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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.

user26857
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    An elementary approach: if $U(R)$ is not finite then pick $x\in m$, $x\ne0$ and consider the elements $ux$, $u\in U(R)$. Then $ux\in m$ and only finitely many of them are distinct, so $u_1x=u_2x=\cdots$. From $(u_1-u_2)x=0$ we conclude $u_1-u_2\in m$ and then $u_1-u_n$ are infinitely many distinct elements of $m$, a contradiction. – user26857 Jul 28 '17 at 16:56
  • @user26857 : why $u_1-u_2 \in m$ ? –  Jul 28 '17 at 17:04
  • Otherwise it's in $U(R)$ and then $x=0$. – user26857 Jul 28 '17 at 17:05
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    @user26857 : ah I see . I have also written an answer –  Jul 28 '17 at 17:31

2 Answers2

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

egreg
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  • what are you suggesting precisely ? –  Jul 28 '17 at 16:19
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    @misao In general, if $S$ is a simple $R$-module, $x\in S$ and $x\ne0$, then $S\cong R/J$, where $J$ is the annihilator of $x$. – egreg Jul 28 '17 at 16:20
  • ah right right , and here the simple $R$-mod is the minimal prime ideal ... –  Jul 28 '17 at 16:24
  • @misao And the annihilator is… ;-) – egreg Jul 28 '17 at 16:29
  • the annihilator $J$ is proper( as $x \ne 0$) , so finite , and so is $R/J$ being isomorphic to a proper ideal ... so we are done –  Jul 28 '17 at 16:34
  • @misao More precisely, the annihilator is $m$, the unique maximal right ideal. So $R/m\cong I$ is finite. – egreg Jul 28 '17 at 16:38
  • How do you know that $R$ has a minimal nonzero ideal? (If $R$ is a noetherian domain then $I^2=I$ and from Nakayama $I=(0)$.) – user26857 Jul 28 '17 at 17:06
  • @user26857 $R$ is obviously Artinian, because $m$ is Artinian (being finite) and $R/m$ is simple. – egreg Jul 28 '17 at 17:14
  • Thanks. (It think this is the way I'd have started such an answer.) – user26857 Jul 28 '17 at 17:18
  • @user26857 Well, my answer was mainly a hint to get the OP started, with details to be filled in. – egreg Jul 28 '17 at 17:19
  • @egreg : I have also written an answer . Could you please verify ? –  Jul 28 '17 at 17:31
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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