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Math Help - [SOLVED] the cardinal number of a finite local ring

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    [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.
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    Quote Originally Posted by xixi View Post
    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}.
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    Quote Originally Posted by NonCommAlg View Post
    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 !
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