# What is the reason for taking $\omega$ to be a primitive $q$-th root unity rather than taking any $q$-th root of unity?

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.

• Why is there a prime $p$ mentioned in problem statement? – enedil Feb 9 at 17:07
• Moreover, if $q$ is prime, every root of unity is primitive (except $\omega = 1$). – enedil Feb 9 at 17:09
• 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? – Dbchatto67 Feb 9 at 17:13
• Yes, that's true. – enedil Feb 9 at 17:15

## 1 Answer

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 $$\sum_{y \in \Bbb {F_q}^*} {\omega}^{y(z+1)}=q-1$$.

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