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Thread: indecomposable module

  1. #1
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    indecomposable module

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

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

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

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

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

    either $\displaystyle \mathfrak{b}=<t^p - 1>$ or $\displaystyle \mathfrak{b}=<t-1>. \ \Box$
    Last edited by NonCommAlg; Jan 24th 2009 at 04:23 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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