Theorem 19.11

Statement

Every finite integral domain is a field.

Proof

Let

(1)
\begin{align} 0,1,a_1,\dotsb,a_n \end{align}

be all the elements of a finite domain $D$. We need to show that for $a\in D$, where $a\neq 0$, there exists $b\in D$ such that $ab=1$. Now consider

(2)
\begin{align} a1,aa_1,\dotsb,aa_n. \end{align}

We claim that all these elements of $D$ are distinct,for $aa_i=aa_j$ implies that $a_i=a_j$,by the cancellation laws that hold in an integral domain.Also,since $D$ has no 0 divisors,none of these elements is 0.Hence by counting,we find that $a1,aa_1,\dotsb,aa_n$ are elements $0,1,a_1,\dotsb,a_n$ in some order,so that either $a1=1$,that is ,$a=1$,or $aa_i=1$ for some $i$. Thus $a$ has a multiplicative inverse.