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Jul 14, 2015 at 15:37 comment added David Hill @Thomas $V$ is not closed under addition. It is just a set.
Jul 14, 2015 at 15:35 comment added Seppo I am a bit confused: Why is $V$ not a representation of $G$ over $\mathbb{F}_q$?
Jul 14, 2015 at 15:31 vote accept Seppo
Jul 13, 2015 at 21:45 comment added David Hill That is a good point.
Jul 13, 2015 at 21:41 comment added Ofir Gorodetsky Note that if $f(x) \in \mathbb{F}_{p}[x]$ is a non-zero polynomial satisfying $f(x)=f(x+\alpha)$ for some $\alpha \in \mathbb{F}_{p}^{\times}$, then necessarily $\deg f \ge p$: This is because if $r$ is a root of $f$ (over the algebraic closure) then its translates $r+i,i\in \mathbb{F}_{p}$ must also be roots.
Jul 13, 2015 at 21:31 history answered David Hill CC BY-SA 3.0