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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?

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  • $\begingroup$ Can you please explain how you construct $M$ ? $\endgroup$ – Georges Elencwajg Feb 6 '14 at 13:18
  • $\begingroup$ @GeorgesElencwajg : He is offline from almost $15$ days and i have no hope that he would respond soon... I am also thinking on same thing but with no success.... :( $\endgroup$ – user87543 Feb 6 '14 at 13:24
  • $\begingroup$ Thanks for answering, @Praphulla. $\endgroup$ – Georges Elencwajg Feb 6 '14 at 17:55
  • $\begingroup$ @GeorgesElencwajg $L:K$ is Galois by the hypotheses. $\endgroup$ – user71815 Feb 6 '14 at 21:31
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    $\begingroup$ @GeorgesElencwajg Okay, my Galois theory is rusty, but let me try: $L$ is the algebraic closure of $K$, so $L:K$ is normal. $[L:K]$ and the fields' characteristic are coprime, so $L:K$ is separable. $\endgroup$ – user71815 Feb 6 '14 at 22:19
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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.

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  • $\begingroup$ Dear Ewan, you implicitly suppose that $i\notin K$. Why is that? $\endgroup$ – Georges Elencwajg Feb 6 '14 at 12:45
  • $\begingroup$ @GeorgesElencwajg Dear Georges, my original presentation indeed gave that mistaken impression. I just corrected that, thanks. In fact, once $M$ has been found (and the OP claims to have already dealt with that so I leave that part aside) $K$ leaves the scene and does not play any role in the rest of the proof. $\endgroup$ – Ewan Delanoy Feb 6 '14 at 13:13
  • $\begingroup$ Ah, thanks Ewan. Now I'll ask the OP how he justifies his claim ! $\endgroup$ – Georges Elencwajg Feb 6 '14 at 13:17

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