## UIC Master’s Exam- Fall 2007 A3

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

Proof: Fix $a\in D, a\neq 0$ and consider the function $\phi(x)=ax$ for all $x\in D$. We wish to show that $\phi$ is injective, and thus $a$ has an inverse.

Suppose that $\phi(x)=\phi(y)$, then $ax=ay$ which implies that $a(x-y)=0$. Since we are in an integral domain we know there are no zero divisors, so either $a=0$ or $x-y=0$. We know that $a\neq 0$ so it must be that $x-y=0$ which implies that $x=y$ and so $\phi$ is injective. Furthermore, since $\phi$ maps $D$ to $D$, and $D$ is finite, we know that $\phi$ is onto $D$. So, there is some $x\in D$ such that $\phi(x)=ax=e$, thus, $x=a^{-1}$.

Thus, $D$ is a field.

$\Box$

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 $D$ is finite, otherwise we wouldn’t be able to say that $\phi$ is onto even though it maps $D$ to $D$ and is one-to-one.

Let $D$ be a finite integral domain and let $d\in D$. If $d=1$ then $d=d^{-1}$, so suppose that $d\neq 1$. Then we know that all powers of $d$ must also be elements of $D$. Since $D$ is finite eventually these powers must repeat, so there exist integers $m\neq n$ such that $d^m=d^n$. Since $D$ is an integral domain we have cancellation and so $d^m=d^n$ implies that $d^{m-n}=1$ (wlog we may assume that $m>n$). Since $d\neq 1$ we know that $m-n>1$ and it follows that $d^{m-n-1}=d^{-1}$. Thus, any non-zero element of $D$ is a unit and so $D$ is a field.
$\Box$