Postgraduate module theory 2013- lectures

1. 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

2. 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 together
with
(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) (rs) • 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 elements
appear 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 M
together 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 • (rs) = (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. MR ) denotes to left (resp. right) R-module M.
Exercise: If the ring R is commutative, then a module M is left R-module if and
only 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 − sm;
(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

3. (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 ideal
of (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 andS is a subring of R with 1S = 1R , then M
is a unitary left S-module as well. For instance the field is an -module, a -module
and a -module.
Proof. Since M is a unitary left R-module, (M,+) is an abelian group and there is a
module multiplication • : R × M → M with 1R • m = m. Define ∗ : S × M → M by
s ∗ 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. Since
1S = 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 -module.
Proof. Let M be any abelian group, let a ∈ M and let n ∈ . Define the module multi-
plication na : × M → M as follows:
If n > 0, then na = a + a + ··· + a (n times); if n = 0, then na = 0; if n < 0, then
na = −((−n)a) = −a − a − ··· − a (| n | times).
Let n,m ∈ and let a ∈ M, thus
(n+m)a = a +a +···+a (n+m times) = (a + a + ··· + a)
n times
+(a + a + ··· + a)
m times
= na +ma.
Also, (nm)a = a + a + ··· + a (nm times)
= (a + a + ··· + a)
m times
+(a + a + ··· + a)
m times
+··· + (a + a + ··· + a)
m times
n times
= ma + ma + ··· + ma
n times
= n(ma).
Let n ∈ and let a,b ∈ M, thus n(a + b) = (a + b) + (a + b) + ··· + (a + b) (n
times) = a + a + ··· + a
n times
+b +b + ··· +b
n times
= na + nb.
Thus the multiplication na : × M → M is a module multiplication and hence M
is a -module. Since 1a = a it follows that M is a unitary -module
3

4. Example 1.11. Let R be a ring with 1 and let n ∈ +. Define Rn = {(a1,a2,··· ,an ) | ai ∈
R, for alli}. Make Rn into a left R-module by componentwise addition and multiplica-
tion by elements of R as follows:
(a1,a2,··· ,an ) + (b1,b2,··· ,bn ) = (a1 + b1,a2 + b2,··· ,an + bn ) and r(a1,··· ,an ) =
(ra1,··· ,ran ), for all r ∈ R and (a1,a2,··· ,an ),(b1,b2,··· ,bn ) ∈ Rn .
Proof. Exercise
Example 1.12. If Mn (R) is the set of n × n matrices over a ring R, then Mn (R) is an
additive abelian group under matrix addition. If (ai j ) ∈ Mn (R) and r ∈ R, then the
operation r(ai j ) = (rai j ) makes Mn (R) into a left R-module. Mn (R) is also a right R-
module under the operation (ai j )r = (ai 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 M
is a left unitary End(M)-module.
Proof. Exercise.
4

5. Submodules
Definition 1.15. Let R be a ring and let M be a left R-module. A left submodule of M
is a subgroup N of M such that r • n ∈ N, for all r ∈ R, and for all n ∈ N, where • is
the module multiplication defined on M. We will use N → M to denote that N is a left
submodule of M.
Proposition 1.16. Let R be a ring, let M be a left R-module and let N be a nonempty
subset 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 of
M. Then N is a left submodule of M if and only if N is a left R-module with the same
addition and module multiplication on M.
Proof. (⇒) Suppose that N is a left submodule of M. Thus (by Definition 1.15) we have
that N is a subgroup of (M,+) such that r • n ∈ N, for all r ∈ R, n ∈ N, where • is the
module multiplication defined on M. Since (M,+) is an abelian group, (N,+) is an
abelian 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 addition
and 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 → N
defined by r ∗ n = r • n, for all r ∈ R, n ∈ N is a module multiplication. Since N is a
subset 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-module
are 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-module
with the same addition and module multiplication on R R. Since I is a subset of R it
follows from Proposition 1.17 that I is a left submodule of a left R-module R R. Hence
every 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 1.16
we have that a −b ∈ N and r • a ∈ N for all a,b ∈ N and for all r ∈ R , where • is the
module 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-module
are exactly the left ideals of a ring (R,+,.).
5

6. Exercise: Find all left submodules of the following left R-modules:
(1) ; (2) F F, where F is a field; (3) ; (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 submodules
of 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 subspaces
that there is no difference between subspaces and submodules of a left F-module M.
Example 1.21. Let M be a left -module. Then the left submodules of a left -module
M 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 submodules
of M.
Define N1 ∩ N2 = {x ∈ M | x ∈ N1 andx ∈ N2} and
N1 + N2 = {y ∈ M | y = a +b witha ∈ 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 ,ra ∈ N1 and a −b ,ra ∈ N2 and this
implies that a −b ,ra ∈ 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 subgroups
of 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 = a1 +b1 and y = a2 +b2, where a1,a2 ∈ N1
and b1,b2 ∈ N2. Thus x − y = a1 + b1 − a2 − b2 = a1 − a2 + b1 − b2 and rx = r(a1 +
b1) = ra1 + rb1. Since N1 → M and N2 → M it follows from Proposition 1.16 that
a1 − a2, ra1 ∈ N1 and b1 − b2, rb1 ∈ N2 and this implies that x − y, rx ∈ N1 + N2. By
Proposition 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∈I
Ni = {x ∈ M | x ∈ Ni for all i ∈ I }. Then
i∈I
Ni 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 submodules
of 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 follows
that x,b ∈ C and hence x −b ∈ C. Since a ∈ A and a = x −b it follows that a ∈ A ∩C
6

7. 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 that
B ∩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 already
have (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 it
follows 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 that
A ⊆ 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 of
M?
7

8. 2 Finitely generated modules, Internal direct sums, di-
rect summands, Quotient modules, Homomorphisms
of 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 M
generated 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 left
submodule of M that contains a subset X.
Proof. By Proposition 1.23, < X > is a submodule of M. Let N be a submodule of M
contains X. By definition of < X > we have that < X >=
A →M
with X⊆A
A. Since N → M and
X ⊆ N it follows that < X >⊆ N. Thus < X > is the smallest submodule of M that
contains a subset X.
Lemma 2.3. Let X be a nonempty subset of a left R-module M and let
N = {
n
i=1
ri xi | ri ∈ R,xi ∈ X,n ∈ +}. Then N → M.
Proof. Exercise
Proposition 2.4. Let X be a subset of a left R-module M. Then
< X >=



