Take the 2-minute tour ×

I am interested in the sums $$g(a,k)=\sum_{n=0}^{p-1}e^{2\pi i a n^k/p}$$ where $p\equiv1\mod k$ is a prime and $a$ is coprime with $p$.

When $k=2$, it is a classical fact that $g(a,2)=\chi(a)g(1,2)$ where $\chi$ is the Legendre symbol. My question is: Is there an analogue of this formula for higher values of $k$? More concretely, is there a Dirichlet character $\chi_k$ such that $$g(a,k)=\chi_k(a)g(1,k)$$ for all $a$?

share|improve this question
These sums are known as Gaussian periods, see en.wikipedia.org/wiki/Gaussian_period –  Seva Jul 8 '14 at 20:13

1 Answer 1

up vote 10 down vote accepted

No, the relation is not as simple in this case. For example, if $k = 3$ and $p = 7$, the three different cubic Gauss sums are roots of $y^{3} - 21y - 7$, and the three roots of this polynomial do not have equal absolute values.

In general, the $k$ different Gauss sums $g(a,k)$ (where $a$ runs over representatives of the cosets of the $k$th powers in $\mathbb{F}_{p}^{\times}$) will be roots of a polynomial of degree $k$, and the coefficient of $x^{k-1}$ in that polynomial will be zero, because $$ \sum_{a \in \mathbb{F}_{p}^{\times}/\left(\mathbb{F}_{p}^{\times}\right)^{k}} g(a,k) = k \sum_{n=0}^{p-1} e^{2 \pi i n / p} = 0. $$ When $k = 2$, this forces both quadratic Gauss sums to be roots of a polynomial of the form $y^{2} - a$ and this makes the quadratic Gauss sums be negatives of each other, but this is no longer the case when $k > 2$.

share|improve this answer

Your Answer


