Timeline for Could the Weil zeroes of curves be evenly distributed?

5 events

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Oct 13, 2021 at 1:13	history	edited	Will Sawin	CC BY-SA 4.0	added 9 characters in body
Oct 12, 2021 at 20:06	comment	added	David E Speyer		Note that we can't expect $f$ to be a permutation of $\mathbb{F}_{q^k}$ for more values of $k$ than this (unless $f$ is a $q^j$-th power of a linear form) because, if we choose $a \in \mathbb{F}_q$ then $\tfrac{f(x)-f(a)}{x-a}$ is a degree $d-1$ polynomial, so it should have a root other than $a$ in some $\mathbb{F}_{q^k}$ with $1 \leq k \leq d-1$.
Oct 12, 2021 at 19:31	comment	added	David E Speyer		Will uses the polynomial $x^q-ax$ where $a$ is a primitive root in $\mathbb{F}_q^{\ast}$, or more generally $\sum_{i=0}^n a_i x^{q^i}$ where $\sum a_i T^i$ is the minimal polynomial of a primitive root in $\mathbb{F}_{q^n}^{\ast}$. I use $x^r$ where $r$ is a prime for which $q$ is a primitive root.
Oct 12, 2021 at 19:28	comment	added	David E Speyer		To summarize Will and my answers: Let $f(x) \in \mathbb{F}_q[x]$ be a polynomial of odd degree $d$ which acts as a permutation on $\mathbb{F}_{q^k}$ for $1 \leq k \leq d-2$. Consider the curve $y^2+y = f(x)$, if we are in characteristic $2$, or $y^2 = f(x)$, if we are in odd characteristic and $f$ is not branched over $0$. Then this curve has genus $\tfrac{d-1}{2}$, and the eigenvalues are spaced by $\tfrac{2 \pi}{d-1}$.
Oct 12, 2021 at 18:10	history	answered	Will Sawin	CC BY-SA 4.0