How I can show that $\sqrt[3]{x}$ is separable over $\mathbb{Z}_{2}(x)$?

Question

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

Take $F = \mathbb{F}_{2}(x)$ 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.

Answer 1

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.

Answer 2

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.