# indecomposable module

• Jan 24th 2009, 04:46 AM
Prometheus
indecomposable module
define $R=\mathbb{Q}G$ where $G=$ 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.
• Jan 24th 2009, 04:50 PM
NonCommAlg
Quote:

Originally Posted by Prometheus

define $R=\mathbb{Q}G$ where $G=$ 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 $\mathfrak{a}=.$ i'll show that $\mathfrak{a}$ is a minimal ideal of $R,$ which solves the problem. it's clear that $R \simeq \frac{\mathbb{Q}[t]}{}.$

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

since $\mathbb{Q}[t]$ is a PID, we have $\mathfrak{b}=,$ for some $f(t) \in \mathbb{Q}[t].$ since $\mathfrak{b} \subseteq ,$ 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 $\mathfrak{b}=$ or $\mathfrak{b}=. \ \Box$
• Jan 25th 2009, 07:18 AM
Prometheus
"it's clear that http://www.mathhelpforum.com/math-he...e0a556a3-1.gif ..."
actually,as clear as it is, I didn't notice it. thanks, that was exactly what I missed