =∏
i∈IM i with M i =M for every i ∈ I .
M (I ) =∐
i∈IM i with M i =M for every i ∈ I .
We call M I the direct product of I copies of M and we call M (I ) the direct sum of Icopies of M .
Definition 4.7. Let {M i }i∈I be a family of left R-modules and let j ∈ I .(1) The natural projection from
∏
i∈IM i onto M j is a mapping πj :
∏
i∈IM i →M j defined
by πj ((a i )i∈I ) = a j , for all (a i )i∈I ∈∏
i∈IM i ;
(2) The natural injection from M j into∏
i∈IM i is a mapping i M j : M j →
∏
i∈IM i defined
by i M j (a j ) = (0, 0, ..., 0, a j , 0, 0, ..., 0), for all a j ∈ M j ;
(3) The natural injection from M j into∐
i∈IM i is a mapping i M j : M j →
∐
i∈IM i defined
by i M j (a j ) = (0, 0, ..., 0, a j , 0, 0, ..., 0), for all a j ∈ M j .
Proposition 4.8. Let {M i }i∈I be a family of left R-modules. Then(1) For each j ∈ I , the natural projection πj :
∏
i∈IM i →M j is a left R-epimorphism;
(2) For each j ∈ I , the mapping πj ρ :∐
i∈IM i →M j is a left R-epimorphism;
where ρ :∐
i∈IM i →∏
i∈IM i is the inclusion mapping;
(3) For each j ∈ I , the natural injections i M j : M j →∏
i∈IM i and i M j : M j →
∐
i∈IM i
are left R-monomorphisms and i M j = ρ i M j , where ρ :∐
i∈IM i →∏
i∈IM i is the inclusion
mapping;
(4) πj i M k =
(
IM k if k = j
0 if k 6= j
Proof. (1) Let (a i )i∈I , (b i )i∈I ∈∏
i∈IM i , let r ∈R and let j ∈ I . Thus
πj ((a i )i∈I +(b i )i∈I ) =πj ((a i +b i )i∈I ) = a j +b j =πj ((a i )i∈I )+πj ((b i )i∈I ).
Alsoπj (r (a i )i∈I ) =πj ((r a i )i∈I ) = r a j = rπj ((a i )i∈I ). Thusπj is a left R-homomorphism,for all j ∈ I .
Let a j ∈M j , thus (0, 0, ..., 0, a j , 0, 0, ..., 0) ∈∏
i∈IM i and πj ((0, 0, ..., 0, a j , 0, 0, ..., 0)) = a j .
Hence πj is a left R-epimorphism, for all j ∈ I .(2) Let j ∈ I . Since πj and ρ are left R-homomorphisms, thus πj ρ is a left R-
homomorphism, for all j ∈ I .Let a j ∈M j , thus (0, 0, ..., 0, a j , 0, 0, ..., 0)∈
∐
i∈IM i and
23
πj ρ((0, 0, ..., 0, a j , 0, 0, ..., 0)) =πj (ρ((0, 0, ..., 0, a j , 0, 0, ..., 0))) =πj ((0, 0, ..., 0, a j , 0, 0, ..., 0)) =a j . Thus πj ρ is a left R-epimorphism, for all j ∈ I .
(3) Let j ∈ I , let a j , b j ∈M j and let r ∈R . Thusi M j (a j+b j ) = (0, 0, ..., 0, a j+b j , 0, 0, ..., 0) = (0, 0, ..., 0, a j , 0, 0, ..., 0)+(0, 0, ..., 0,b j , 0, 0, ..., 0) =i M j (a j )+ i M j (b j ) andi M j (r a j ) = (0, 0, ..., 0, r a j , 0, 0, ..., 0) = r (0, 0, ..., 0, a j , 0, 0, ..., 0) = r i M j (a j ). Thus i M j is aleft R-homomorphism, for all j ∈ I .
