Ring Theory and Vector Calculus
UNIT-I
Binary Operation: Let 𝑆 be a non-empty set. A map (bop) ⋆ : 𝑆 × 𝑆 → 𝑆, (𝑎, 𝑏) → 𝑎 ⋆ 𝑏
is called a binary operation on 𝑆. So ⋆ takes 2 inputs 𝑎, 𝑏 from 𝑆 and produces a single
output 𝑎 ⋆ 𝑏 ∈ 𝑆. In this situation we may say that 𝑆 is closed under ⋆ .
Example:
1. Let 2, 3 ∈ R ⟹ 2 + 3 ∈ 𝑅, addition (+) is a binary operation on R or closed on R.
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2. Let 2, 4 ∈ 𝑁 ⟹ 4 = 2 ∉ 𝑁, division (÷) is not a binary operation on N or not closed on
N.
3. + is not a binary operation on the set 𝑆 = {0, 1}. We have 1 ∈ 𝑆 but 1 + 1 = 2 ∉ 𝑆.
Algebraic Structure: The ordered pair (𝐺, ⋆) is called an algebraic structure or algebraic
system, if ⋆ is a binary operation on G .
Example: (𝑁, +) and (𝑁, ∙) are both algebraic systems, but (𝑁, −) and (𝑁, ÷) are not
algebraic systems. Since −, ÷ are not a binary operations on 𝑁.
Group: A non-empty set 𝐺 with a binary operation ⋆ is called a group if 𝐺 satisfies the
following conditions.
(i). 𝑎 ⋆ 𝑏 ∈ 𝐺, for all 𝑎, 𝑏 ∈ 𝐺. (Closure law)
(ii). 𝑎 ⋆ (𝑏 ⋆ 𝑐 ) = ( 𝑎 ⋆ 𝑏) ⋆ 𝑐 for all 𝑎, 𝑏, 𝑐 ∈ 𝐺 (Associative law)
(iii). There exists an element 𝑒 ∈ 𝐺 such that 𝑎 ⋆ 𝑒 = 𝑒 ⋆ 𝑎 = 𝑎 for all 𝑎 ∈ 𝐺, where 𝑒 is
called the identity in 𝐺 (Existence of Identity)
(iv). For each 𝑎 ∈ 𝐺, there exists an element 𝑏 ∈ 𝐺 such that 𝑎 ⋆ 𝑏 = 𝑏 ⋆ 𝑎 = 𝑒, where 𝑏 is
called the inverse of 𝑎 in 𝐺. (Existence of Inverse).
Example:
(i). (𝑍𝑛 , ⨁) is finite group of integers over addition modulo 𝑛, where ⨁ addition modulo
binary operation.
(ii). The systems (𝑍, +), (𝑄, +) , (𝑅, +), (𝐶, +) are all infinite groups under usual addition.
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Remark: (𝑁, +) is not a group, where 𝑁 is the set of all Natural numbers, since the axioms
(iii) and (iv) (in the definition of group) are not satisfied in 𝑁.
Abelian group: A group (𝐺, ⋆) is called abelian or commutative group if 𝑎 ⋆ 𝑏 = 𝑏 ⋆ 𝑎 for
all 𝑎, 𝑏 ∈ 𝐺.
Example: The groups (𝐼, +), (𝑄, +) and (𝑅, +) are abelian.
Remark: The set of all square matrices with respect to usual multiplication does not form an
abelian group. Since matrices does not satisfy commutative property with respect to
multiplication.
Ring: Let R be a non-empty set and +, ∙ be two binary operations defined on R. Then (𝑅, +,
∙) is said to be a ring, if
(i). (𝑅, +) is a commutative group
(ii). (𝑅, ∙) is a semi group and
(iii). Distributive laws hold
( or )
Let R be a non-empty set and +, ∙ be two binary operations on R. Then (𝑅, +, ∙) is said
to be a ring if
Addition
(A1) Associativity: for all 𝑎, 𝑏, 𝑐 ∈ 𝑅 we have 𝑎 + (𝑏 + 𝑐) = (𝑎 + 𝑏) + 𝑐
(A2) Identity: there exists 0 ∈ 𝑅 such that for all 𝑎 ∈ 𝑅 we have 𝑎 + 0 = 0 + 𝑎 = 𝑎
(A3) Inverse: for any 𝑎 ∈ 𝑅 there exists −𝑎 ∈ 𝑅 such that 𝑎 + (−𝑎) = (−𝑎) + 𝑎 = 0
(A4) Commutativity: for all 𝑎, 𝑏 ∈ 𝑅 we have 𝑎 + 𝑏 = 𝑏 + 𝑎
Multiplication
(M1) Associativity: for all 𝑎, 𝑏, 𝑐 ∈ 𝑅, we have 𝑎 · (𝑏 · 𝑐) = (𝑎 · 𝑏) · 𝑐
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Addition and Multiplication together (Distributive laws)
(D) For all 𝑎, 𝑏, 𝑐 ∈ 𝑅,
Left Distributivity: 𝑎 · (𝑏 + 𝑐) = 𝑎 · 𝑏 + 𝑎 · 𝑐
Right Distributivity: (𝑎 + 𝑏) · 𝑐 = 𝑎 · 𝑏 + 𝑏 · 𝑐
Note:
1. The properties 𝐴1, 𝐴2, 𝐴3 and 𝐴4 states that (R, +) is a commutative group.
2. The properties 𝑀1, 𝑀2 states that (R, ·) is a semi group and D states that distributive laws
in R.
Example:
1. (𝑍, +,·), (𝑄, +,·) , (𝑅, +,·), (𝐶, +,·) are all rings.
2. The set N of natural numbers is not a ring w.r.to usual additions and multiplication,
because (N, +) is not a group (additive identity 0 ∉ 𝑁, so identity law fails).
3. The set of all irrational numbers is not a ring w.r.to +, · , because there is no zero
element.
Note: We sometimes say 𝑅 is a ring, taken it as given that the ring operations are denoted
+ and · . As in ordinary arithmetic we shall frequently suppress · and write 𝑎𝑏 instead of 𝑎 ·
𝑏.
Commutative ring: If a ring (𝑅, +, ·) satisfies commutative property w.r.to multiplication,
then (𝑅, +, ·) is called commutative ring. i.e., ∀ 𝑎, 𝑏 ∈ 𝑅 ⇒ 𝑎. 𝑏 = 𝑏. 𝑎.
Example:
1. (𝑍, +,·), (𝑄, +,·) , (𝑅, +,·), (𝐶, +,·) are the commutative rings.
2. The set of all square matrices with respect to usual addition and usual multiplication does
not form a commutative ring, since matrices does not satisfy commutative law w.r.to
multiplication.
Ring with unity: If a ring (𝑅, +, ·) satisfies the multiplicative identity property, then (𝑅, +,
·) is said to be ring with unity. i.e.,
∀ 𝑎 ∈ 𝑅, then ∃ an identity element e ∈ R ∋ 𝑎. 𝑒 = 𝑒. 𝑎 = 𝑎
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Example:
1. (𝑍, +, ·), (𝑄, +, ·), (𝑅, +, ·), (𝐶, +, ·) are all the rings with unity.
2. (2𝑍, +, ·) is a ring without unity, since 0 ∉ 2𝑍.
3. The set of all integers w.r.to usual addition and multiplication is a commutative ring with
unity. Since
(i) (Z,+) Is a commutative group
(ii) (Z, ·) is a semigroup
(iii) Distributive laws are holds under +, · .
Properties of a ring:
1. The zero element of R is unique and a + 0 = a, for every element a 𝜖 𝑅.
2. For every element a 𝜖 𝑅, the additive inverse –a 𝜖 𝑅 is unique and a + ( - a ) = 0.
3. For a 𝜖 𝑅, - ( - a ) = a.
4. For 0 𝜖 𝑅, - 0 = 0.
5. For a, 𝑏 𝜖 𝑅, - ( a + b ) = - a - b.
6. For a, 𝑏, 𝑐 𝜖 𝑅, 𝑎 + 𝑏 = 𝑎 + 𝑐 ⟹ 𝑏 = 𝑐 and
𝑏+𝑎 =𝑐+𝑎 ⟹ 𝑏 =𝑐
7. For a, 𝑏, 𝑥 𝜖 𝑅, the equations 𝑎 + 𝑥 = 𝑏 and 𝑥 + 𝑎 = 𝑏 have unique solutions.
Notation: If 𝑅 is a ring and 0, a, b 𝜖 𝑅, we have
1. a + ( - b ) can be written as a – b
2. 𝑎 + 𝑎 written as 2𝑎
3. 𝑎. 𝑎 written as a 𝑎2
Theorem: If 𝑅 is a ring and 0, a, b, c in R, then
(i) 0𝑎 = 𝑎0 = 0
(ii) 𝑎(−𝑏) = (−𝑎)𝑏 = −(𝑎𝑏)
(iii) (−𝑎)(−𝑏) = 𝑎𝑏
(iv) 𝑎(𝑏 − 𝑐 ) = 𝑎𝑏 − 𝑎𝑐
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Theorem: If ( R, + , ·) is a ring with unity, then multiplicative identity is unique.
