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Let $E$ be a subfield of $\mathbb{C}$ and Let $\overline{E}=\{\overline{z} \, |\, z \in E \}$ with $\overline{z}$ being the complex conjugate of $z$. Let $K$ be a subfield of $\mathbb{C}$ with $\overline{K}=K$ and $w\in \mathbb{C}$ and $w^2 \in K$. Is (when is) $\overline{K(w)}=K(w)$?

I have no idea how to show it and would be thankful for hints (please no solutions at this point). Some things I know: Let $w\notin K$. Since $w\in \mathbb{C}$ and $w^2 \in K \Rightarrow [K(w):K]=2 \Rightarrow$ Minimalpolynomial of $w$ over $K$ has degree $2$. I also know that $\mathbb{C}$ is algebraic closed.

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Hint(s): You know that $[K(\omega):K]=2$- what does that mean for what the elements of $K(\omega)$ will look like?- also, notice that the function that takes conjugates has some nice algebraic properties. Finally, if $\omega$ is a root of a quadratic over $K$, say, $x^2+ax+b$ (with $a$ and $b\in K$) what can you say about $\overline\omega$?
(extra hint below)

What field can you find $\overline\omega$ in?- look at the specific fact that $\omega^2\in K$ and find a condition on $\omega$'s components

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First of all thank you for taking the time and helping:

1.) $[K(w):K]=2 \Rightarrow \{1,w\}$ is $K$-basis of $K(w)$. Every element of $K(w)$ can be written as $a+bw$ with $a,b\in K$.

2.) The function that takes conjugates is a homomorphism of field. It's easy to verify that $\overline{a+b}=\overline{a}+\overline{b}$ and $\overline{a\cdot b}=\overline{a}\cdot \overline{b}$. This might also mean that the function implicates an extension from $K(w)\rightarrow \mathbb{C}$ because $\overline{K}=K$?

3.) If $w$ is a root of that polynomial, isn't $\overline{w}$ also a root since because $a,b \in K$ and $\overline{K}=K$?

From here on I just don't know what to do...

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    $\begingroup$ For item 3: Consider the case $K=\Bbb{Q}(i)$. Let $w=\sqrt{2+i}$ (to be specific, let's use the square root in the first quadrant). Then $w^2\in K$. Furthermore, $w$ is a zero of the polynomial $x^2-(2+i)$ (which has coefficients in $K$). But $\overline{w}$ is not a zero of that polynomial. Instead, it is a zero of the polynomial $x^2-(2-i)$. $\endgroup$ – Jyrki Lahtonen Jan 17 at 6:43

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