Also, if i M j (a j ) = i M j (b j ), then (0, 0, ..., 0, a j , 0, 0, ..., 0) = (0, 0, ..., 0,b j , 0, 0, ..., 0) ⇒ a j =b j and hence i M j is a left R-monomorphisms, for all j ∈ I .
Similarly, we can prove that i M j : M j →∐
i∈IM i is a left R-monomorphism (H.W.)
Let a j ∈M j , thusρ i M j (a j ) = ρ(i M j (a j )) = ρ((0, 0, ..., 0, a j , 0, 0, ..., 0)) = (0, 0, ..., 0, a j , 0, 0, ..., 0) = i M j (a j ).Hence ρ i M j = i M j .
(4) Let a k ∈M k .If k = j , then (πj i M k )(a k ) = (πk i M k )(a k ) = πk (i M k (a k )) = πk ((0, 0, ..., 0, a k , 0, 0, ..., 0)) =a k = IM k (a k ) and hence πj i M k = IM k .If k 6= j , then (πj i M k )(a k ) = πj (i M k (a k )) = πj ((0, 0, ..., 0, a k , 0, 0, ..., 0)) = 0 (since j 6= k )and hence πj i M k = 0.
Connection between the internal and external direct sums
The following theorem gives a connection between the internal and external directsums of modules.
Theorem 4.9. Let {M i }i∈I be a family of left R-modules. Then∐
i∈IM i =⊕
i∈IM
′
i where
M′
i = {(0, 0, ..., 0, a i , 0, 0, ..., 0) | a i ∈M i } for all i ∈ I and M′
i∼=M i ,
in other words, the external direct sum of the modules M i is equal to the internal directsum of the submodules M
′
i of∐
i∈IM i isomorphic to M i .
Proof. Let i ∈ I . Define αi : M i → M ′i by αi (a i ) = (0, 0, ..., 0, a i
︷ ︸︸ ︷
i th component
, 0, 0, ..., 0),
for all a i ∈M i . Thus αi is a left R-isomorphism (H.W.) and hence M′
i∼=M i for all i ∈ I .
Let M =∑
i∈IM ′
i and let x ∈ M , thus x ∈∑
i∈IM ′
i and hence x ∈<⋃
i∈IM ′
i >. Thus x =∑
j∈I ′
I ′⊆I and I ′ finite
a j where a j ∈M ′j .
For all j ∈ I ′, let a j = (0, 0, ..., 0,b j , 0, 0, ..., 0) with b j ∈ M j . Since I ′ is finite, thusx ∈∐
i∈IM i and hence∑
i∈IM ′
i ⊆∐
i∈IM i .
Let 0 6= (a i )i∈I ∈∐
i∈IM i and let a j1 6= 0, a j2 6= 0, ..., a jn 6= 0 where j1, j2, ..., jn ∈ I ,
whereas a i = 0 for all other i ∈ I , thus it follows that (a i )i∈I = (0, 0, ..., 0, a j1 , 0, 0, ..., 0) +(0, 0, ..., 0, a j2 , 0, 0, ..., 0)+ ...+(0, 0, ..., 0, a jn , 0, 0, ..., 0)∈M ′
j1+M ′
j2+ ...+M ′
jn⊆∑
i∈IM ′
i . Thus
24
∐
i∈IM i ⊆∑
i∈IM ′
i and hence∐
i∈IM i =∑
i∈IM ′
i . Since for each j ∈ I we have that
M ′j ∩∑
i∈Ii 6=j
M ′i = 0 (H.W.) thus M =
⊕
i∈IM ′
i and hence∐
i∈IM i =⊕
i∈IM
′
i .
Remark 4.10. From now we will denote to the external (or internal) direct sum of thefamily {M i }i∈I of left R-modules by
⊕
i∈IM i and is called the direct sum of the family
{M i }i∈I of left R-modules.
