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.

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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 .

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(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 )

︸ ︷︷ ︸

n times

+(a +a + · · ·+a )︸ ︷︷ ︸

m times

= na+m a .

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

=m a +m a + · · ·+m a︸ ︷︷ ︸

n times

= n (m 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

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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.

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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 ,+, .).

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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

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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 ?

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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 > .

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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}.

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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.

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 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.

Since ker(ϕ) = {x ∈M 1 |ϕ(x ) = 0((M 1+M 2)/M 2) =M 2}= {x ∈M 1 | x +M 2 =M 2}= {x ∈M 1 | x ∈M 2}=M 1 ∩M 2, thus M 1/(M 1 ∩M 2)∼= (M 1+M 2)/M 2.

Corollary 2.46. If M =M 1⊕M 2, then M/M 2∼=M 1.

Proof. Exercise

15

3 Exact and split sequences of modules

Definition 3.1. A sequence M 1f→M

g→M 2 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 : · · · →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 )).

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 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 ).

Proof. (1) Suppose thatβ 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→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)).

19

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 Bα // 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 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

αj : M j →N j is epimorphism.

26

Let b j ∈ N j , thus (0, 0, ..., 0, b j︷ ︸︸ ︷

j th component

, 0, 0, ..., 0) ∈⊕

i∈INi . Since⊕

i∈Iαi is epimor-

phism, there is (a i )i∈I ∈⊕

i∈IM i such that (

⊕

i∈Iαi )((a i )i∈I ) = (0, 0, ..., 0,b j , 0, 0, ..., 0). Thus

(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)) =

(0, 0, ..., 0,αj1(a j1), 0, 0, ..., 0,αj2(a j2), 0, 0, ..., 0,αjn (a jn ), 0, 0, ..., 0) = (b i )i∈I , thus⊕

i∈Iαi is epi-

morphism.(5) (⇒) Suppose that

∏

i∈Iαi is isomorphism, thus

∏

i∈Iαi 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 {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]).

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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.

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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.

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