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where zeta_8 is the 8th root of unity over Q and p is an odd prime integer.

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You can write $\zeta_8$ explicitly – Thomas Andrews Nov 7 '12 at 22:14

You can write down all the quadratic subfields of $\mathbb{Q(\zeta_8)}$ and show that none of them are $\mathbb{Q}(\sqrt{p})$.

More explicitly, this field contains $i = \zeta_8^2$. It also contains $\sqrt{2} = \zeta_8 + \frac{1}{\zeta_8}$, where I've used that $$ \zeta_8 = \frac{\sqrt{2}}{2}(1 + i) $$

And it contains $\sqrt{-2}$ by using the above formula and the fact that it contains $i$. So our field contains $\mathbb{Q}(\sqrt{2}), \mathbb{Q}(i)$, and $\mathbb{Q}(\sqrt{-2})$. Since our field is Galois over $\mathbb{Q}$ of degree 4, its Galois group must be cyclic or Klein 4; we've found three subgroups of order 2, so it's the Klein 4 group and in particular we've found all the quadratic subfields.

(If you know some algebraic number theory, this argument can be made simpler by appealing to the fact that no odd primes ramify in $\mathbb{Q}(\zeta_8)$.)

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