# 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 such that $a^{j}=a^{k}$. Thus, $a^{k}-a^{j}=0$. Since $j 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. ∎

Title alternative proof that a finite integral domain is a field AlternativeProofThatAFiniteIntegralDomainIsAField 2013-03-22 16:21:54 2013-03-22 16:21:54 Wkbj79 (1863) Wkbj79 (1863) 6 Wkbj79 (1863) Proof msc 13G05