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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

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  • $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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