{
n
i=1
ri xi | ri ∈ R,xi ∈ X,n ∈ +}
0
,if X = φ
,if X = φ
Proof. Suppose that X = φ. Since {0} → M and X = φ ⊆ {0}, thus
A →M
with φ⊆A
A = 0.
Since < φ >=
A →M
with φ⊆A
A ⇒< X >=< φ >= 0.
Suppose that X = φ and let N = {
n
i=1
ri xi | ri ∈ R,xi ∈ X,n ∈ +}.
We will prove that < X >= N. Let a ∈ N ⇒ a =
m
i=1
si yi with si ∈ R, yi ∈ X, m ∈ +.
Let A be any submodule of M contains X. Since yi ∈ X and si ∈ R ⇒ si yi ∈ A,∀i =
1,2,...,m ⇒ a =
m
i=1
si yi ∈ A ⇒ N ⊆ A, for all submodule A of M contains X ⇒ N ⊆
A →M
with X⊆A
A.
Since < X >=
A →M
with X⊆A
A ⇒ N ⊆< X > .
8

9. By Lemma 2.2, < X > is the smallest left submodule of M contains X. Since N is a
left submodule of M (by Lemma 2.3) and X ⊆ N ((H.W) Why?) it follows that < X >⊆ N.
Thus < X >= N = {
n
i=1
ri xi | ri ∈ R,xi ∈ X,n ∈ +}.
Definition 2.5. (1) A left R-module M is said to be finitely generated if it generated by a
finite 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} contains
one element only, that is M =< {a} >.
Remarks 2.6. (1) If M is a finitely generated left R-module generated by a subset X =
{a1,a2,...,an } for some n ∈ +, then we will write M =< a1,a2,...,an >. Also, from
Proposition 2.4 we get that M =< a1,a2,...,an >= {
n
i=1
ri ai | ri ∈ R}.
(2) If M is a cyclic left R-module generated by a subset X = {a}, then we will write
M =< a >. Also, from Proposition 2.4 we get that M =< a >= {ra | 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∈I
Ni >=



{
i∈I
ai | ai ∈ Ni , I ⊆ I withI is finite}
0
,if I = φ
,if I = φ
That is, in the case I = φ, <
i∈I
Ni > is the set of all finite sums ai with ai ∈ Ni .
Proof. If I = φ, then
i∈I
Ni = φ. By Proposition 2.4, <
i∈I
Ni >= 0.
If I = φ, then
i∈I
Ni = φ. By Proposition 2.4, <
i∈I
Ni >= {
n
j =1
rj xj | rj ∈ R,xj ∈
i∈I
Ni , n ∈ +}. Let x ∈<
i∈I
Ni >, thus x =
n
j =1
rj xj , for some rj ∈ R, xj ∈
i∈I
Ni and n ∈ +.
If we bring together all summands rj xj which lie in a fixed Ni to form a sum xi and if
we treat with the remaining summands similarly then it follows that x =
n
j =1
rj xj =
i∈I
xi .
Thus <
i∈I
Ni >⊆ {
i∈I
ai | ai ∈ Ni , I ⊆ I withI is finite}.
Conversely, let x ∈ {
i∈I
ai | ai ∈ Ni , I ⊆ I withI is finite}, thus x =
j ∈I
bj with bj ∈
Nj , I ⊆ I withI is finite. We can write I = {1,2,...,n}, for some n ∈ +. Thus x =
n
j =1
bj
with bj ∈
i∈I
Ni ⇒ x ∈<
i∈I
Ni > and hence {
i∈I
ai | ai ∈ Ni , I ⊆ I withI is finite} ⊆<
i∈I
Ni >. Thus <
i∈I
Ni >= {
i∈I
ai | ai ∈ Ni , I ⊆ I withI is finite}.
9

10. 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∈I
Ni to denote the sum of the submodules {Ni }i∈I
and defined by
i∈I
Ni =<
i∈I
Ni >.
Remark 2.10. Let {N1,N2,...,Nn } be a family of submodules of a left R-modules M. Then
n
i=1
Ni = {
n
i=1
ai | ai ∈ Ni }.
Internal direct sums and direct summands
Definition 2.11. (Internal direct sum) Let M be a left R-module and let {Ni }i∈I be a
family 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∈I
Ni if the following two conditions are hold:
(1) M =
i∈I
Ni and
(2) Nj ∩
i∈I
i=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 let
N1 = {(x,0,0) | x ∈ R}, N2 = {(0,y,0) | y ∈ R} and N3 = {(0,0,z) | z ∈ R}.
Prove that:
(1) N1,N2 andN3 are left submodules of M;
(2) M is the internal direct sum of the submodules N1,N2 andN3.
Proof. (1) Exercise.
(2) Since N1,N2 andN3 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}andN2 + N3 =
{(0,y,z) | y,z ∈ R}. Thus N1 ∩ (N2 + N3) = 0, N2 ∩ (N1 + N3) = 0, andN3 ∩ (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 trivial
submodules 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 andN2.
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) Nj ∩
i∈I
i=j
Ni = 0, for all j ∈ I .
(2) For every x ∈ M the representation x =
i∈I
bi with bi ∈ Ni , I ⊆ I , I finite, is
unique in the following sense:
If x =
i∈I
bi =
i∈I
ci with bi , ci ∈ Ni , then bi = ci for all i ∈ I .
10

