5.2 Direct Products and Sums of Modules; Free Modules

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Let {M_, }_2A be a family of R-modules. Assume we are given a family

(P, __)_2A consisting of an R-module P, and R-module homomorphisms

__ : P ! M_, _ 2 A, satisfying the following universal property: If

(P0, _0

_)_2A is another family consisting of an R-module P0 and R-module

homomorphisms _0

_ : P0 ! M_, _ 2 A, then there exists a unique R-module

homomorphism _ : P0 ! P, making the triangle below commute, for each

_ 2 A.

P0 M_

P

-

@

@

@R 􀀀

􀀀

􀀀_

_0

_

_ __

Then the family (P, __)_2A is called a (direct) product of the R-modules

M_, _ 2 A. Sometimes we simplify the language a bit by simply calling

P a direct product of the modules M_, without explicitly referring to the

mappings __.

The usual sort of “abstract nonsense” shows that if (P, __)_2A and

(P0, _0

_)_2A are both products of the family M_, _ 2 A, then P _= P0.

This leaves the question of existence of a product; however this is already

afforded by the ordinary cartesian product:

P =

Y

_2A

M_.

To give this a module structure, recall first the definition of the cartesian

product:

Y

_2A

M_ = {f : A !

[

_2A

M_| f(_) 2 M_ for each _ 2 A}.

Now define addition and R-scalar multiplication in

Y

_2A

M_ pointwise: (f +

g)(_) = f(_) + g(_), (r · f)(_) = r · f(_), _ 2 A, f, g 2

Y

_2A

M_, r 2 R.

108 CHAPTER 5. MODULE THEORY

The “projection maps” __ :

Y

_2A

M_ ! M_, _ 2 A are defined by setting

__(f) = f(_), _ 2 A.

Theorem 5.2.1 The family (

Y

_2A

M_, __) is a product of the family {M_}_2A.

Dual to the above notion is that of the direct sum of the family {M_}_2A.

The family (D, μ_)_2A is said to be a direct sum of the family {M_}, if

there exist module homomorphisms μ_ : M_ ! D, satisfying the following

universal criterion: If D0 is any other module, with homomorphisms μ0

_ :

M_ ! D0, then there exists a unique homomorphism _ : D ! D0, such that

for each _ 2 A, the triangle below

M_ D

D0

-

@

@

@R

􀀀

􀀀

􀀀        

μ_

__ _

commutes. Again, as in the case of the direct product, if the direct sum of

the family {M_}_2A, exists, it is unique up to isomorphism.

Using the direct product

Y

_2A

M_, one can construct a direct sum, as

follows. Namely, set

D = {f 2

Y

_2A

M_| f(_) = 0 for all but finitely many _ 2 A}.

Note that D is clearly an R-submodule of

Y

_2A

M_. Next we define R-module

homomorphisms μ_ : M_ ! D by setting

μ_(m_)(_) =

_

m_ if _ = _,

0 if _ 6= _.

Then one has the following:

5.2. DIRECT PRODUCTS AND SUMS OF MODULES 109

Proposition 5.2.2 The direct sum of a family {M_} exists and is constructed

as above.

We denote the direct sum of the family {M_} by

M

_2A

M_. Note that

if μ_ : M_ !

M

_2A

M_, are as above, then every element of

M

_2A

M_ can be

uniquely written as a sum

X

_2A

μ_(m_), m_ 2 M_.

There is also an “internal” version of direct sum. Let M be an Rmodule,

and assume that {M_} is a family of submodules. For each _ 2 A,

let i_ : M_ ! M be the inclusion map. If (M, i_)_2A satisfies the universal

criterion above, we say that M is the internal direct sum of the submodules

M_, _ 2 A, and write M =

M

_2A

M_.

Fortunately, there is a simple criterion for M to be an internal direct

sum of submodules M_, _ 2 A.

Proposition 5.2.3 Let M be an R-module, and let {M_}, _ 2 A be a

family of submodules. Then M =

M

_2A

M_, if and only if

(i) M =

X

_2A

M_, and

(ii) for each _ 2 A, M_ \

X

_6=_

M_ = 0.

