I need to calculate a couple of Gauss sums to solve a problem I'm working on, but I keep getting the wrong answer because the absolute value of what I calculate is impossible for such a sum. Can anyone spot my mistake?

Suppose $F$ is a field with four elements, as $0, 1, a, a^2$, and $\chi$ a nontrivial character on $F$. Thus we must have $\chi(0) = 0, \chi(1) = 1$, and we can assume $\chi(a) = \zeta$ and $\chi(a^2) = \zeta^2$, where $\zeta = e^{2\pi i /3}$. This is because $\chi$ sends a nonzero member of $F$ to a third root of unity, so the only other homomorphism (which would be $\chi^2$) would interchange where $a$ and $a^2$ are sent.

Now the Gauss sum of $\chi$ is given by $g(\chi) = \sum\limits_{t \in F} \chi(t)\zeta^{tr(t)}$, where $tr$ is the trace function, defined by $tr(t) = t + t^2$. The trace function has its range equal to $\mathbb{Z}_2 \subset F$, each member of which is interpreted as the corresponding $0$ or $1$ in $\mathbb{C}$.

We have $tr(0) = 0, tr(1) = 1 + 1^2 = 0$. Also $tr(a) = a + a^2 = 1$, and $tr(a^2) = a^2 + a^4 = a^2 + a = tr(a) = 1$.

Thus the Gauss sum is $\chi(0)\zeta^{tr(0)} + \chi(1)\zeta^{tr(1)} + \chi(a)\zeta^{tr(a)} + \chi(a^2)\zeta^{tr(a^2)} = 0 + 1 + \zeta \zeta + \zeta^2\zeta = 1 + \zeta^2 + 1 = 2 + \zeta^2$.

But $\zeta^2 = e^{2 \pi i (2/3)} = Cos(4 \pi / 3) + i Sin(4 \pi / 3) = -1/2 - i\sqrt{3}/2$, whence $g(\chi) = 1.5 - \sqrt{3}i/2$, meaning $|g(\chi)| = \sqrt{2.25 + 3/4} = \sqrt{3}$.

The Gauss sum is supposed to have absolute value $\sqrt{|F|} = 2$. What have I done wrong here?

  • $\begingroup$ Without background and/or links, books, papers, etc., I think it is going to be hard someone will understand what's going wrong with your calculations, if at all, since all that looks fine... $\endgroup$ – DonAntonio Jan 28 '13 at 2:32
  • $\begingroup$ Besides the above, and as far as I remember, if $\,\chi\,$ is a primitive character, as in your case, the absolute module of its Gauss sum should be, I thing ,$\,\sqrt 4=2\,$ and not $\,\sqrt 2\,$ , as you say...check this. $\endgroup$ – DonAntonio Jan 28 '13 at 2:34
  • $\begingroup$ yes, it should be square root of two. umm as far as links to the subject, these notes I think cover it modular.math.washington.edu/edu/2010/414/projects/wen_wang.pdf $\endgroup$ – D_S Jan 28 '13 at 3:32
  • $\begingroup$ Well, proposition 3.1 in that paper, which is exactly the same, including the proof, as in Ireland-Rosen's book, talks of working on the prime field $\,\Bbb F_p\,$ , which is not your case, and I'm not sure we can expect to have the same bound... $\endgroup$ – DonAntonio Jan 28 '13 at 13:52

A Gauss sum involves a multiplicative character and an additive character. The additive character on the field of $4$ elements does not involve $\zeta=e^{2\pi i/3}$; it involves $-1$.

  • 1
    $\begingroup$ Holy derp. ${}$ $\endgroup$ – anon Jan 28 '13 at 5:24
  • $\begingroup$ I'm silly. thank you $\endgroup$ – D_S Jan 28 '13 at 5:27

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