# Does the Gauss sum attached to $\chi$ ever belong to $\mathbb{Q}(\chi)$?

Let $$p$$ be a prime number and $$\chi$$ be a primitive Dirichlet character of conductor $$p$$. We let $$g(\chi)=\sum_{a=1}^{p}{\chi(a)e^{2i\pi a/p}}$$ be the Gauss sum attached to $$\chi$$. Is this known that $$g(\chi)$$ does not belong to $$\mathbb{Q}(\chi)$$, the algebraic extension of $$\mathbb{Q}$$ generated by the values of $$\chi$$?

Many thanks!

• Take a quadratic Gauss sum for instance. Mar 8 '19 at 22:45
• If $\chi$ is the quadratic character, then the Gauss sum has degree two over the rationals. Mar 8 '19 at 22:47
• I've slightly edited the title; when reading it I have understood the question as asking for something trivially false. Mar 8 '19 at 22:51
• @SylvainJULIEN $\mathbb Q(\chi)$ is itself a cyclotomic field, since $\chi$ has takes values in roots of unity. What you probably imply is that the Gauss sum belongs to $\mathbb Q(\chi,e^{2\pi i/p})$, which is true and obvious. The question is whether it belongs to some particular smaller field. Mar 8 '19 at 23:53
• there is an element of Galois group fixing ${\mathbb Q}(\chi)$ and sending $e^{2\pi i/p}$ to $e^{2\pi i b/p}$ (here $b$ is a number relatively prime with $p$). Applying this element to $g(\chi)$ we get $\chi(b^{-1})g(\chi)$. Thus $g(\chi) \not \in {\mathbb Q}(\chi)$ if $\chi$ is nontrivial. Mar 9 '19 at 0:48