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I would like to know if $\mathbb Q \left[\sqrt {2+\sqrt 2}\right ]: \mathbb Q$ is normal . The roots of the minimal polynomail is $\pm\sqrt {2\pm\sqrt 2}$ .

Now the thing that i have really tried and have no idea to get is to write $\sqrt {2-\sqrt 2}$ as the combination of the powers of $\sqrt {2+\sqrt 2}$

If at all it is possible ??

What are the possible ways of finding the coefficients ?

Thanks for you help .

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3 Answers 3

up vote 7 down vote accepted

$Q(\sqrt{2+\sqrt{2}})$ contains $\sqrt{2}$ and

$\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt{2}}=\sqrt{2}$ so $\sqrt{2-\sqrt{2}}\in Q(\sqrt{2+\sqrt{2}})$

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As an incidental remark, this field (which is a normal extension of $\mathbb Q$, as the other answers show) is in fact the maximal totally real subfield of $\mathbb Q(\zeta_{16})$ (where $\zeta_{16}$ denotes a primitive $16$th root of unity).

In fact, $\zeta_{16} + \zeta_{16}^{-1} = \sqrt{2 + \sqrt{2}},$ and so the field in question is indeed equal to $\mathbb Q(\zeta_{16} + \zeta_{16}^{-1}).$

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Well, $\sqrt{2+\sqrt2}\cdot\sqrt{2-\sqrt2} = \sqrt{4-2}=\sqrt2$.

Let $\alpha:= \sqrt{2+\sqrt2}$, then $\sqrt2=\alpha^2-2$, so $$\sqrt{2-\sqrt2}=\frac{\alpha^2-2}{\alpha} = \alpha-\frac2\alpha $$ Can you express $\frac2\alpha$? (Use the minimal polynomial.)

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