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.


(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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.