## Compute a quotient module

Let $R$ be a principal ideal domain, let $a \in R$ be nonzero, and let $M = R/(a)$. Given a prime $p \in R$, say $a = p^nq$, where $p$ does not divide $q$. Prove that $(p^k)M/(p^{k+1})M$ is module-isomorphic to $R/(p)$ if $k < n$ and to $0$ if $k \geq n$.

We begin with some lemmas.

Lemma 1: Let $C \subseteq A,B$ be ideals of a ring $R$, and consider $A/C$ as an $R$-module. Then $B(A/C) = (BA)/C$. Proof: $(\subseteq)$ If $x \in B(A/C)$, then $x = \sum b_i(a_i + C) = \sum (b_ia_i + C)$ $= (\sum b_ia_i) + C$ $\in (BA)/C$. $(\supseteq)$ If $x \in (BA)/C$, then $x = (\sum b_ia_i) + C$ $= \sum (b_ia_i + C)$ $= \sum b_i(a_i+C)$. Thus $x \in B(A/C)$. $\square$

Lemma 2: Let $R$ be a principal ideal domain and let $a,b,c \in R$ be nonzero with $b|c$. Note that $(ac) \subseteq (ab)$. Prove that $(ab)/(ac) \cong_R (b)/(c)$. Proof: Let $\psi: (b) \rightarrow (ab)/(ac)$ be given by $bx \mapsto \overline{abx}$. (This is well-defined since $R$ is a domain, and is clearly an $R$-module homomorphism.) Certainly $\psi$ is surjective. Now if $bx \in \mathsf{ker}\ \psi$, then $abx \in (ac)$, so that $bx \in (c)$. Conversely, if $bx \in (c)$, then $bx = cy$ for some $y$, and so $\overline{abx} = \overline{acy} = 0$. By the First Isomorphism Theorem, $(b)/(c) \cong_R (ab)/(ac)$. $\square$

Lemma 3: Let $R$ be a principal ideal domain and $a,b,c \in R$ such that $b|c$. If $(a,c/b) = (1)$, then $(a)((b)/(c)) = (b)/(c)$. Proof: Say $c = bd$. Note that $ax+dy = 1$ for some $x,y \in R$, so that $abx + cy = b$. Now $b+(c) = abx + (c)$ $= a(bx+(c))$ $\in (a)((b)/(c))$, so that $(a)((b)/(c)) = (b)/(c)$. $\square$

If $k < n$, then $p^{k+1}|a$, so that $(a) \subseteq (p^k), (p^{k+1})$. By Lemma 1, we have $(p^k)(R/(a))/(p^{k+1})(R/(a)) = ((p^k)/(a))/((p^{k+1})/(a))$ $\cong_R (p^k)/(p^{k+1})$ by the Third Isomorphism Theorem. Using Lemma 2, this is isomorphic to $(1)/(p) \cong_R R/(p)$.

If $k \geq n$ with (say) $k = n+t$, then $((p^k)(R/(a)))/((p^{k+1})/(R/(a)) = ((p^t)((p^n)/(p^nq)))/((p^{t+1})((p^n)/(p^nq)))$ $= ((p^n)/(p^nq))/((p^n)/(p^nq))$ $= 0$, using Lemma 3 and the Third Isomorphism Theorem.