Proof: Suppose 1, 11 are two multiplicative identities in a ring R. Then
1. 𝑥 = 𝑥. 1 = 𝑥, ∀ x ∈ 𝑅 and 11 . 𝑥 = 𝑥. 11 = 𝑥, ∀ x ∈ 𝑅
For the identity 1, we have 11 . 1 = 1. 11 = 11 ------(1)
For the identity 11 , we have 1. 11 = 11 . 1 = 1 ------(2)
From (1) and (2), we get 1 = 11 .
Note: If R is a ring with a unity element 1 and a∈ 𝑅, then
(i). ( - 1 ) a = - a and (ii). ( - 1 ) ( - 1 ) = 1
Boolean Ring: In a ring R, if 𝑎2 = 𝑎, ∀ a ∈ 𝑅. Then R is called a Boolean ring.
Example:
1. 0 and 1 are idempotent elements, since 02 = 0 and 12 = 1, ∀ 0, 1 ∈ 𝑅.
2. The ring (𝑍2 = {0, 1}, +2 , ×2 ) is Boolean.
Theorem: If R is a Boolean ring, then
(i). 𝑎 + 𝑎 = 0, ∀ 𝑎 ∈ 𝑅
(ii). 𝑎 + 𝑏 = 0 ⟹ 𝑎 = 𝑏
(iii). R is a commutative under multiplication
(or)
Prove that every Boolean ring is abelian
Proof:
(i) For 𝑎 ∈ 𝑅 ⟹ 𝑎 + 𝑎 ∈ 𝑅
Since 𝑎2 = 𝑎, ∀ 𝑎 ∈ 𝑅, we have
(𝑎 + 𝑎)2 = 𝑎 + 𝑎
⟹ (𝑎 + 𝑎)(𝑎 + 𝑎) = 𝑎 + 𝑎
⟹ 𝑎 (𝑎 + 𝑎 ) + 𝑎 (𝑎 + 𝑎 ) = 𝑎 + 𝑎
⟹ (𝑎2 + 𝑎2 )(𝑎2 + 𝑎2 ) = 𝑎 + 𝑎
⟹ (𝑎 + 𝑎) + (𝑎 + 𝑎) = (𝑎 + 𝑎) + 0 (since R is a Boolean ring, so 𝑎2 = 𝑎 )
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By using left cancelation law of group (R, +), we get 𝑎 + 𝑎 = 0.
(ii) For 𝑎, 𝑏 ∈ 𝑅, assume 𝑎 + 𝑏 = 0
⟹ 𝑎 + 𝑏 = 𝑎 + 𝑎 ( from (𝑖) )
⟹ 𝑏 = 𝑎 (by left cancellation law)
(iii) For a, 𝑏 ∈ 𝑅, we have
𝑎 + 𝑏 ∈ 𝑅 and (𝑎 + 𝑏)2 = 𝑎 + 𝑏
⟹ (𝑎 + 𝑏)(𝑎 + 𝑏) = 𝑎 + 𝑏
⟹ 𝑎 (𝑎 + 𝑏 ) + 𝑎 (𝑎 + 𝑏 ) = 𝑎 + 𝑏
⟹ (𝑎2 + 𝑎𝑏) + (𝑏𝑎 + 𝑏2 ) = 𝑎 + 𝑏
⟹ (𝑎 + 𝑎𝑏) + (𝑏𝑎 + 𝑏) = 𝑎 + 𝑏
⟹ (𝑎 + 𝑏) + (𝑎𝑏 + 𝑏𝑎) = (𝑎 + 𝑏) + 0
⟹ 𝑎𝑏 + 𝑏𝑎 = 0 (𝑏𝑦 𝐿. 𝐶. 𝐿 )
⟹ 𝑎𝑏 = 𝑏𝑎 ( 𝑏𝑦 (𝑖𝑖 ) )
Solved problem(s):
1. Prove that the power set of every non-empty set is a Boolean ring.
Solution: Let S be a non-empty set and R = P(S) = power set of S.
Take 𝑆 = {𝑎, 𝑏}. Then
𝑅 = 𝑃 (𝑆 )
= 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑠𝑢𝑏𝑠𝑒𝑡𝑠 𝑜𝑓 𝑆
⇒ 𝑅 = {∅, {𝑎}, {𝑏}, 𝑆}
For A, B ∈ 𝑅, we define
𝐴 + 𝐵 = (𝐴 ∪ 𝐵) − (𝐴 ∩ 𝐵) = (𝐴 − 𝐵)𝑈(𝐵 − 𝐴) and
𝐴. 𝐵 = 𝐴 ∩ 𝐵
Clearly (R, +, .) is a ring.
Now we prove that R is a Boolean ring.
(1) Let {a} ∈ 𝑅. Then 𝑎2 = 𝑎. 𝑎 = 𝑎 ∩ 𝑎 = 𝑎.
(2) Similarly, for {b} ∈ 𝑅, 𝑏2 = 𝑏. 𝑏 = 𝑏 ∩ 𝑏 = 𝑏.
(3) For {∅}∈ 𝑅, ∅2 = ∅. ∅ = ∅ ∩ ∅ = ∅.
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(4) For {𝑎, 𝑏} ∈ 𝑅, {𝑎, 𝑏}2 = {𝑎, 𝑏}. {𝑎, 𝑏} = {𝑎, 𝑏} ∩ {𝑎, 𝑏} = {𝑎, 𝑏}.
2. 𝑅 is a boolean ring and for 𝑎 ∈ 𝑅, 2𝑎 = 0 ⟹ 𝑎 = 0. Then prove that 𝑅 = {0}.
Solution: Let 𝑎 ∈ 𝑅, 𝑅 𝑖𝑠 𝑎 𝑏𝑜𝑜𝑙𝑒𝑎𝑛 𝑟𝑖𝑛𝑔. This implies that
⟹ 𝑎 + 𝑎 = 0, ∀ 𝑎 ∈ 𝑅
⟹ 2𝑎 = 0, ∀ 𝑎 ∈ 𝑅
⟹ 𝑎=0
∴ 𝑎 = 0, ∀𝑎 ∈ 𝑅 ⟹ 𝑅 = {0}.
3. If R is a ring with identity element 1 and 1 = 0, then R = {0}.
Solution: Let 𝑥 ∈ 𝑅. Then
𝑥 = 1. 𝑥
⟹ 𝑥 = 0. 𝑥
⟹𝑥=0
Therefore R = { 0 }.
Note: The above example shows that a ring R with unity has at least 2 elements, if R ≠ { 0 }.
4. Prove that the set of all even integers is a commutative ring without unity under usual
addition and multiplication of integers.
Solution: Let R = the set all of even integers= {2𝑥 ∶ 𝑥 ∈ 𝑍}.
Let 𝑎, 𝑏, 𝑐 ∈ 𝑅 ⟹ 𝑎 = 2𝑚, 𝑏 = 2𝑛, 𝑐 = 2𝑝 , where 𝑚, 𝑛, 𝑝 ∈ 𝑍
To prove that (Z, +) is a commutative group
Closure law: 𝑎 + 𝑏 = 2𝑚 + 2𝑛 = 2(𝑚 + 𝑛) ∈ 𝑅 , since 𝑚 + 𝑛 ∈ 𝑍
Associative law: 𝑎 + (𝑏 + 𝑐 ) = 2𝑚 + (2𝑛 + 2𝑝)
= 2𝑚 + 2(𝑛 + 𝑝)
= (2𝑚 + 2𝑛) + 2𝑝
= (𝑎 + 𝑏) + 𝑐
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Identity law: Let 𝑎 = 2𝑚 ∈ 𝑅, then there exists 𝑒 = 2. 0 ∈ 𝑅 such that
𝑎 + 𝑒 = 2𝑚 + 2. 0 = 2(𝑚 + 0) = 2𝑚 = 𝑎 and
𝑒 + 𝑎 = 2.0 + 2𝑚 = 2(0 + 𝑚) = 2𝑚 = 𝑎
This shows that 𝑎 + 𝑒 = 𝑒 + 𝑎 = 𝑎 and 𝑒 = 2 . 0 is the identity element in 𝑅.
Inverse law: For every element 𝑎 = 2𝑚 ∈ 𝑅, then there exists −𝑎 = −2𝑚 ∈ 𝑅 such that
𝑎 + (−𝑎) = 2𝑚 + (−2𝑚) = 0 and
(−𝑎) + 𝑎 = (−2𝑚) + 2𝑚 = 0
This shows that 𝑎 + (−𝑎) = (−𝑎) + 𝑎 = 0 = 𝑒 and −𝑎 = −2𝑚 is the inverse
element of 𝑎 = 2𝑚 ∈ 𝑅.
