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For the finite field $\mathbb{F}_7$, find an additive Fourier transform of nontrivial multiplicative characters.

If the field contains $q$ elements, there are $q-2$ nontrivial values.

Where do I even start?

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Are you trying to define the Fourier transform for $\mathbb{F}_7$, or are you trying to trying to find the Fourier transform of a multiplicative character on $\mathbb{F}_7^*$? My guess is the latter based on the fact that there are $q-2$ nontrivial multiplicative characters on $\mathbb{F}_q$. –  JavaMan Apr 22 '11 at 19:07
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I would edit it if I could, but technically the term "multiplicative character on $\mathbb{F}_7^*$" is redundant. It should read either multiplicative character of $\mathbb{F}_7$ or character of $\mathbb{F}_7^*$. –  JavaMan Apr 22 '11 at 19:29
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1 Answer 1

Not so much a hint, but a push in the right direction:

Let $\mathbb{F}_q^d$ denote the $d$-dimensional vector space over the finite field with $q$ elements. Recall that for a function $f : \mathbb{F}_q^d \to \mathbb{C}$, the Fourier transform of $f$ is given by

$$ \widehat{f}(m) = q^{-d} \sum_{x \in \mathbb{F}_q^d} f(x) \chi(x \cdot m) $$

where $\chi$ is a nontrivial additive character on $\mathbb{F}_q$. It is worth noting that this definition of $\widehat{f}$ may or may not agree with the definition you use. In particular, your definition might not include the normalization factor $q^{-d}$.

Now, when $q$ is prime, we can actually take $\chi(z) = \exp(2 \pi i z/q)$. Furthermore, when $q$ is prime, we can identify $\mathbb{F}_q$ with $\mathbb{Z}_q = \{0,1, \dots , q-1\}$.

For this particular problem, we have $q = 7$ and $d = 1$. Let $\psi$ denote any nontrivial multiplicative character. Write:

$$ \widehat{\psi}(m) = \frac{1}{7} \sum_{x \in \mathbb{Z}_7} \psi(x) \exp(2 \pi i xm/7).... $$

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