Let $p$ be an odd prime and $n \in \mathbb{N}$. Let $a,b,c$ be arbitrary integers such that $ab \neq 0$. We write $p^{\alpha}A = a$ and $p^{\beta}B = B$ for some $\alpha, \beta \in \mathbb{N}_0$ and $A, B \in \mathbb{Z}$ are such that $(AB,p)=1$. Further, we assume that $c$ is chosen such that $p^{\alpha} \mid c$.

Observe that \begin{align*} 4a(ax^2+bxy+cy^2) &= (2ax+by)^2 + (4ac-b^2)y^2\\ p^{\alpha}(ax^2+bxy+cy^2) &\equiv (4A)^{-1}(2ax+by)^2+(4A)^{-1}(4ac-b^2)y^2 \mod{p^n}. \end{align*}

Let $X := 2p^{\alpha}Ax + by$ and let $e(q):= e^{2\pi \imath q}$ for any rational number $q$. Then consider the double sum \begin{align*} \frac{1}{p^{2\alpha}}& \sum_{x,y=0}^{p^{n+\alpha}-1} e\left(\frac{p^{\alpha}(ax^2+bxy+cy^2)}{p^{n+\alpha}}\right)\\ &= \frac{1}{p^{2\alpha}}\sum_{y=0}^{p^{n+\alpha}-1} e\left(\frac{(4A)^{-1}(4ac-b^2)y^2}{p^{n+\alpha}}\right) \sum_{x=0}^{p^{n+\alpha}-1} e\left(\frac{(4A)^{-1}X^2}{p^{n+\alpha}}\right). \end{align*}

It can be shown that if $\beta < \alpha$ then \begin{align*} \sum_{x=0}^{p^{n+\alpha}-1} e\left(\frac{(4A)^{-1}X^2}{p^{n+\alpha}}\right) = 0. \end{align*}

Let $R:= \mathbb{Z}/p^{n+\alpha}\mathbb{Z}$. From my previous question (see here), I can define an $R$-module endomorphism $\phi: R \times R \rightarrow R \times R$ by $\phi(x,y) = (2p^{\alpha}Ax+by,y)$. Under this endomorphism, we have that $\phi(R \times R) \simeq \mathbb{Z}/p^{n}\mathbb{Z} \times R$.

I want to now argue that \begin{align} \sum_{x=0}^{p^{n+\alpha}-1} e\left(\frac{(4A)^{-1}X^2}{p^{n+\alpha}}\right) = \sum_{X=0}^{p^{n+\alpha}-1} e\left(\frac{(4A)^{-1} p^{2\alpha}(2Ax+by)^2}{p^{n+\alpha}}\right). (1) \end{align}

I think I need to use the fact that $p^{\alpha}(2Ax)+by$ is an $R$-module and then use the 1st and 3rd isomorphism theorems. I am not sure how to proceed though. I may have it wrong, but I think that $\phi$ is an onto map, and the elements of the first component $R$ will get mapped to $\mathbb{Z}/p^{n}\mathbb{Z}$, and will yield $p^{\alpha}$ copies. I basically need an argument so that I can replace the index $x$ by $X$ in my sum.

I can obtain my desired result by considering the two cases $\alpha \leq \beta$ and $\alpha > \beta$, so I know that (1) will hold (with possibly an extra square factor $p^{2\alpha}$ missing.) But I'm pretty sure there's some clever way to show this holds using the properties of $R$-module endomorphisms. I would really like to avoid specifying cases for $\beta$.

Any suggestions on things to try or any corrections would be really helpful! Thanks!

  • $\begingroup$ I think it will work if we can argue that $<p^{\alpha}>(\mathbb{Z}/p^{n+\alpha} \mathbb{Z}) \simeq R$ or something along those lines. We need the ideal $ <p^{\alpha}>$ to pop out after our endomorphism $\endgroup$ – Greg Doyle Aug 7 '15 at 21:52

I think I have it figured out. Define \begin{align*} P^{\alpha} := p^{\alpha}R = \{p^{\alpha}r | r \in R\} \end{align*} and it should follow that $P^{\alpha}$ is an $R$-submodule and is isomorphic to $\mathbb{Z}/p^n\mathbb{Z}$


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