# Find minimal polynomial of root squared

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.

• Isn't $\alpha\mapsto\alpha^2$ the Frobenius automorphism? – Angina Seng Sep 30 '20 at 18:01
• 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 $1, \beta, \beta^2, \beta^3, \beta^4$. – Brian Moehring Sep 30 '20 at 18:10

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 $$b^3=b^4+1; b^4+b^3+1=0$$
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 $$\Bbb F_2$$.