I want to give an example of a non-separable, non-normal, finite field extension. So I have to find an extension of a field, which is not perfect. I would suggest $K=\mathbb{F}_2(t)$. An extension, that could work is $L=\mathbb{F}_2(t)[X]/(X^6-t)$. I already know, that $L/K$ is finite and that it is not separable. But I can't show that it is not normal. If I suppose that the polynomial $Z^6-t$ factors in $L[Z]$, then I can show that all sixth roots of unity are in $L$. But I cannot get a contradiction from this.


Using your example, let $K = \mathbb{F}_2(t), L = \mathbb{F}_2(t^{1/6})$. Consider the equation $$x^3 - t = 0$$ then it has a root in $L$, namely $t^{1/3}$. Then all the roots of this equation will be $t^{1/3}, \zeta t^{1/3}, \zeta^2 t^{1/3}$, where $\zeta$ is a primitive third root of unity. If $\zeta t^{1/3} \in L$, we must have $\zeta \in \mathbb{F}_2$, however $\mathbb{F}_2$ does not have a primitive third root of unity.

Hence $x^3-t = 0$ does not have all its root in $L$, note also that $x^3-t$ is irreducible over $K[x]$. This shows the extension $L/K$ is not normal.

  • $\begingroup$ Thank you. Could you please explain, why $\zeta\in\mathbb{F}_2$. I only see why it is in $L$ $\endgroup$ – tiefi Jan 28 '18 at 12:41
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    $\begingroup$ @tiefi If $\zeta \in L$ is a rational function $$\zeta = \frac{f(t^{1/6})}{g(t^{1/6})}$$ with coprime $f,g \in \mathbb{F_2}[t]$, and $\deg f$ or $\deg g \geq 1$. Then $\zeta^3 \neq 1$. Hence $\zeta$ must be a "pure constant", i.e., $\zeta \in \mathbb{F}_2$. $\endgroup$ – pisco Jan 28 '18 at 12:43

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