Additional Terminology and Notation. Let {M_}_2A be a family of

R-modules. If (

Y

_2A

M_, __)_2A is a product, we frequently call the mappings

__ :

Y

_2A

M_ ! M_ projection mappings. Correspondingly, if (__2AM_, μ_)

is a sum, we frequently call the mappings μ_ : M_ ! __2AM_ coordinate

mappings.

Next suppose that we have a collection of R-module homomorphisms

p_ : P ! M_, _ 2 A. Then we use the notation

{p_}_2A : P −!

Y

_2A

M_

110 CHAPTER 5. MODULE THEORY

for the induced mapping. In particular, if p1 : P ! M1, p2 : P ! M2 is

a pair of R-module homomorphisms into R-modules M1, M2, we have the

induced mapping

{p1, p2} : P −! M1 ×M2.

When we have a collection of R-module homomorphisms i_ : M_ ! D, _ 2

A, the we use the notation

hi_i_2A : __2AM_ −! D

for the induced mapping. In particular, where we have a pair of maps

i1 : M1 ! D, i2 : M2 ! D, the induced mapping is denoted

hi1, i2i : M1 _M2 −! D.

Finally, let {M_}_2A, {M0

_}_2A be families of R-modules, indexed by

the same index set A. If we have R-module homomorphisms __M_ !

M0

_, _ 2 A, there there is a naturally induced map

Y

__ :

Y

_2A

M_ −!

Y

_2A

M0

_

induced by the composite maps

Q

_2AM_

__! M_

__! M0

_, _ 2 A. In an

entirely analogous fashion, we get naturally induced homomorphisms

___ :

M

_2A

M_ −!

M

_2A

M0

_ .

There is another universal construction, reminiscent of that for free

groups. Let M be an R-module, and let S be a set. Say that M is free

on the set S if there exists a map _ : S ! M, satisfying the following universal

property. If N is any R-module, and if _ : S ! N is any map, then

there is a unique R-module homomorphism _ : M ! N such that

S M

N

-

@

@

@R

􀀀

􀀀

􀀀        

_

_ _

5.2. DIRECT PRODUCTS AND SUMS OF MODULES 111

commutes.

The following is easily anticipated.

Proposition 5.2.4 If S is any set, then there exists a free module M on

the set S, which is unique up to isomorphism.

In fact, the above construction is based on the direct sum construction,

as follows. Given the set S and the ring R, let Rs = R regarded as a left

R-module, let M =

M

s2S

Rs, and let μs0 : Rs0 !

M

s2S

Rs be the coordinate

mappings. Define _ : S !

M

s2S

Rs by setting _(s) = μs(1), s 2 S. Then M is

the desired free module.

Again, there is an “internal” criterion for freeness, as follows. Let M be

and R-module, and let B _ M. If B spans M and is R-linearly independent,

then B is called a basis for M. It need not happen that the R-module M

admits a basis. A good example is the additive group (i.e. Z-module)

Q of rational numbers. Note first that Q is not a cyclic group and so it

cannot have a basis consisting of one element. Next, let r1 = a1

b1

, r2 =

a2

b2

2 Q, with r1 6= r2. Then a2b1r1 − a1b2r2 = 0, and so the set {r1, r2}

is Z-linearly dependent. Therefore, it follows that any subset B of Q of

cardinality greater than 1 is Z-linearly dependent.

The significance of having a basis is as follows.

Proposition 5.2.5 The R-module M is free if and only if it has a basis.

In case M is a free module over a commutative ring R, we can actually

say more. Indeed, if J _ R is a maximal ideal then R/J is a field, and

it’s easy to see that the quotient module M/JM is acually an R/J-module,

i.e., is an R/J-vector space. Furthermore, if {m_| _ 2 A} is a basis for

M, it is easy to check that the set {m_ + JM| _ 2 A} is a vector space

basis for M/JM. Since any two bases of a fixed vector space have the same

cardinality, we conclude the following result:

Proposition 5.2.6 If M is a free module over the commutative ring R,

then any two bases have the same cardinality.