Commutative law:
𝑎 + 𝑏 = 2𝑚 + 2𝑛
= 2(𝑚 + 𝑛 )
= 2(𝑛 + 𝑚 ) (Since 𝑍 is Commutative)
= 2𝑛 + 2𝑚
= 𝑏+𝑎
∴ (𝑅, +) is an abelian group
To prove that (R, .) is a semi group
Closure law: 𝑎. 𝑏 = (2𝑚)(2𝑛)
= 2(2𝑚𝑛)
= 2𝑙 ∈ 𝑅 , where 𝑙 = 2𝑚𝑛 ∈ 𝑍
Associative law: 𝑎. (𝑏. 𝑐 ) = (2𝑚)((2𝑛)(2𝑝))
= 8𝑚(𝑛𝑝)
= 8(𝑚𝑛)𝑝
= 𝑎(𝑏. 𝑐)
Distributive laws:
(1) 𝑎. (𝑏 + 𝑐 ) = (2𝑚). (2𝑛 + 2𝑝)
= 2𝑚. 2𝑛 + 2𝑚. 2𝑝
= 𝑎. 𝑏 + 𝑎. 𝑐
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(2) Similarly as (1), we get (𝑏 + 𝑐 ). 𝑎 = 𝑏. 𝑎 + 𝑐. 𝑎
Therefore we conclude that (R,+, .) is a ring. Since the multiplicative identity 1 is not
even integer (1∉ 𝑅) and hence the set of all even integers 𝑅 = 2𝑍 is a commutative ring
without unity under usual addition and multiplication of integers.
5. If 𝑅 is a commutative ring, then prove that (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2 .
Solution: (𝑎 + 𝑏)2 = (𝑎 + 𝑏)(𝑎 + 𝑏)
= 𝑎2 + 𝑎𝑏 + 𝑏𝑎 + 𝑏2
= 𝑎2 + 𝑎𝑏 + 𝑎𝑏 + 𝑏2 , since R is commutative
= 𝑎2 + 2𝑎𝑏 + 𝑏2
6. Is it the set of pure imaginary numbers {𝑖𝑦/𝑦 ∈ 𝑅} is a ring R w.r.to addition &
multiplication of complex numbers?
Solution: L𝑒𝑡 𝐴 = {𝑖𝑦/𝑦 ∈ 𝑅}.
Take 𝑖𝑦1 , 𝑖𝑦2 ∈ 𝑅, where 𝑦1 , 𝑦2 ∈ 𝑅.
Now, 𝑖𝑦1 . 𝑖𝑦2 = 𝑖 2 𝑦1 . 𝑦2 = −𝑦1 . 𝑦2 ∉ 𝐴, since −𝑦1 . 𝑦2 ∈ 𝑅.
Therefore the multiplication is not a binary operation on 𝐴 = {𝑖𝑦/𝑦 ∈ 𝑅} and hence
(𝐴, +, . ) is not a ring.
Problem 4:Prove that𝑄√2 = {𝑎 + 𝑏√2/ 𝑎, 𝑏€𝑄 }is a ring with ordinary addition and
multiplication:
Proof :To show (𝑄√2,+) abelian group
Now let 𝑥 = 𝑎1 + 𝑏1 √2
𝑦 = 𝑎2 + 𝑏2 √2
𝑧 = 𝑎3 + 𝑏2 √2
1.closure law :x,y € 𝑄√2
Consider 𝑥 + 𝑦 ∈ 𝑄√2
x+y ={(𝑎1 +𝑏1 √2) + (𝑎2 + 𝑏2 √2)
=𝑎1 +𝑎2 + (𝑏1 + 𝑏2 )√2
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=𝑎 + 𝑏√2 ∈ 𝑄√2 is closed
2.Associative law : 𝑥, 𝑦, 𝑧 ∈ 𝑄√2
Consider ,(𝑥 + 𝑦) + 𝑧 = ⦏(𝑎1 + 𝑏√2) + (𝑎2 + 𝑏√2)⦐ + 𝑎3 + 𝑏3 √2)
=(𝑎1 + 𝑎2 + 𝑎3 ) + (𝑏1 + 𝑏2 + 𝑏3 )√2
=𝑎1 + (𝑎2 + 𝑎3 ) + (𝑏1 + 𝑏2 )√2
=(𝑎1 + 𝑏1 √2) + ⦏(𝑎2 + 𝑎3 ) + 𝑏2 + 𝑏3 √2
=𝑥 + (𝑦 + 𝑧) ∈ 𝑄√2
3.Identity law :0 = 0 + 0√2 ∈ Q√2
Conisider 𝑥 + 0 = 𝑎1 + 𝑏1 √2 +0+0√2
= (𝑎1 + 0) + (𝑏1 +0 )√2
= 𝑎1 + 𝑏 √2
= 𝑥
∴ 0 + 0√2 is identity element.
4.Inverse law :𝑥 = 𝑎1 + 𝑏√2 where a,b,∈ 𝑄
⇒ −𝑎, −𝑏 ∈ 𝑄
∴ −𝑎1 + (−𝑏1 )√2 ∈ 𝑄√2
Consider 𝑎1 + 𝑏1 √2 + (−𝑎1 ) + (−𝑏1 )√2
∴ 𝑎1 − 𝑎1 +(𝑏1 − 𝑏1 ) √2 = 0 + 0√2 which is the identity element in Q√2
The inverse element of 𝑎1 + 𝑏1 √2 is (−𝑎1 ) + (−𝑏1 )√2 (or) x is –x
Commutative law: 𝑥 + 𝑦 = (𝑎1 + 𝑏1 √2) + (𝑎2 + 𝑏2 √2)
= 𝑎1 +𝑎2 + (𝑏1 + 𝑏2 )√2
= 𝑎2 +𝑎1 + (𝑏2 + 𝑏1 )√2
= (𝑎2 + 𝑏2 √2) + (𝑎1 + 𝑏1 √2)
= 𝑦 + 𝑥 ∈Q√2
∴ (Q√2, +) is an abelian group
Semi group: (Q√2, . )
(i) Closure law: (𝑎1 + 𝑏1 √2) + (𝑎2 + 𝑏2 √2)
= 𝑎1 𝑎2 + 𝑎1 𝑏2 √2 + 𝑏1 𝑎2 √2 + 2𝑏1 𝑏2
= 𝑎1 𝑎2 + (𝑎1 𝑏2 + 𝑏1 𝑎2 )√2 + 2𝑏1 𝑏2
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= (𝑎1 𝑎2 + 2𝑏1 𝑏2 ) + (𝑎1 𝑏2 + 𝑏1 𝑎2 )√2
= 𝑥 + 𝑦√2 is closed under multiplication
(ii) Associative law: 𝑥(𝑦𝑧) = (𝑥𝑦)𝑧
(𝑥. 𝑦). 𝑧 = [(𝑎1 𝑎2 + 2𝑏1 𝑏2 ) + (𝑎1 𝑏2 + 𝑏1 𝑎2 )√2].( 𝑎3 + 𝑏3 √2)
= (𝑎1 𝑎2 𝑎3 + 2𝑏1 𝑏2 𝑎3 )+𝑎1 𝑎2 𝑏3 √2)+2𝑏1 𝑏2 𝑏3 √2 + 𝑎1 𝑏2 𝑎3 √2 + 𝑏1 𝑎2 𝑎3 √2
+2𝑎1 𝑏2 𝑏3 + 2𝑏1 𝑎2 𝑎3
= 𝑎1 𝑎2 𝑎3 + 2𝑏1 𝑏2 𝑎3 + 2√2𝑏1 𝑏2 𝑏3 + ⋯
= 𝑥. (𝑦. 𝑧)
Associative law holds.
(iii) Distributive law: 𝑎(𝑏 + 𝑐 ) = 𝑎. 𝑏 + 𝑎. 𝑐
𝑎𝑛𝑑 (𝑏 + 𝑐 ). 𝑎 = 𝑏. 𝑎 + 𝑐. 𝑎
Distributive law holds
Hence,(𝑄√2, +, . ) is a Ring .
Problem: If Z is the set of Integers then Prove that (Z, ∗,𝑜) is a commutative ring with unity
.where ∗,𝑜 are defined by 𝑎 ∗ 𝑏 = 𝑎 + 𝑏 − 1 𝑎𝑛𝑑 𝑎 𝑜 𝑏 = 𝑎 + 𝑏 − 𝑎𝑏 ∀ 𝑎, 𝑏 ∈ Z .
Solution: Z is the set of Integers
1.Now prove that (Z, ∗) is an abelian group
∗ is the operation considered on Z as 𝑎 ∗ 𝑏 = 𝑎 + 𝑏 − 1 for 𝑎, 𝑏 ∈ Z.
