# Estimating certain short Kloosterman sums

Recall that for the classical Kloosterman sum $$K(a,b,p^t):= \sum_{x \in (\mathbb{Z}/ p^t \mathbb{Z})^* } \psi \left(\frac{ax+bx^{-1}}{p^t} \right),$$ where $$\psi(x)=e^{2\pi ix}$$, $$a,b,t$$ are natural numbers and $$p$$ is a fixed prime number. We have the Weil bound, i.e. $$\vert K(a,b,p^t) \rvert \leq (t+1) \sqrt{(\gcd(a,b,p^t))} \sqrt{p^t}.$$ Now if we consider the following short Kloosterman sum $$K'(a,b,p^t):= \sum_{x \in A} \psi \left(\frac{ax+bx^{-1}}{p^t} \right),$$ where $$A:=\left \{ x \rvert x \equiv 1 \mod p^m,\;x \in (\mathbb{Z}/ p^t \mathbb{Z})^* \right \}.$$ Here m is a fixed positive integer and we can also assume that $$t>>m$$. My question is that can we also achieve an analogy of Weil bound for the above short Kloosterman sum $$K'(a,b,p^t)$$? Hopefully, I expect the following bound $$\vert K'(a,b,p^t) \rvert \leq A_m \cdot (t+1) \sqrt{(\gcd(a,b,p^t))} \sqrt{p^t}.$$ Here $$A_m$$ is a positive constant only depend on the choice of $$p$$ and $$m$$.

Since we assume that $$t>>m$$, then the number of elements in the finite set $$A$$ is greater than $$\sqrt{p^t}$$. Then from the “short Kloosterman sums" in Wikipedia, we may achieve an analogy of Weil bound for the above short Kloosterman sum. However, I cannot find any references. So any ideas or references for the bound of above short Kloosterman sum are welcome.

If $$x=1+p^my$$ ($$1\le y\le p^{t-m}$$) then $$\psi \left(\frac{ax+bx^{-1}}{p^t} \right)=\psi \left(\frac{a(1+p^my)+b(1+p^my)^{-1}}{p^t} \right)=\psi \left(\frac{a+b}{p^t} \right)\psi \left(\frac{f(y)}{p^{t-m}} \right),$$ where $$f(y)=ay+b(-y+y^2p^m-y^3p^{2m}+\cdots)$$. The sum $$\sum_{y=1}^{p^{n}} \psi \left(\frac{f(y)}{p^{n}} \right)$$ can be calculated explicitly, because $$f(y)$$ has a nice form (almost all coefficients are divisible by $$p$$), see Lemma 2.1 in Generalized Twisted Kloosterman Sum Over ℤ[i] by S. Varbanets. The main idea is to take $$y=y_0+y_1p^{n-1}$$, where $$1\le y_0\le p^{n-1}$$, $$1\le y_1\le p.$$ The sum becomes linear over $$y_1$$. Apply this idea twice.

For $$n\ge 1$$ $$f(y)\equiv f(y_0)+f'(y_0)y_1p^{n-1}\equiv f(y_0)+(a-b)y_1p^{n-1}\pmod{ p^n}.$$ So $$S_n(a,b)=\sum_{y=1}^{p^n}e_{p^n}(f(y))=\sum_{y_0=1}^{p^{n-1}}\sum_{y_1=1}^{p}e_{p^n}(f(y_0)+(a-b)y_1p^{n-1})=p\delta_p(a-b)\sum_{y_0=1}^{p^{n-1}}e_{p^n}(f(y_0)),$$ where $$e_N(x)=e^{2\pi ix/N}$$ and $$\delta_q(x)=\begin{cases} 1,& \text{ if }q\mid x;\\ 0,& \text{ if }q\nmid x. \end{cases}$$ This sum does not vanish if $$a\equiv b\pmod{p }$$. We may also assume that $$a\equiv b\not \equiv0\pmod{p }$$ because otherwise original sum can be simplified: for $$a=pa_1$$, $$b=pb_1$$ $$\sum_{x \in (\mathbb{Z}/ p^n \mathbb{Z})^*}e_{p^n}(ax+bx^{-1})=p\sum_{x \in (\mathbb{Z}/ p^{n-1} \mathbb{Z})^*}e_{p^{n-1}}(a_1x+b_1x^{-1}).$$ Let $$a=b+p^\alpha a_1$$, $$\alpha\ge 1$$, $$(a_1,p)=1$$. Then
$$f(y)=a_1p^\alpha y+b(y^2p^m-y^3p^{2m}+\ldots)$$, and $$S_n(a,b)=p \sum_{y=1}^{p^{n-1}}e_{p^{n-1}}(f(y)p^{-1}).$$

If $$\alpha\ge m$$ then $$S_n(a,b)=p \sum_{y=1}^{p^{n-1}}e_{p^{n-m}}(g(y))=p^m \sum_{y=1}^{p^{n-m}}e_{p^{n-m}}(g(y)),$$ where $$g(y)=a_1p^{\alpha-m}y+b(y^2-y^3p^{m}+\ldots),$$ and one can apply Lemma 2.1 from the cited article.

If $$\alpha then $$S_n(a,b)=p^\alpha \sum_{y=1}^{p^{n-\alpha}}e_{p^{n-\alpha}}(g(y)),$$ where $$g(y)=a_1y+b(y^2p^{m-\alpha}-y^3p^{2m-\alpha}+\ldots).$$ Again $$y=y_0+y_1p^{n-\alpha-1}$$, $$1\le y_0\le p^{n-\alpha-1}$$, $$1\le y_1\le p$$ $$g(y)\equiv g(y_0)+g'(y_0)y_1p^{n-\alpha-1}\equiv g(y_0)+a_1y_1p^{n-\alpha-1}\pmod{ p^{n-\alpha}}.$$ $$S_n(a,b)= \sum_{y=1}^{p^{n-\alpha}}e_{p^{n-\alpha}}(g(y))= \sum_{y_0=1}^{p^{n-\alpha-1}}\sum_{y_1=1}^{p}e_{p^{n-\alpha}}(g(y_0)+a_1y_1p^{n-\alpha-1})=0,$$ because sum over $$y_1$$ vanishes.

• Thanks a lot! Is there any similar result for $p$ equals 2? – JACK Feb 21 at 15:41
• I think maybe we cannot apply Lemma 2.1 in your reference. Since Lemma 2.1 requires that the coefficient of $y^2$ should be relatively prime to $p$. However, the polynomial in your calculation does not satisfy this condition.@Alexey Ustinov – JACK Feb 21 at 21:35
• @JACK You may repeat the same calculations for $p=2$. Lemma 2.1 not always applied directly. I've added more details in my answer. – Alexey Ustinov Feb 22 at 2:46
• Thanks! @Alexey Ustinov I think for $p$ equals 2 case, we may not repeat the same calculations since Lemma 2.1 cannot apply to this case. So maybe we need some modifications. – JACK Feb 23 at 2:09
• @JACK For $p=2$ take $y=y_0+y_1p^{n-\alpha-2}$. It works for $m>1$. For $m=1$ take $y=y_0+y_1p^{n-\alpha-3}$. – Alexey Ustinov Feb 23 at 2:38