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Let $p$ be a prime and $n$ be a positive integer such that $p^n > 2$. Let $R:= \mathbb{Z}/p^n \mathbb{Z}$ and suppose $(x,y) \in R \times R$, where addition and multiplication are defined component-wise, and are given by addition and multiplication modulo $p^n$.

Suppose we define a map on $R \times R$ by $\phi(x,y) \mapsto (px,y)$. Then this map will be an $R$-module homomorphism,. The kernel is given by $ker(\phi) = \{(0,0), (p^{n-1},0), (2p^{n-1}, 0), \ldots, ((p-1)p^{n-1},0)\}$

I want to now look at the isomorphism given by $(R \times R)/\ker(\phi) \simeq \phi(R \times R)$. If I can write $\ker(\phi)$ as a direct product $A \times B$, then by Exercise 11 of page 350 of Dummit and Foote, we will have that $(R \times R)/\ker(\phi) \simeq R/A \times R/B$.

I think that $A = \mathbb{Z}/p\mathbb{Z}$ and $B = \{1\}$ but I am unable to show this. Does anyone see a way to do this? Or if I've made a mistake, if you could point it out. (It's been a couple years since I've worked with modules so my memory is a little fuzzy.)

EDIT

Based on the answer from Alex G, I expect the following will hold.

Let $\alpha, \beta \in \mathbb{N}$ such that $\alpha, \beta < n$. Let $w_1$ and $w_2$ be arbitrary integers. Define two $R$-module endomorphisms $\phi_1$ and $\phi_2$ such that \begin{align*} \phi_1(x,y) \mapsto (p^{\alpha}x+w_1y, y) \text{ and } \phi_2(x,y) \mapsto (x,p^{\beta}y+w_2x). \end{align*}

Then $\ker(\phi_1) = <p^{n-\alpha}> \times \{0\}$ and $\ker(\phi_2) = \{0\} \times <p^{n-\beta}>$.

Define $\phi:= \phi_1 \circ \phi_2$ and I would expect that \begin{align*} \ker(\phi) \simeq <p^{n-\alpha}> \times <p^{n-\beta}> \end{align*}

and so by the $R$-module homomorphism theorem and the exercise by Dummit and Foote, we get \begin{align*} \phi(R \times R)/\ker(\phi) \simeq \mathbb{Z}/p^{n-\alpha}\mathbb{Z} \times \mathbb{Z}/p^{n-\beta}\mathbb{Z} \end{align*}

Further, I think that if we define $X_1 := p^{\alpha}x + w_1y$ and $X_2:= p^{\beta}y+w_2x$ then $\phi_1$ and $\phi_2$ are onto maps, such that for fixed $y$, as $x$ takes all values of $R$, $X_1$ will take all values of $\displaystyle \mathbb{Z}/p^{n-\alpha}\mathbb{Z}$ and will yield $p^{\alpha}$ copies. Similarly for fixed $x$ as $y$ runs over $R$, repeated $p^{\beta}$ times

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  • $\begingroup$ It might also be that $A = \{1\}$ and $B = \mathbb{Z}/p\mathbb{Z}$. I just need to show that by mapping one of my components to $p$, I'll effectively be working over $\mathbb{Z}/p^{n-1}\mathbb{Z} \times \mathbb{Z}/p^n \mathbb{Z}$. $\endgroup$ – Greg Doyle Aug 6 '15 at 17:50
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You've done most of the work already. Just observe that $\ker(\phi) = \langle p^{n-1}\rangle \times \{0\} \subset R\times R$. So indeed $$(R\times R) / \ker(\phi) \,\,\,\,\simeq \,\,\,\,(R/\langle p^{n-1}\rangle) \times R\,\,\,\, \simeq \,\,\,\,\Bbb Z/p^{n-1} \Bbb Z \times \Bbb Z / p^n \Bbb Z$$

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