2. Free Modules
Math 540, Lecture notes
Dr. Alshammari
2. Free Modules
2.1. Direct Sums and Products
Let I be some indexing set. Suppose for each i Î I , M i is an R-module.
 The direct product of the family {M i }i Î I , written P i Î I M i , is defined by
P i Î I M i = {(x i )i Î I : x i Î M i },
i.e. it is just their Cartesian product.
 The direct product P i Î I M i can be made into an R-module, where


Addition is defined componentwise (xi )i Î I + (xi¢)i Î I = (xi + xi¢)i Î I .
Scalar multiplication is defined by r (x i )i Î I := (rx i )i Î I , for any r Î R and
(x i )i Î I Î P i Î I M i . i.e. r multiplied by each component.
 Note that, for each j Î I , the natural projection p j : P i Î I M i ® M j : p j (x i )i Î I = x j is an Rhomomorphism.
 The direct sum of the family {M i }i Î I , denoted Åi Î I M i , is defined as the set of all elements
(x i )i Î I Î Õ i Î I M i having only finitely many nonzero components, that is
Åi Î I M i = {(x i )i Î I Î P i Î I M i : x i = 0, for all but finit ely many i Î I } .
 Åi Î I M i is in fact an R-submodule of P i Î I M i .
 Note that, for each j Î I , the natural inclusion i j : M i ® Åi Î I M i : i j (x ) = (xi )i Î I , where
x i = 0 for i ¹ j and x j = x is a homomorphism. Thus each M j is embedded as a submodule of
Åi M i .
 We also note that for any y Î Åi Î I M i , y = (x i )i Î I where x i Î M i x i = 0 for all but a finite
number of i Î I . Then å i Î I i i (x i ) is really a finite sum and is equal to y.
2.1.1. Universal Property of Products
Given an R-module N, there is a one to one correspondence between
f
f (n ) = (f (n )i )i Î I , f (n )i = fi (n )
{maps f
: N ® P iÎ I Mi
¬ ¾¾
{ fi }i Î I
«
}
f
{sequences of maps {fi }i Î I , fi
¾ ¾®
: N ® Mi
}
{ fi }i Î I , fi = pi o f
Since addition and scalar multiplication is componentwise. It follows that g is a homomorphism iff each
gi is a homomorphism. This says that P i Î I M i has the property that for any R-module N with
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2. Free Modules
Math 540, Lecture notes
Dr. Alshammari
homomorphisms fi : N ® M i , then there exists a unique homomorphism f : N ® P i Î I M i such that
pi o f = fi . i.e. such that the following diagram is commutative
N
fk
fj
f
Mk
k
i I Mi
Mj
j
This property is called the universal property of the direct product.
Theorem 1. The universal property determines P i Î I M i up to isomorphism.
Proof. Let M be an R-module with maps a i : M ® M i for each i so that M has the same property.
Since P i Î I M i satisfies the universal property, there exists a unique homomorphism g : M ® P i Î I M i
such that the following diagram commutes.
M
k
j
g
Mk
Mj
i I Mi
k
j
On the other hand since M also satisfies the universal property, there exists a unique
homomorphism h : P i Î I M i ® M such that the following diagram commutes.
M
k
j
g
Mk
Mj
h
k
i I Mi
j
Notice that the homomorphism g o h : P i Î I M i ® P i Î I M i satisfies p j o g o h = a j o h = p j for all j Î I .
Since the identity id is the unique such map  g o h = id on P i Î I M i . Similarly, h o g = id on M,
hence g and h are isomorphisms.