i) Closure law:
For 𝑎, 𝑏 ∈ Z ⟹ 𝑎 + 𝑏 ∈ Z, 1 ∈ Z
⟹𝑎+𝑏−1 ∈Z
⟹𝒂∗𝑏 ∈Z
∴ Closure law holds
ii) Associativity: Let 𝑎, 𝑏, 𝑐 ∈ 𝑍
Consider (𝑎 ∗ 𝑏) ∗ 𝑐 = (𝑎 + 𝑏 − 1) ∗ 𝑐
=𝑎+𝑏−1+𝑐−1
=𝑎+𝑏+𝑐−2
Consider 𝑎 ∗ (𝑏 ∗ 𝑐 ) = 𝑎 ∗ (𝑏 + 𝑐 − 1)
= 𝑎 + (𝑏 + 𝑐 − 1) − 1
=𝑎+𝑏+𝑐−2
∴ (𝑎 ∗ 𝑏 ) ∗ 𝑐 = 𝑎 ∗ (𝑏 ∗ 𝑐 )
∴ Associative law holds
iii) Existence of identity:
Let 𝑎 ∈ 𝑍
Suppose 𝑎 ∗ 𝑒 = 𝑒 ∗ 𝑎 = 𝑎
𝑎+𝑒−1 =𝑎
𝑒−1 =0
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𝑒=1∈𝑧
iv) Existence of inverse :
Let 𝑎 ∈ 𝑧,
Suppose 𝑎 ∗ 𝑏 = 𝑏 ∗ 𝑎 = 𝑒
⇒𝑎 + 𝑏 = 2
⇒𝑏 = 2 − 𝑎
=is an integer
∈𝑧
∴ the inverse of 𝑎 is (2 − 𝑎)
v) Commutative law :
𝑎∗𝑏 =𝑎+𝑏−1
=𝑏+𝑎−1
=𝑏 ∗ 𝑎
𝑧 is a commutative hold
∴ (𝑧,∗) is an abelian group
1 To show (𝑧, 𝑜) is semi group :
1. Closure law : Let 𝑎, 𝑏 ∈ 𝑧
𝑎𝑜𝑏 = 𝑎 + 𝑏 − 𝑎𝑏
∈𝑧
∴ Closure law holds
2. Associative law : Let 𝑎, 𝑏, 𝑐 ∈ 𝑧
Consider ,(𝑎𝑜𝑏)𝑜𝑐 = (𝑎 + 𝑏 − 𝑎𝑏) 𝑜 𝑐
= (𝑎 + 𝑏 − 𝑎𝑏 + 𝑐) − (𝑎 + 𝑏 − 𝑎𝑏)𝑐
= (𝑎 + 𝑏 − 𝑎𝑏 + 𝑐 − 𝑎𝑐 − 𝑏𝑐 + 𝑎𝑏𝑐)
= 𝑎 + 𝑏 + 𝑐 − 𝑎𝑏 − 𝑏𝑐 − 𝑎𝑐 + 𝑎𝑏𝑐) → 1
( )
Again consider 𝑎𝑜 𝑏𝑜𝑐 = 𝑎𝑜(𝑏 + 𝑐 − 𝑏𝑐)
=(𝑎 + 𝑏 + 𝑐 − 𝑏𝑐 − 𝑎(𝑏 + 𝑐 − 𝑏𝑐)
=(𝑎 + 𝑏 + 𝑐 − 𝑏𝑐 − 𝑎𝑏 − 𝑎𝑐 + 𝑎𝑏𝑐)
=(𝑎 + 𝑏 + 𝑐 − 𝑎𝑏 − 𝑏𝑐 − 𝑎𝑐 + 𝑎𝑏𝑐) → 2
From 1&2 (𝑎𝑜𝑏)𝑜𝑐 = 𝑎𝑜(𝑏𝑜𝑐)
∴ associative holds
3.Distributive law : consider 𝑎𝑜(𝑏 ∗ 𝑐 ) = (𝑎𝑜𝑏) ∗ (𝑎𝑜𝑐)
Clarely distributive law is satisfised.
4.commutative ring: 𝑎𝑜𝑏 = 𝑎 + 𝑏 − 𝑎𝑏
=𝑏 + 𝑎 − 𝑎𝑏
=𝑏𝑜𝑎
∴ (𝑧, 𝑜 ∗) is commutative ring
Since 1∈ 𝑧 then 𝑧 is commutative ring with unity
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Zero divisors of a ring (ring with zero divisors): Two non-zero elements a, b of a ring R
are said to be zero divisiors, if 𝑎 𝑏 = 0 where 0 ∈ 𝑅 is the zero element. i.e.,
𝑎 ≠ 0 𝑎𝑛𝑑 𝑏 ≠ 0 ⇒ a b = 0
No zero divisors: (ring with no zero divisors): A ring R has non zero divisors,
if 𝑎 ≠ 0 𝑎𝑛𝑑 𝑏 ≠ 0 ⇒ a b ≠ 0. i.e.,
𝑎, 𝑏 are no zero divisors ⇔ ab = 0 ⇒ 𝑎 = 0 𝑜𝑟 𝑏 = 0
Example: In the ring (𝑍6 , +, .), 𝑍6 = {0, 1, 2, 3, 4, 5},
2×6 3 = 0 and 4 ×6 3 = 0, here 2 ≠ 0, 3 ≠ 0 and 4 ≠ 0.
Therefore 2, 3, 4 are zero divisors of a ring (𝑍6 ,+6 , ×6 ).
Solved problem(s):
1. Find the zero divisors in the ring (𝑍12 , +12 , ×12 ), 𝑍12 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.
Solution: We have 2×12 6 = 0, 4×12 6 = 0, 4×12 9 = 0, 3×12 4 = 0, 3×12 8 = 0, 6×12 10 = 0.
So 2, 3, 4, 6, 8, 9, 10 are zero divisors and all are non zeros (𝑎𝑏 = 0 ⟹ 𝑎 ≠ 0, 𝑏 ≠ 0).
Therefore the zero divisors of the ring (𝑍12 , +12 , ×12 ) are 2, 3, 4, 6, 8, 9, 10.
Note: The ring of integers Z has no zero divisors. i.e., for every𝑎, 𝑏 ∈ 𝑅,
ab = 0 ⟹ 𝑒𝑖𝑡ℎ𝑒𝑟 𝑎 = 0 𝑜𝑟 𝑏 = 0.
2. Find the zero divisors in the ring (𝑍10 , +10 , ×10 ), where 𝑍10 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
Solution: Two elements 𝑎, 𝑏 are said to be zero divisors in 𝑍10 , if 𝑎 ≠ 0, 𝑏 ≠ 0 and 𝑎𝑏 = 0.
5 ≠ 0, 4 ≠ 0 ⇒ 5 ×10 4=0
2 ≠ 0, 5 ≠ 0 ⟹ 2 ×10 5 = 0
5 ≠ 0, 6 ≠ 0 ⟹ 5 ×10 6 = 0
5 ≠ 0, 8 ≠ 0 ⟹ 5 ×10 8 = 0
Therefore the zero divisors of 𝑍10 are 2, 4, 5, 6, 8.
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3. Is it the ring of all 2× 2 matrices has zero divisors.
1 1 −1 −1 0 0
Solution: Let A= [ ], 𝐵 = [ ], 𝑂 = [ ] are 2× 2 matrices, where A and B
1 1 1 1 0 0
are non zeros.
Here 𝐴. 𝐵 = 0, i.e., 𝐴 ≠ 0, 𝐵 ≠ 0 ⟹ 𝐴. 𝐵 = 0.
Cancellation laws: In a ring R and for a, b, c ∈ 𝑅, 𝑖𝑓
𝑎 ≠ 0, 𝑎𝑏 = 𝑎𝑐 ⟹ 𝑏 = 𝑐 &
𝑎 ≠ 0, 𝑏𝑎 = 𝑐𝑎 ⟹ 𝑏 = 𝑐
Then we say that the cancellation laws hold in R.
Theorem: A ring R has no zero divisors if the cancellation laws hold in R.
Proof:
Case 1: Let the ring R has no zero or without zero divisors.
Now we have to prove that the cancellation laws hold in R.
Since R has no zero divisors, i.e., ab = 0 ⟹ 𝑎 = 0 or 𝑏 = 0.
Let a, b, c ∈ 𝑅 & 𝑎 ≠ 0, ab = ac
⟹ 𝑎𝑏 − 𝑎𝑐 = 0
⟹ a (b – c) = 0
⟹ 𝑏 − 𝑐 = 0 (∵ 𝑎 ≠ 0)
⟹𝑏=𝑐
Similarly we can prove that, for 𝑎 ≠0 and 𝑏𝑎 = 𝑐𝑎 ⟹ 𝑏 = 𝑐.
Therefore the cancellation laws holds in R.
Case 2: Consider the cancellation laws hold in R.
Now we have to prove that R has no zero divisors.
Suppose that R has zero divisors, then there exists 𝑎, 𝑏 𝜖 𝑅 ∋ a ≠ 0, b ≠ 0 and ab = 0.
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For ab = 0, we have a𝑏 = 𝑎. 0
⟹𝑏=0 (𝑏𝑦 𝐿. 𝐶. 𝐿)
Which is a contradiction to b ≠ 0.
Therefore a ≠ 0, b ≠ 0 and ab = 0 is not true in R and hence R has no zero divisors.
Solved Problem(s):
1. Solve the equation 𝑥 2 - 5x + 6 in the ring 𝑍12 .
Solution: In the ring of integers Z the given polynomial has two solutions 2, 3. Since
𝑥 2 -5x + 6 = 0 ⟹ ( x – 2 ) ( x – 3 ) = 0 ⟹ x = 2 and x = 3, where 2, 3 ϵ Z.
But in 𝑍12 , for x = 6 ⟹ ( 6 – 2 ) ×12 ( 6 – 3 ) = 4 ×12 3 = 0,
for x = 11 ⟹ ( 11 – 2 ) ×12 ( 11 – 3 ) = 9 ×12 8 = 0
Therefore the given polynomial has four solutions, namely 2, 3, 6 & 11 in the ring
𝑍12 .
Note: If P is a prime, then 𝑍𝑃 has no zero divisors.
Integral Domain: A commutative ring with unity containing non zero divisors is called an
integral domain and it is denoted with the symbol D.
