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.)


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

  • $\begingroup$ 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

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 $$(R\times R) / \ker(\phi) \,\,\,\,\simeq \,\,\,\,(R/\langle p^{n-1}\rangle) \times R\,\,\,\, \simeq \,\,\,\,\Bbb Z/p^{n-1} \Bbb Z \times \Bbb Z / p^n \Bbb Z$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.