Q:

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. 

A:

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 $\alpha<m$
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.

