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Let $R$ be a principal ideal domain but not a field, and let $M$ be an $R$-module. Show the following:

(i) Let $p \in R$ be an irreducible element and $r \in R \setminus \{0\}$. Then $(R/ \langle r \rangle)[p] \cong R/ \langle p^n \rangle$, where $n=\max\{k \in \mathbb N_0 : p^k\mid r\}$.

(ii) $M$ is simple iff $\exists \space p \in R$ irreducible such that $M \cong R/ \langle p \rangle$.

I am pretty stuck on both items.

In (i) I've tried to prove it by induction on $\mathbb N_0$, but I could only prove it for the base case $n=0$: If $n=0$, then $R/ \langle p^n \rangle=R/ \langle 1 \rangle=0$. Now, $$(R/ \langle r \rangle)[p]=\{\overline{a} \in R/ \langle r \rangle : p^m\overline{a}=0 \space \text{for some } m \in \mathbb N\}=\{a \in R : p^ma \in \langle r \rangle \space \text{for some } m \in \mathbb N \}$$

If I call this set $S$ (which is also a submodule), then I would like to conclude $S=\langle r \rangle$. The inclusion $\langle r \rangle \subset S$ is immediate. Now take $s \in S$, then $p^ms=rq$. Using the fact that $R$ is a UFD, one can deduce that $p^m \sim q$, then $p^ms=rup^m$ for some $u \in \mathcal U(R)$, from here it follows $s=ru \in \langle r \rangle$.

I couldn't prove the induction step, maybe induction is not the best way attack this problem.

As for (ii), I could show that $M \cong R/ \langle p \rangle \implies M$ is simple: since $p$ is irreducible, it is also prime, as we are in a PID, this implies $\langle p \rangle$ is maximal, so $R/ \langle p \rangle$ is simple, it immediately follows $M$ is simple.

I would appreciate suggestions to prove (i) and the other implication in (ii). Thanks in advance.

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(i) Write $r=p^ns$ with $\gcd(p,s)=1$. From $p^ma\in(r)$ we get $p^ma=rb$, that is, $p^ma=p^nsb$. Now consider two cases: (a) $m\leq n$, and then $a=p^{n-m}sb\in(s)$, or (b) $m>n$, and then $p^{m-n}a=sb$. Since $s$ and $p$ are coprime we get $b=p^{m-n}c$ and therefore $a=sc\in(s)$.
Conclusion: $(R/(r))[p]$ is $(s)/(r)$, the ideal of $R/(r)$ generated by $s$. From $(s)/(r)=(s)/(p^ns)$ we get $(R/(r))[p]\simeq R/(p^n)$.

(ii) $M$ simple, then $M$ is cyclic, so $M\simeq R/I$. But $R/I$ should not have proper $R$-submodules, that is, proper ideals, so $I$ is maximal. Since $R$ is a PID, $I$ is principal and being maximal it is generated by an irreducible element.

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  • $\begingroup$ Oh, how silly, what I meant was: why is it that $(s)/(p^ns) \cong R/(p^n)$ (sorry about the confusion)? $\endgroup$ – user16924 Nov 13 '14 at 11:37
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    $\begingroup$ Think of this like simplifying the fractions (by $s$). Formally: $R\to(s)$ by $a\mapsto as$, and compose this with the canonical projection $(s)\to (s)/(p^ns)$. Both maps are surjective and the kernel is $\{a\in R:as\in(p^ns)\}=\{a\in R:a\in(p^n)\}=(p^n)$. $\endgroup$ – user26857 Nov 13 '14 at 11:43

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