# [SOLVED] the cardinal number of a finite local ring

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• May 29th 2010, 09:33 AM
xixi
[SOLVED] the cardinal number of a finite local ring
If $R$ is a finite local commutative ring having residue field(= $R/M$ where $M$ is the unique maximal ideal of $R$) consisting of $p^n$ elements ( $p=$ prime integer) then prove that the cardinal number of $R$ is $p^{nm}$ i.e. $|R|=p^{nm}$ where $m$ denotes the length of the (Artinian) ring $R$.
• May 29th 2010, 03:24 PM
NonCommAlg
Quote:

Originally Posted by xixi
If $R$ is a finite local commutative ring having residue field(= $R/M$ where $M$ is the unique maximal ideal of $R$) consisting of $p^n$ elements ( $p=$ prime integer) then prove that the cardinal number of $R$ is $p^{nm}$ i.e. $|R|=p^{nm}$ where $m$ denotes the length of the (Artinian) ring $R$.

every simple module over a commutative ring $R$ is isomorphic with $R/I,$ where $I$ is some maximal ideal of $R.$ now your ring has only one maximal ideal $M$ and so if $N$ is a simple $R$ module, then

$N \cong R/M.$ In particular $|N|=|R/M|=p^n.$ now let $(0)=M_0 < M_1 < \cdots < M_{m-1}=M < M_m=R$ be a composition series for $R.$ then $|M_i/M_{i-1}|=p^n,$ for all $1 \leq i \leq m,$ because every

$M_i/M_{i-1}$ is a simple $R$ module. therefore $|R|=|M_m|=\prod_{i=1}^m |M_i|/|M_{i-1}|=\prod_{i=1}^m |M_i/M_{i-1}|=\prod_{i=1}^m p^n=p^{mn}.$
• May 30th 2010, 04:55 AM
xixi
Thanks
Quote:

Originally Posted by NonCommAlg
every simple module over a commutative ring $R$ is isomorphic with $R/I,$ where $I$ is some maximal ideal of $R.$ now your ring has only one maximal ideal $M$ and so if $N$ is a simple $R$ module, then

$N \cong R/M.$ In particular $|N|=|R/M|=p^n.$ now let $(0)=M_0 < M_1 < \cdots < M_{m-1}=M < M_m=R$ be a composition series for $R.$ then $|M_i/M_{i-1}|=p^n,$ for all $1 \leq i \leq m,$ because every

$M_i/M_{i-1}$ is a simple $R$ module. therefore $|R|=|M_m|=\prod_{i=1}^m |M_i|/|M_{i-1}|=\prod_{i=1}^m |M_i/M_{i-1}|=\prod_{i=1}^m p^n=p^{mn}.$

Thank YOU very much !