ADVANCED ALGEBRA UNIT-5:MODULES

Dr. SHIVANGI UPADHYAY

ACADEMIC CONSULTANT UTTRAKHAND OPEN UNIVERSITY HALDWANI(UTTRAKHAND) [email protected]

Dr. Shivangi Upadhyay Advanced Algebra 1 / 19 Overview

1 Modules left R-module Unital Module and Right R-module Examples Elementary Properties 2 Sub-modules Definition Theorem 3 Direct Sum Definition Theorem 4 Quotient Module 5 Module Homomorphism

Dr. Shivangi Upadhyay Advanced Algebra 2 / 19 Modules Modules

Definition Let (R, +,.) be a ring and let M be anon empty set. Then M is calles a left R-module or simply a left module over R if the following are satisfied: (M, +) is an abelian group ∀ r ∈ R and ∀ m ∈ M =⇒ rm ∈ M This law satisfies the following conditions: (1) r(m + n) = rm + rn (2) (r + s)m = rm + rn (3) (rs)m = r(sm) ∀ r, s ∈ R and all m, n ∈ M.

Dr. Shivangi Upadhyay Advanced Algebra 3 / 19 Modules

Unital Module If R has unity element 1 such that 1 m = m∀m ∈ M, then M is called a Unital Module

Right R-module If in the definition of Left R-module we replace r m by m r, then we have right R-module.

If R is a field, then a unital module M is a vector space over R.

Dr. Shivangi Upadhyay Advanced Algebra 4 / 19 Modules Examples

Example 1 A ring R is an R-module over its subring Sol. Let S be a subring of R. Since R is a ring. Therefore it is an additive abelian group. Taking the multiplicative in R is scalar multiplication we can see that ∀ m ∈ R and ∀ r ∈ S =⇒ rm ∈ R further (i) (r(m + n) = rm + rn for all m, n ∈ R, r ∈ S (∵ left distributive law in R) (ii) (r + s)m = rm + sm for all m ∈ R, r, s ∈ S (∵ Right distributive law in R) (iii) (rs)m = r(sm) for all r, s ∈ S and m ∈ R (∵ Associativity in R) Hence R is an R-module over S.

Dr. Shivangi Upadhyay Advanced Algebra 5 / 19 Modules Elementary Properties Elementary Properties

Theorem 1 Let R be a ring and M be an R− module. Then (i) r0 = 0 ∀ r ∈ R (ii) 0m = 0 ∀ m ∈ M (iii) (−r)m = −(rm) = r(−m) ∀ r ∈ R m ∈ M (iv) r(m − n)m = rm − rn ∀ r ∈ R m, n ∈ M (v) (r − s)m = rm − sm ∀ r, s ∈ R m ∈ M

Proof (i)Since 0 + 0 = 0, 0 is the identity of M. =⇒ r(0 + 0) = r0 ∀ r ∈ R =⇒ r0 + r0 = r0 =⇒ r0 + r0 = r0 + 0 =⇒ r0 = 0 (∵ cancellation law in M)

Dr. Shivangi Upadhyay Advanced Algebra 6 / 19 Modules Elementary Properties

(ii)Since 0 + 0 = 0, 0 is the identity of M. =⇒ (0 + 0)m = 0m ∀ m ∈ M =⇒ 0m + 0m = 0m =⇒ 0m + 0m = 0m + 0 =⇒ 0m = 0 (∵ cancellation law in M)

Dr. Shivangi Upadhyay Advanced Algebra 7 / 19 Modules Elementary Properties

(iii) For r ∈ R, r + (−r) = 0 =⇒ [r + (−r)]m = 0m ∀ m ∈ M =⇒ rm + (−r)m = 0 (Using (ii)) =⇒ (−r)m = −(rm) Similarly m + (−m) = 0 =⇒ r[m + (−m)] = r0 =⇒ r[m + (−m)] = 0 =⇒ rm + r(−m) = 0 =⇒ r(−m) = −(rm) Hence r(−m) = −(rm) = (−r)m ∀ r ∈ R ∀ m ∈ M

Dr. Shivangi Upadhyay Advanced Algebra 8 / 19 Modules Elementary Properties

(iv) For r ∈ R, m, n ∈ M r(m − n) = r[m + (−n)] = rm + r(−n) = rm − rn (from (iii)) (v) For rs ∈ R, m ∈ M (r − s)m = [r + (−s)]m = rm + (−s)m = rm − sm (from (iii))

Dr. Shivangi Upadhyay Advanced Algebra 9 / 19 Sub-modules Sub-modules

Definition Let M be an R-module over ring R A non void subset N of M is said to be Sub-modules of M if N itself is an R module under the operation of addition and scalar multiplication given for M restricted to N.

Every R-module M has two submodules M and {0}. These are called Improper Submodule. An R-module M is said to be Irreducible Sub-module if its only submodule are {0} and M.

