TITLE: Local ring with finite maximal ideal is finite
QUESTION [8 upvotes]: Let $(R,  m)$ be a commutative local ring which is not a field such that $m$ is finite. Then is it true that $R$ is finite ? 

I can see that $R$ has finitely many ideals and all proper ideals are finite; so in particular $R$ is Artinian. Moreover $m=R\setminus U(R)$ is finite where $U(R)$ denotes the group of units of $R$ . To show $R$ is finite it would be enough to show either $U(R)$ is finite or that $R/m$ is finite. But I am unable to conclude either. Is the claim at all true ? 
Please help. Thanks in advance.

REPLY [3 votes]: Let $R$ be not a field ; then $\exists 0 \ne x \in R\setminus U(R)$ . Then $Rx$ and $ ann(x)$ are both proper ideals of $R$ , hence both of them are finite . And obviously $Rx \cong R/ann(x)$  as $R$-modules ; hence $R/ann(x)$ is also finite . Thus $R$ is finite