A field is an integral domain, but the converse is not true. In this post, we will prove this fact. Before we prove, let us recall what are fields and integral domains.
| Field: A field is a non-trivial commutative ring with unity where each non-zero element is a unit. For example, (ℝ, +, ⋅) is a field. Integral Domain: An integral domain is a non-trivial commutative ring with unity containing no zero divisors. For example, (ℝ, +, ⋅) is an integral domain. |
Table of Contents
Prove that a Field is an Integral Domain
Proof:
Let F be a field. So F is a nontrivial commutative ring with unity. We will show that F is an integral domain, that is, it does not contain zero divisors.
Let a ∈ F be non-zero. Then as F is a field, its multiplicative inverse a-1 exists in F and we have that
a⋅a-1 = a-1⋅a = 1.
We claim that a is not a zero divisor. If possible assume that a is so. Thus there exists b ∈ F such that
a⋅b = 0
⇒ a-1⋅(a⋅b) = a-1⋅0
⇒ (a-1⋅a)⋅b = 0 as the multiplication is associative.
⇒ 1⋅b = 0
⇒ b = 0
Hence, a is not a left divisor of zero. Similarly, we can show that a is not a right divisor of zero. In other words, a is not a zero divisor.
So F contains no zero divisors.
As F is a nontrivial commutative ring with unity having no zero divisors, we conclude that F is an integral domain. This proves that every field is an integral domain.
But the Converse is not True
An integral domain need not be a field.
For example, the ring (ℤ, +, ⋅) is an integral domain, but it is not a field.
Related Topics: Introduction to Ring Theory
Idempotent and Nilpotent Elements
Prove that Every Finite Integral Domain is a Field
FAQs
Answer: True. A field is always an integral domain.
