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

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    [SOLVED] the cardinal number of a finite local ring

    If $\displaystyle R$ is a finite local commutative ring having residue field(=$\displaystyle R/M$ where $\displaystyle M$ is the unique maximal ideal of $\displaystyle R$) consisting of $\displaystyle p^n$ elements ($\displaystyle p=$ prime integer) then prove that the cardinal number of $\displaystyle R$ is $\displaystyle p^{nm}$ i.e. $\displaystyle |R|=p^{nm}$ where $\displaystyle m$ denotes the length of the (Artinian) ring $\displaystyle R$.
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    Quote Originally Posted by xixi View Post
    If $\displaystyle R$ is a finite local commutative ring having residue field(=$\displaystyle R/M$ where $\displaystyle M$ is the unique maximal ideal of $\displaystyle R$) consisting of $\displaystyle p^n$ elements ($\displaystyle p=$ prime integer) then prove that the cardinal number of $\displaystyle R$ is $\displaystyle p^{nm}$ i.e. $\displaystyle |R|=p^{nm}$ where $\displaystyle m$ denotes the length of the (Artinian) ring $\displaystyle R$.
    every simple module over a commutative ring $\displaystyle R$ is isomorphic with $\displaystyle R/I,$ where $\displaystyle I$ is some maximal ideal of $\displaystyle R.$ now your ring has only one maximal ideal $\displaystyle M$ and so if $\displaystyle N$ is a simple $\displaystyle R$ module, then

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

    $\displaystyle M_i/M_{i-1}$ is a simple $\displaystyle R$ module. therefore $\displaystyle |R|=|M_m|=\prod_{i=1}^m |M_i|/|M_{i-1}|=\prod_{i=1}^m |M_i/M_{i-1}|=\prod_{i=1}^m p^n=p^{mn}.$
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    Quote Originally Posted by NonCommAlg View Post
    every simple module over a commutative ring $\displaystyle R$ is isomorphic with $\displaystyle R/I,$ where $\displaystyle I$ is some maximal ideal of $\displaystyle R.$ now your ring has only one maximal ideal $\displaystyle M$ and so if $\displaystyle N$ is a simple $\displaystyle R$ module, then

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

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