**Problem Statement:** Show that a finite integral domain is a field.

**Proof: **Fix and consider the function for all . We wish to show that is injective, and thus has an inverse.

Suppose that , then which implies that . Since we are in an integral domain we know there are no zero divisors, so either or . We know that so it must be that which implies that and so is injective. Furthermore, since maps to , and is finite, we know that is onto . So, there is some such that , thus, .

Thus, is a field.

**Reflection:** This is a classic qual problem. It’s very straightforward and there are many ways to approach the proof. The key rests in the fact that is finite, otherwise we wouldn’t be able to say that is onto even though it maps to and is one-to-one.

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*Related*

As mentioned in the reflection above, there are many ways to approach this proof. Below I have written up the proof using another approach.

Proof:

Let be a finite integral domain and let . If then , so suppose that . Then we know that all powers of must also be elements of . Since is finite eventually these powers must repeat, so there exist integers such that . Since is an integral domain we have cancellation and so implies that (wlog we may assume that ). Since we know that and it follows that . Thus, any non-zero element of is a unit and so is a field.