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Math Help - indecomposable R-module

  1. #1
    Newbie dangkhoa's Avatar
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    indecomposable R-module

    i) Let p in R be a prime of a PID R. If A ad B are submodules of R/(p^n), n >=1 .
    Show that either A belongs to B or B belongs to A. Deduce that R/(p^n) is an indecomposable R-module.
    (ii) Let R be a PID and d is non_zero, non_unit. If R/(d) is indecomposable, show that d~p^n for some prime p in R.

    Can you give me some hints how to do this question please?

    Thank you so much
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  2. #2
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    Quote Originally Posted by dangkhoa View Post
    i) Let p in R be a prime of a PID R. If A ad B are submodules of R/(p^n), n >=1 .
    Show that either A belongs to B or B belongs to A. Deduce that R/(p^n) is an indecomposable R-module.
    an R-submodule of S=\frac{R}{<p^n>} is in the form \frac{I}{<p^n>}, where I is an ideal of R containing <p^n>. so A=\frac{<a>}{<p^n>}, \ B=\frac{<b>}{<p^n>}, for some a,b \in R and a \mid p^n, \ b \mid p^n. thus a=p^i u, \ b = p^j v,

    for some i,j \leq n, and units u,v. it's clear now that if i \leq j, then B \subseteq A and if i \geq j, then A \subseteq B. it's now obvious that S is indecomposable because if S=A \oplus B, for some submoules A,B,

    then we'll have A \cap B=(0). but, as we just showed either A \cap B = A or A \cap B=B. so we must have either A=(0) or B=(0).

    (ii) Let R be a PID and d is non_zero, non_unit. If R/(d) is indecomposable, show that d~p^n for some prime p in R.

    Can you give me some hints how to do this question please?

    Thank you so much
    suppose d has at least two distinct prime factors. let p be a prime factor of d. so we have d=p^n a, where \gcd(p,a)=1. therefore <p^n> + <a>=R. see that \frac{<p^n>}{<d>} \cap \frac{<a>}{<d>} = (0) and so

    \frac{R}{<d>}=\frac{<p^n>}{<d>} \oplus \frac{<a>}{<d>}, which contradicts indecomposablity of \frac{R}{<d>}. thus p is the only prime factor of d and hence d and p^n are associates.
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