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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!

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    $\begingroup$ Take a quadratic Gauss sum for instance. $\endgroup$ – Wojowu Mar 8 at 22:45
  • $\begingroup$ If $\chi$ is the quadratic character, then the Gauss sum has degree two over the rationals. $\endgroup$ – Gerry Myerson Mar 8 at 22:47
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    $\begingroup$ I've slightly edited the title; when reading it I have understood the question as asking for something trivially false. $\endgroup$ – Wojowu Mar 8 at 22:51
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    $\begingroup$ @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. $\endgroup$ – Wojowu Mar 8 at 23:53
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    $\begingroup$ 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. $\endgroup$ – Victor Ostrik Mar 9 at 0:48

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