11. Proof. (1) ⇒ (2). Let x ∈ M such that x =
i∈I
bi =
i∈I
ci with bi , ci ∈ Ni I ⊆ I , I finite ⇒
i∈I
bi −
i∈I
ci = 0.
Thus for all j ∈ I we have that bj +
i∈I
i=j
bi − (cj +
i∈I
i=j
ci ) = 0 ⇒ bj − cj =
i∈I
i=j
(ci −bi ).
Since
i∈I
i=j
(ci −bi ) ∈
i∈I
i=j
Ni ⇒ bj −cj ∈
i∈I
i=j
Ni . Since bj −cj ∈ Nj ⇒ bj −cj ∈ Nj ∩
i∈I
i=j
Ni .
Since Nj ∩
i∈I
i=j
Ni ⊆ Nj ∩
i∈I
i=j
Ni and since Nj ∩
i∈I
i=j
Ni = 0 ⇒ Nj ∩
i∈I
i=j
Ni = 0 ⇒ bj −cj = 0 ⇒
bj = cj , for all j ∈ I .
(2) ⇒ (1). Let b ∈ Nj ∩
i∈I
i=j
Ni , thus b = bj ∈ Nj andb ∈
i∈I
i=j
Ni ⇒ there is a finite subset
I ⊆ I with j /∈ I and b = bj =
i∈I
bi , bi ∈ Bi .
If we add to the left-hand side the summands 0 ∈ Bi , i ∈ I and to the right-hand
side the summand 0 ∈ Bj , then the same finite index set I ∪{j } appears on both sides
and from uniqueness in (2) it follows that b = bj = 0. Thus Nj ∩
i∈I
i=j
Ni = 0.
Definition 2.16. (Direct summand) A submodule N of a left R-module M is said to be
a 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 = as a left -module. Prove that < 0 > and are the only direct
summands of M = .
Proof. Assume that N is a direct summand of M with N = 0 and N = . Thus N =<
n > with n ∈ and n = 0, n = 1, n = −1. Since N is a direct summand of M, there is
a submodule K =< m > of M for some m ∈ such that =< n > ⊕ < m >⇒< n >
∩ < m >= 0. Since nm ∈< n > ∩ < m >⇒ nm = 0. Since n = 0 and Z is an integral
domain ⇒ m = 0. Since =< n > + < m >⇒ =< n >. Since either =< 1 > or
=< −1 >⇒ either n = 1 or n = −1 and this is a contradiction. Thus < 0 > and are
the only direct summands of M = .
11

12. 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 submodule
of M. The (additive, abelian) quotient group M/N can be made into a left R-module by
defining a module multiplication • : R × M/N → M/N by r • (x + N) = (rx) + N, for all
r ∈ 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) if
for all a,b ∈ N and r ∈ R, then f (a +b) = f (a)+ f (b) and f (ra) = r f (a).
(2) A left R-module homomorphism is a monomorphism if it is injective and is an
epimorphism if it is surjective. A left R-module homomorphism is an isomorphism if
it 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 of
f ). The R-module M/im(f ) is called the cokernel of f . The cokernel of f will be denoted
by coker(f ).
(4) Define HomR (N,M) to be the set of all left R-module homomorphisms from N
into 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 from
Proposition 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) that
im(f ) is a left submodule of N.
12

13. Example 2.27. [Hungerford, p. 170] For any modules the zero map 0 : A → B given by
0(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 iN : N → M defined by iN (x) = x, for all x ∈ N is a left R-monomorphism.
Proof. Let x,y ∈ N and let r ∈ R. Thus iN (x + y ) = x + y = iN (x) + iN (y ) and iN (rx) =
rx = r iN (x). Hence inclusion mapping iN is a left R-homomorphism. Since iN is an
injective mapping ⇒ iN is a left R-monomorphism.
Example 2.29. Let M be a left R-module. The identity mapping IM : M → M defined by
IM (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. Then
the 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(rx) = rx + N = r(x + N) = rπ(x). Thus π : M → M/N is a left
R-homomorphism. Since π is a surjective mapping ⇒ π is an R-epimorphism.
Also, ker(π) = {x ∈ M | π(x) = 0} = {x ∈ M | x + N = N} = {x ∈ M | x ∈ N} = N.
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 -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 ∈ , 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 -homomorphism.
(⇐) Suppose that f : G1 → G2 is -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 a
function. Then f is a left F-homomorphism if and only if it is linear transformation
over F.
Proof. Exercise.
Examples 2.33. Let f : → defined by f (x) = 2x, for all x ∈ . Then f is a left Z-
module homomorphism but it is not ring homomorphism, since f (1) = 2 = 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 → M2×2(R) by f (x) =
x 0
0 x
for all x ∈ R.
Then f is a left R-monomorphism.
Proof. Exercise.
13

14. Proposition 2.35. Let N be a left R-submodule of a left R-module M. Then A is a left
R-submodule of a left R-module M/N if and only if there is a unique left R-submodule
B of M such that N ⊆ B and A = B/N.
Proof. Exercise.
Proposition 2.36. Let α : N → M be a left R-homomorphism. Then α is R-monomorphism
if and only if ker(α) = {0N }.
Proof. Exercise.
Proposition 2.37. Let f : N → M and g : M → K be left R-homomorphisms. Then
g ◦ 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 )(ra) = g (f (ra)) = g (r f (a)) = r g (f (a)) = r((g ◦ f )(a)). Thus g ◦ f : N → K
is 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, thus ϕ((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(rx +ker(f )) = f (rx) = r f (x) = rϕ(x +ker(f )). Hence ϕ is
a well-defined left R-homomorphism.
Let y ∈ im(f ) ⇒ ∃x ∈ M f (x) = y . Since x +ker(f ) ∈ M/ker(f ) and ϕ(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 is
denoted 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 a
ring R.
Proof. Exercise.
Proposition 2.42. A left R-module M is cyclic if and only if M ∼= R/annR (m) for some
m ∈ M.
14

