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}



$\alpha^3 + \alpha + 1=0$ 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 $\frac{1}{\beta}=\beta^2 + \beta$.

  • $\begingroup$ 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

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 ).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.