TITLE: Homework: No field extension is "degree 4 away from an algebraic closure"
QUESTION [5 upvotes]: Question: Suppose $[L:K]=4$ and char$K \neq 2$ 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 have successfully found such $M$. Would somebody please give me some hints to the last part?

REPLY [2 votes]: There is a $i\in M$ such that $i^2=(-1)$. Since
$[L:M]=2$, there is a $d\in M$ such that $L=M(\sqrt{d})$. 
Since $L$ is algebraically closed, there is a $\alpha\in L$
such that $\alpha^2=\sqrt{d}$. 
You can write $\alpha=x+y\sqrt{d}$ with $x,y\in M$. Then
$\alpha^2=(x^2+dy^2)+(2xy)\sqrt{d}$, and
$x^2+dy^2=0, 2xy=1$ (why?).
Now, $y\neq 0$ (why?), so $(\frac{x}{y})^2=-d$, $(i\frac{x}{y})^2=d$.
Notice that $i\frac{x}{y} \in M$, and deduce a contradiction from this.