Note: D is an integral domain iff (i) D is a ring (ii) D is a commutative
(iii) D has unity element (iv) D has no zero divisors
Example: The ring of integers Z is an integral domain. Since,
(1) 1 ϵ Z is the unity element (identity)
(2) ∀ a, b ϵ Z, we have ab = ba (commutativity) and
(3) ab = 0 ⇒ a = 0 or b = 0 (no zero divisors)
Example: Q, R, C w.r.to +, ∙ are integral domains.
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Exercise problem(s):
1. Prove that the set of all Gaussian integers Z(i) = { a + ib : a, b ϵ Z} is an integral
domain.
2. Prove that the Equivalence class modulo P, 𝑍𝑃 is integral domain if P is a prime.
3. Prove that the set of all 2 × 2 matrices is not an integral domain.
( it is not commutative and has zero divisors)
Invertible / Multiplicative Inverse: Let R be a ring with unity element 1. A non zero
element a ϵ R is said to be invertible under multiplication, if ∃ b ϵ R such that a b = b a = 1, b
ϵ R is called multiplicative inverse of a.
Example: 1 is the multiplicative identity element in the ring of integers Z.
For -1, 1 ϵ Z ⇒ (1) (1) = 1 and (-1) (-1) = 1.
Therefore -1, 1 are the invertible elements in Z.
Unit of a ring: Let R be a ring with unity. An element u ϵ R is said to be unity of R, if it has
multiplicative inverse in R.
1 1 1
Example: In the ring (Q, +, ∙ ), 2 ϵ Q and there exist ½ ϵ Q such that 2 ∙ 2 = 2 ∙ 2 = 1. Also 2
1
is called the multiplicative inverse of 2 and 2, 2 are called units.
Note: Zero element of a ring is not an unit.
Theorem: Every finite integral domain is a field.
Proof: Let 0, 1, 𝑎1 ,𝑎2 , ..., 𝑎𝑛 be the (n+1) distinct elements in the finite integral domain D.
Since, every integral domain is a commutative ring with unity and having no zero
divisors.
To prove the theorem, it is enough to prove that every non zero element of D has
multiplicative inverse.
Let a ≠ 0 ϵ D. Then the (n+1) products a.0, a.1, a𝑎1 , …, 𝑎𝑎𝑛 belongs to D. First we
prove that these are distinct.
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Suppose that a𝑎𝑖 = 𝑎𝑎𝑗 for i ≠ j.
⇒ a𝑎𝑖 − 𝑎𝑎𝑗 = 0
⇒ a(𝑎𝑖 − 𝑎𝑗 ) = 0
Since a ≠ 0 and by cancellation laws, we have
𝑎𝑖 − 𝑎𝑗 = 0
⇒ 𝑎𝑖 = 𝑎𝑗
This is contradiction, since 𝑎𝑖 ≠ 𝑎𝑗 for i ≠ j.
∴ a.0, a.1, a𝑎1 , …, 𝑎𝑎𝑛 are (n+1) distinct elements in D and are equal to 0, 1, 𝑎1 ,𝑎2 , ..., 𝑎𝑛 in
D in some order.
Since 1 ϵ D, so that a = 1 or a 𝑎𝑖 = 1 for some i.
For a ≠ 0 ϵ D, then there exists b = 𝑎𝑖 ϵ D such that ab=1
This shows that, for a ≠ 0 ϵ D has multiplicative inverse b in D.
Therefore we conclude that every finite integral domain is a field.
Idempotent element of a ring: In a ring R, if 𝑎2 = 𝑎 for 𝑎 ϵ R then 𝑎 is called an
idempotent element of R w.r.to multiplication.
Example: 0, 1 are the only idempotent elements in the ring of integers Z.
Note: The only idempotent elements in an integral domain are 0, 1 only.
Nilpotent element: Let R be a ring and a≠ 0 ϵ R, if there exists n ϵ N such that 𝑎𝑛 = 0 (zero
element). Then “a” is called nilpotent element in R.
Example: 2, 4 are the nilpotent elements in (𝑍8 , +8 , ×8 ), because 23 = 0 & 42 = 0.
Note: 𝑇here is no nilpotent elements in the ring (𝑍6 , +6 , ×6 ).
Theorem: An integral domain has no nilpotent elements other than zero.
Proof: Let R be an integral domain and 𝑎 ≠ 0 ∈ 𝑅.
We have 𝑎1 = 𝑎 ≠ 0, 𝑎2 = 𝑎. 𝑎 ≠ 0, since R has no zero divisors.
For n ∈ N, 𝑎𝑛 ≠ 0 and then 𝑎𝑛+1 = 𝑎𝑛 . 𝑎 ≠ 0.
Therefore by induction, 𝑎𝑛 ≠ 0 for every n ∈ N and hence the proof.
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Characteristic of a ring: The characteristic of a ring R is defined as the least positive integer
𝑝 such that 𝑝𝑎 = 0 for all 𝑎 ∈ 𝑅 and it is denoted by ch R In case such a positive integer
does not exist, then we say that the characteristic of R is zero or infinite.
Example: R = {0, 1, 2, 3, 4, 5, 6} be the ring under addition & multiplication modulo7.
Here the zero element in R is 0.
Now ∀ 𝑎 ∈ 𝑅 ⟹ 7𝑎 = 0 (𝑚𝑜𝑑 7)
⟹ 7𝑎 = 0 ∀ 𝑎 ∈ 𝑅
Further for 1 ∈ 𝑅, 𝑝. 1 = 𝑝 ≠ 0, where 𝑝 ≠ 0 and 0 < 𝑝 < 7.
∴ ∀ 𝑎 ∈ 𝑅, then there exists a least positive integer 7 such that 7𝑎 = 0.
Hence the characteristic of the ring R = 7.
Note:
1. The characteristic of the ring (Z, +, .) is zero. Because, for every 𝑎 ∈ 𝑍 there exists
no positive integer 𝑛 such that 𝑛𝑎 = 0.
2. The characteristic of an integral domain D is zero or a positive integer according as the
order of any non zero element of R regarded as a member of the additive group (R, +).
Theorem: The characteristic of an integral domain is either a prime (or) zero.
Proof: Let (R, +, .) be an integral domain with characteristic p (p > 0). Now we have to prove
that p prime.
Suppose that p is not a prime. Then
𝑝 = 𝑚 𝑛, where 1 < 𝑚, 𝑛 < 𝑝.
For 𝑎 ≠ 0 ∈ 𝑅 ⇒ 𝑎. 𝑎 = 𝑎2 ≠ 0 ∈ 𝑅.
Since ch R= 𝑝, then 𝑝𝑎2 = 0
⇒ (𝑚𝑛)𝑎2 = 0
⇒ (𝑚𝑎)(𝑛𝑎) = 0
⇒ (𝑚𝑎) = 0 or (𝑛𝑎) = 0, since R has no zero divisors.
Let 𝑚𝑎 = 0. For any 𝑥 ∈ 𝑅, (𝑚𝑎)𝑥 = 0 ⇒ 𝑎(𝑚𝑥 ) = 0 ⇒ 𝑚𝑥 = 0, since 𝑎 ≠ 0.
This is absurd as 1 < 𝑚 < 𝑝 and characteristic of 𝑅 = 𝑝. Therefore 𝑚𝑎 ≠ 0.
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Let 𝑛𝑎 = 0 and for any 𝑥 ∈ 𝑅, similarly we get 𝑛𝑎 ≠ 0.
i.e., for (𝑚𝑎)(𝑛𝑎) = 0 ⇒ 𝑚𝑎 ≠ 0 or 𝑛𝑎 ≠ 0. This is contradiction to 𝑅 is an
integral domain. Hence the proof follows.
Solved Problem(s):
1. Prove that the characteristic of a Boolean ring is 2.
Solution: Since 𝑎2 = 𝑎, ∀ 𝑎 ∈ 𝑅, we have (𝑎 + 𝑎)2 = 𝑎 + 𝑎
⟹ (𝑎 + 𝑎)(𝑎 + 𝑎) = (𝑎 + 𝑎)
⟹ 𝑎 (𝑎 + 𝑎 ) + 𝑎 (𝑎 + 𝑎 ) = (𝑎 + 𝑎 )
⟹ ( 𝑎 2 + 𝑎 2 ) + (𝑎 2 + 𝑎 2 ) = 𝑎 + 𝑎
⟹ (𝑎 + 𝑎) + (𝑎 + 𝑎) = (𝑎 + 𝑎)+0
⟹𝑎+𝑎 =0
⟹ 2𝑎 = 0
This proves that for every 𝑎 ∈ 𝑅, we have 2𝑎 = 0. Further, for 𝑎 ≠ 0, 1. 𝑎 = 𝑎 ≠ 0.
Therefore 2 is the least positive integer such that 2𝑎 = 0, ∀ 𝑎 ∈ 𝑅 and hence the
characteristic of R is 2.
2. If the characteristic of a ring is 2 and the elements a, b of the ring are commute. Then
prove that (𝑎 + 𝑏)2 = 𝑎2 +𝑏2 = (𝑎 − 𝑏)2 .
Solution: Since the characteristic of the ring R, Ch R = 2.
⟹ 2𝑥 = 0, ∀ 𝑥 ∈ 𝑅
Since 𝑎, 𝑏 ∈ 𝑅 are commute, then 𝑎𝑏 = 𝑏𝑎.
