Every Finite Integral Domain is a Field: Proof

If R is a finite integral domain, then it must be a field. In this article, we will prove that every finite integral domain is a field.

Every finite integral domain is field

Theorem: Prove that every finite integral domain is a field

Proof:

Let R be a finite integral domain with n elements 1, a₁, a₂, a₃, …, a_n-1, 1 being the unity in R. We will prove that R is a field.

That is, we will show that every non-zero element of R is a unit. Let a_r ∈ R be a non-zero element.

Let us consider the products a_r⋅1, a_r⋅a₁, a_r⋅a₂, a_r⋅a₃, …, a_r⋅a_n-1. All these elements belong to R. As R is an integral domain, it does not contain any zero divisors. This forces that none of these products will be zero.

We claim that all these elements are distinct. If possible suppose that ar⋅ai = ar⋅aj for some 1 ≤ i, j ≤ n-1 with i ≠ j. ⇒ ar⋅(ai – aj) = 0 ⇒ ai – aj = 0 as R has no zero divisors. ⇒ ai = aj.

This proves our claim. As a_r⋅a₁, a_r⋅a₂, a_r⋅a₃, …, a_r⋅a_n-1 are (n-1) distinct non-zero elements of R, one of them must be unity.

⇒ a_r⋅a_s = 1 for some 1 ≤ s ≤ n-1.

⇒ a_r⋅a_s = a_s⋅a_r = 1, since R is a commutative ring.

This shows that a_r is a unit.

Since a_r is an arbitrary non-zero element of R, and it is a unit, we deduce that each non-zero element of R is a unit. Therefore, R is a field. This completes the proof of the fact that each finite integral domain is a field.

Related Topics: Introduction to Ring Theory

Units of a Ring

Characteristic of a Ring

Zero divisors of a ring

Idempotent and Nilpotent Elements

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