25
Homomorphisms of direct products and sums
Proposition 4.11. Let {M i }i∈I and {Ni }i∈I be two families of left R-modules and let αi :M i →Ni be left R-homomorphisms, for all i ∈ I . Then:(1) The mapping
∏
i∈Iαi :∏
i∈IM i →∏
i∈INi defined by (
∏
i∈Iαi )((a i )i∈I ) = (αi (a i ))i∈I , for all
(a i )i∈I ∈∏
i∈IM i , is a left R-homomorphism;
(2) The mapping⊕
i∈Iαi :⊕
i∈IM i →⊕
i∈INi defined by (
⊕
i∈Iαi )((a i )i∈I ) = (αi (a i ))i∈I , for all
(a i )i∈I ∈⊕
i∈IM i , is a left R-homomorphism.
Proof. (1) Let (a i )i∈I , (b i )i∈I ∈∏
i∈IM i and let r ∈R . Thus
∏
i∈Iαi ((a i )i∈I +(b i )i∈I ) =
∏
i∈Iαi ((a i +b i )i∈I ) = (αi (a i +b i ))i∈I = (αi (a i )+αi (b i ))i∈I =
(αi (a i ))i∈I +(αi (b i ))i∈I =∏
i∈Iαi ((a i )i∈I )+∏
i∈Iαi ((b i )i∈I ).
Also,∏
i∈Iαi (r (a i )i∈I ) =∏
i∈Iαi ((r a i )i∈I ) = (αi (r a i ))i∈I = (rαi (a i ))i∈I = r ((αi (a i ))i∈I )
= r (∏
i∈Iαi ((a i )i∈I )). Thus
∏
i∈Iαi is a left R-homomorphism.
(2) By similar way. Exercise.
Proposition 4.12. Let {M i }i∈I and {Ni }i∈I be two families of left R-modules and let αi :M i →Ni be left R-homomorphisms, for all i ∈ I . Then:
(1)∏
i∈Iαi is monomorphism if and only if for each i ∈ I , αi is monomorphism;
(2)⊕
i∈Iαi is monomorphism if and only if for each i ∈ I , αi is monomorphism;
(3)∏
i∈Iαi is epimorphism if and only if for each i ∈ I , αi is epimorphism;
(4)⊕
i∈Iαi is epimorphism if and only if for each i ∈ I , αi is epimorphism;
(5)∏
i∈Iαi is isomorphism if and only if for each i ∈ I , αi is isomorphism;
(6)⊕
i∈Iαi is isomorphism if and only if for each i ∈ I , αi is isomorphism.
Proof. (1) (⇒) Suppose that∏
i∈Iαi is monomorphism. Let j ∈ I and let a j ,b j ∈M j such
that αj (a j ) = αj (b j ). Put a i = b i = 0 for all i ∈ I and i 6= j , thus αi (a i ) = αi (b i ) = 0 forall i ∈ I and i 6= j and hence (αi (a i ))i∈I = (αi (b i ))i∈I .Thus (∏
i∈Iαi )((a i )i∈I ) = (∏
i∈Iαi )((b i )i∈I ). Since
∏
i∈Iαi is monomorphism, (a i )i∈I = (b i )i∈I
and hence a j =b j . Therefore αj is monomorphism, for each j ∈ I .(⇐) Suppose that αi is monomorphism for all i ∈ I . Let (a i )i∈I , (b i )i∈I ∈
∏
i∈Iαi such
that (∏
i∈Iαi )((a i )i∈I ) = (∏
i∈Iαi )((b i )i∈I ), thus (αi (a i ))i∈I = (αi (b i ))i∈I and hence
αi (a i ) =αi (b i ), for all i ∈ I . Since αi is monomorphism for all i ∈ I , thus a i = b i , for alli ∈ I and hence (a i )i∈I = (b i )i∈I . Thus
∏
i∈Iαi is monomorphism.