Corollary 2.
2.1.2. Universal Property of direct sums
Let N be an R-module. Then there is a 1-1 correspondence between
{
homomorphisms f : Åi Î I M i ® N
}« {
sequences of homomorphisms { fi }i Î I
where fi :M i ® N
9
}
2. Free Modules
Math 540, Lecture notes
Dr. Alshammari
Given f, then { fi } is defined by fi = f o ii . Given a sequence { fi } of homomorphisms fi : M i ® N ,
then f is define by f ((x i )i Î I ) = å i Î I fi x i . This sum makes sense because only a finite number of
components of (x i )i Î I are nonzero.
f : f ((x i )i Î I ) = å
¬ ¾¾
i Î I fi x i
{
homomorphisms f : Åi Î I M i ® N
}
{ fi }i Î I
{
«
sequences of homomorphisms {fi }i Î I
where fi :M i ® N
¾ ¾®
f
{ fi }i Î I : fi = f o i i
This correspondence can be rephrased as a property of the direct sum: If N is any R-module with
homomorphisms fi : M i ® N for all i Î I , then there exists a unique homomorphism
f : Åi Î I M i ® N such that fi = f o ii i.e. such that the following diagram of maps commutes:
k
Mk
j
i I Mi
f
fk
Mj
fj
N
This property is called the universal property of the direct sum.
Theorem 3. The universal property characterizes Åi Î I M i up to isomorphism.
Proof. Exercise.
Consider I = {1, 2} , then the sum and product properties are described explicitly in the following
diagram
N
n
f1
f
m1
1
M1 M2
(m 1 , m 2 )
1
M1
M2
2
M2
f2
m2
f
f2 (n )
m1
2
m2
(0, m 2 )
(m 1 , m 2 )
( f1 ( n ) , f2 ( n ))
M1
(m 1 , 0)
M1
f2
n
f1 (n )
m1
n
M2
f1
f1 ( m 1 )
f1 (m 1 )
m2
f2 ( m 2 )
N
f2 (m 2 )
Exercise 4. Let M1, M2 be submodules of an R-module M. Prove that M 1 Å M 2 @ M 1 + M 2 iff
M1 Ç M 2 = 0 .
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}
2. Free Modules
Math 540, Lecture notes
Dr. Alshammari
2.2. Free Modules
Let S be a nonempty subset of an R-module M. Then the set of all finite linear combinations of elements
of S is denoted S i.e.
S = {r1s1 + r2s2 + K + rt st : ri Î R , si Î S , for i = 1, 2, K , t }
which is a submodule called the module generated by S. The set S is called linearly independent if for any
finite linear combination of the form
r1s1 + r2s2 + K + rt st = 0 Þ ri = 0 , for all i.
NOTE. If S = {x i }i Î I for some indexing set I, then elements S
assume ri = 0 for all but finitely many i Î I .
are expressed as S i Î I ri x i where we
Definition 5. An R-module M is said to be free if M = 0 or M = S , for some nonempty linearly
independent set, and in this case we call S a basis for M over R. Similarly we call M a finitely generated Rmodule if M = S where S is finite.
Example 6. R n is a finitely generated free R-module since R n = {e1, e2 , K , en } , where
e1 = (1, 0, 0, K , 0) , e2 = (0, 1, 0, K , 0) , …, en = (0, 0, K , 0, 1) .
In general, if I is an indexing set, then the direct sum Åi Î I R is free with basis {ei }i Î I .
Example 7. The quotient R I is finitely generated over R, since R I = {1 + I } , but not free
unless I = 0 .
Example 8. R [x ] is free over R as R [x ] = {1, x , x 2 , x 3 , K } . But not finitely generated, since any
finite set of polynomials must have bounded degree.
Lemma.9. If M is free R-module with basis {xi }i Î I , then each m
Î M is uniquely represented as a
linear combination.
Proof. Suppose there are two families {ai }i Î I and {bi }i Î I of elements of R so that
å
iÎ I
ai xi =
å
bx
iÎ I i i
Þ
å
iÎ I
(ai - bi )x i = 0 Þ (ai - bi ) = 0 , for all i.

Therefore, the two families must be the same.
Theorem 10. Let M is a free R-module with basis {x i }i Î I . If N is an R-module and {y i }i Î I is a
family of elements of N. Then there exists a unique homomorphism f : M ® N such that f (x i ) = yi
for all i.
Proof. Let x be an element of M. There is a unique family {ai }i Î I of elements of R such that
x = å
iÎ I
ai x i . Define
f (x ) =
å
iÎ I
11
ai yi
2. Free Modules
Math 540, Lecture notes
Dr. Alshammari
It is easy to see that f is a homomorphism with f (x i ) = yi for all i. To see that it is unique follows
because it must satisfy f (x ) = å i Î I ai f (x i ) .

Corollary 11. If in the theorem we assume {y i }i Î I is a basis for N, then f is an isomorphism.
Proof. Here we get a homomorphism g : N ® M such that g (yi ) = x i for all i. So f o g is the
identity map N and g o f is the identity map on M. Hence f and g are isomorphisms.