Let (𝑎 + 𝑏)2 = (𝑎 + 𝑏)(𝑎 + 𝑏)
= 𝑎(𝑎 + 𝑏) + 𝑏(𝑎 + 𝑏)
= 𝑎𝑎 + 𝑎𝑏 + 𝑏𝑎 + 𝑏𝑏
= 𝑎2 + 𝑎𝑏 + 𝑎𝑏 + 𝑏2
= 𝑎2 + 2𝑎𝑏 + 𝑏2 , since 𝑅 is commutative
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= 𝑎2 + 𝑏2 , since Ch R=2
Similarly, we can prove that (𝑎 − 𝑏)2 = 𝑎2 + 𝑏2 .
Exercise Problem: If the characteristic of a ring is 3 and the elements a, b of the ring are
commute. Then prove that (𝑎 + 𝑏)3 = 𝑎3 +𝑏3 = (𝑎 − 𝑏)3 .
Divisor (or) Factor: Let R be a commutative ring and 𝑎 ≠ 0, 𝑏 ∈ 𝑅. If there exists 𝑞 ∈ 𝑅
such that 𝑏 = 𝑎𝑞, then 𝑎 is said to be divide b and it is denoted by 𝑎|𝑏.
Example: In the ring (Z,+, .), 3 × 5 = 15. Then 3|15.
But, in the ring Q of rational numbers; 3/7 exist because there exist (7/3)∈ 𝑄 such that
7=3.(7/3)
Unit: Let R be a commutative ring with unity .An element 𝑎 ∈ 𝑅 is said to be a unit in R if
there exist an element 𝑏 ∈ 𝑅 such that 𝑎𝑏 = 1 in R. However the unity element 1 is also a
unit because 1.1=1
Associates: Let R be a commutative ring with unity .Two element 𝑎, 𝑏 ∈ 𝑅 are said to be
associates if 𝑏 = 𝑢𝑎 for some unit u in R.
Example: in the ring Z of integers,the units are 1,-1 only.
For,𝑎 ≠ 0 ∈ 𝑅 we have 𝑎. 1 = 𝑎 and 𝑎 = (−𝑎)(−1) only
Hence ,𝑎 ∈ 𝑍 has only two associates,namely 𝑎, −𝑎.
Examples(1):Find all the associates of (2 − 𝑖)in the ring of Gaussian integers.
Sol: We have 2 − 𝑖 = (2 − 𝑖 ). 1; (2 − 𝑖 ) = (−2𝑖 − 1). 𝑖
2 − 𝑖 = (−2 + 𝑖 )(−1) ; (2 − 𝑖 ) = (2𝑖 + 1). (−𝑖)
∴ 2 − 𝑖, −2 + 𝑖, −2𝑖 − 1 𝑎𝑛𝑑 2𝑖 + 1 are the associates.
Example(2): Find all units of 𝑍14
Sol: 𝑍14 = {0,1,2,3,4,5,6,7,8,9,10,11,12,13}
Unity element =1 is unit. since 14 is even, even number in 𝑍14 can not be unit
For, 3 ∈ 𝑍14 we have 3.5 = 15 = 1 ⇒ 3,5 are units
For, 9 ∈ 𝑍14 we have 9.11 = 99 = 1 ⇒ 9,11 are units
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For, 13 ∈ 𝑍14 we have 13.13 = 169 = 1 ⇒ 13 is a unit
Exercise Problem: If the characteristic of a ring is 3 and the elements a, b of the ring are
commute. Then prove that (𝑎 + 𝑏)3 = 𝑎3 +𝑏3 = (𝑎 − 𝑏)3 .
Subring: Let (R, +, .) be a ring S be a non-empty subset of R if (S, +, .) is also a ring
with respect to the two operations +, . in R then (S, +, .) is a sub ring of R.
Example:
(1) The set of even integers is a subring of ring (Z, +, .).
𝑎
(2) Let (Q, +, .) be a ring of rational number then S={ 2 /a∈ 𝑧} is a non empty subset of
1 1 1 1
Q and (S, +) is a subgroup (Q, +). But for ∈ 𝑆, we have (2 )( 2 )= 4 ∉ 𝑆 and hence
2
‘.’ is not a binary operation in S. Thus (S,+,.) is not a Subring of (Q,+,.)
(3) Let (R, +, .) be a ring and 0 ∈ 𝑅 be the zero element of R, then S={0} is a non
empty subset of R so that (S, +, .) is itself a ring.
∴ (𝑆, +, . ) is a subring of R,
Also ({0}, +, .) is called improper subring of R.
For each positive integer n, the set nz ={0, ±𝑛, ±2𝑛, ±3𝑛, . ..} is a subring of Z
(4) The set of Gaussian integers Z[i] = { a+ib : a, b∈ 𝑍, 𝑖 2 = −1 } is a subring of
complex numbers field ‘C’.
Subfield & Subdomain: Let (F, +, .) be a field and (S, +, .) be a subring of F.
If (S, +, .) is a field then we say that S is a subfield of F.
If (S,+,.) is an integral domain then we say that S is a subdomain of F.
Example: If (S, +, .) is a subfield of the field (F, +, .) then (S, +) is a subgroup and (S-
{0},.) is a subgroup of (F-{0},.).
Necessary and Sufficient Condition for a non empty set to be a subring.(subring
test): Let S be a non empty subset of a ring R. Then S is a subring of R if and only if
a-b ∈ 𝑆 𝑎𝑛𝑑 𝑎. 𝑏 ∈ 𝑆, ∀ 𝑎, 𝑏 ∈ 𝑆.
Proof: Let S be a non empty subset of a ring (R,+, . ).
Now we have to prove that S is a subring of R if and only if a-b ∈ 𝑆 𝑎𝑛𝑑
𝑎. 𝑏 ∈ 𝑆, ∀ 𝑎, 𝑏 ∈ 𝑆.
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Case 1: Let S be a subring of R. Then 𝑆 is itself is a ring.
For 𝑎, 𝑏 ∈ 𝑆 ⟹ 𝑎, −𝑏 ∈ 𝑆
⟹ 𝑎 + (−𝑏) = 𝑎 − 𝑏 ∈ 𝑆 (𝑤. 𝑟, 𝑡𝑜 ′ + ′) and
for 𝑎, 𝑏 ∈ 𝑆 ⟹ 𝑎𝑏 ∈ 𝑆 (w.r.to ‘.’)
Case 2: Let 𝑎 − 𝑏 ∈ 𝑆 ---(1) & 𝑎𝑏 ∈ 𝑆 ∀ 𝑎, 𝑏 ∈ 𝑆 ---(2).
For 𝑎, 𝑎 ∈ 𝑆, then 𝑎 − 𝑎 = 0̅ ∈ 𝑆 from (1).
Also, for 0̅, 𝑎 ∈ 𝑆 then 0̅ − 𝑎 = −𝑎 ∈ 𝑆
For 𝑎, 𝑏 ∈ 𝑆, then 𝑎, −𝑏 ∈ 𝑆 ⟹ 𝑎 + 𝑏 = 𝑎 − (−𝑏) ∈ 𝑆
From (2), 𝑏 ∈ 𝑆 ∀ 𝑎, 𝑏 ∈ 𝑆 .
This implies that 𝑆 satisfies all the properties of a ring w.r.to the operations
defined on the ring 𝑅. Hence 𝑆 is a subring of 𝑅.
NOTE: every subring contains at least zero element of the ring.
Theorem: The intersection of two subrings of a ring R is a subring of R.
Proof: Let 𝑆1 and 𝑆2 be two subrings of R. Let 0̅ ∈ 𝑅 be a zero element in 𝑅.
Since every subring contains at least zero element of the ring, then
0̅ ∈ 𝑆1 and 0̅ ∈ 𝑆2 ⟹ 0̅ ∈ 𝑆1 ∩ 𝑆2 .
This implies that 𝑆1 ∩ 𝑆2 ≠ ∅ & 𝑆1 ∩ 𝑆2 ⊂ 𝑅.
Let 𝑎, 𝑏 ∈ 𝑆1 ∩ 𝑆2 then 𝑎, 𝑏 ∈ 𝑆1 𝑎𝑛𝑑 𝑎, 𝑏 ∈ 𝑆2 .
For 𝑎, 𝑏 ∈ 𝑆1 and 𝑆1 is a subring of R. Then 𝑎 − 𝑏 ∈ 𝑆1 & 𝑎𝑏 ∈ 𝑆1 ---- (1)
For 𝑎, 𝑏 ∈ 𝑆2 and 𝑆2 is a subring of R. Then 𝑎 − 𝑏 ∈ 𝑆2 & 𝑎𝑏 ∈ 𝑆2 ---- (2)
From (1) & (2), −𝑏 ∈ 𝑆1 ∩ 𝑆2 & 𝑎𝑏 ∈ 𝑆1 ∩ 𝑆2 , for all 𝑎, 𝑏 ∈ 𝑆1 ∩ 𝑆2 .
∴ 𝑆1 ∩ 𝑆2 is a subgroup of R.
𝑎 𝑏]
Problem: Show that the set of all matrices [ is a subring of the ring of 2× 2
0 𝑐
matrices whose elements are integrals.
