# A question about proof (-Module over a PID)

I can understand the final part of the proof of the Lemma 6.8, page 248, Algebra, Hungerford.

Lemma6.8
Let $$A$$ be a module over a principal ideal domain $$R$$ such that $$\ p^{n}A=0, p^{n-1}A \neq 0$$ for some prime $$p\in R$$ and positive integer $$n$$. Let $$a$$ be an element of $$A$$ of order $$\ p^{n}$$

(ii) There is a submodule $$C$$ of $$A$$ such that $$A = Ra \oplus C$$.

in case of $$A\neq Ra$$, Roughly speaking , the author constructs the maximal submodule $$C$$ s.t $$Ra \cap C = 0.......(1)$$ (by using Zorn's Lemma where $$C$$ is a maximal submodule which satisfies (1)). And then, the author ascertains that the quotient group $$A/C$$ is cyclic, which is generated $$a+C$$, i.e $$A/C=R(a+C)$$. And, finally

... $$A/ C$$ is the cyclic $$R$$-module generated by $$a + C$$ (that is, $$A/C = R(a + > C)$$). Consequently, A = Ra + C, whence A=Ra $$\oplus$$ C
(Q.E.D)

But I cannot understand that $$A/C=R(a+C)$$ implies $$A = Ra+ C$$.

If $$A/C=R(a+C)$$, then as sets of cosets we have that $$\{a'+C:a'\in A\}=\{ra+C:r\in R,c\in C\}$$ since $$r(a+C)=ra+C$$ for $$r\in R$$ by the definition of the action on the quotient module.

Again as a set, for any $$a'\in A$$ we have that $$a'+C=\{a'+c:c\in C\}$$

Since $$0\in C$$, for any $$a'\in A$$ we have $$a'=a'+0\in a'+C=ra+C$$ for some $$r\in R$$. Then $$a'=ra+c\in Ra+C$$ for some $$c\in C$$.

This shows that $$A\subseteq Ra+C$$. Clearly $$Ra+C\subseteq A$$, so $$A=Ra+C$$. By the definition of $$C$$ we have that $$Ra\cap C=0$$, so the sum is direct and $$A=Ra\oplus C$$.

(Edited for clarity)