(2) By similar way of (1) above. Exercise.(3) By similar way of (4) below. Exercise.(4) (⇒) Suppose that
⊕
i∈Iαi is epimorphism and let j ∈ I . We will prove that
(αi (a i ))i∈I = (0, 0, ..., 0,b j , 0, 0, ..., 0) and hence αj (a j ) = b j with a j ∈M j . Thus αj : M j →N j is epimorphism, for all j ∈ I .(⇐) Suppose that αi : M i →Ni is epimorphism, for all i ∈ I .
Let (b i )i∈I ∈⊕
i∈INi . If (b i )i∈I = (0Ni )i∈I . Since
⊕
i∈Iαi is left R-homomorphism, thus
(⊕
i∈Iαi )((0M i )i∈I ) = (0Ni )i∈I .
If (b i )i∈I 6= (0Ni )i∈I , then (b i )i∈I = (0, 0, ..., 0,b j1 , 0, 0, ..., 0,b j2 , 0, 0, ..., 0,b jn , 0, 0, ..., 0) withb j i 6= 0 and b j i ∈ N j i (i = 1, 2, ..., n). Since αj i : M j i → N j i is epimorphism, thus there isa j i ∈M j i such thatαj i (a j i ) =b j i . Since (0, 0, ..., 0, a j1 , 0, 0, ..., 0, a j2 , 0, 0, ..., 0, a jn , 0, 0, ..., 0)∈⊕
i∈IM i and (⊕
i∈Iαi )((0, 0, ..., 0, a j1 , 0, 0, ..., 0, a j2 , 0, 0, ..., 0, a jn , 0, 0, ..., 0)) =
morphism. By (1) and (3) above we have that αi is monomorphism and epimorphism,for all i ∈ I . Thus αi is isomorphism, for all i ∈ I .(⇐) Exercise.(6) Exercise.
Corollary 4.13. Let {M i }i∈I and {Ni }i∈I be two families of left R-modules and letαi : M i →Ni be left R-homomorphisms, for all i ∈ I . Then:
(1)∏
i∈Iαi is monomorphism if and only if
⊕
i∈Iαi is monomorphism;
(2)∏
i∈Iαi is epimorphism if and only if
⊕
i∈Iαi is epimorphism;
(3)∏
i∈Iαi is isomorphism if and only if
⊕
i∈Iαi is isomorphism.
Proof. Exercise.
Proposition 4.14. Let {M i }i∈I and {Ni }i∈I be two families of left R-modules and letαi : M i →Ni be left R-homomorphisms, for all i ∈ I . Then:
(1) ker(∏
i∈Iαi )∼=∏
i∈Iker(αi ); (2) ker(
⊕
i∈Iαi )∼=⊕
i∈Iker(αi );
(3) im(∏
i∈Iαi )∼=∏
i∈Iim(αi ); (4) im(
⊕
i∈Iαi )∼=⊕
i∈Iim(αi ).
Proof. Exercise. See [Kasch, Lemma 4.3.2, p. 86]).
27
5 Free modules and Finitely presented modules
Definition 5.1. Let M be a left R-module. A subset X of M is said to be linear indepen-
dent if for every x1,x2, ...,xn ∈ X such thatn∑
i=1ri x i = 0, with ri ∈ R for all i = 1, 2, ..., n,
then ri = 0, for all i = 1, 2, ..., n.
Definition 5.2. Let M be a left R-module. A subset X of M is said to be a basis for M if
(i) M =< X > (in other words, ∀m ∈M , m =n∑
i=1ri x i with ri ∈ R and x i ∈ X for all
i = 1, 2, ..., n).(ii) X is linear independent.
Proposition 5.3. Let M be a left R-module and let X be a generating set for M , then X
is a basis for M if and only if for each m ∈M the representation m =n∑
i=1ri x i with ri ∈ R
and x i ∈X for all i = 1, 2, ..., n, is unique.
Proof. (⇒) Suppose that X is a basis for M . Let m ∈M such that m =n∑
i=1ri x i =
n∑
i=1t i x i
with ri , t i ∈R and x i ∈X are distinct elements for all i = 1, 2, ..., n . Thusn∑
i=1(ri−t i )x i = 0.