Corollary 12. Every free module M over R is isomorphic to a direct sum Åi Î I R .
Proof. If M has the basis {m i }i Î I , then consider the R-module Åi Î I R with basis {ei }i Î I . The result

follows from Corollary12.
Corollary 13. M is finitely generated over iff M is homomorphic image of a free finitely generated Rmodule.
Proof. Suppose M is generated by the set S = {m 1, m 2, K , m k } . Consider the free module Åki= 1R
which is generated by the basis {ei }ik= 1 . By Theorem10, there exists a unique homomorphism
j : Åki= 1R ® M with j (ei ) = m i . Since S generates M, it follows that j is surjective.

2.3. The existence of free modules
Let S be a nonempty set. We construct for any ring R a free R-module generated by S. Define
S := all functions j : S ® R with j (x ) = 0 , for all but finitely many x Î S .
Then S is an abelian group, if for j , y Î S , we define (j + y )(x ) = j (x ) + y (x ). S is an
R-module, if we define (r j )(s ) = r j (s ) , for all r Î R , j Î S . For each x Î S , let l x Î S be
the function l x (x ) = 1 and l x (y ) = 0 for y ¹ x . Then the family {l x }x Î S forms a basis for S :
Let l x i Î {l x }x Î S , so that
r1l x1 + r2l x 2 + K + rm l x m = 0 ,
then
(r1l x1 + r2l x 2 + K + rm l x m )(x i ) = 0 (x i )
r1l x1 (x i ) + r2l x 2 (x i ) + K + rm l x m (x i ) = 0
ri = 0
Let j Î S be zero outside the set {x1, x 2, K , x m } Í S . If j (x i ) = ri , for i = 1, 2, K , m , then
j = r1l x1 + r2l x 2 + K + rm l m .

Corollary 14. Any R-module is a homomorphic image of a free module.
Proof. If M is an arbitrary R-module we can choose S := M , so that M is free with basis
{l m }m Î M . From Theorem 10 there is a homomorphism f : M
12
® M such that f (l m ) = m .

Math 540, Lecture notes
2. Free Modules
Dr. Alshammari
Exercise 15. If R is a commutative ring, M an R-module, and I an ideal of R, then show that:
a. IM = {å ai m i : ai Î I , m i Î M } is a submodule of M.
b. M IM is a module over R I by setting (r + I )(x + IM ) := rx + IM , for r + I Î R I
and x + IM Î M IM .
Theorem 16. If M is a free R-module and R is commutative, then any two bases have same
cardinality.
If J is a maximal ideal, then R J is a field and M JM is a vector space over R J . Now given a basis
{x i }i Î I for M. Using the projection p : M ® M JM , we claim that the family {p (x i )}i Î I is a basis
for the vector space M JM over R J : Clearly {p (x i )}i Î I generates M JM over R J since {x i }i Î I
generates M over R. Suppose å ri p (x i ) = 0 in M JM , then å ri x i Î IM . Since {x i }i Î I generates
M, then å ri x i = å ai x i for some ai Î J . By unique representation ri = ai for all i , that is ri Î J
for all i and hence ri = 0 in the field R J . Since any two bases of the vector space M JM have the
same cardinality, hence any two bases for M also do.

Definition 17. Over a commutative ring R, the rank of a free R-module is defined to be the cardinality
of a basis.
If R is not commutative then Theorem16 dose not hold, the following proposition contains what we can
say about a free module in general.
Proposition 18. Let R be a ring. The following are equivalent for a free R-module F:
(i)
(ii)
(iii)
F is finitely generated.
F has a finite basis.
Every basis of F is finite.
Proof. Clearly if F has a finite basis then it is finitely generated. Conversely, suppose F is generated by
T = {f1, f2, K , fn }, and let S be a basis of F. Then for each i, fi = å s Î S ri ,ss , for some ri ,s Î R ,
where the set of s Î S with ri ,s ¹ 0 must be finite, call it S i . Now the finite set S ¢ = Uni= 1 S i is
linearly independent (being a subset of S ) and generates F since it generates T. Since S is linearly
independent, we must have S = S ¢ which shows that S is finite.
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