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Let $\mathbb{F}_{16} = \mathbb{F}_2[x]/(x^4 + x^3 + 1)$ and let $\alpha $ be a root of $x^4 + x^3 +1$. Compute the minimal polynomial of $\alpha^2$ over $\mathbb{F}_2$ in $\mathbb{F}_{16}$.

I have to find $g(x)$ s.t. $g(\alpha^2) = 0$ where $g$ has minimal degree and is monic.

Can anyone give me a hint how to start here.

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    $\begingroup$ Isn't $\alpha\mapsto\alpha^2$ the Frobenius automorphism? $\endgroup$ – Angina Seng Sep 30 '20 at 18:01
  • $\begingroup$ The most elementary route I can imagine is to set $\beta = \alpha^2$ and then compute $\beta^2, \beta^3, \beta^4$ in terms of the $\mathbb{F}_2$-basis $1, \alpha, \alpha^2, \alpha^3$. Then find a linear dependence among $1, \beta, \beta^2, \beta^3, \beta^4$. $\endgroup$ – Brian Moehring Sep 30 '20 at 18:10
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People have suggested various methods. Here is something a bit simpler than computing powers of the root.

Suppose $a^4+a^3+1=0$ and $b=a^2$ then $$b^2+ab+1=0$$ and $$ab=b^2+1$$(characteristic $2$). Square this to get $$a^2b^2=b^3=b^4+2b^2+1=b^4+1$$so $$b^3=b^4+1; b^4+b^3+1=0$$

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For my own sake, I'm going to let $\beta=\alpha^2$ here. Then, presuming I didn't make any mistakes, $\beta^3=\alpha^3+\alpha^2+\alpha+1$ and $\beta^4=\alpha^3+\alpha^2+\alpha$. Thus $\beta^4+\beta^3+1=0$, so $\beta$ is a root of $x^4+x^3+1$, and this is the minimal polynomial over $\Bbb F_2$.

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