Since X is linear independent, thus ri − t i = 0, and hence ri = t i for all i = 1, 2, ..., n .
Therefore, for each m ∈M the representation m =n∑
i=1ri x i with ri ∈ R and x i ∈ X for all
i = 1, 2, ..., n , is unique.(⇐)We must prove that X is a basis for M . Since X is a generating set for M (by hy-
pothesis) thus we need only prove that X is linear independent. Suppose thatn∑
i=1ri x i =
0 with ri ∈ R and x i ∈ X for all i = 1, 2, ..., n . Sincen∑
i=10x i = 0 thus
n∑
i=10x i =
n∑
i=1t i x i . Since
the representation of each element in M is unique, thus ri = 0 for all i = 1, 2, ..., n andhence X is linear independent. Thus X is a basis for M .
Definition 5.4. A left R-module M is said to be free if it has a basis.
Examples 5.5. (1) Every ring R with identity 1 is free left R-module.
Proof. Let X = {1}. It is clear that R R =< {1}>=<X >.Let r 1= 0 with r ∈R , thus r = 0 and hence X = {1} is linear independent. Hence X is abasis for a left R-module R . Thus R is a free left R-module.
(2) Z is a free Z-module.(3) Zn is a free Zn -module, for every n ∈Z+.(4) Let R be a ring with identity 1.Then Rn is a free left R-module for every n ∈Z+.
Proof. Let X = {(1, 0, 0, ..., 0)︸ ︷︷ ︸
n−times
, (0, 1, 0, ..., 0)︸ ︷︷ ︸
n−times
, ..., (0, 0, 0, ..., 0, 1)︸ ︷︷ ︸
n−times
}. We will prove that X is a ba-
sis for Rn as left R-module.Let r ∈Rn , thus r = (r1, r2, ..., rn ), with ri ∈R for all i = 1, 2, ..., n .Hence r = (r1, 0, ..., 0) + (0, r2, 0, ..., 0) + · · ·+ (0, 0, ..., 0, rn ) = r1(1, 0, ..., 0) + r2(0, 1, 0, ..., 0) +· · ·+rn (0, 0, ..., 0, 1) and this implies that Rn =<X >.
28
Let s1, s2, ..., sn ∈R such that s1(1, 0, ..., 0)+ s2(0, 1, 0, ..., 0)+ · · ·+ sn (0, 0, ..., 0, 1) = 0, thus(s1, s2, ..., sn ) = (0, 0, ..., 0) and hence s i = 0 for all i = 1, 2, ..., n . Thus X is linear indepen-dent and hence X is a basis for a left R-module Rn . Therefore, Rn is a free left R-modulefor every n ∈Z+.
(5) Let R be a ring with identity 1 and let n ∈Z+. Then the matrix ring Mn×n(R) is a free
left R-module.
Proof. (H.W.)
(6) Let F be a field. Then every left F -module (F-vector space) is free left F -module.
Proof. (H.W.)
(7) Z4 is not free left Z-module.
Proof. It is clear that {1}, {3}, {0, 1}, {0, 3}, {1, 2}, {1, 3}, {2, 3}, {0, 1, 2}, {0, 1, 3}, {0, 2, 3},{1, 2, 3} and Z4 = {0, 1, 2, 3} are all generating sets of Z4 as left Z-module (Why?).If X = {1} is a basis of Z4 as left Z-module. Since 3. 1= 7. 1= 3, thus 3= 7 (by Proposi-tion 5.3) and this is a contradiction. Thus X = {1} is not a basis of Z4 as left Z-module.If X = {3} is a basis of Z4 as left Z-module. Since 1. 3= 5. 3= 3, thus 1= 5 (by Proposi-tion 5.3) and this is a contradiction. Thus X = {1} is not a basis of Z4 as left Z-module.Similarly, we can prove that {0, 1}, {0, 3}, {1, 2}, {1, 3}, {2, 3}, {0, 1, 3}, {1, 2, 3} and Z4 ={0, 1, 2, 3} are not a basis of Z4 as left Z-module. Hence Z4 is not free left Z-module.