Therefore, if M is a free module over the commutative ring R, we may

speak of the rank of this module. In general, if R is a ring whose free

112 CHAPTER 5. MODULE THEORY

modules have well-defined ranks, we often say that R has IBN (invariant

basis number); therefore, commutative rings have IBN. One can show that,

more generally, any left Noetherian ring has IBN. (See Joseph Rotman, An

Introduction to Homological Algebra, Academic Press, 1979, Theorem 4.9,

page 111.)

Lemma 5.2.7 Let R be a commutative ring and let M be a finitely generated

R-module. If J _ R is an ideal and M=JM, then (1 − x)M = 0 for

some x 2 J.

Lemma 5.2.8 Let R be a commutative ring and let M be a finitely generated

R-module. If the R-module homomorphism f : M ! M is surjective,

then it is injective (and hence is an isomorphism).

Note that the above generalizes Exercise 3 of Section 4.

The following shows again the rough similarity between free modules

over a commutative ring and vector spaces.

Theorem 5.2.9 Let R be a commutative ring and letM be a free R-module

of finite rank r. If {m1,m2, . . . ,mr} generates M, then it is a basis of M.

Exercises

1. Let

0 ! M0 μ

! M _! M00 ! 0

be an exact sequence of left R-modules. Show that the following two

conditions are equivalent:

(a) There exists a module homomorphism _ : M ! M0 such that

_ _ μ = 1M0 .

(b) There exists a module homomorphism _ : M00 ! M such that

_ _ _ = 1M00 .

Show that if either of the above two cases hold, then M _= M0 _M00.

When this happens, we say that the exact sequence 0 ! M0 μ

! M _!

M00 ! 0 splits.

5.2. DIRECT PRODUCTS AND SUMS OF MODULES 113

2. Let M be an R-module. Prove that M is free if and only if M is

isomorphic to the direct sum of copies of R.

3. Prove that any R-module is the homomorphic image of a free Rmodule.

4. Give an example of a free R-module M and a submodule N such that

N is not free.

5. Prove that the direct sum of a family of free R-modules is also free.

6. Let F1 be a free R-module on the set S1, and let F2 be a free R-module

on the set S2. If S1, S2 have the same cardinality, prove that F1

_= F2.

7. Consider the diagram of R-modules and R-module homomorphisms:

A0 B0

A B

-

-

_ ? ?_

_0

_

From the above diagram, construct the sequence:

A

μA0_+μB_

−! A0 _ B

h−_0,_i

−! B0,

where μA0 : A0 ! A0 _ B, μB : B ! A0 _ B are the coordinate maps.

Show that the above square is commutative if and only if the above

sequence is differential, i.e., h−_0, _i _ (μA0_ + μB_) = 0.

8. Let {M_}_2A be a family of R-modules. For each pair of indices _, _ 2

A, define homomorphisms p__ : M_ ! M_ by setting p__ = 1M_, if

_ = _ and p__ = 0 : M_ ! M_, if _ 6= _. Therefore, by universality

oYf direct product, we get induced homomorphisms {p__}_ : M_ !

_2A

M_, _ 2 A. In turn, we get an induced map

h{p__}_i_ :

M

_2A

M_ −!

Y

_2A

M_.

Analogously, obtain induced maps

{hp__i_}_ :

Y

_2A

M_ −!

M

_2A

M_.

114 CHAPTER 5. MODULE THEORY

Show that the composition of the two maps

M

_2A

M_ !

M

_2A

M_ is the

identity, by

(a) using the explicit constructions of

M

_2A

M_ and

Y

_2A

M_, and

(b) using the universality properties.

9. Let N

μ!

M _! N be R-module homomorphisms with _μ an automorphism

of N. Prove that M = μN _ ker _.

10. Let F be a field and let R be the ring

R =

8<

:

2

4

a 0 0

0 b 0

c d e

3

5| a, b, c, d, e 2 F

9=

;,

let M be the left R-module

M =

8<

:

2

4

x

y

z

3

5| x, y, z 2 F

9=

;,

and let N _ M be the submodule

N =

8<

:

2

4

0

0

z

3

5| z 2 F

9=

;.

Prove that M is not the direct sum of two proper submodules, but

that the quotient M/N _= M1 _M2 for nontrivial submodules M1 and

M2.

5.3. MODULES OVER A PRINCIPAL IDEAL DOMAIN 115

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