I want to calculate Gauss Sum of the field with 8 elements. I studied in Lidl & Niederreiter's book.

$\chi(c)=e^{2 \pi i Tr(c)/p}$ for all $c\in F_q$, defines an additive character of $F_q$, where $Tr$ is from $F_q$ to $F_p$. The characteristic of $F_q$ is $p$. And,

$\psi(g^k)=e^{{2\pi i k}/{q-1}}$ for $k=0,1,2,...,q-2$ defines a multiplicative character of $F_q$, where $g$ is a primitive element of $F_q$.

$G(\chi,\psi)=\sum_{c\in F_q^*}\chi(c)\psi(c)$ is the Gauss sum.

For $p=2$ and $q=8$, $F_8^*=\{a^0, a^1, a^2, a^3, a^4, a^5, a^6\}$ and their trace are $\{1,0,0,1,0,1,1\}$ respectively. Then

$G(\chi,\psi)=(e^{{2\pi i Tr(1)}/{2}})(e^{{2\pi i0}/{7}})+(e^{{2\pi i Tr(a)}/{2}})(e^{{2\pi i1}/{7}})+(e^{{2\pi i Tr(a^2)}/{2}})(e^{{2\pi i2}/{7}})+(e^{{2\pi i Tr(a^3)}/{2}})(e^{{2\pi i3}/{7}})+(e^{{2\pi i Tr(a^4)}/{2}})(e^{{2\pi i4}/{7}})+(e^{{2\pi i Tr(a^5)}/{2}})(e^{{2\pi i5}/{7}})+(e^{{2\pi i Tr(a^6)}/{2}})(e^{{2\pi i6}/{7}})$ $=\zeta^{1/2}.1+1.\zeta^{1/7}+1.\zeta^{2/7}+\zeta^{1/2}.\zeta^{3/7}+1.\zeta^{4/7}+\zeta^{1/2}.\zeta^{5/7}+\zeta^{1/2}.\zeta^{6/7}$

where I take $\zeta=e^{2\pi i}$ and so $\zeta^{1/2}=e^{\pi i}=-1$. Thus I get


What is wrong with my calculations?

Additionally, do the additive characters always take $p^{th}$ roots of unity? Which is $2^{nd}$ roots of unity $=\{-1,+1\}$ in my example.

Thanks for any help.

  • $\begingroup$ You want to compute $G(\chi,k)=\sum_{n=0}^{q-1} \chi(n) e^{2i \pi n k / q}$ i.e. the discrete Fourier transform of $\chi$, and with the change of variable $m = nk$ if $gcd(k,q) = 1$ you get $G(\chi,k)=\sum_{n=0}^{q-1} \overline{\chi(k)}\chi(nk) e^{2i \pi n k / q}=\overline{\chi(k)}\sum_{m=0}^{q-1} \chi(m) e^{2i \pi m / q}= \overline{\chi(k)} G(\chi,1)$. Conclude by showing $G(\chi,k) = 0$ if $gcd(k,q) > 1$ $\endgroup$ – reuns Nov 30 '16 at 5:27
  • $\begingroup$ Thanks for your reply. Since I do not understand well this subject, I do not understand what do you mean. Could you explain your idea simply please? And where the calculations are wrong and my question? $\endgroup$ – Kenan DOGAN Dec 1 '16 at 13:34
  • $\begingroup$ Do you know the discrete Fourier transform ? And what do you not understand ? $\endgroup$ – reuns Dec 1 '16 at 13:41

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