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I need to decide if this following statement is true or false: $\forall K \text{ field (not algebraically closed) } \exists L,K\leq L: [L:K]=2$

For finite fields this statement should be true, but I have no idea if this is true for all $K$.

mast
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1 Answers1

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It is false.

The complex constructible numbers are not algebraically closed, but they are quadratically closed (everything has a square root).

mr_e_man
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  • Oh cool, thanks! – mast Mar 28 '22 at 17:15
  • More generally, if $K$ is any field (let's work in characteristic $\neq2$, so that quadratic extensions are necessarily Galois), you can form the quadratic closure $K^{quad}$ of $K$ inside a separable closure by the analogous procedure, which has no quadratic extensions itself. The Galois group $\mathrm{Gal}(K^{quad}/K)$ is the maximal pro-$2$ quotient of the absolute Galois group $\mathrm{Gal}K$. Thus, $\mathrm{Gal}{K^{quad}}$ possesses no $2$-group quotients, which translates to $K^{quad}$ having no quadratic extensions. – Thorgott Mar 28 '22 at 17:21