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I have seen this problem asked by another user but it isn't completely solved in the answers. I'm trying to do it, but I can't.

Question: Suppose $[L:K]=4$ and $charK≠2$ and $L$ is algebraically closed. Show that there is an intermediate field M such that $[L:M]=2$ and that $X^2+1$ splits over $M$. Show that this leads to a contradiction.

I can't show that this M exists, and for this reason, I can't follow with the other parts.

Thanks in advanced.

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  • $\begingroup$ Can you give the link to the other Question? It might help us to isolate a more specific question that you need help answering. $\endgroup$ – hardmath Feb 6 '14 at 12:44
  • $\begingroup$ What does it mean to say that $[L:K]=4$? $\endgroup$ – John Habert Feb 6 '14 at 12:50
  • $\begingroup$ @JohnHabert : I am not able to write properly.. I know that by $[L:K]=4$ $L$ is a $4$ dimensional vector space when viewed with base field $K$ but then i am not able to write precisely in terms of minimal polynomials or galois theory terms... $\endgroup$ – user87543 Feb 6 '14 at 13:20
  • $\begingroup$ This is the link. math.stackexchange.com/questions/615604/… I don't know if the question is correct. Maybe not. I'm with you John. I don't have enought hipotesis... you can't use normality, separability, finite field... nothing. $\endgroup$ – Carles Pérez Feb 6 '14 at 15:42
  • $\begingroup$ And there are other problem. It's impossible to give a counterexample because the last question is "show that hypothesis are impossible"... $\endgroup$ – Carles Pérez Feb 6 '14 at 15:51
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Case 1: $x^2+1$ does not split over $K$. Simply take $M$ as the splitting field of $x^2+1$.

Case 2: $x^2+1$ splits over $K$, so I only need to find some extension $M/K$ of degree $2$. $L/K$ is a finite Galois extension because $L$ is algebraically closed. $\operatorname{Gal}(L/K)$ has order $4$, hence it has an element $\sigma$ of order two. You can take $M$ as the field fixed by $\sigma$.

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