# Some easy questions about multiplicative characters and Jacobi sums.

First I want to give you some context. Then I will ask my questions. I think that my questions are easy and fast to answer, so I've decided to put them together in one question here.

Context

Consider the equations $$a_1x_1^{l_1} + \dots + a_rx_r^{l_r} = b$$ with $$a_1, \dots , a_r \in F^{*}$$, where $$F$$ is a finite field. Let us say that $$F$$ has $$m$$ elements. $$N$$ is the number of solutions. Now I know that if $$b \neq 0$$ :

$$N = m^{r-1} + \sum \chi_1\chi_2\cdots\chi_r(b) \chi_1(a_1^{-1})\chi_2(a_2^{-1})\cdots\chi_r(a_r^{-1})J(\chi_1,\dots,\chi_r)$$. The summation is over $$r$$-tuples of characters $$\chi_1, \dots, \chi_r$$, where $$\chi_i^{l_i} = \varepsilon$$ and $$\chi_i \neq \epsilon$$ for $$i = 1, \dots, r$$.

I want to specializing this to $$x^2 + y^4 = 1$$. Obviously $$r=2$$. So I want to sum over all $$2$$-tuples of characters $$\chi_1, \chi_2$$ where $$\chi_i^{l_i} = \varepsilon$$ and $$\chi_i \neq \epsilon$$ for $$i = 1,2$$.

So the $$N = m + J(\rho,\chi) + J(\rho,\chi^2) + J(\rho,\chi^3)$$, where $$\rho$$ is a character of order $$2$$ and $$\chi$$ is a character of order $$4$$.

Questions

1) Why $$\chi^2 = \rho$$ ? I need that to say $$J(\rho,\chi^2) = -1$$.

2) We know $$\chi^4 = \varepsilon$$, but why this is enough to say that $$\chi^3 = \bar{\chi}$$ ?

3) Why $$J(\rho,\bar{\chi}) = \overline{J(\rho,\chi)}$$ ?

4) Now let us say that $$\pi = -J(\rho,\chi)$$. We know that $$\rho$$ takes the values $$\pm1$$ and $$\chi \pm1$$, $$\pm i$$. Why is this enougth to say that $$\pi = a+bi$$?

5)If my four questions are answered I can say that $$N = m - 1 - \pi - \bar{\pi}$$ . I know that $$a^2 + b^2 = \pi \bar{\pi} = m$$. Why can I say that $$N = m - 1 - 2a$$ ?

I hope that my question are clear. Sorry that they might be trivial for you. I'm an absolute beginner.

$$\chi^2$$ has order $$2$$ and $$\rho$$ is the only character of order $$2$$.

$$\chi^3=\chi^{-1}=\overline\chi$$.

$$\overline{J(\rho,\chi)}=J(\overline\rho,\overline\chi)=J(\rho,\overline\chi)$$.

$$J(\rho,\chi)$$ is the sum of terms all of which have the form $$\pm1$$ or $$\pm i$$.

$$\pi+\overline\pi=(a+bi)+(a-bi)=2a.$$

• Thank you for your answer. I get it now. Happy new year :). – Memories Jan 1 at 18:34
• Sorry for disturbing you again, but can I ask you one final question? My question seems so simple that it wouldnt be necessary to use " ask Question". I showed now $\rho(a) -1 \equiv 0 (2)$ and $\chi(a) -1 \equiv 0(1+i)$. Combine this to get $(\rho(a) -1)(\chi(b) -1) \equiv 0(2+2i)$. Thus $\sum_{a+b=1} (\rho(a) -1)(\chi(b) -1) \equiv 0(2+2i)$. I have expanded this term and got $\sum_a \rho(a)$ and $\sum_b \chi(b)$. Now I need that these two sum are zero. But I don't see why. – Memories Jan 2 at 16:46
• @Memories It's a general fact about characters that the values of a non-trivial character sum to zero. – Lord Shark the Unknown Jan 2 at 16:51