Let $F$ be a field of characteristic 2 and $E/F$ be separable field extension of degree 2. Prove that there exists $a\in E$ s.t. $E=F(a)$ and $a^2+a\in F$.

I know that since $E/F$ has degree 2, for all $a\in E-F$, $m_a$ has degree 2.

So I think $x^2+x+1$ is minimal polynomial for some $a$ but I don't know how to do it.


1 Answer 1


Choose $a\in E-F$. Then the minimal polynomial of $a$ is of degree two. Since you are in a field of characteristic $2$, it has to be of the type $x^2+\alpha x + \beta $ where $ \alpha,\beta\in F$. The possibility $\alpha=0$ contradicts the separability of $E/F$, hence $\alpha\neq 0$. Then the element $a/\alpha$ does the job and we are done.

  • $\begingroup$ why $\alpha ,\beta$ is 0 or 1?? $\endgroup$
    – Arturo
    Nov 14, 2013 at 14:19
  • $\begingroup$ @JinyongGo Sorry I did not realize, the above proof is incorrect. I mistakenly assumed that you were working in a field with two elements. $\endgroup$
    – user90041
    Nov 14, 2013 at 14:21
  • $\begingroup$ @JinyongGo But may be the proof can be repaired, I guess we just need that $\alpha$ is non-zero. Then the element $a/\alpha$ does the job ? $\endgroup$
    – user90041
    Nov 14, 2013 at 14:23

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