(8) In general, Zn is not free left Z-module, for all n ≥ 2.
Theorem 5.6. Let F be a left R-module. Then F is free left R-module if and only ifF ∼=R (I ) as a left R-module for some index I .
Proof. Exercise. (See [Kasch, Lemma 4.4.1, p.88])
Theorem 5.7. The following statements are equivalent for a left R-module M :(1) M is finitely generated left R-module;(2) there is a left R-epimorphism α : Rn →M for some n ∈Z+.
Proof. (1)⇒ (2). Suppose that M is finitely generated left R-module, thus M =< x1,x2, ...,xn >for some n ∈Z+. Define α : Rn →M by α((r1, r2, ..., rn )) = r1x1+ r2x2+ ...+ rn xn , for all(r1, r2, ..., rn ) ∈ Rn . It is clear that α is a left R-epimorphism (H.W.). Thus there is a leftR-epimorphism α : Rn →M for some n ∈Z+.(2)⇒ (1). Suppose that there is a left R-epimorphism α : Rn →M for some n ∈Z+.
Since Rn =< (1, 0, 0, ..., 0), (0, 1, 0, ..., 0), ..., (0, 0, 0, ..., 0, 1)> thus we can prove thatM =<α((1, 0, 0, ..., 0)),α((0, 1, 0, ..., 0)), ...,α((0, 0, 0, ..., 0, 1))> (Why?) and hence M is finitelygenerated left R-module.
Lemma 5.8. Let I be an index set. Then a free left R-module R (I ) is finitely generated ifand only if I is finite set. In other words: a free left R-module R (I ) is finitely generated ifand only if R (I ) =Rn , for some n ∈Z+.
Proof. Exercise.
29
Corollary 5.9. The following statements are equivalent for a left R-module M :(1) There is a left R-epimorphism α : Rn →M for some n ∈Z+;(2) M is a homomorphic image of a finitely generated free left R-module.
Proof. (1)⇒ (2). This is obvious (H.W.).(2)⇒ (1). Suppose that M is a homomorphic image of a finitely generated free left
R-module, thus there is a left R-epimorphism ϕ : F →M with F is a finitely generatedfree left R-module. By Theorem 5.6 and Lemma 5.8, F ∼= Rn , for some n ∈ Z+ andhence there is an isomorphism β : Rn → F . Put α = βϕ : Rn →M . It is clear that α isa left R-epimorphism and hence there is a left R-epimorphism α : Rn → M for somen ∈Z+
Exercise: Show that whether the quotient module of a free left R-module is free ornot and why?
30
Finitely presented modules
Definition 5.10. A left R-module M is said to be finitely presented if there is an exact
sequence 0→ Kf→ F
g→M → 0 of left R-modules, where F is finitely generated and free
and K is finitely generated. Such a sequence will be called a finite presentation of M.
Proposition 5.11. Let n ∈Z+ and let N be a left submodule of a left R-module Rn . if Nis finitely generated, then the left R-module Rn/N is finitely presented.
Proof. By Theorem 5.6, Rn is a free left R-module and by Lemma 5.8 we have that Rn
is finitely generated. Since the sequence 0→Ni→Rn π→Rn/N → 0 is exact and since N
is finitely generated it follows that Rn/N is a finitely presented left R-module.
Corollary 5.12. Let M be a left R-module. Then M is finitely presented if and only ifM ∼= Rn/N for some n ∈ Z+ and for some finitely generated left submodule N of a leftR-module Rn .
Proof. Exercise.
Examples 5.13. (1) For every n , m ∈Z+ the Z -module Z m/ < n > is finitely presented.(2) For every n ∈Z+ the Z -module Zn is finitely presented.
Theorem 5.14. The following statements are equivalent for a left R-module M .(1) M is finitely presented.
