Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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
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
add comment

1 Answer

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:

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

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.