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.

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