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Definition : Let $K$ be a field with $char(K)=0$ , let's define $ K^{quad}\subset \overline{K}$ by: $$ K^{quad} = \bigcup\limits_{i \geqslant 1} {K_i } $$ where $$ K_1 = K $$ $$ K_{i + 1} = K_i(\sqrt{K_i}) $$

and where $K_i(\sqrt{K_i})$ means the field generated over $K_i$ by all the roots of $K_i$ i.e elements in an algebraic closure of $K_i$ such that $x^2 \in K_i $

Prove that if $K\subset \Bbb C$ is such that it's closed under conjugation, then $K^{quad}$ is also closed under conjugation.

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Induction on $i$? –  Jyrki Lahtonen Dec 6 '12 at 15:43
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1 Answer 1

If $K$ is closed by conjugation, so is $K(\sqrt K)$, because for any $x \in K$, the conjugates of the square roots of $x$ are precisely the square roots of the conjugates of $x$ :

$y^2 - \overline{x} = \overline{\overline{y}^2 - x}$, hence $y$ is a root of $X^2- \overline{x}$ if and only if $\overline{y}$ is a root of $X^2 - x$.

The elements of $K(\sqrt{K})$ whose conjugates are still in $K(\sqrt{K})$ form a subfield of $K(\sqrt{K})$ containing $K$ (because it is closed by conjugation) and $\sqrt{K}$ (from the argument above), thus it is $K(\sqrt{K})$

Thus, by induction, every $K_i$ is closed by conjugation, and thus so is their reunion, $K^{quad}$

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