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March 29th, 2017, 02:51 AM   #1
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Question 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...
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June 2nd, 2017, 01:15 PM   #2
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$\displaystyle \mathbb{F}_{p}\subset \mathbb{F}_{p^h}.$

$\displaystyle f$ 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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