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I'd like to ask how to prove or disprove the following identity (which I have to use on some computation regarding $\epsilon$-constant for constructing certain local Langlands parameter):

Let $\psi$ be any non-trivial additive character on $\mathbb{F}_q$, the field of $q$ elements. Then

$$\sum_{x\in\mathbb{F}_{q^d}}\left(\frac{x}{q^d}\right)\psi(\text{Tr}_{\mathbb{F}_{q^d}/\mathbb{F}_q}x)=\left(\sum_{x\in\mathbb{F}_{q}}\left(\frac{x}{q}\right)\psi(x)\right)^d,$$

where $\left(\frac{\cdot}{q^d}\right)$ denote the quadratic symbol on $\mathbb{F}_{q^d}^*$. The identity is definitely correct up to sign, and I come up with it just via numerical checking. Thanks!

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1 Answer 1

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I think this is just the Davenport-Hasse Theorem, which Wikipedia calls the Hasse-Davenport lifting relation.

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