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March 29th, 2017, 02:51 AM  #1 
Senior Member Joined: Mar 2012 From: Belgium Posts: 653 Thanks: 11  minimal polynomials over a finite field
Suppose $\displaystyle f$ is a minimal polynomial for a certain element $\displaystyle \alpha \in \mathbb{F}_{p^h}$. Suppose $\displaystyle g(\alpha)=0$ for a $\displaystyle g \in \mathbb{F}_{p}[x] $. Does this imply that $\displaystyle f \mid g$ ? I believe this is not true when were not working in a finite field but i can't really wrap my head around the case where it is a finite field... 
June 2nd, 2017, 01:15 PM  #2 
Newbie Joined: May 2017 From: Russia Posts: 19 Thanks: 1 
$\displaystyle \mathbb{F}_{p}\subset \mathbb{F}_{p^h}.$ $\displaystyle f$ divides $\displaystyle g$ because there exists division algorithm for polynomials over any field (in one variable). $\displaystyle {\rm deg}\ f\leq {\rm deg }\ g$ because $\displaystyle f$ is the minimal polynomial. $\displaystyle g(x)=q(x)\cdot f(x)+r(x)$, $\displaystyle \ \ \ \ \ {\rm deg}\ r\lt {\rm deg}\ f $ or $\displaystyle r(x)\equiv0$. $\displaystyle g(\alpha)=q(\alpha)\cdot f(\alpha)+r(\alpha)$. So, $\displaystyle r(\alpha)=0$. If r(x) were not 0, we would get $\displaystyle \ \ {\rm deg}\ r\lt {\rm deg}\ f $ and $\displaystyle r(\alpha)=0$. But it is impossible because $\displaystyle f$ is the minimal polynomial. The only variant is $\displaystyle r(x)\equiv 0$. Last edited by ABVictor; June 2nd, 2017 at 01:27 PM. 

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field, finite, minimal, polynomials 
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