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

1. ## [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 $R$.

2. 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 $|R|=|M_m|=\prod_{i=1}^m |M_i|/|M_{i-1}|=\prod_{i=1}^m |M_i/M_{i-1}|=\prod_{i=1}^m p^n=p^{mn}.$

3. ## Thanks

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 !