Definitions and Examples of Rings and Fields

DEFINITIONS AND EXAMPLES OF RINGS AND FIELDS :

Definition 1: Ring           [A.U M/J 2014]

An algebraic system (S, +, .) is called a ring if the binary operations + and on S satisfy the following three properties :

1. (S,+) is an abelian group

2. (S, .) is a semigroup

3. The operation . is distributive over +; that is, for any a, b, c Є S,

a. (b + c) = a.b + a.c and (b + c).a = = b.a + c.a

Examples:

1. The set of all integers Z, the set of all rational numbers R⁺, the set of all real numbers R are rings under the usual addition and usual multiplication.

2. The set of all n ˟ n matrices M_n is a ring under the matrix addition and matrix multiplication.

Definition 2: Integral domain.

A commutative ring (S, +, .) with identity and without divisors of zero is called can integral domain.

Definition 3: Field

A commutative ring (S, +, .) which has more than one element such that every non-zero element of S has a multiplicative inverse in S is called a field.

Definition 4: Sub ring.

A subset RCS where (S, +, .) is a ring is called a subring if (R, +, .) is itself with the operations + and restricted to R.

Examples:

1. The ring of integers Z is a subring of the ring of all rational numbers Q.

2. In Z the ring of all integers the set of all even integers is a subring.

2. The ring Q of all rational numbers, and the ring R of real numbers are fields.

3. The ring (Z₇, +₇, ×₇) is a field.

4. The ring (Z₁₀, +₁₀, ×₁₀) is not an integral domain. (as 5 ×₁₀ 2 = 0, yet 5≠0, 2≠0 in Z₁₀).

5. The ring Z of all integers is an integral domain but not a field.

Definition 6. Commutative Ring :

A ring (R, +, codt) is said to be commutative

if a.b = b.a ∀ a, b Є R

Theorem 1: Every finite integral domain is a field.

Proof : Let (R, +, .) be a finite integral domain.

To prove (R - {0}, .) is a group

i.e., to prove

(i) there exists an element 1 Є R such that

1. a = a. 1 = a, for all a Є R (1 Є R is an identity)

(ii) for every element of 0 ≠ a Є R, there exists an element a^-1 Є R such that

a.a^-1 =  a^-1.a = 1

Let R - {0} = {a₁, a₂, a₃, …. a_n}

Let a Є R - {0}, then the elements aa₁, aa₂, … aa_n are all in R - {0} and they are all distinct.

(i.e.,) If a.a_i = a.a_j, i ≠ j

Then  a.(a_i - a_j) = 0

Since R is an integral domain and a ≠ 0, we must have a_i - a_j =0,

(i.e.,) a_i = a_j which is a contradiction.

R - {0} has exactly n elements, and R is a commutative ring with cancellation law

we get a = a.i₀, for some i₀ (since a Є R- {0})

i.e., a.ai₀ = ai₀.a (Since R is commutative)

Thus, let x = a. a_i for same a_i Є R - {0}, and

y.ai₀ = a.ai₀ = (a_i. a) ai₀ = a_i.a = a. a_j = y

Hence ai₀ is an unity R - {0}. We write it as 1.

Since 1 Є R {0}, therefore there exists an element aa_k Є R-{0}

such that

aak = 1

ba = ab = 1 (let a_k = b)

b is the inverse of a, and conversely.

Hence (R, +, .) is a field.

Thereom 2: Every field is an integral domain, but the converse need not be true.

Proof :

Let (F, +, .) is a field.

(i.e.,) F is a commutative ring with unity.

To prove F is an integral domain it is enough to show that it has non zero divisor.

Let a, b Є F, such that a .b = 0

Let a ≠ 0, then a^-1 Є F

a.b = 0

⇒ a^-1. (a.b) = a^-1.0

⇒ 1.b = 0

⇒ b = 0

Hence the proof.

Example 1: Define ring and give an example of an ring with zero-divisors.   [A.U. N/D, 2005]

Solution :    

Definition: Ring

An algebraic system (S, +, .) is called a ring if the binary operations + and . on S satisfy the following three properties.

1. (S, +) is an abelian group.

2. (S, .) is a semi group.

3. The operation . is distributive over + ; that is, for any a, b, c Є S.

a. (b + c) = a.b + a.c and

(b + c).a = b.a + c.a

Example: The ring (Z₁₀, +₁₀, ×₁₀) is not an integral domain.

Since 5 × 10² = 0, yet 5 ≠ 0, 2 ≠ 0 in Z₁₀.

Theorem 3:

Let (R, +, .) be a ring. Then

1. a.0 = 0.a = 0, for every a Є R

2. a.(-b) = - (a.b) = (-a). b, for all a, b Є R

3. a.(b-c) = a.b - a.c, for all a, b, c Є R

4. (a - b).c = a.c - b.c, for all a, b, c Є R

Proof :

1. Let a Є R. Then

a.0 = a. (0+0) as 0 + 0 = 0

= a.0+ a.0 by left distributive law

So a.0 = 0, by the cancellation law for the group (R, +)

Similarly 0.a = 0

2. Let a, b Є R then

0 = a.0= a. (b + (-b))

= a. b + a. (-b)    by left distributive law

It follows that a. (-b) = - (a.b)

Similarly, (-a). (b) = - (a.b)

3. a. (b-c) = a. (b + (−c))

= a.b + a. (-c)

= a.b - a.c     by (2)

4. (a - b).c = (a + (-b)).c

= a.c + (-b).c

= a.c - b.c

Theorem 4: A commutative ring (R, +, .) is an integral domain if and only if the cancellation laws holds in R.

