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 +.
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:
- Closure property under addition: For all a, b ∈ R, we have a + b ∈ R.
- Associative property under addition: For all a, b, c ∈ R, we have ( a + b ) + c = a + ( b + c )
- Existence of additive identity: For all a ∈ R, there exists 0 ∈ R such that a+ 0 = a = 0 + a
- Existence of additive inverse: For each a ∈ R, there exists a ∈ R such that a + (-a) = 0 = (-a) + a
- Commutative property: For all a, b ∈ R, we have a + b = b + a
- Closure property under multiplication: For all a, b ∈ R, we have ab ∈ R
- Associative property under multiplication: For all a, b, c ∈ R, we have a(bc) = (ab)c
- 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.
A ring R is called a field if it is
- Has unit element,
- 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.
A ring R is called an integral domain if it is
- Has unit element
- 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.
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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