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Consider the extensions $\Bbb F_2(\alpha)$ and $\Bbb F_2(\beta)$ over $\Bbb F_2$ where $\alpha$ and $\beta$ are roots of the polynomials $f(x)=x^3+x+1$ and $g(x)=x^3+x^2+1\in\Bbb F_2[x]$ respectively

The problem is to establish by finding an explicit isomorphism that $\Bbb F_2(\alpha)\cong\Bbb F_2(\beta)$

We have: $$\alpha^3+\alpha+1=0\implies\alpha^3=-(\alpha+1)=\alpha+1\in\Bbb F_2$$ and $$\beta^3+\beta^2 +1=0\implies\beta^3=-(\beta^2 +1)=\beta^2 +1\in\Bbb F_2$$

I don't really know what to do with that.

So going a different way, we observe that $\deg f=\deg g=3$ and they are bith irreducible in $\Bbb F_2$ (that is given to us). So $[\Bbb F_2(\alpha):\Bbb F_2]=[\Bbb F_2(\beta):\Bbb F_2]=3$ so looking at those extensions as vector spaces of dimension $3$ over a field of cardinal $2$, they are both of cardinal $2^3=8$ so they are isomorphic since all fields of the same finite cardinal are isomorphic.

Now to explicitly give the isomorphism, and I'm not sure about this, but does this work:

\begin{equation} \phi(u) = \begin{cases} u & \text{if $u\in\Bbb F_2$}\\ {\beta u\over\alpha}=\beta\alpha^{-1}u & \text{if $u\notin\Bbb F_2$}\\ \end{cases} \end{equation}

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HINT:

$\alpha^3 + \alpha + 1=0$ implies $1 + \frac{1}{\alpha^2} + \frac{1}{\alpha^3}=0$, so $\frac{1}{\alpha}$ is a root of $1 + x^2 + x^3$. So the isomorphism should take $\frac{1}{\alpha}$ to $\beta$, and so $\alpha$ to $\frac{1}{\beta}$. Note that $\frac{1}{\beta}=\beta^2 + \beta$.

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  • $\begingroup$ Why ${1\over\beta}=\beta^2+\beta$? $\endgroup$ – John Cataldo May 6 '18 at 11:57
  • $\begingroup$ @John:Cataldo Oh, because $\beta(\beta^2 + \beta)=1$ $\endgroup$ – Orest Bucicovschi May 6 '18 at 11:59
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As you already said, since the degree of both minimal polynomials $p,q$ is $3$,
$\mathbb{F}[X]/(p)$ and $\mathbb{F}[X]/(q)$ are both $\cong$ $\mathbb{F}_{2^3}$. So you have finite dimensional vector spaces over $\mathbb{F}_2$ and an isomorphism $$\phi:\mathbb{F}_2[X]/(p) \rightarrow \mathbb{F}_2[X]/(q)$$ between them is completely determined when explained on their basis $\{1,\alpha,\alpha^2\},\{1,\beta,\beta^2\}$ respectively (like in Linear Algebra). You can then easily see that $\phi|\mathbb{F}_2\equiv \text{id}$ (this is a general fact of field extensions, any automorphism $\phi \in Aut(K)$, where $K$ is a field extension of a prime field $P$ leaves $P$ fixed ).

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  • $\begingroup$ What do you mean by $\phi\mid\Bbb F_2$? $\endgroup$ – John Cataldo May 6 '18 at 11:12
  • $\begingroup$ Oh ok I get it now $\endgroup$ – John Cataldo May 6 '18 at 11:13
  • $\begingroup$ no worries, I should have been more explicit: I meant the restriction of $\phi$ to $\mathbb{F}_2$ $\endgroup$ – Simonsays May 6 '18 at 11:16

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