(i.e.,) a.b = a.c and

a≠0 ⇒ b = c, for all a, b, c Є R

Proof :

Let R be an integral domain and

a.b = a.c and a ≠ 0, for all a, b, c Є R

we have a.b – a.c = 0 ⇒ a. (b-c) = 0

since R is an integral domain and a ≠ 0, then b - c = 0. (R has no zero divisor)

b = c. Hence the cancellation law holds.

Conversely: Assume that the cancellation law holds in a ring R.

Let a.b = 0, for a ≠ 0 and b Є R

We have ab = 0 a0

b = 0

Thus ab = 0 in R ⇒ a = 0 or b = 0

R has no zero divisors.

R is an integral domain.

Theorem 5: If R be a commutative ring with unity whose ideals are {0} and R, then R is a field.

Proof :

We have to show that for any 0 ≠ a Є R there exists an element 0 ≠ b Є R such that

ab = 1

Let 0≠a Є R

Define Ra = {ra | r Є R}

Claim: Ra is an ideal in Ra.

Let u, v Є Ra, then u = r₁ a, v = r₂ a, for some r₁, r₂ Є R.

Now u – v = r₁ a – r₂ a = (r₁ – r₂­) a, Є R a

Ra is a group under addition.

Now, let r Є R, and u Є R.

Then ru =  r (r₁a) (rr₁) a Є Ra

and ru = ur Є R a (since R is commutative)

Hence (Ra, +, .) is an ideal in R.

Since e Є R ⇒ ea ⇒ Ra

⇒ a Є Ra

Ra ≠ {0} (since a ≠ 0)

By the hypothesis of the theorem

Ra = R

This means that every element of R is a multiple of 'a' by some element of R.

(ie.,) ∀ x Є R, x = ra, for some r Є R

For I Є R

1 = ba, for some 0 ≠ b Є R

ab = 1

Hence the theorem.

Example 1: Show that (Z, +, ×) is an integral domain where Z is the set of all integers.    [A.U N/D 2010]

Solution :

(Z, +) is an abelian group.

(Z, ×) is monoid.

The operation × is distributive over +

(Z, ×) is commutative.

(Z, +, ×): without zero divisors.

 (Z, +, ×) is an integral domain where Z is the set of all integers.

Example 2: Prove that the set Z₄ = {[0], [1], [2], [3]} is a commutative ring with respect to the binary operation addition modulo and multiplication modulo +₄, ×₄   [AU N/D 2012]

Solution :

The composition tables for additive modulo 4 multiplicative modulo 4 are shown below.

Composition table for additive modulo 4.

From tables we get

(i) All the entries in both the tables belong to Z₄.

Z₄ is closed under the operations +₄ and ×₄.

 (ii) In both the tables,

Entries in the first row = Entries in the first column.

Entries in the second row = Entries in the second column.

Entries in the third row = Entries in the third column.

Entries in the fourth row = Entries in the fourth column,

The operations +₄ and ×₄ are commutative in Z₄.

(iii) Also, for any a, b, c Є Z₄, we have

a +₄ (b +₄ c) = (a +₄ b) +₄ c

and a ×₄ (b ×₄ c) = (a ×₄ b) ×₄ c

Since, 0 +₄ (1 +₄ 2) = 0 +₄ 3 = 3

and (0 +₄ 1) +₄ 2 = 1 +₄ 2 = 3

0 +₄ (1 +₄ 2) = (0 +₄ 1) +₄ 2

Also, 1 ×₄ (2 ×₄ 3) = (1 ×₄ 2) = 2

and (1 ×₄ 2) ×₄ 3 = 2 ×₄ 3 = 2

1 ×₄ (2 ×₄ 3) = (1 ×₄ 2) ×₄ 3

Thus, the operations +₄ and ×₄ are associative in Z₄

(iv) 0 is the additive identity of Z₄ and 1 is the multiplicative identity

of Z₄.

(v) Additive inverse of 0, 1, 2, 3 are respectively 0, 3, 2, 1.

Multiplicative inverses of the non-zero elements 1, 2 and 3 are 1, 2 and 3, respectively.

(vi) If a, b, c Є Z₄, then

a ×₄ (b +₄ c) = (a ×₄ b) +₄ (a ×₄ c).

and (a +₄ b) ×₄ c = (a ×₄ c) +₄ (b ×₄ c)

Thus, the operation ×₄ is distributed over +₄ in Z₄.

Hence, (Z₄, +₄, ×₄) is a commutative ring with unity.

Example 3: Give an example of a ring which is not a field. [A.U N/D 2013]

Solution :

The ring Z of all integers is an integral domain but not a field.