(2) There exists an exact sequence Rm α→ Rnβ→ M → 0 of left R-modules for some
m , n ∈Z+(3) There exists an exact sequence 0→ F1→ F0→M → 0 of left R-modules, where F1
and F0 are finitely generated free R-modules.
Proof. (1)⇒ (2) Suppose that M is a finitely presented left R-module, thus there is an
exact sequence 0→ Kf→ F
g→M → 0 of left R-modules, where F is finitely generated
and free and K is finitely generated. By Lemma 5.8, F ∼= Rn for some n ∈ Z+. Supposethat K is generated by m elements, thus from Theorem 5.7 we have that there is a leftR-epimorphism λ : Rm → K .Put α= f λ and β = g , thus im(α) = im( f λ) = f (im(λ)) (by Proposition 3.8(2))
= f (K ) = im( f ) = ker(g ) = ker(β ). Hence the sequence Rm α→ Rnβ→ M → 0 of left
R-modules is exact. Hence there exists an exact sequence Rm α→ Rnβ→ M → 0 of left
R-modules for some m , n ∈Z+
(2)⇒ (1) Suppose that the sequence Rm α→ Rnβ→M → 0 of left R-modules is exact
and let K = ker(β ). Thus the sequence 0→ Ki→ Rn
β→M → 0 is exact. Since im(α) =
ker(β ) it follows that im(α) = K and hence α(Rn ) = K . Since Rn is finitely generated itfollows from Theorem 5.7 that K is finitely generated. Thus we get an exact sequence
0→ Ki→Rn
β→M → 0 of left R-modules with Rn is finitely generated free left R-module
and K is finitely generated module and hence M is finitely presented.(2)⇔ (3) Exercise.
31
Corollary 5.15. Every finitely presented left R-module is finitely generated.
Proof. Let M be a finitely presented left R-module. By Theorem 5.14, there exist m , n ∈Z+ such that the sequence Rm α→ Rn
β→M → 0 of left R-modules is exact and hence β
is an epimorphism. Thus Theorem 5.7 implies that M is finitely generated.
Proposition 5.16. Every finitely generated free left R-module is finitely presented.
Proof. Let M be a finitely generated free left R-module, thus M ∼= Rn for some n ∈ Z+(by Lemma 5.8). Let λ1 : R→Rn+1 be the injection mapping defined byλ1(a ) = (a , 0, 0, ..., 0︸ ︷︷ ︸
n times
), for all a ∈ R and let ρ1 : Rn+1 → Rn be the projection mapping
defined by ρ1(a 1, a 2, ..., a n+1) = (a 2, a 3, ..., a n+1), for all (a 1, a 2, ..., a n+1)∈Rn+1. It is clearthat λ1 is a left R-monomorphism and ρ1 is a left R-epimorphism (H.W.).
It is an easy to prove that the sequence 0→ Rλ1→ Rn+1
ρ1→ Rn → 0 of left R-modules isexact (H.W.). Since Rn+1 is finitely generated free left R-module and R is finitely gen-erated left R-module, Rn is a finitely presented left R-module and hence M is finitelypresented.
32
References
[Anderson and Fuller] F. W. Anderson and K. R. Fuller, Rings and Categories of Mod-ules, Springer-Verlag, 1992.
[Bland] P. E. Bland, Rings and Their Modules, Walter de Gruyter GmbH and Co. KG,Berlin/New York, 2011.
[Grillet] P. A. Grillet, Abstract Algebra, Springer Science and Business Media, 2007.
[Hungerford] Thomas W. Hungerford, Algebra, Springer-Verlag, New York, 1974.
[Kasch] F. Kasch, Modules and Rings, Academic Press, London, New York, 1982.
[Lam] T. Y. Lam, Lectures on Modules and Rings, Springer-Verlag, New York, 1999.
[Plyth] T. S. Plyth, Module theory an approach to linear algebra, 1977.