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.

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## 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, a_{1}, a_{2}, a_{3}, …, 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}⋅a_{1}, a_{r}⋅a_{2}, a_{r}⋅a_{3}, …, 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 a_{r}⋅a_{i} = a_{r}⋅a_{j} for some 1 ≤ i, j ≤ n-1 with i ≠ j.⇒ a _{r}⋅(a_{i} – a_{j}) = 0⇒ a _{i} – a_{j} = 0 as R has no zero divisors.⇒ a _{i} = a_{j}. |

This proves our claim. As a_{r}⋅a_{1}, a_{r}⋅a_{2}, a_{r}⋅a_{3}, …, 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

Idempotent and Nilpotent Elements

## FAQs

**Q1: Every finite integral domain is a field but converse is not true.**

Answer: Yes, every finite integral domain is a field, but the converse is not true. For example, (ℝ, +, ⋅) is a field, but not a finite integral domain.