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alternative proof that a finite integral domain is a field
Proof. Let $R$ be a finite integral domain and $a \in R$ with $a \neq 0$ . Since $R$ is finite, there exist positive integers $j$ and $k$ with $j<k$ such that $a^j=a^k$ . Thus, $a^k-a^j=0$ . Since $j<k$ and $j$ and $k$ are positive integers, $k-j$ is a positive integer. Therefore, $a^j(a^{k-j}-1)=0$ . Since $a \neq 0$ and $R$ is an integral domain, $a^j \neq 0$ . Thus, $a^{k-j}-1=0$ . Hence, $a^{k-j}=1$ . Since $k-j$ is a positive integer, $k-j-1$ is a nonnegative integer. Thus, $a^{k-j-1} \in R$ . Note that $a \cdot a^{k-j-1}=a^{k-j}=1$ . Hence, $a$ has a multiplicative inverse in $R$ . It follows that $R$ is a field. 
alternative proof that a finite integral domain is a field is owned by Warren Buck.
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