Let $p$ be a prime and $n \in \mathbb{N}$ is such that $p^n > 2$. We let $\alpha \in \mathbb{N}$ be such that $0 < \alpha < n$. Let $R := \mathbb{Z}_{p^n}$ denote the ring of integers modulo $p^n$ and let $<p^{\alpha}>$ denote the principal ideal given by the set $\{p^{\alpha}r | r \in R\}$.

Suppose $\phi: R \times R \rightarrow R \times R$ is an $R$-module endomorphism given by $\phi(x,y) = (p^{\alpha}x+y,y)$. It is clear that $\ker(\phi) = <p^{n-\alpha}> \times <0>$.

By the 1st isomorphism theorem, and as product modules act component-wise, we can deduce that \begin{align*} \phi(R \times R) \simeq R/<p^{n-\alpha}> \times R \simeq <p^{\alpha}> \times R. \end{align*}

Set $e(q):=e^{2\pi \imath q}$ for any rational number $q$. Suppose we consider the sum \begin{align*} \sum_{y=0}^{p^n-1} e\left(\frac{y^2}{p^n}\right) \sum_{x=0}^{p^n-1} e\left(\frac{(p^{\alpha}x+y)^2}{p^n}\right). \end{align*}

I want to argue that because $\phi(R \times R) \simeq <p^{\alpha}> \times R$, I can just replace $p^{\alpha}x+y$ by $p^{\alpha}x$ in my sum. Basically, as $x$ and $y$ run through complete residue system modulo $p^n$, the corresponding coordinate points will act like $(p^{\alpha}x+y,y)$ modulo $p^n$. As the image of these coordinate points is isomorphic to $<p^{\alpha}> \times R$, I can replace the coordinate points $(p^{\alpha}x+y,y)$ by $(p^{\alpha}x,y)$.

Where I'm getting stuck is the fact that, $\phi(x,y) \in <p^{\alpha}> \times R$ whenever $y \in <p^{\alpha}>$. If $y \not\in <p^{\alpha}>$, then $p^{\alpha}x+y$ doesn't "behave" like $p^{\alpha}x$ modulo $p^n$. Am I really justified in just writing $p^{\alpha}x$ for $p^{\alpha}x+y$ in my sum? Does an $R$-module isomorphism on product modules means the first coordinate behaves like an $R$-module homomorphism on $R$ with a similar kernel?

This is either obvious, or I'm missing something.

(I have previous questions of this nature here and here.)

EDIT: It can be shown that if $n \geq 2\alpha$ we have \begin{align*} \sum_{\substack{y=0\\y \equiv 0 \mod {p^{\alpha}}}}^{p^n-1} e\left(\frac{y^2}{p^n}\right) = \sum_{y=0}^{p^n-1} e\left(\frac{y^2}{p^n}\right) \end{align*} but I would really like this $R$-module endomorphism idea to work so I don't need to divide $y$ up into cases depending on $p^{\alpha}$.


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