15. Proof. (⇒) Suppose that a left R-module M is cyclic, thus there is m ∈ M such that
M =< m >. Define f : R → M by f (r) = r m for every r ∈ R. It is clear that f is a left
R-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), thus
M ∼= 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 M1 and M2 are left sub-
modules of a left R-module M such that M1 ⊆ M2, then M2/M1 is a left submodule of
M/M1 and (M/M1)/(M2/M1) ∼= M/M2.
Proof. Exercise.
Example 2.44. Since < 4 >→< 2 >→ thus Theorem 2.43 implies that
Z/ < 2 >∼= (Z/ < 4 >)/(< 2 > / < 4 >).
Theorem 2.45. (Third Isomorphism Theorem for Modules) If M1 and M2 are left sub-
modules of a left R-module M, then M1/(M1 ∩ M2) ∼= (M1 + M2)/M2.
Proof. Define ϕ : M1 → (M1 + M2)/M2 by ϕ(x) = x + M2, for all x ∈ M1. We can prove
that ϕ is a left R-epimorphism (H.W.). By Corollary 2.39, M1/ker(ϕ) ∼= (M1 + M2)/M2.
Since ker(ϕ) = {x ∈ M1 | ϕ(x) = 0((M1+M2)/M2) = M2} = {x ∈ M1 | x + M2 = M2} = {x ∈
M1 | x ∈ M2} = M1 ∩ M2, thus M1/(M1 ∩ M2) ∼= (M1 + M2)/M2.
Corollary 2.46. If M = M1 ⊕ M2, then M/M2
∼= M1.
Proof. Exercise
15

16. 3 Exact and split sequences of modules
Definition 3.1. A sequence M1
f
→ M
g
→ M2 of left R-modules and left R-homomorphism
is said to be exact at M if im(f ) = ker(g ).
Definition 3.2. A sequence of left R-modules and left R-homomorphism of the form
S : ··· → Mn−1
fn−1
→ Mn
fn
→ Mn+1 → ···, n ∈ ,
is said to be an exact sequence if it is exact at Mn between a pair of R-homomorphisms
for each n ∈ .
Proposition 3.3. Let A, B and C be left R-modules. Then
(1) The sequence 0 → A
f
→ B is exact (at A) if and only if f is injective.
(2) The sequence B
g
→ C → 0 is exact (at C) if and only if g is surjective.
Proof. (1) (⇒) Suppose that the sequence 0 → A
f
→ 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 → A
f
→ B is exact (at A).
(2) Exercise.
Corollary 3.4. The sequence 0 → A
f
→ 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 → A
f
→ B
g
→ C → 0 of left R-modules is exact,
thus the sequences 0 → A
f
→ 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 → A
f
→ B, A
f
→ B
g
→ C and B
g
→ C → 0 are exact. Thus the
sequence 0 → A
f
→ B
g
→ C → 0 of left R-modules is exact.
Definition 3.5. The exact sequence of the form 0 → A
f
→ 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 → N
i
→ M
π
→ M/N → 0 is a short exact sequence, where i is the inclusion mapping
and 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 → N
i
→
M
π
→ 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

17. 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
M
f
→ N
π
→ N/f (M) is exact. Hence the sequence 0 → ker(f )
i
→ M
f
→ N
π
→ N/f (M) → 0
is 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)).
Conversely, let x ∈ α−1(ker(β)) ⇒ α(x) ∈ ker(β) ⇒ β(α(x)) = 0K ⇒ (βα)(x) = 0K ⇒
x ∈ ker(βα) ⇒ α−1(ker(β)) ⊆ ker(βα). Thus ker(βα) = α−1(ker(β)).
(2) Let x ∈ im(βα) ⇒ ∃a ∈ M (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) y = β(x) ⇒ ∃a ∈ M 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).
Proof. (1) Suppose that β is an R-monomorphism, thus ker(β) = 0N (by Proposition 2.36).
By Proposition 3.8(1), ker(βα) = α−1(ker(β)) = α−1(0N ) = ker(α).
(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 → M1
α
→ M
f
→ N → 0 and S2 : 0 → N
g
→ M2
β
→ M3 → 0 be
short exact sequences of left R-modules. Then the sequence S3 : 0 → M1
α
→ M
g f
→ M2
β
→
M3 → 0 is exact.
Proof. Exercise.
Definition 3.11. The diagram
A B
f
//
C D
β
//
A
C
α
B
D
g
of left R-modules and left R-homomorphisms is said to be commutative if g f = βα.
17

18. Theorem 3.12. (The four lemma) Suppose that the diagram of left R-modules and left
R-homomorphisms
A B
f
// B C
g
// C D
h //
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 left
R-homomorphisms
A B
f
// B C
g
// C D
h // D E
l //
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) if α1 is an epimorphism and α2 and α4 are monomorphisms, then α3 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. By
applying Theorem 3.12(1) (The four lemma) to the left-hand three squares we see that
α3 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 that
α3 is an epimorphism.
(3) Suppose that α1, α2, α4 and α5 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 and
left R-homomorphisms
0 B// B C
f
// 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

19. Proof. Take A = A = E = E = 0 in Theorem 3.13 (The five lemma).
Proposition 3.15. Let the diagram
A B
f
// 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 A
f
→ B
g
→ C is
exact if and only if the sequence A
f
→ B
g
→ C is exact.
Proof. (⇒) Suppose that the sequence A
f
→ 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 monomorphism
if im(α) is a direct summand in N.
Definition 3.17. A left R-epimorphism β : N → K is said to be split epimorphism if
ker(β) is a direct summand in N.
Examples 3.18. (1) Let i : 2 → ZZ6 be the inclusion Z-homomorphism. Then i is a
split 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 B
α // 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)).
19

20. Proof. (1) Since λ = βα, thus im(λ) = im(βα) = β(im(α)) (by Proposition 3.8(2)) ⇒
β−1(im(λ)) = β−1(β(im(α))) = im(α)+ ker(β) (by Lemma 3.19(1)).
(2) ker(λ) = ker(βα) = α−1(ker(β)) (by Proposition 3.8(1)).
Thus α(ker(λ)) = α(α−1(ker(β))) = im(α) ∩ ker(β) (by Lemma 3.19(2)).
Corollary 3.21. Let the diagram
A B
α // B
C
β

A
C
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 that λ 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. Let α : M → N be a left R-homomorphism. Then α 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 direct
summand 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) by
α0(a) = α(a), for all a ∈ M. Thus α0 is an isomorphism (H.W.)?.
Put β = α−1
0 π : N → M. Since α−1
0 and π are left R-homomorphisms, thus β is a left
R-homomorphism.
For all a ∈ M we have that (βα)(a) = β(α(a)) = β(α0(a)) = α−1
0 (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 such
that 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 α
is 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 be
the inclusion mapping. Define α0 : A → N by α0(a) = α(a), for all a ∈ A. It is clear that
α0 is a left R-homomorphism (H.W.)?.
Let a,b ∈ A such that α0(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

