Let $p$ and $q$ be two distinct odd primes. Let $\omega$ be a primitive $q$-th root of unity. Consider the sum

$$S = \sum_{x \in \Bbb {F_q}^*} \left ( \frac x q \right ) {\omega}^x.$$ Prove that

$$S^2 = q \left (\frac {-1} {q} \right ).$$

I have proved the above as follows $:$

$$\begin{align} S^2 & = \sum_{x,y \in \Bbb {F_q}^*} \left ( \frac {xy} {q} \right ) {\omega}^{x+y}. \end{align}$$

Putting $x = yz$ we have

$$\begin{align} S^2 & = \sum_{y,z \in \Bbb {F_q}^*} \left (\frac {y^2 z} {q} \right ) {\omega}^{y(z+1)} \\ & = \sum_{y,z \in \Bbb {F_q}^*} \left (\frac z q \right ){\omega}^{y(z+1)} \\ & = \sum_{y \in \Bbb {F_q}^*} \left ( \frac {-1} {q} \right ) + \sum_{z \neq -1} \left ( \frac z q \right ) \sum_{y \in \Bbb {F_q}^*} {\omega}^{y(z+1)} \\ & = \left ( \frac {-1} {q} \right )(q-1) + (-1) \sum_{z\neq -1} \left (\frac z q \right ) \\ & = q \left ( \frac {-1} {q} \right ) + (-1) \sum_{z \in \Bbb {F_q}^*} \left (\frac z q \right ) \\ & = q \left (\frac {-1} {q} \right ). \end{align}$$

But I don't find any proper reason as to why should I take $\omega$ as a primitive $q$-th root of unity instead of taking any ordinary $q$-th root of unity. Would anybody please point out that where have I used this fact implicitly?

Thank you very much.

  • $\begingroup$ Why is there a prime $p$ mentioned in problem statement? $\endgroup$ – enedil Feb 9 at 17:07
  • 1
    $\begingroup$ Moreover, if $q$ is prime, every root of unity is primitive (except $\omega = 1$). $\endgroup$ – enedil Feb 9 at 17:09
  • $\begingroup$ Yeah @enedil you are correct. If we don't take the primitive $q$-th root of unity then we would have $S=0$. Isn't it so? $\endgroup$ – Dbchatto67 Feb 9 at 17:13
  • $\begingroup$ Yes, that's true. $\endgroup$ – enedil Feb 9 at 17:15

The step where you simplify $\sum_{y \in \Bbb {F_q}^*} {\omega}^{y(z+1)}$ to $-1$ is only valid if $\omega^{z+1}\neq 1$. It seems you mean to write $z\neq -1$ in the outer summation instead of $z\neq 1$, so that $\omega^{z+1}\neq 1$ for all such $z$ as long as $\omega$ is a primitive $q$th root of unity. On the other hand, if $\omega=1$, then $\omega^{z+1}=1$ for all $z$ and we get $\sum_{y \in \Bbb {F_q}^*} {\omega}^{y(z+1)}=q-1$.

  • $\begingroup$ Yes. Exactly it is. I have corrected my error now @Eric Wofsey. $\endgroup$ – Dbchatto67 Feb 9 at 17:20

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