TITLE: Is the following field extension normal $\mathbb Q(\!\sqrt {2+\sqrt 2}): \mathbb Q$
QUESTION [5 upvotes]: I would like to know if $\mathbb Q \left[\sqrt {2+\sqrt 2}\right ]: \mathbb Q$ is normal .

The roots of the minimal polynomial 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.

REPLY [2 votes]: 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 $\mathbb Q(\zeta_{16} + \zeta_{16}^{-1}).$