22
𝑎𝑏]
𝑺𝒐𝒍: Let 𝑅 = { [ /𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝑍} be the ring of 2× 2 matricec and
0𝑐
𝑎 𝑏]
𝑆 = {[ /0, 𝑏, 𝑐, 𝑑 ∈ 𝑍} then 𝑆 ≠ ∅ & 𝑆 ⊂ 𝑅.
0 𝑐
𝑎1 𝑏1 𝑎 𝑏2
Let A, B ∈ 𝑆, where 𝐴 = [ ] & B= [ 2 ], 0, 𝑎1 , 𝑏1 , 𝑐1 , 𝑎2 , 𝑏2 , 𝑐2 ∈ 𝑍
0 𝑐1 0 𝑐2
𝑎1 𝑏1 𝑎 𝑏2 𝑎 − 𝑎2 𝑏1 − 𝑏2
Now 𝐴 − 𝐵 = [ ]−[ 2 ]=[ 1 ]and
0 𝑐1 0 𝑐2 0 𝑐1 − 𝑐2
𝑎 𝑏1 𝑎2 𝑏2 𝑎 𝑎 𝑎1 𝑏2 + 𝑏1 𝑐2
AB= [ 1 ][ ]=[ 1 2 ]
0 𝑐1 0 𝑐2 0 𝑐1 𝑐2
Since 𝑎1 − 𝑎2 , 𝑎1 𝑎2 , 𝑐1 𝑐2 , 𝑐1 − 𝑐2 , 𝑎1 𝑏2 + 𝑏1 𝑐2 ∈ 𝑍, then A-B ∈ 𝑆 & 𝐴𝐵 ∈ 𝑆, for all
A, B ∈ 𝑆.
Therefore 𝑆 is a subring of R.
Problem: If R is a ring and 𝐶 (𝑅) = {𝑥 ∈ 𝑅: 𝑥𝑎 = 𝑎𝑥 ∀ 𝑎 ∈ 𝑅} then prove that C(R) is
a subring of R.
Sol: Let 0 ∈ 𝑅, the zero element of the ring R, we have 0. 𝑎 = 𝑎. 0 ∀𝑎 ∈ 𝑅.
By the definition of C(R), 0 ∈ 𝐶 (𝑅). Then 𝐶 (𝑅) ≠ ∅ and 𝐶 (𝑅) ⊂ 𝑅.
Let 𝑥, 𝑦 ∈ 𝐶 (𝑅), then
𝑥𝑎 = 𝑎𝑥, 𝑦𝑎 = 𝑎𝑦 ∀ 𝑎 ∈ 𝑅 -------(1)
Consider 𝑎(𝑥 − 𝑦) = 𝑎𝑥 − 𝑎𝑦 = 𝑥𝑎 − 𝑦𝑎 = (𝑥 − 𝑦)𝑎
Also ∀ 𝑎 ∈ 𝑅,
𝑎(𝑥𝑦) = (𝑎𝑥 )𝑦 = (𝑥𝑎)𝑦
= 𝑥(𝑎𝑦)
= 𝑥(𝑦𝑎)
= (𝑥𝑦)𝑎 (from (1))
Therefore, we conclude that 𝑥 − 𝑦, 𝑥𝑦 ∈ 𝐶 (𝑅) ∀ 𝑥, 𝑦 ∈ 𝐶 (𝑅) and hence C(R) is
a subring of R.
Exercise problems:
1. If R is a division Ring then show that 𝐶 (𝑅) = {𝑋 ∈ 𝑅⁄𝑥𝑎 = 𝑎𝑥 ∀ 𝑎 ∈ 𝑅} then
prove that C(R) is a Field
2. Let R the ring of 2X2 matrices whose elements are real numbers . then prove
𝑎 0
that the set 𝑆 = {( )⁄𝑎 , 𝑏 ∈ 𝑅} is a sub ring of R.
𝑏 0
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Left Ideal: A non empty subset U of a ring R is called a left ideal if
(i) 𝑎, 𝑏 ∈ 𝑈 ⇒ 𝑎 − 𝑏 ∈ 𝑈
(ii) 𝑎 ∈ 𝑈, 𝑟 ∈ 𝑅 ⇒ 𝑟𝑎 ∈ 𝑈
Right Ideal: A non empty subset U of a ring R is called a right ideal if
(i) 𝑎, 𝑏 ∈ 𝑈 ⇒ 𝑎 − 𝑏 ∈ 𝑈
(ii) 𝑎 ∈ 𝑈, 𝑟 ∈ 𝑅 ⇒ 𝑎𝑟 ∈ 𝑈
Ideal: Let (𝑅, +, . ) be a Ring. A non empty subset U of a ring R is called a two sided ideal
or ideal if
(i) 𝑎, 𝑏 ∈ 𝑈 ⇒ 𝑎 − 𝑏 ∈ 𝑈 and
(ii) 𝑎 ∈ 𝑈, 𝑟 ∈ 𝑅 ⇒ 𝑎𝑟, 𝑟𝑎 ∈ 𝑈
Note: An ideal is both a left and a right ideal.
Examples:
1. If R is a ring, then R itself is an ideal of R. So, R is called unit ideal or improper ideal.
2. In a ring R, 0 ∈ 𝑅 is the zero element in R. Then U={0} is an ideal of R. Also U is called
Null ideal or zero ideal or trivial ideal.
3. Let R= the ring of all integers & U = the set of all even integers. Since, the set of all even
integers form a subring of R. Hence, U is a subring of R. Also for 𝑎 ∈ 𝑈, 𝑟 ∈ 𝑅 ⇒ 𝑎𝑟, 𝑟𝑎 ∈
𝑈. Therefore U is an ideal of R.
Theorem: If U is an ideal of a ring R with unity element 1∈ 𝑈 then 𝑈 = 𝑅.
Proof: By the def. of an ideal, U⊆ 𝑅 − − − (1).
Also, for 𝑥 ∈ 𝑅 and 1 ∈ 𝑈, we have 𝑥. 1 ∈ 𝑈 ⇒ 𝑥 ∈ 𝑈
i.e., if 𝑥 ∈ 𝑅 then 𝑥 ∈ 𝑈, and then 𝑅 ⊆ 𝑈 − − − (2).
From (1) & (2), 𝑈 = 𝑅.
Theorem: A field has no proper non trivial ideals.
(or)
The ideals of a field F are only {0} and F only.
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Proof: Let U be an ideal of a field F such that 𝑈 ≠ {0}.
Now we have prove that U=F.
By the definition of an ideal, U⊆ 𝐹 ------- (1)
Let 𝑎 ≠ 0 ∈ 𝑈 be an arbitrary element. For 𝑎 ≠ 0 ∈ 𝐹 (𝑠𝑖𝑛𝑐𝑒 U ⊆ 𝐹 ) then there
exists 𝑎−1 ∈ 𝐹 such that 𝑎𝑎−1 = 1.
For 𝑎 ∈ 𝑈, 𝑎−1 ∈ 𝐹 and U is an ideal of 𝐹, then
𝑎𝑎−1 = 1 ∈ 𝑈
For 𝑥 ∈ 𝐹, 1 ∈ 𝑈 and 𝑈 is an ideal of 𝐹, then
𝑥. 1 ∈ 𝑈
⇒ 𝑥∈𝑈
i.e., for each 𝑥 ∈ 𝐹 we have 𝑥 ∈ 𝑈 and then 𝐹 ⊆ 𝑈 ------ (2)
From (1) & (2), we have 𝐹 = 𝑈. Therefore, we conclude that the ideals of a field
F are either {0} or F.
Theorem: Prove that the intersection of two ideals of a ring R is an ideal of R.
Proof: Let 𝑈1 and 𝑈2 be two ideals of a ring R.
For the zero element 0 ∈ 𝑅, then 0 ∈ 𝑈1 & 0 ∈ 𝑈2 .
This implies that 0 ∈ 𝑈1 ∩ 𝑈2 and 𝑈1 ∩ 𝑈2 ≠ ∅.
Let 𝑎, 𝑏 ∈ 𝑈1 ∩ 𝑈2 𝑎𝑛𝑑 𝑟 ∈ 𝑅. Then 𝑎, 𝑏 ∈ 𝑈1 & 𝑎, 𝑏 ∈ 𝑈2 .
For 𝑎, 𝑏 ∈ 𝑈1 , 𝑟 ∈ 𝑅 and 𝑈1 is an ideal, then
𝑎 − 𝑏 ∈ 𝑈1 𝑎𝑛𝑑 𝑎𝑟, 𝑟𝑎 ∈ 𝑈1 ------(1)
For 𝑎, 𝑏 ∈ 𝑈2 , 𝑟 ∈ 𝑅 𝑎𝑛𝑑 𝑈2 is an ideal, then
𝑎 − 𝑏 ∈ 𝑈2 𝑎𝑛𝑑 𝑎𝑟, 𝑟𝑎 ∈ 𝑈2 -------(2)
From (1) & (2), we have 𝑎 − 𝑏 ∈ 𝑈1 ∩ 𝑈2 𝑎𝑛𝑑 𝑎𝑟, 𝑟𝑎 ∈ 𝑈1 ∩ 𝑈2 . And therefore
𝑈1 ∩ 𝑈2 is an ideal of R.