21. Let b ∈ N. Since α : M → N is an epimorphism, thus there is a ∈ M such that
α(a) = b. Since M = ker(α) ⊕ A ⇒ a = a1 +b1 with a1 ∈ A, b1 ∈ ker(α). Thus b = α(a) =
α(a1 +b1) = α(a1) + α(b1) = α(a1) + 0N = α(a1) = α0(a1). Hence α0 is an epimorphism
and thus α0 : A → N is an isomorphism.
Put β = i α−1
0 : N → M. Since i and α−1
0 are left R-homomorphisms, thus β is a left
R-homomorphism.
For all x ∈ N we have that (αβ)(x) = α(β(x)) = α(i α−1
0 (x)) = α(α−1
0 (x)) = α0(α−1
0 (x)) =
x = IN (x). Thus αβ = IN and hence there is a left R-homomorphism β : N → M such
that 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. Thus
α : M → N is a split epimorphism.
Definition 3.24. An exact sequence A
f
→ 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 A1
f1
→ A2
f2
→ A3 → ··· → An
fn
→ An+1 is said to
be split on whether im(fi ) = ker(fi+1) is a direct summand in Ai+1 for all i = 1,2,...,n −1.
Corollary 3.26. Let 0 → A
f
→ 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 → A
f
→ B
g
→ C → 0 is split;
(2) the exact sequence A
f
→ 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

22. 4 Direct sum and product of modules, Homomorphisms
of direct products and sums
Definition 4.1. Let {Mi }i∈I be a family of left R-modules. The direct product of {Mi }i∈I
is the cartesian product:
i∈I
Mi = {(ai )i∈I | ai ∈ Mi for all i ∈ I }
in which (ai )i∈I = (bi )i∈I if and only if ai = bi for all i ∈ I .
Proposition 4.2. Let {Mi }i∈I be a family of left R-modules.
(1) Define addition on
i∈I
Mi as follows:
(ai )i∈I + (bi )i∈I = (ai +bi )i∈I , for all (ai )i∈I , (bi )i∈I ∈
i∈I
Mi . Then (
i∈I
Mi ,+) is an
abelian group.
(2)
i∈I
Mi is a left R-module.
Proof. (1)Exercise.
(2) Define • : R ×
i∈I
Mi →
i∈I
Mi by r • (ai )i∈I = (rai )i∈I , for all r ∈ R and for
all (ai )i∈I ∈
i∈I
Mi . Then • is a module multiplication, since for all r,s ∈ R and for all
(ai )i∈I , (bi )i∈I ∈
i∈I
Mi we have that r • ((ai )i∈I + (bi )i∈I ) = r • ((ai + bi )i∈I ) = (r(ai +
bi ))i∈I = (rai + rbi )i∈I = (rai )i∈I + (rbi )i∈I = r • (ai )i∈I + r • (bi )i∈I and
(r + s) • (ai )i∈I = ((r + s)ai )i∈I = (rai + sai )i∈I = (rai )i∈I + (sai )i∈I = r • (ai )i∈I + s •
(ai )i∈I .
Also, (rs)•(ai )i∈I = ((rs)ai )i∈I = (r(sai ))i∈I = r •(sai )i∈I = r •(s •(ai )i∈I ). Thus • is a
module multiplication and hence
i∈I
Mi is a left R-module.
Definition 4.3. Let {Mi }i∈I be a family of left R-modules. The external direct sum of
{Mi }i∈I is denoted by
i∈I
Mi and defined as follows:
i∈I
Mi = {(xi )i∈I ∈
i∈I
Mi | xi = 0Mi
for all i but finite many i ∈ I }.
Proposition 4.4. Let {Mi }i∈I be a family of left R-modules. Then
i∈I
Mi is a left submod-
ule of a left R-module
i∈I
Mi .
Proof. Since 0Mi
∈ Mi for all i ∈ I , thus 0 = (0Mi
)i∈I ∈
i∈I
Mi and hence
i∈I
Mi = φ. It is
clear that
i∈I
Mi ⊆
i∈I
Mi . Let (xi )i∈I , (yi )i∈I ∈
i∈I
Mi and let r ∈ R, thus xi = 0Mi
for all i
but finite many i ∈ I and yi = 0Mi
for all i but finite many i ∈ I and (xi )i∈I − (yi )i∈I =
(xi − yi )i∈I = (zi )i∈I with zi = 0 for all i but finite many i ∈ I and hence (zi )i∈I ∈
i∈I
Mi .
Thus (xi )i∈I − (yi )i∈I ∈
i∈I
Mi .
Also, r •(xi )i∈I = (rxi )i∈I = (wi )i∈I with wi = 0Mi
for all i but finite many i ∈ I . Thus
(wi )i∈I ∈
i∈I
Mi and hence r • (xi )i∈I ∈
i∈I
Mi . Therefore,
i∈I
Mi is a left submodule of a
left R-module
i∈I
Mi .
Corollary 4.5. If the index set I is finite, then
i∈I
Mi =
i∈I
Mi .
22

