Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Induction on $i$? – Jyrki Lahtonen Dec 6 '12 at 15:43

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, $K^{quad}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.