Theorem: If 𝑈1 𝑎𝑛𝑑 𝑈2 are two ideals of a ring R then 𝑈1 ∪ 𝑈2 is an ideal of R if and
only if 𝑈1 ⊆ 𝑈2 or 𝑈2 ⊆ 𝑈1 .
Proof: Let 𝑈1 ∪ 𝑈2 be an ideal of R. Now we prove that 𝑈1 ⊆ 𝑈2 or 𝑈2 ⊆ 𝑈1 .
Suppose that 𝑈1 ⊈ 𝑈2 and 𝑈2 ⊈ 𝑈1 .
For 𝑈1 ⊈ 𝑈2 , taking an element 𝑎 ∈ 𝑈1 𝑎𝑛𝑑 𝑎 ∉ 𝑈2 . Similarly, for 𝑈2 ⊄ 𝑈1 , taking
an element 𝑏 ∈ 𝑈2 𝑎𝑛𝑑 𝑏 ∉ 𝑈1 .
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For 𝑎 ∈ 𝑈1 and 𝑏 ∈ 𝑈2 , then 𝑎, 𝑏 ∈ 𝑈1 ∪ 𝑈2 .
Taking 𝑎, 𝑏 ∈ 𝑈1 ∪ 𝑈2 and 𝑈1 ∪ 𝑈2 is an ideal, then
𝑎 − 𝑏 ∈ 𝑈1 ∪ 𝑈2 .
This implies that 𝑎 − 𝑏 ∈ 𝑈1 or 𝑎 − 𝑏 ∈ 𝑈2 .
For 𝑎 − 𝑏 ∈ 𝑈1 and 𝑎 ∈ 𝑈1 , then
𝑎 − (𝑎 − 𝑏) = 𝑏 ∈ 𝑈1 ------ (1)
Similarly, for 𝑎 − 𝑏 ∈ 𝑈2 and 𝑏 ∈ 𝑈2 , then
𝑏 + (𝑎 − 𝑏) = 𝑎 ∈ 𝑈2 ------ (2)
(1) & (2) contradicts to 𝑎 ∉ 𝑈2 , 𝑏 ∉ 𝑈1 .
Therefore our assumption that 𝑈1 ⊈ 𝑈2 and 𝑈2 ⊈ 𝑈1 is wrong and hence 𝑈1 ⊆ 𝑈2 or
𝑈2 ⊆ 𝑈1 .
Conversely, suppose that 𝑈1 ⊆ 𝑈2 or 𝑈2 ⊆ 𝑈1 .
Then 𝑈1 ∪ 𝑈2 = 𝑈2 𝑜𝑟 𝑈1 , 𝑈2 and 𝑈1 ideals and therefore 𝑈1 ∪ 𝑈2 is an ideal of
𝑅.
Principal ideal: Let R be a commutative ring with unity and 𝑎 ∈ 𝑅. Then the ideal {𝑟𝑎 ∕
𝑟 ∈ 𝑅} of all multiples of 𝑎 is called the principal ideal generated by 𝑎 and it is denoted
by {𝑎} or < 𝑎 >.
An ideal U of the ring R is a principal ideal, then 𝑈 =< 𝑎 >= {𝑟𝑎 ∕ 𝑟 ∈ 𝑅} for
some 𝑎 ∈ 𝑅.
Example: The set of all integers Z is a commutative ring with unity element. The
principal ideal generated by the element 2 ∈ 𝑍 = < 2 > = {2𝑧⁄𝑛 ∈ 𝑍} = the set of all
even integers
Quotient Rings or Factor Rings: Let R be a ring and U be an ideal of R. Then the set
𝑅⁄ = {𝑥 + 𝑈⁄𝑥 ∈ 𝑅} with respect to induced operations of addition and multiplication
𝑈
of Cosets defined by (𝑎 + 𝑈) + (𝑏 + 𝑈) = (𝑎 + 𝑏) + 𝑈; (𝑎 + 𝑈). (𝑏 + 𝑈) = (𝑎. 𝑏) + 𝑈
for (𝑎 + 𝑈), (𝑏 + 𝑈) ∈ 𝑅⁄𝑈 is a ring. This ring (𝑅⁄𝑈 , + , . ) is called Quotien ring or
Factor ring of residue class ring R.
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Theorem: If U is an ideal of a ring R then the set 𝑅⁄𝑈 = {𝑥 + 𝑈⁄𝑥 ∈ 𝑅} is a ring with
respect to the induced operations of addition and multiplication of cosets defined as
(𝑎 + 𝑈) + (𝑏 + 𝑈) = (𝑎 + 𝑏) + 𝑈 and
Proof: Since (R, +) is commutative group, then the quotient group (𝑅⁄𝑈 , +) is commutative.
In order to show that (𝑅⁄𝑈 , + , . ) is a ring. We prove that
1) Addition and Multiplication of cosets is well defined
2) Multiplication is associative
3) Distributive laws holds.
To Prove that Addition and Multiplication of cosets is well defined:
Let (𝑎 + 𝑈) = 𝑎1 + 𝑈 and (𝑏 + 𝑈) = 𝑏1 + 𝑈. Then
𝑎𝑏 = (𝑎1 + 𝑢1 )(𝑏1 + 𝑢2 ) = 𝑎1 𝑏1 + 𝑎1 𝑢2 + 𝑢1 𝑏1 + 𝑢1 𝑢2
Since U is an ideal , 𝑎1 𝑢2 , 𝑢1 𝑏1 , 𝑢1 𝑢2 ∈ 𝑈
∴ 𝑎𝑏 − 𝑎1 𝑏1 ∈ 𝑈 and hence 𝑎𝑏 + 𝑈 = 𝑎1 𝑏1 + 𝑈
⇒ (𝑎 + 𝑈. )(𝑏 + 𝑈) = (𝑎1 + 𝑈). (𝑏1 + 𝑈)
There fore multiplication of cosets is well defined.
To Prove that Multiplication is Associative:
Let 𝑎 + 𝑢, 𝑏 + 𝑈, 𝑐 + 𝑈 ∈ 𝑅⁄𝑈 , then
(𝑎 + 𝑈)[(𝑏𝑐) + 𝑈] = (𝑎 + 𝑈)[(𝑏 + 𝑈)(𝑐 + 𝑈)]
∴ multiplication is associative holds
To prove that Distributive Laws hold:
Let 𝑎 + 𝑈 𝑏 + 𝑈, 𝑐 + 𝑈 ∈ 𝑅⁄𝑈, then
(𝑎 + 𝑈). [(𝑏 + 𝑈) + (𝑐 + 𝑈)] = (𝑎 + 𝑈). [𝑏 + 𝑐 ) + 𝑈] = 𝑎(𝑏 + 𝑐 ) = 𝑈
=𝑎(𝑏 + 𝑐 ) = 𝑈
=(𝑎𝑏 + 𝑎𝑐) = 𝑈
=(𝑎𝑏 + 𝑈)(𝑎𝑐 + 𝑈)
=(𝑎 + 𝑈). (𝑏 + 𝑈) + (𝑎 + 𝑈. (𝑐 + 𝑈)
Similarly we can prove that
[(𝑏 + 𝑈) + (𝑐 + 𝑈)]. (𝑎 + 𝑈) = (𝑏 + 𝑈). (𝑎 + 𝑈) + (𝑐 + 𝑈). (𝑎 + 𝑈)
Hence (𝑅⁄𝑈, + ,∗) is a ring
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Theorem: If 𝑅⁄𝑈 is a quotient ring then prove that
(i) 𝑅⁄𝑈 is commutative if R is commutative and
(ii) 𝑅⁄𝑈 has unity element if R has unity element.
Proof :
(i) Let R is commutative , then 𝑎𝑏 = 𝑏𝑎, ∀ 𝑎, 𝑏 ∈ 𝑅
Let 𝑎 + 𝑈, 𝑏 + 𝑈 ∈ 𝑅⁄𝑈 , then
(𝑎 + 𝑈)(𝑏 + 𝑈) = 𝑎𝑏 + 𝑈 = 𝑏𝑎 + 𝑈 = (𝑏 + 𝑈)(𝑎 + 𝑈)
∴ 𝑅⁄𝑈 is commutative .
(𝑖𝑖 ) Let R has unity element, i.e., there exist 1 ∈ 𝑅 such that
𝑎. 1 = 1. 𝑎 = 𝑎, ∀ 𝑎 ∈ 𝑅
Let (𝑎 + 𝑈) ∈ 𝑅⁄𝑈 . Then, for 1 ∈ 𝑅, wehave 1 + 𝑈 ∈ 𝑅⁄𝑈.
Now we prove that 1+U is the unity element in 𝑅⁄𝑈.
Taking (𝑎 + 𝑈)(1 + 𝑈) = 𝑎. 1 + 𝑈 = 𝑎 + 𝑈 and
(1 + 𝑈)(𝑎 + 𝑈) = 1. 𝑎 + 𝑈 = 𝑎 + 𝑈, ∀ 𝑎 ∈ 𝑅.
i.e., (𝑎 + 𝑈)(1 + 𝑈) = (1 + 𝑈)(𝑎 + 𝑈) = 𝑎 + 𝑈.
∴ 1+U is the unity element in 𝑅⁄𝑈
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