23. Proof. Let I = {1,2,...,n}, for some n ∈ +. We will prove that
n
i=1
Mi =
n
i=1
Mi .
By Proposition 4.4,
n
i=1
Mi ⊆
n
i=1
Mi . Let x ∈
n
i=1
Mi = M1 × M2 × ... × Mn , thus x =
(x1,x2,...,xn ) with xi ∈ Mi for all i ∈ I . Since xi = 0 for all i but finite many i ∈ I , thus
x ∈
n
i=1
Mi and hence
i∈I
Mi ⊆
i∈I
Mi . Thus
i∈I
Mi =
i∈I
Mi .
Notation 4.6. Let I be a non-empty set and let M be a left R-module. Then let
MI =
i∈I
Mi with Mi = M for every i ∈ I .
M(I ) =
i∈I
Mi with Mi = M for every i ∈ I .
We call MI the direct product of I copies of M and we call M(I ) the direct sum of I
copies of M.
Definition 4.7. Let {Mi }i∈I be a family of left R-modules and let j ∈ I .
(1) The natural projection from
i∈I
Mi onto Mj is a mapping πj :
i∈I
Mi → Mj defined
by πj ((ai )i∈I ) = a j , for all (ai )i∈I ∈
i∈I
Mi ;
(2) The natural injection from Mj into
i∈I
Mi is a mapping iMj
: Mj →
i∈I
Mi defined
by iMj
(a j ) = (0,0,...,0,a j ,0,0,...,0), for all a j ∈ Mj ;
(3) The natural injection from Mj into
i∈I
Mi is a mapping iMj
: Mj →
i∈I
Mi defined
by iMj
(a j ) = (0,0,...,0,a j ,0,0,...,0), for all a j ∈ Mj .
Proposition 4.8. Let {Mi }i∈I be a family of left R-modules. Then
(1) For each j ∈ I , the natural projection πj :
i∈I
Mi → Mj is a left R-epimorphism;
(2) For each j ∈ I , the mapping πj ρ :
i∈I
Mi → Mj is a left R-epimorphism;
where ρ :
i∈I
Mi →
i∈I
Mi is the inclusion mapping;
(3) For each j ∈ I , the natural injections iMj
: Mj →
i∈I
Mi and iMj
: Mj →
i∈I
Mi
are left R-monomorphisms and iMj
= ρ iMj
, where ρ :
i∈I
Mi →
i∈I
Mi is the inclusion
mapping;
(4) πj iMk
=
IMk
if k = j
0 if k = j
Proof. (1) Let (ai )i∈I , (bi )i∈I ∈
i∈I
Mi , let r ∈ R and let j ∈ I . Thus
πj ((ai )i∈I + (bi )i∈I ) = πj ((ai +bi )i∈I ) = a j +bj = πj ((ai )i∈I ) + πj ((bi )i∈I ).
Also πj (r(ai )i∈I ) = πj ((rai )i∈I ) = ra j = rπj ((ai )i∈I ). Thus πj is a left R-homomorphism,
for all j ∈ I .
Let a j ∈ Mj , thus (0,0,...,0,a j ,0,0,...,0) ∈
i∈I
Mi 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 ∈ Mj , thus (0,0,...,0,a j ,0,0,...,0) ∈
i∈I
Mi and
23

24. 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 , bj ∈ Mj and let r ∈ R. Thus
iMj
(a j +bj ) = (0,0,...,0,a j +bj ,0,0,...,0) = (0,0,...,0,a j ,0,0,...,0)+(0,0,...,0,bj ,0,0,...,0) =
iMj
(a j )+ iMj
(bj ) and
iMj
(ra j ) = (0,0,...,0,ra j ,0,0,...,0) = r(0,0,...,0,a j ,0,0,...,0) = r iMj
(a j ). Thus iMj
is a
left R-homomorphism, for all j ∈ I .
Also, if iMj
(a j ) = iMj
(bj ), then (0,0,...,0,a j ,0,0,...,0) = (0,0,...,0,bj ,0,0,...,0) ⇒ a j =
bj and hence iMj
is a left R-monomorphisms, for all j ∈ I .
Similarly, we can prove that iMj
: Mj →
i∈I
Mi is a left R-monomorphism (H.W.)
Let a j ∈ Mj , thus
ρ iMj
(a j ) = ρ(iMj
(a j )) = ρ((0,0,...,0,a j ,0,0,...,0)) = (0,0,...,0,a j ,0,0,...,0) = iMj
(a j ).
Hence ρ iMj
= iMj
.
(4) Let ak ∈ Mk .
If k = j , then (πj iMk
)(ak ) = (πk iMk
)(ak ) = πk (iMk
(ak )) = πk ((0,0,...,0,ak ,0,0,...,0)) =
ak = IMk
(ak ) and hence πj iMk
= IMk
.
If k = j , then (πj iMk
)(ak ) = πj (iMk
(ak )) = πj ((0,0,...,0,ak ,0,0,...,0)) = 0 (since j = k)
and hence πj iMk
= 0.
Connection between the internal and external direct sums
The following theorem gives a connection between the internal and external direct
sums of modules.
Theorem 4.9. Let {Mi }i∈I be a family of left R-modules. Then
i∈I
Mi =
i∈I
Mi where
Mi = {(0,0,...,0,ai ,0,0,...,0) | ai ∈ Mi } for all i ∈ I and Mi
∼= Mi ,
in other words, the external direct sum of the modules Mi is equal to the internal direct
sum of the submodules Mi of
i∈I
Mi isomorphic to Mi .
Proof. Let i ∈ I . Define αi : Mi → Mi by αi (ai ) = (0,0,...,0, ai
ith component
,0,0,...,0),
for all ai ∈ Mi . Thus αi is a left R-isomorphism (H.W.) and hence Mi
∼= Mi for all i ∈ I .
Let M =
i∈I
Mi and let x ∈ M, thus x ∈
i∈I
Mi and hence x ∈
i∈I
Mi . Thus x =
j ∈I
I ⊆I and I finite
a j where a j ∈ Mj .
For all j ∈ I , let a j = (0,0,...,0,bj ,0,0,...,0) with bj ∈ Mj . Since I is finite, thus
x ∈
i∈I
Mi and hence
i∈I
Mi ⊆
i∈I
Mi .
Let 0 = (ai )i∈I ∈
i∈I
Mi and let a j1
= 0,a j2
= 0,...,a jn
= 0 where j1, j2,..., jn ∈ I ,
whereas ai = 0 for all other i ∈ I , thus it follows that (ai )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) ∈ Mj1
+Mj2
+...+Mjn
⊆
i∈I
Mi . Thus
24

25. i∈I
Mi ⊆
i∈I
Mi and hence
i∈I
Mi =
i∈I
Mi . Since for each j ∈ I we have that
Mj ∩
i∈I
i=j
Mi = 0 (H.W.) thus M =
i∈I
Mi and hence
i∈I
Mi =
i∈I
Mi .
Remark 4.10. From now we will denote to the external (or internal) direct sum of the
family {Mi }i∈I of left R-modules by
i∈I
Mi and is called the direct sum of the family
{Mi }i∈I of left R-modules.
25

