# No field extension is “degree 4 away from an algebraic closure”

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.

• What does it mean to say that $[L:K]=4$? – John Habert Feb 6 '14 at 12:50
• @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... – user87543 Feb 6 '14 at 13:20
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 $\sigma$.