How can I evaluate the sum $$\sum\_{a+b=1 ;\ a,b \in \mathbb F\_p}\left(\frac{a}{p}\right) \chi(b)$$
where $\chi$ is a multiplicative character of $\mathbb F\_p^{\*}$ of order three?


Thanks!



How can I evaluate the sum $\sum_{a+b=1;\ a,b \in \mathbb F_p} \left(\frac{a}{p}\right) \chi(b)$ where $\chi$ is a multiplicative character of $\mathbb F_p^{*}$ of order three?
If I've understood the question correctly, this is just a Jacobi sum. The question only seems to make sense for $p$ a prime congruent to 1 mod 3. In this case, the order 3 character and the quadratic character are distinct, and not inverses of each other either, so the absolute value of the Jacobi sum is going to be $\sqrt{p}$ (see [Wikipedia](http://en.wikipedia.org/wiki/Jacobi_sum)). It's also clearly an algebraic integer, so it is one of the factors of $p$ in the factorization of $p$ into two primes in $\mathbb{Z}[\omega]$ with $\omega$ a cube root of unity. Now one gets into the usual thorny questions about signs, but the bottom line is that there may be a relatively simple way of expressing the sum in terms of $A$ and $B$ once you have written $p=A^2+AB+B^2$, the main problem with making this precise is that $A$ and $B$ are not unique, and you'll have to get things right mod 3 before you know exactly which way is up.