26. Homomorphisms of direct products and sums
Proposition 4.11. Let {Mi }i∈I and {Ni }i∈I be two families of left R-modules and let αi :
Mi → Ni be left R-homomorphisms, for all i ∈ I . Then:
(1) The mapping
i∈I
αi :
i∈I
Mi →
i∈I
Ni defined by (
i∈I
αi )((ai )i∈I ) = (αi (ai ))i∈I , for all
(ai )i∈I ∈
i∈I
Mi , is a left R-homomorphism;
(2) The mapping
i∈I
αi :
i∈I
Mi →
i∈I
Ni defined by (
i∈I
αi )((ai )i∈I ) = (αi (ai ))i∈I , for all
(ai )i∈I ∈
i∈I
Mi , is a left R-homomorphism.
Proof. (1) Let (ai )i∈I , (bi )i∈I ∈
i∈I
Mi and let r ∈ R. Thus
i∈I
αi ((ai )i∈I + (bi )i∈I ) =
i∈I
αi ((ai +bi )i∈I ) = (αi (ai +bi ))i∈I = (αi (ai )+ αi (bi ))i∈I =
(αi (ai ))i∈I + (αi (bi ))i∈I =
i∈I
αi ((ai )i∈I ) +
i∈I
αi ((bi )i∈I ).
Also,
i∈I
αi (r(ai )i∈I ) =
i∈I
αi ((rai )i∈I ) = (αi (rai ))i∈I = (rαi (ai ))i∈I = r((αi (ai ))i∈I )
= r(
i∈I
αi ((ai )i∈I )). Thus
i∈I
αi is a left R-homomorphism.
(2) By similar way. Exercise.
Proposition 4.12. Let {Mi }i∈I and {Ni }i∈I be two families of left R-modules and let αi :
Mi → Ni be left R-homomorphisms, for all i ∈ I . Then:
(1)
i∈I
αi is monomorphism if and only if for each i ∈ I , αi is monomorphism;
(2)
i∈I
αi is monomorphism if and only if for each i ∈ I , αi is monomorphism;
(3)
i∈I
αi is epimorphism if and only if for each i ∈ I , αi is epimorphism;
(4)
i∈I
αi is epimorphism if and only if for each i ∈ I , αi is epimorphism;
(5)
i∈I
αi is isomorphism if and only if for each i ∈ I , αi is isomorphism;
(6)
i∈I
αi is isomorphism if and only if for each i ∈ I , αi is isomorphism.
Proof. (1) (⇒) Suppose that
i∈I
αi is monomorphism. Let j ∈ I and let a j ,bj ∈ Mj such
that αj (a j ) = αj (bj ). Put ai = bi = 0 for all i ∈ I and i = j , thus αi (ai ) = αi (bi ) = 0 for
all i ∈ I and i = j and hence (αi (ai ))i∈I = (αi (bi ))i∈I .
Thus (
i∈I
αi )((ai )i∈I ) = (
i∈I
αi )((bi )i∈I ). Since
i∈I
αi is monomorphism, (ai )i∈I = (bi )i∈I
and hence a j = bj . Therefore αj is monomorphism, for each j ∈ I .
(⇐) Suppose that αi is monomorphism for all i ∈ I . Let (ai )i∈I , (bi )i∈I ∈
i∈I
αi such
that (
i∈I
αi )((ai )i∈I ) = (
i∈I
αi )((bi )i∈I ), thus (αi (ai ))i∈I = (αi (bi ))i∈I and hence
αi (ai ) = αi (bi ), for all i ∈ I . Since αi is monomorphism for all i ∈ I , thus ai = bi , for all
i ∈ I and hence (ai )i∈I = (bi )i∈I . Thus
i∈I
αi is monomorphism.
(2) By similar way of (1) above. Exercise.
(3) By similar way of (4) below. Exercise.
(4) (⇒) Suppose that
i∈I
αi is epimorphism and let j ∈ I . We will prove that
αj : Mj → Nj is epimorphism.
26

27. Let bj ∈ Nj , thus (0,0,...,0, bj
j th component
,0,0,...,0) ∈
i∈I
Ni . Since
i∈I
αi is epimor-
phism, there is (ai )i∈I ∈
i∈I
Mi such that (
i∈I
αi )((ai )i∈I ) = (0,0,...,0,bj ,0,0,...,0). Thus
(αi (ai ))i∈I = (0,0,...,0,bj ,0,0,...,0) and hence αj (a j ) = bj with a j ∈ Mj . Thus αj : Mj →
Nj is epimorphism, for all j ∈ I .
(⇐) Suppose that αi : Mi → Ni is epimorphism, for all i ∈ I .
Let (bi )i∈I ∈
i∈I
Ni . If (bi )i∈I = (0Ni
)i∈I . Since
i∈I
αi is left R-homomorphism, thus
(
i∈I
αi )((0Mi
)i∈I ) = (0Ni
)i∈I .
If (bi )i∈I = (0Ni
)i∈I , then (bi )i∈I = (0,0,...,0,bj1
,0,0,...,0,bj2
,0,0,...,0,bjn
,0,0,...,0) with
bji
= 0 and bji
∈ Nji
(i = 1,2,...,n). Since αji
: Mji
→ Nji
is epimorphism, thus there is
a ji
∈ Mji
such that αji
(a ji
) = bji
. Since (0,0,...,0,a j1
,0,0,...,0,a j2
,0,0,...,0,a jn
,0,0,...,0) ∈
i∈I
Mi and (
i∈I
αi )((0,0,...,0,a j1
,0,0,...,0,a j2
,0,0,...,0,a jn
,0,0,...,0)) =
(0,0,...,0,αj1
(a j1
),0,0,...,0,αj2
(a j2
),0,0,...,0,αjn
(a jn
),0,0,...,0) = (bi )i∈I , thus
i∈I
αi is epi-
morphism.
(5) (⇒) Suppose that
i∈I
αi is isomorphism, thus
i∈I
αi is monomorphism and epi-
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 {Mi }i∈I and {Ni }i∈I be two families of left R-modules and let
αi : Mi → Ni be left R-homomorphisms, for all i ∈ I . Then:
(1)
i∈I
αi is monomorphism if and only if
i∈I
αi is monomorphism;
(2)
i∈I
αi is epimorphism if and only if
i∈I
αi is epimorphism;
(3)
i∈I
αi is isomorphism if and only if
i∈I
αi is isomorphism.
Proof. Exercise.
Proposition 4.14. Let {Mi }i∈I and {Ni }i∈I be two families of left R-modules and let
αi : Mi → Ni be left R-homomorphisms, for all i ∈ I . Then:
(1) ker(
i∈I
αi ) ∼=
i∈I
ker(αi ); (2) ker(
i∈I
αi ) ∼=
i∈I
ker(αi );
(3) im(
i∈I
αi ) ∼=
i∈I
im(αi ); (4) im(
i∈I
αi ) ∼=
i∈I
im(αi ).
Proof. Exercise. See [Kasch, Lemma 4.3.2, p. 86]).
27

