alternative proof that a finite integral domain is a field

Proof.

Let $R$ be a finite integral domain and $a\in R$ with $a\ne 0$. Since $R$ is finite, there exist positive integers $j$ and $k$ with such that $a^j=a^k$. Thus, $a^k-a^j=0$. Since and $j$ and $k$ are positive integers, $k-j$ is a positive integer. Therefore, $a^j(a^{k-j}-1)=0$. Since $a\ne 0$ and $R$ is an integral domain, $a^j\ne 0$. 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
Canonical name	AlternativeProofThatAFiniteIntegralDomainIsAField
Date of creation	2013-03-22 16:21:54
Last modified on	2013-03-22 16:21:54
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	6
Author	Wkbj79 (1863)
Entry type	Proof
Classification	msc 13G05