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Let $E$ be a field with characteristic not equal to $2$. Let $F$ be an extension field of degree $2$. Prove that $F=E(\alpha)$ with $Irr(\alpha,E)=x^2-b$. Show that this is false for some field $E$ of characteristic $2$.

My thought is let $p(x) = Irr(\alpha,E)$ and $x^2-b$ should be equal to $(x-\alpha)q(x)$ so $q(x)$ must has degree $1$. If the characteristic of E is not equal to $2$ then $p'(x) \neq 0$ so $\alpha$ is separate over F. Am I correct? And what's next? I was stuck here. Can someone please help me? Thanks.

Bérénice
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jacky123456789
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  • Hint: $\alpha$ satisfies a quadratic, which in fields of characteristic not 2 can be solved by the quadratic formula. Think of the formula. – J.G Jun 01 '16 at 02:39
  • It would improve your Question to explain (define) the notation $Irr (\alpha,E)$. It seems to mean the minimal (irreducible) polynomial over $E$ satisfied by $\alpha$. – hardmath Jun 01 '16 at 04:21

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If the characteristic of $E$ is not $2$, let $b \in F - E$, then $\{1,b\}$ is an $E$-basis of $F$. So $b$ satisfies a degree two polynomial $b^2+ub+v=0$ where $u,v \in E$. The characteristic of $E$ is not $2$, so $\frac{u}{2}$ makes sense and $(b+\frac{u}{2})^2=b^2+ub+\frac{u^2}{4}=\frac{u^2}{4}-v$, let $a=b+\frac{u}{2}\in F - E$, $a^2 \in E$ and $F=E(a)$. So $F=E(a)$ and $Irr(a,E)=x^2-a^2$ with $a^2 \in E$.

For the counter example, consider $\mathbb{F}_2[X]/(X^2+X+1)$ extension of $\mathbb{F}_2$, as $(u+vX)^2=(u^2+v^2)+v^2X$ is in $\mathbb{F}_2$ if and only if $v=0$. You should be able to conclude from there.

Bérénice
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