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I'm interested if I solved this somewhat correctly, and would like to be set straight if it is wrong. This is an exercise from an introductory text.

(Unless stated otherwise, all theorems and exercises are from Hungerford's Algebra.)

Lemma IV.6.8: Let $A$ be a module over a principal ideal domain $R$ such that $p^nA = 0$ and $p^{n-1}A \neq 0$ for some prime $p \in R$. Let $a$ be an element of $A$ of order $p^n$. There is a submodule $C$ of $A$ such that $A = Ra \oplus C$.

Here order of $a$ is defined as the generator of the annihilator of cyclic submodule $Ra$. The proof is given in elementary form, and additonally as an exercise involving injective modules and Baer's criterion:

Lemma IV.3.8: Let $R$ be a unital ring. $R$-module $A$ is injective if and only if for every left ideal $L$ of $R$ and homomorphism $f$ of $L$ into $A$, $f$ extends to a homomorphism of $R$ into $A$.

One of the exercises made the criterion a bit easier to handle. Here is an equivalent statement:

Exercise IV.3.2: Let $R$ be a unital ring. $R$-module $A$ is injective if and only if for every left ideal $L$ of $R$ and homomorphism $f$ of $L$ into $A$, there exists $a \in A$ such that $f(r) = ra$ for every $r \in L$.

And here is the sequence of exercises.

Exercise IV.6.7: Let $A$ and $a \in A$ satisfy the hypotheses of Lemma IV.6.8.

(i) Every $R$-submodule of $R/(p^n)$-module with action $(r+(p^n))a = ra$. Conversely, every $R/(p^n)$-submodule of $A$ is an $R$-submodule by pullback along $R \rightarrow R/(p^n)$.

(ii) The submodule $Ra$ is isomorphic to $R/(p^n)$.

(iii) The only proper ideals of the ring $R/(p^n)$ are the ideals generated by $p^i + (p^n), (i= 1,2, \ldots , n-1)$

(iv) $R/(p^n)$ is injective as $R/(p^n)$-module.

(v) There exists $C$ such that $A = C \oplus Ra$.

(ii) $Ra \cong R/(p^n)$ by map $R \rightarrow Ra$ and Isomorphism theorems.

(iii) If $q \neq p$ is any prime in $R$ then $gcd(p,q) = gcd(p^n, q) = 1$ hence $xp^n + yq = 1$ so $q$ is invertible in $R/(p^n)$, hence the only nonunits in $R/(p^n)$ are $p^i (i=2,\ldots, n-1)$ and only these can generate proper (necessarily principal) ideals.

(iv) $L$ is any of the possible ideals of $R/(p^n)$ and $f:L \rightarrow R/(p^n)$ an $R/(p^n)$-module homomorphism. Image of $f$ is an $R/(p^n)$-submodule, so one of the ideals. Then since $L = \{cp^i + (p^n) \vert c \in R\}$, $f[cp^i + (p^n)] = (c+p^i)f[p^i+(p^n)]$, action of $f$ is determinated by where it sends $p^i + (p^n)$.

Let $f[p^i + (p^n)] = rp^j + (p^n)$ for some $r \in R$. If $j \geq i$ then $f[cp^i + (p^n)] = (c+(p^n))f[p^i+(p^n)] = crp^j + (p^n) = (cp^i + (p^n))(rp^{j-i} + (p^n))$. Hence the image of $L$ would be of the required form, and $R/(p^n)$ would be injective.

If $i > j$, then $0 + (p^n) = f[p^n + (p^n)] = (p^{n-i}+(p^n))f[p^i + (p^n)] = p^{n-i+j} + (p^n) \neq 0 + (p^n)$, contradiction.

(v) We have an exact sequence of $R/(p^n)$-modules $0 \rightarrow Ra \rightarrow A \rightarrow A/Ra \rightarrow 0$, where $Ra \cong R/(p^n)$ hence $Ra$ is injective and therefore the sequence splits, hence $A = Ra \oplus A/Ra$. These are $R$-modules again by pullback along canonical projection $R \rightarrow R/(p^n)$. $\blacksquare$

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