indecomposable R-module

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

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

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

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

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