Exercise on modules over PID involving injective modules, Baer's criterion.

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.

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:

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:

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.

Let $A$ and $a \in A$ satisfy the hypotheses.

(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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Thanks for reading it. I sort of thought it was too long for anybody to care enough to get through it. Indeed author leads the reader into a solution, but I need some confirmation I somewhat understood the material correctly, as similar concepts occur frequently later in the book. – Malman Jul 11 '12 at 9:59
@navigetor Exercise is from Hungerford. It is self-study, I complement it with Robert Ash's text Basic Abstract Algebra. – Malman Jul 11 '12 at 16:16