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

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

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)


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