# indecomposable R-module

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• Dec 10th 2009, 04:37 AM
dangkhoa
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
• Dec 10th 2009, 06:17 PM
NonCommAlg
Quote:

Originally Posted by dangkhoa
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 $\displaystyle S=\frac{R}{<p^n>}$ is in the form $\displaystyle \frac{I}{<p^n>},$ where $\displaystyle I$ is an ideal of $\displaystyle R$ containing $\displaystyle <p^n>.$ so $\displaystyle A=\frac{<a>}{<p^n>}, \ B=\frac{<b>}{<p^n>},$ for some $\displaystyle a,b \in R$ and $\displaystyle a \mid p^n, \ b \mid p^n.$ thus $\displaystyle a=p^i u, \ b = p^j v,$

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

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

Quote:

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

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