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Give an example of extension $K/F$ that is neither separable nor purely inseparable.

Take $F = \mathbb{Z}_{2}(x)$ and $K = F(\sqrt[6]{x})$. We can write $K = F(\sqrt{x},\sqrt[3]{x})$. Thus $(\sqrt{x})^{2^{1}} = x \in F$ so, $\sqrt{x}$ is purely inseparable over $F$. Also, $(\sqrt[3]{x})^{2^{n}} = x^{\frac{2^{n}}{3}}$ and $\frac{2^{n}}{3} \not\in \mathbb{N}$ so, $\sqrt[3]{x}$ is not purely inseparable. How I can show that $\sqrt[3]{x}$ is separable over $F$? I couldn't determine the minimal polynomial.

After that, $F(\sqrt{x})/F$ is purely inseparable and $F(\sqrt[3]{x})/F$ is separable, then $K/F$ is neither separable nor purely inseparable.

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  • $\begingroup$ The element $\root3\of x$ is a zero of $P(t)=t^3-x$. This is irreducible by Eisenstein (the prime $x$), so it is the minimal polynomial. The other zeros of $P(t)$ are $\alpha\root3\of x$ and $\alpha^2\root3\of x$, where $\alpha\in\Bbb{F}_4\setminus\Bbb{F}_2$ is a primitive third root of unity. $\endgroup$ – Jyrki Lahtonen Jun 23 '18 at 5:29
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The minimal polynomial of $\sqrt[3]{x}$ is $t^3-x$. This is separable since the degree, $3$, is prime to the characteristic, $2$.

Note that if $\sqrt[3]{x}$ satisfied a quadratic polynomial, you could write (with constants in $GF(2)(x)$) $\sqrt[3]{x^2}=A\sqrt[3]{x}+B$, so $x=A\sqrt[3]{x^2}+B\sqrt[3]{x}=(A^2+B)\sqrt[3]{x}+AB$ which, lest $\sqrt[3]{x}\in GF(2)(x)$ results in $A^2=B$ and $AB=x$ in $GF(2)(x)$. But then $\sqrt[3]{x}=A\in GF(2)(x)$ which is preposterous.

Note that I am writing $GF(2)$ for the field with two elements, which is I assume what you are asking about. The $2$-adic integers $\Bbb{Z}_2$ are not a field, and they have characteristic zero, so I assume that is not what you had in mind.

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The minimal polynomial of $\sqrt[3]x$ is $t^3-x$, which is irreducible (by an elementary argument), and its derivative $3t^2 \ne 0$, so it is separable.

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  • $\begingroup$ You don’t even need to show $t^3-x$ is irreducible, because it is coprime to its derivative $t^2$. $\endgroup$ – egreg Jun 23 '18 at 5:52
  • $\begingroup$ On the contrary, the OP really does need that the minimal polynomial has degree greater than 1. Otherwise he couldn't stack a purely inseparable extension on top of it and get an extension that is neither separable nor purely inseparable. $\endgroup$ – C Monsour Jun 23 '18 at 6:23

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