indecomposable module

**Prometheus** · January 24th 2009, 03:46 AM

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.

**NonCommAlg** · January 24th 2009, 03:50 PM

Originally Posted by Prometheus

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 $\mathfrak{b}=<t-1>. \ \Box$

**Prometheus** · January 25th 2009, 06:18 AM

"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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