> [!NOTE] Theorem > A [[Finite Set|finite]] [[Integral Domain|integral domain]] is a [[Field (Algebra)|field]]. ###### Proof \[MA268\] Let $R$ be a finite integral domain and let $0\neq a\in R$. The sequence $a,a^2,a^3,\dots$ must have a repetition. Thus there are $n>m$ such that $a^m=a^n$. Thus $a^m(a^{n-m}-1)=0$. As $a\neq{0}$ and $R$ is an integral domain, $a^{m-n}=1$. But $m-n\geq{1}$, so $a$ has an inverse in $R$, namely $a^{m-n-1}$. ###### Proof Let $R$ be a finite integral domain whose [[Ring Unity|unity]] is $1$ and [[Ring Zero Element|zero]] is $0.$ Let $0\neq a\in R.$ Consider the [[Ideal of Ring|ideal]] $aR=\{ ar \mid r\in R \}.$ For any $r_{1},r_{2}\in R,$ $ar_{1}=ar_{2}\implies a(r_{1}-r_{2})=0\implies r_{1}=r_{2}.$ Thus the elements of $aR$ are distinct therefore $aR$ has the same number of elements as $R.$ Since $aR$ is a subset of $R,$ we have $aR=R.$ Since $1\in R=aR,$ there exists $b\in R$ such that $ab=1.$ # Applications This proof generalises the fact that [[Integers modulo prime is a field and integers modulo composite is not|the integers modulo prime form a field]].