28. 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 that
n
i=1
ri xi = 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=1
ri xi with ri ∈ R and xi ∈ 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=1
ri xi with ri ∈ R
and xi ∈ 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=1
ri xi =
n
i=1
ti xi
with ri , ti ∈ R and xi ∈ X are distinct elements for all i = 1,2,...,n. Thus
n
i=1
(ri −ti )xi = 0.
Since X is linear independent, thus ri − ti = 0, and hence ri = ti for all i = 1,2,...,n.
Therefore, for each m ∈ M the representation m =
n
i=1
ri xi with ri ∈ R and xi ∈ 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 that
n
i=1
ri xi =
0 with ri ∈ R and xi ∈ X for all i = 1,2,...,n. Since
n
i=1
0xi = 0 thus
n
i=1
0xi =
n
i=1
ti xi . Since
the representation of each element in M is unique, thus ri = 0 for all i = 1,2,...,n and
hence 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 a
basis for a left R-module R. Thus R is a free left R-module.
(2) is a free -module.
(3) n is a free n -module, for every n ∈ +.
(4) Let R be a ring with identity 1.Then Rn is a free left R-module for every n ∈ +.
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

29. 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 si = 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-module
for every n ∈ +.
(5) Let R be a ring with identity 1 and let n ∈ +. Then the matrix ring M
n×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) 4 is not free left -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 4 = {0,1,2,3} are all generating sets of 4 as left -module (Why?).
If X = {1} is a basis of 4 as left -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 4 as left -module.
If X = {3} is a basis of 4 as left -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 4 as left -module.
Similarly, we can prove that {0,1}, {0,3}, {1,2}, {1,3}, {2,3}, {0,1,3}, {1,2,3} and 4 =
{0,1,2,3} are not a basis of 4 as left -module. Hence 4 is not free left -module.
(8) In general, n is not free left -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 if
F ∼= 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 ∈ +.
Proof. (1) ⇒ (2). Suppose that M is finitely generated left R-module, thus M = x1,x2,...,xn
for some n ∈ +. 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 left
R-epimorphism α : Rn → M for some n ∈ +.
(2) ⇒ (1). Suppose that there is a left R-epimorphism α : Rn → M for some n ∈ +.
Since Rn = (1,0,0,...,0),(0,1,0,...,0),...,(0,0,0,...,0,1) thus we can prove that
M = α((1,0,0,...,0)),α((0,1,0,...,0)),...,α((0,0,0,...,0,1)) (Why?) and hence M is finitely
generated left R-module.
Lemma 5.8. Let I be an index set. Then a free left R-module R(I ) is finitely generated if
and only if I is finite set. In other words: a free left R-module R(I ) is finitely generated if
and only if R(I ) = Rn , for some n ∈ +.
Proof. Exercise.
29

30. 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 ∈ +;
(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 generated
free left R-module. By Theorem 5.6 and Lemma 5.8, F ∼= Rn , for some n ∈ + and
hence there is an isomorphism β : Rn → F. Put α = βϕ : Rn → M. It is clear that α is
a left R-epimorphism and hence there is a left R-epimorphism α : Rn → M for some
n ∈ +
Exercise: Show that whether the quotient module of a free left R-module is free or
not and why?
30

31. Finitely presented modules
Definition 5.10. A left R-module M is said to be finitely presented if there is an exact
sequence 0 → K
f
→ 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 ∈ + and let N be a left submodule of a left R-module Rn . if N
is 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 → N
i
→ 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 if
M ∼= Rn /N for some n ∈ + and for some finitely generated left submodule N of a left
R-module Rn .
Proof. Exercise.
Examples 5.13. (1) For every n,m ∈ + the Z-module Zm / n is finitely presented.
(2) For every n ∈ + 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
→ Rn
β
→ M → 0 of left R-modules for some
m, n ∈ +
(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 → K
f
→ 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 ∈ +. Suppose
that K is generated by m elements, thus from Theorem 5.7 we have that there is a left
R-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
→ Rn
β
→ M → 0 of left
R-modules is exact. Hence there exists an exact sequence Rm
α
→ Rn
β
→ M → 0 of left
R-modules for some m, n ∈ +
(2) ⇒ (1) Suppose that the sequence Rm
α
→ Rn
β
→ M → 0 of left R-modules is exact
and let K = ker(β). Thus the sequence 0 → K
i
→ 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 it
follows from Theorem 5.7 that K is finitely generated. Thus we get an exact sequence
0 → K
i
→ 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

32. 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 ∈
+ 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 ∈ +
(by Lemma 5.8). Let λ1 : R → Rn+1 be the injection mapping defined by
λ1(a) = (a,0,0,...,0
n times
), for all a ∈ R and let ρ1 : Rn+1 → Rn be the projection mapping
defined by ρ1(a1,a2,...,an+1) = (a2,a3,...,an+1), for all (a1,a2,...,an+1) ∈ Rn+1. It is clear
that 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 → R
λ1
→ Rn+1
ρ1
→ Rn → 0 of left R-modules is
exact (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 finitely
presented.
32

33. 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.
33