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.

  • $\begingroup$ 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

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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.