Determining the kernel of a module homomorphism

Question

Let $p$ be a prime and let $n$ be a positive integer such that $p^n > 2$. Set $R:= \mathbb{Z}_{p^n}$, that is, the residue ring with binary operations of addition and multiplication modulo $p^n$. Let $\alpha_1,\alpha_2,\alpha_3 \in \mathbb{N}_0$ be such that $\alpha_1 > \alpha_2$ and assume $n$ is sufficiently large that $\alpha_1 + \alpha_2 + \alpha_3 < n$. Let $A_1, A_2, A_3 \in \mathbb{Z}$ be arbitrary integers coprime to $p$.

Define a map $\phi: R^3 \rightarrow R^3$ by

\begin{align*} \phi(x_1,x_2,x_3) = (p^{\alpha_1}A_1x_1 + p^{\alpha_2}A_2x_2+x_3,p^{\alpha_3}A_3x_2+x_3,x_3) \end{align*}

Note that this map can be represented by the $3x3$ linear transformation matrix \begin{align*} \left(\begin{array}{ccc} p^{\alpha_1}A_1 & p^{\alpha_2}A_2 & 1\\ 0 & p^{\alpha_3}A_3 & 1\\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{c} x_1\\x_2\\x_3\end{array}\right) \end{align*}

I would like to determine the kernel of $\phi$. By definition, we have that \begin{align*} \ker(\phi) &= \{(x_1,x_2,x_3) \in R^3 | \phi(x_1,x_2,x_3) = (0,0,0)\} \end{align*} This means that $x_1, x_2$ and $x_3$ must satisfy the system of congruences given by \begin{align*} p^{\alpha_1}A_1x_1 + p^{\alpha_2}A_2x_2 + x_3 &\equiv 0 (\mod{p^n})\\ p^{\alpha_3}A_3x_2 + x_3 &\equiv 0 (\mod {p^n})\\ x_3 &\equiv 0 (\mod {p^n}). \end{align*}

Thus, if $(x_1,x_2,x_3)$ is in the kernel, we must have $x_3 =0$ and $\displaystyle x_2 \equiv 0 (\mod{p^{n-\alpha_3}})$. Hence, every such element $x_2$ will be given by $x_2:= p^{n-\alpha_3}x_2'$, where $x_2' = 0, \ldots, p^{\alpha_2}-1$.

Our kernel can now be written as \begin{align*} &\{(x_1,p^{n-\alpha_3}x_2',0) | x_1 \in R; x_2' = 0, \ldots, p^{\alpha_2}-1; p^{\alpha_1}A_1x_1 + p^{n+\alpha_2-\alpha_3}A_2x_2' \equiv 0 (\mod{p^n})\}. \end{align*} We have that \begin{align*} p^{\alpha_1}&\left(A_1x_1 + p^{n+\alpha_2-\alpha_3-\alpha_1}A_2x_2'\right)\equiv 0 (\mod {p^n})\\ &\rightarrow A_1x_1 + p^{n+\alpha_2-\alpha_3-\alpha_1}A_2x_2' \equiv 0 (\mod {p^{n-\alpha_1}})\\ &\rightarrow x_1 \equiv -A_1^{-1}A_2p^{n+\alpha_2-\alpha_3-\alpha_1}x_2' (\mod {p^{n-\alpha_1}}). \end{align*}

Thus, our the kernel of $\phi$ is given by the set \begin{align*} &\{(x_1,p^{n-\alpha_3}x_2',0) | x_1 \in R; x_2' = 0, \ldots, p^{\alpha_2}-1; x_1 \equiv -A_1^{-1}A_2p^{n+\alpha_2-\alpha_3-\alpha_1}x_2' (\mod {p^{n-\alpha_1}})\}. \end{align*}

I want to express the kernel as a direct product of $3$ sub-modules (i.e. ideals) of $R$. I know that the kernel of the second component will be given by the ideal $<p^{n-\alpha_3}>$ and the last component is $<0>$. Hence, our kernel resembles something like \begin{align*} \ker(\phi) = I \times <p^{n-\alpha_3}> \times <0>. \end{align*}

The argument is that since multiplication is distributed component-wise (i.e. scalar), and addition is component-wise, the collection of all the first-components in the kernel of $\phi$ form an ideal. (Or that submodules of rings are ideals.) It also may happen that this should be properly expressed as an isomorphism; $\ker(\phi)\simeq I \times <p^{n-\alpha_3}> \times <0>$.

Since $R$ is a principal ideal ring, I know that the ideal $I$ will be given by $<p^{\beta}>$ for some $\beta$. If $\alpha_3 = 0$ then the ideal is clearly given by $<p^{n-\alpha_1}>$. However, for $\alpha_3 > 0$ it is not obvious to me about how to deduce the generator for $I$ given what's been stated above.

For example, let $\alpha_2=0$ and $\alpha_3=1$. Then the kernel of $\phi$ is given by \begin{align*} &\{(x_1,p^{n-\alpha_3}x_2',0) | x_1 \in R; x_2' = 0, \ldots, p-1; x_1 \equiv -A_1^{-1}A_2p^{n-1-\alpha_1}x_2' (\mod {p^{n-\alpha_1}})\}. \end{align*} Since $x_1$ can only represent $p$ total elements, I suspect that the elements of the first component of the kernel will be given by $<p^{n-1}>$, or are isomorphic to such a thing. But that means $\alpha_1$ doesn't really have an effect?

I may have things wrong, but any clarification, hints, or a counterexample would be greatly appreciated!