Prove that Every Field is an Integral Domain

In this article, we will discuss and prove that every field in the algebraic structure is an integral domain. A field is a non-trivial ring R with a unit. If the non-trivial unitary ring is commutative and each non-zero element of R is a unit, so the non-empty set F forms a field with respect to two binary operations. and +. 

Ring: 

Let R be a non-empty set with two binary operations, addition, and multiplication, then the algebraic structure ( R, +, ∗ ) is called a ring if it satisfies the following conditions:  

1. Closure property under addition: For all a, b ∈ R, we have a + b ∈ R.
2. Associative property under addition: For all a, b, c ∈ R, we have ( a + b ) + c = a + ( b + c )
3. Existence of additive identity: For all a ∈ R, there exists 0 ∈ R such that a+ 0 = a = 0 + a
4. Existence of additive inverse: For each a ∈ R, there exists a ∈ R such that a + (-a) = 0 = (-a) + a
5. Commutative property: For all a, b ∈ R, we have a + b = b + a
6. Closure property under multiplication: For all a, b ∈ R, we have ab ∈ R
7. Associative property under multiplication: For all a, b, c ∈ R, we have a(bc) = (ab)c
8. Distributive property: For all a, b, c ∈ R, we have a ( b + c ) = a . b + a . c  

Commutative ring: A ring R for which a . b = b . a for all a, b ∈ R is called a commutative ring. 

Field: 

A ring R is called a field if it is 

1. Commutative
2. Has unit element,
3. And each non-zero elements possess a multiplicative inverse.

Example of Field: The set R of all real numbers is a field as R is a commutative ring with unity and each non-zero element has a multiplicative inverse. 

Integral domain:

A ring R is called an integral domain if it is

1. Commutative
2. Has unit element
3. And has no zero divisors. 

Example: The set Z of all integers is an integral domain as Z is a commutative ring with unity and also does not possess zero divisors.

Proof:

Let F be any field. We know that field F is a commutative ring with unity. So, in order to prove that every field is an integral domain, we have to show that F has no zero divisors. 

Let a & b be elements of F with a ≠ 0 such that ab = 0.

Now, a ≠ 0 implies that a^-1 exists.

For ab = 0,
multiply a-1 to both sides,
(ab)a-1  = (0)a-1
(a.a-1)b = 0
(1)b = 0
⇒ b = 0

Therefore, a ≠ 0, ab = 0 implies that b = 0

Similarly, let ab = 0 and b ≠ 0

Now, b≠0 implies that b^-1 exists.

For ab = 0,
multiply b-1 to both sides,
(ab)b-1  = (0)b-1
(b.b-1)a = 0
(1)a = 0
⇒ a = 0

Therefore, b ≠ 0, ab = 0 implies that a = 0

In field F, 

ab = 0 ⇒ a = 0 or b = 0

Therefore, F has no zero divisors.

Hence proved, Field is an integral domain.

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