Help finding the root of a polynomial in a field extension which adjoins the root of a separate polynomial.

Question

If $g(x)=x^3+x^2+1, f(x)=x^3+x+1, f(\alpha)=0, g(\beta)=0$, then an extension including $\alpha $ is defined as $\Bbb F_2(\alpha):=\{a+b\alpha+c\alpha^2|a,b,c\in \Bbb F_2\}$. g(x) has some root in this extension but I cant figure out what it is. Obviously its not any $a \in \Bbb F_2(\alpha)$ as $g(x) $ had no root in $\Bbb F_2$ but I cant figure out what element to use. I've tried trial and error by splitting it into the 6 possible elements of $\Bbb F_2 $ but none of these seem to be roots. Any suggestions ?

Answer 1

Remember that in $\Bbb{F}_2(\alpha)$ one has $(a+b)^2=a^2+b^2$ and that $\alpha^3=\alpha+1$. So

$$\begin{align} g(\alpha^2+1)=&(\alpha^2+1)^3+(\alpha^2+1)^2+1\\ =&(\alpha^4+1)(\alpha^2+1)+(\alpha^4+1)+1\\ =&(\alpha^4+1)\alpha^2+1\\ =&\alpha^6+\alpha^2+1\\ =&(\alpha+1)^2+\alpha^2+1=0 \end{align}$$

The hint of Will Jagy is very powerfull. One has

$$x^3g({1\over x})=f(x)$$

By substituting $\alpha$ one gets

$$g({1\over \alpha})=0$$

Now $\alpha^3+\alpha+1=0$ and dividing by $\alpha^2$, one gets

$${1\over\alpha}=\alpha^2+1$$

Answer 2

What is $$ x^3 \; g \left( \frac{1}{x} \right) \; \; ? $$