Local ring with finite maximal ideal is finite

Question

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.

Answer 1

Let $I$ be a minimal right ideal; let $x\in I$, $x\ne 0$. What's the annihilator of $x$?

Answer 2

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