Consider the following question:

(a) The following conditions on a field K are equivalent :So, Prove them

(i) every irreducible polynomial in K[x] is separable ;

(ii) every algebraic closure K of K is Galois over K;

(iii) every algebraic extension field of K is separable over K;

(iv) either char K = 0 or char K = p and $K = K^p$.

(i)=>(ii) This part I am not able to do. What i am missing is how to prove that for all $\sigma \in Aut_K{\bar K}$ only y$\in K$ gets fixed and rest all are not fixed.

(ii)=> (iii) I have done as if an algebraic extension is galois over K and then the extension is also separable over K ( This is a theorem).

(iii)=>(iv) I tried by assuming that char K $\neq 0$, so, As field is integral domain, so Characterstic must be prime and so, let char K=p but then I am not able to move towards proving $K=K^{p}$.

(iv)=>(i) I have solved . So, I don't require help with that.

Kindly shed some light on this

  • $\begingroup$ What are you allowed to use? That $f$ is separable iff $gcd(f,f')=1$? $\endgroup$ Jan 22, 2021 at 14:43
  • $\begingroup$ See also the answer of this post. $\endgroup$ Jan 22, 2021 at 14:50
  • $\begingroup$ @DietrichBurde Yes, I am allowed to use that. $\endgroup$
    – Avenger
    Oct 17, 2021 at 15:59

1 Answer 1


For (i)=>(ii): Let $L$ be an algebraic closure of $K$. Then $L/K$ is by definition normal and by assumption separable. Depending on your definition of Galois-extension we might be done here. If your definition of Galois extension is $K=L^{Aut_KL}$, then look at the answer to one of your previous questions.
For (iii)=>(iv): If $a\in K$ there is some $b\in L$ in an extension field with $b^p=a$. Now look at the irreducible factors (over $K$) of the polynomial $x^p-a=(x-b)^p$ and use your assumption.

  • $\begingroup$ Why is in $(iii) \to (iv) $ there exists some $b\in L$ in an extension field with $b^p =a$? can you please tell? $\endgroup$
    – Avenger
    Oct 17, 2021 at 6:17

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.