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Math Help - indecomposable module

  1. #1
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    indecomposable module

    define R=\mathbb{Q}G where G=<g> is a cyclic group of prime order p, then R is a module over itself (it's a ring). If x=\sum g^i then I need to show that the set \{y\in R \mid xy=0\} is an indecomposable module.
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  2. #2
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    Quote Originally Posted by Prometheus View Post

    define R=\mathbb{Q}G where G=<g> is a cyclic group of prime order p, then R is a module over itself (it's a ring). If x=\sum g^i then I need to show that the set \{y\in R \mid xy=0\} is an indecomposable module.
    let \mathfrak{a}=\{y \in R: \ xy=0 \}. note that \mathfrak{a} is the augmentation ideal of R and \mathfrak{a}=<g-1>. i'll show that \mathfrak{a} is a minimal ideal of R, which solves the problem. it's clear that R \simeq \frac{\mathbb{Q}[t]}{<t^p - 1>}.

    now since \mathfrak{a} \simeq \frac{<t-1>}{<t^p-1>}, we only need to prove that if \mathfrak{b} is an ideal of \mathbb{Q}[t] such that <t^p-1> \subseteq \mathfrak{b} \subseteq <t-1>, then either \mathfrak{b}=<t^p - 1> or \mathfrak{b}=<t-1>. this is very easy to see:

    since \mathbb{Q}[t] is a PID, we have \mathfrak{b}=<f(t)>, for some f(t) \in \mathbb{Q}[t]. since \mathfrak{b} \subseteq <t-1>, we have f(t)=(t-1)g(t), for some g(t) \in \mathbb{Q}[t]. but t^p - 1 \in \mathfrak{b}, which means t^p - 1 =f(t)h(t), for

    some h(t) \in \mathbb{Q}[t]. thus t^p - 1 = (t-1)g(t)h(t), and hence t^{p-1} + \cdots + t + 1 = g(t)h(t). but since t^{p-1} + \cdots + t + 1 is irreducible over \mathbb{Q}, we must have either g(t)=1 or h(t)=1. thus

    either \mathfrak{b}=<t^p - 1> or \mathfrak{b}=<t-1>. \ \Box
    Last edited by NonCommAlg; January 24th 2009 at 05:23 PM. Reason: typo!
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  3. #3
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    "it's clear that ..."
    actually,as clear as it is, I didn't notice it. thanks, that was exactly what I missed
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