Introduction to Modules
Lecture Notes: Week-3
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Direct sum Contd...
Definition
If {Mi }i∈I is a family of submodules
P of M, then M is called a direct sum
of family of submodules
if M = i∈I Mi and every m ∈ M can be written
P
uniquely as m = i∈I mi , mi ∈ Mi , mi = 0 except for finitely many i 0 s.
We write M = ⊕i∈I Mi .
Remark
Q
1
It is easy to see that ⊕i∈I Mi ⊆
finite.
2
Direct product of the Mi ’s is an R-module and the direct sum of the
Mi ’s is a submodule of their direct product.
i∈I Mi
and equality holds if I is a
Example
Rn = R
| ⊕R ⊕
{z· · · ⊕ R} .
n times
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Direct summand
Definition
A submodule N of a module M is said to be a direct summand or simply a
summand of M if there exists a submodule N 0 of M such that
M = N ⊕ N 0 . Such a N 0 is called a supplement of N.
Remark
Supplement of a summand need not be unique.
Example
R3 = R(0, 0, 1) ⊕ R(0, 1, 0) ⊕ R(1, 0, 0)
R3 = R(0, 0, 1) ⊕ R(0, 1, 1) ⊕ R(1, 1, 1)
In this example, summand R(0, 0, 1) has two supplements.
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Free Module
Definition
1
Linear independence: Let M be an R-module. A subset N of M is
said to be linearly independent over R if for any finite subset
k
X
{n1 , · · · , nk } ⊆ N,
ri ni = 0, ri ∈ R implies r1 = r2 = · · · = rk = 0.
2
Free module: An R-module M is called a free module, if M has a
basis B. i.e., a linearly independent subset B of M such that M is
spanned by B over
PR, i.e., every element: m ∈ M can be written
uniquely as m = b∈B λb .b, λb ∈ R, λb = 0 except for finitely many
b’s, i.e., m is finite linear combination of elements in B, the scalars
being unique for m.
i=1
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Examples of Free modules
Example
1
2
For any ring R with 1, the left R-module M = R is free with basis
{1}.
Let R be a ring with 1. Then R n = R
{z· · · × R} is free module.
| ×R ×
n times
The set B = {(1, 0, · · · , 0), (0, 1, · · · , 0), · · · , (0, 0, · · · , 1)} is a basis
for R n , called usual basis or standard basis.
3
Every vector space V over F is a free module.
4
Q is not a free Z-module.
m2
1
Let p1 = m
n1 , p2 = n2 be two elements of Q, where m1 , m2 , n1 , n2 ∈ Z
then (m2 n1 )p1 + (−m1 n2 )p2 = 0. i.e. no two elements of Q are
linearly independent.
5
Any finite abelian group M is not a free Z-module.
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Examples contd...
Suppose M is free and B is a basis of M over Z. Let x ∈ M be such that
x 6= 0. Let nx = 0, n ≥ 2 for some n ∈ N such that mx 6= 0 for m < n.
since B is a basis of M, so there exist b1 , b2 , · · · , br ∈ B and
n1 , n2 , · · · , nr ∈ Z such that
x = n1 b1 + n2 b2 + · · · + nr br
⇒ nx = nn1 b1 + nn2 b2 + · · · + nnr br = 0
⇒ nn1 = 0, nn2 = 0, ......, nnr = 0
⇒ n1 = 0, n2 = 0, ......, nr = 0 (Since n 6= 0)
⇒ x = 0, which is a contradiction
Hence M is not free over Z.
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Semi-simple module
Definition
A left R-module M is said to be semi-simple module if it is direct sum of a
family of simple modules.
Example
1
Every simple module is semi-simple.
2
Every matrix ring over a division ring D is a semi-simple module.
Theorem
For an R-module M, the followings conditions are equivalent(i)
M is a sum of family of simple submodules.
(ii)
M is a direct sum of family of simple submodules.
(iii)
Every submodule of M is a direct summand.
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Proof
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Contd...
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