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I want to show:

Let $F$ be a field with characteristic 2. If $E/F$ is a separable extension with degree 2, then $E= F(\alpha)$ for some $\alpha\in F$ such that $\alpha^2+\alpha\in F$.

My attempt: As $E/F$ is a finite extension, let $\alpha\in E\setminus F$ so that $E = F(\alpha)$. Consider the minimal polynomial $m_\alpha$. Since $m_\alpha$ splits in $E$, $m_\alpha = (x-\alpha)(x-(m\alpha+n))$ where $m,n\in F$. As $m_\alpha\in F[x]$, $x^2-(m\alpha+\alpha+n)x+\alpha(m\alpha+n)\in F[x]$ so in particular, $(m+1)\alpha+n\in F$ so that $m=-1$. Hence, the form should be $(x-\alpha)(x-(\alpha+n))$. Here's where I'm stuck. How can I show $n=1$ is possible? Any comment will be appreciated.

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Since $\alpha$ is separable, $n\neq 0$. Now set $\beta=\dfrac{\alpha}{n}$. Then $E=F(\alpha)=F(\beta)$, and $\beta^2+\beta=\dfrac{\alpha^2+n\alpha}{n^2}=\dfrac{\alpha(\alpha+n)}{n^2}\in F$.

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  • $\begingroup$ Oh, that's a nice trick! thanks $\endgroup$ – love_sodam Oct 16 '20 at 7:53

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