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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. ∎

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## Mathematics Subject Classification

13G05*no label found*

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