Linear exponential sum over $(\mathbb{Z}/(p^t))^*$

Question

Let $p$ be a prime and $t$ a natural number. Let us denote $(\mathbb{Z}/(p^t))^*$ to be the group of units of $\mathbb{Z}/(p^t)$. I have the following exponential sum $$ S = \sum_{w \in (\mathbb{Z}/(p^t))^*} e^{2 \pi i a w/ p^t}, $$ which I am convinced is $0$ if $a \not \equiv 0 (\mod p^t)$. However, I am not too sure how to proceed to prove this statement. I would greatly appreciate any hints or comments or explanations! Thanks!

Answer 1

Write that sum as $S(a,t)$. Here are some thoughts:

• If $\gcd(a,p)=1$, then $S(a,t)=S(1,t)$ because raising a primitive $p^t$-root of unity to $a$ still gives you a primitive $p^t$-root of unity.

• If $\gcd(a,p)>1$, then $S(a,t)=m S(a',t')$ for some $m\in \mathbb N$ and $a'<a$ with $\gcd(a',p)=1$ and $t'<t$.

So, the only sum that you need to care about is $S(1,t)$.

For $t=1$, we have $$ S(1,1) = \theta+\theta^2+\cdots+\theta^{p-1}=\dfrac{\theta^p-1}{\theta-1}-1=-1 $$ For $t=2$, we have $$ \begin{eqnarray} S(1,2) &=&1+\theta+\theta^2+\cdots+\theta^{p^2-1} -(1+\theta^p+\theta^{2p}+\cdots+\theta^{(p-1)p}) \\ & = & 0-(1+\theta^p+\theta^{2p}+\cdots+\theta^{(p-1)p}) =-\dfrac{\theta^{p^2}-1}{\theta^p-1}=0 \end{eqnarray} $$