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Weil's character bound states that: Let $\mathbb{F}_{q}$ be a finite field of size $q$. Let $\chi$ be a multiplicative character of order $m$. Let $f(x)$ be a polynomial of degree $d$ such that $f(x) \neq c . g(x)^{m}$ for any $c \in \mathbb{F}_{q}$ and $g(x) \in \mathbb{F}_{q}[x]$. Then $$ \left| \sum_{x \in \mathbb{F}_{q}} \chi(f(x)) \right| \leq (d-1)\sqrt{q} $$

The Riemann Hypothesis over finite fields, or rather the Hasse-Weil bound is concerned with the number of rational points on curves. To put it simply, again suppose we have a finite field $\mathbb{F}_{q}$ and an absolutely irreducible polynomial $h(x,y)$ of total degree $d$ over $\mathbb{F}_{q}$, (a simple version of) it states that $$ \left| N - q \right| \leq O(d^2) \sqrt{q} $$ where $N$ is the number of rational points on the variety defined by the polynomial $h$ i.e. $$N = \left|\left\lbrace (a,b) \in \mathbb{F}_{q}^2 : h(a,b)=0 \right\rbrace \right|$$

Now if $\chi$ is the quadratic character defined by $\chi(a) = (\frac{a}{q})$ where $(\frac{a}{q})$ is the Legendre symbol (whether $a$ is quadratic residues or not modulo $q$), then since $\chi$ takes only 1/-1 values, it is easy to see that the number of rational points on the curve $y^2 - f(x) = 0$ can be used to count the sum $\left| \sum_{x \in \mathbb{F}_{q}} \chi(f(x)) \right|$. Most references simply state that the quadratic character bound, and the general character sum bound are special cases of counting points on varieties and the Riemann hypothesis. But how are the two results related in the general case, where $\chi$ no longer takes only 1/-1 values? Is there a simple correspondence like in the quadratic case? Thanks.

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    $\begingroup$ You need a little bit help from $L$-functions to bridge the gap in the general case $m>2$. Basically you use Hasse-Weil of the curve $y^m-f(x)=0$. You can easily write the number of solutions of that in terms of character sums related to $\chi$ and its powers. The $L$-function allows you to separate the effects of distinct powers of $\chi$. To that end you need to use Hasse-Weil to that equation not only over $\Bbb{F}_q$ but over all its extension fields as well. At least that's how you can do it with relatively elementary tools. Number theorists have more powerful tools at their disposal. $\endgroup$ – Jyrki Lahtonen Jan 20 '14 at 6:42
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    $\begingroup$ So I get the impression that one needs to wade into a lot more involved theory to "see" the connection between character sums and rational points on curves. I am a coding theorist rather than a number theorist. Would you recommend any references just to get some working intuition and big picture connections, without getting into the nitty-gritties (which I do want to explore slowly at my own pace)? Thanks. $\endgroup$ – BharatRam Jan 21 '14 at 3:12
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This can be explained in a pretty elementary way.

For $0 \leq j < m$, let $\gamma_j$ be the character sum $$\gamma_j := \sum_{x \in \mathbb{F}_q} \chi^j(f(x)).$$ Let $X$ be the curve $y^m = f(x)$. Note that the number of points on $X$ over $\mathbb{F}_p$ is $$\# X(\mathbb{F}_q) = \sum_{x \in \mathbb{F}_q} \# \{m\mbox{-th roots of f(x)} \}= \sum_{x \in \mathbb{F}_q} \sum_{j=0}^{m-1} \chi^j(f(x)) = \sum_{j=0}^{m-1} \gamma_j$$

Let $g$ be a primitive element in $\mathbb{F}_q$, so $\chi(g)$ is a primitive $m$-th root of unity. Define $\zeta = \chi(g)$. For $0 \leq i < m$, let $X_i$ be the curve $y^m = g^i f(x)$. Then $$\# X_i(\mathbb{F}_q) = \sum_{x \in \mathbb{F}_q} \sum_{j=0}^{m-1} \chi^j(g^i f(x)) = \sum_{j=0}^{m-1} \zeta^{ij} \gamma_j.$$

So the character sums and the number of points on the curves $X_i(\mathbb{F}_q)$ are related by the finite Fourier transform. The Riemann hypothesis $\# X_i(\mathbb{F}_q) = q + O(\sqrt{q})$ transforms to $\gamma_0 = q$ and $\gamma_j = O(\sqrt{q})$ for $j \neq 0$.

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