Dr. Shivangi Upadhyay Advanced Algebra 10 / 19 Sub-modules Theorem Theorem

Theorem 2 Let R be a ring and let M be an R-module. A non-void subset N of M is submodule of M iff (i) x − y ∈ N, for all x, y ∈ N (ii) rx ∈ N, for all x ∈ N, r ∈ R

Proof If N is a submodule of N. Then N is an abelian group under addition and is closed under scalar multiplication Therefore (i) and (ii) hold.

Dr. Shivangi Upadhyay Advanced Algebra 11 / 19 Sub-modules Theorem

Converse Let N be non empty subset of M sucht that (i) and (ii) hold. Now (i) =⇒ N is additive subgroup of M. Therefore N itself is an abelian group under addition. (ii) =⇒ N is closed under scalar multiplication The remaining axioms for scalar multiplication hold for all elements in N as they hold in M. Hence N is a submodule of M.

Dr. Shivangi Upadhyay Advanced Algebra 12 / 19 Direct Sum Direct Sum

Definition

Let M be an R-module and let M1, M2,...... ,Mn be submodule of M. Then M is called the direct sum of M1, M2,...... ,Mn if every element m ∈ M is uniquely expressed as m = m1 + m2 + ..... + mn where m1 ∈ M1, m2 ∈ M2, ...... , mn ∈ Mn M = M1 ⊗ M2 ⊗ ...... Mn Symbollically it is denoted as

Dr. Shivangi Upadhyay Advanced Algebra 13 / 19 Direct Sum Theorem Theorem

Theorem 3

Let M be an R-module and let N1, N2,...... ,Nk be submodule of M. Then the following statements are equivalent: (i) M = N1 ⊗ N2 ⊗ ...... Nk (ii) If n1 + n2 + ..... + nk = 0, then n1 = n2 = ..... = nk = 0 for ni ∈ Ni (iii) Ni ∩ {N1 + ...... + Ni + Ni+1 + ...... + Nk } = {0}

Proof. (i) =⇒ (ii) Let M is the direct sum of N1, N2,...... ,Nk and n1 + n2 + ..... + nk = 0 for ni ∈ Ni , i = 1, 2, ..., k Since every element of M has unique expression 0 ∈ M which is written as 0=0+0+...... +0 implies that n1 = n2 = ..... = nk = 0.

Dr. Shivangi Upadhyay Advanced Algebra 14 / 19 Direct Sum Theorem

(ii) =⇒ (iii) Let x ∈ (Ni ∩ {N1 + ...... + Ni + Ni+1 + ...... + Nk }). So there exist n1 ∈ N1, n2 ∈ N2, ...... , ni−1 ∈ Ni−1, ni+1 ∈ Ni+1, ...... nk ∈ Nk such that x = n1 + n2 + .... + ni−1 + ni + ..... + nk =⇒ n1 + n2 + .... + ni−1 + (−x) + ni + ..... + nk = 0 =⇒ n1 = n2 = .... = ni−1 = x = ni = ..... = nk = 0. Since x ∈ Ni =⇒ −x ∈ N − i i.e., −x is the ith element in the sum. Hence x = 0 and Ni ∩ {N1 + ...... + Ni + Ni+1 + ...... + Nk } = {0}

Dr. Shivangi Upadhyay Advanced Algebra 15 / 19 Direct Sum Theorem

(iii) =⇒ (i) Let us assume that for m ∈ M, we have two different representations m = m1 + m2 + ..... + mk = n1 + n2 + ..... + nk Then 0 = (m1 − n1) + (m2 − n2) + ...... + (mk − nk ) =⇒ (m1 − n1) = (n1 −m1)+...... +(ni−1 −mi−1)+(ni −mi )+(ni+1 −mi+1)+...+(nk −mk ) Now (mi − ni ) ∈ Ni and [(n1 − m1) + ...... + (ni−1 − m1) + (ni−1 − mi−1) + (ni − mi ) + (ni+1 − mi+1) + ... + (nk − mk )] ∈ (N1 + ...... + Ni−1 + Ni + Ni+1 + ...... + Nk ) So (mi − ni ) ∈ (N1 + ...... + Ni−1 + Ni + Ni+1 + ...... + Nk ) = {0} =⇒ mi = ni ∀ i. Thus m ∈ M has unite representations Hence M is the direct sum of N1, N2,...... ,Nk .

Dr. Shivangi Upadhyay Advanced Algebra 16 / 19 Quotient Module Quotient module

Quotient Module Let M be an R-module over ring R and let N be submodule of M. then the set

M/N = {N + x : x ∈ M} is called quotient module.

Dr. Shivangi Upadhyay Advanced Algebra 17 / 19 Module Homomorphism Module Homomorphism

Module Homomorphism 0 Let M and M be R-modules. 0 A mapping f : M → M is called an R-module homomorphism if (i) f (x + y) = f (x) + f (y), for all x, y ∈ M and (ii) f (rx) = rf (x) for all r ∈ R, x ∈ M and if R is a ring with unity then we can combine (i) and (ii) as

f (rx + sy) = rf (x) + sf (y)

Dr. Shivangi Upadhyay Advanced Algebra 18 / 19 Module Homomorphism

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Dr. Shivangi Upadhyay Advanced Algebra 19 / 19