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Let $\zeta \in \mathbb{C}$ a primitive $p^{th}$ root of unit ($\zeta^p=1$ holds and no smaller power works) with $p$ an odd primeand assume $p > 2$. Consider $E = \mathbb{Q}(\zeta)$. This cyclotomic extension is a Galois extension with $$ G = \mathrm{Gal}(E/\mathbb{Q}) = \left( \mathbb{Z} / p \mathbb{Z} \right)^{\times}$$ so $G$ is the multiplicative group $$ G = \left( \{ 1 , 2 , ... , p-1 \} , \cdot \bmod p \right)$$. Let $H$ be a subgroup of $G$ of order 2. Define $$ \alpha = \sum_{i \in H} \zeta^i \quad \mbox{ and } \quad \beta = \sum_{i \in G-H} \zeta^i \ .$$ I showed that $\alpha$ and $\beta$ are fixed under $H$ and that $\alpha$ and $\beta$ are the roots of $x^2 + x + \alpha \beta \in \mathbb{Q}[x]$.

I want to calculate $\alpha \beta$ and from that deduce that the fixed field of $H$ (that is, $E^H$) is $E^H = \mathbb{Q}(\sqrt{p})$ when $p = 1 \bmod 4$ and $E^H = \mathbb{Q}(\sqrt{-p})$ when $p = 3 \bmod 4$.

We have that $H = \{ \pm 1 \}$ since $(p-1)^2 \stackrel{p}{\equiv} (-1)^2 = 1$ and thus it is an element of order $2$. I have calculated $\alpha = \zeta + \zeta^{p-1} = \zeta + \zeta^{-1}$ and $\beta = \zeta^2 + ... + \zeta^{p-2}$ by multiplying them and thus use the formula for geometric series but did not reach a simple expression, which relates to the second part of the question.

I would appriciate help.

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I don't think that $\alpha \beta \in \Bbb Q$. Are you sure you want to pick $H$ of order $2$ and not of index $2$ ? – mercio Dec 8 '12 at 15:10
This is how the problem appears. Note that $\alpha + \beta = -1$ and that for any $j \in G-H$ we have $\alpha^j = \beta$ and $\beta^j = \alpha$. Since $$\forall k \in G : (\alpha \beta)^k = \alpha \beta$$ it must be in the fixed field $E^G = \mathbb{Q}$. – LinAlgMan Dec 8 '12 at 15:44
As mercio says, there is some serious confusion going on in your post between "index" and "order" of a subgroup. I recommend that you first review the statement of the Galois correspondence and then your post. – Alex B. Dec 10 '12 at 19:17

The easiest way I know of uses ramification. Since $\Bbb Q(\zeta_p)$ is totally ramified at $p$, any sub-extension (including the unique, quadratic sub-field) is ramified only at $p$. Then since we know that the ramification is completely controlled by the discriminant, and that the discriminant of $K=\Bbb Q(\sqrt m)$ for $m$ square-free (without any loss of generalization we may assume this is the case) is

$$\Delta_K =\begin{cases} m & m\equiv 1\mod 4 \\ 4m & m\equiv 2,3\mod 4\end{cases}$$

Then since $2$ is ramified for any $m\not\equiv 1\mod 4$ we see that $p^*=(-1)^{(p-1)/2}p\equiv 1\mod 4$, so that the quadratic sub-field is $\Bbb Q(\sqrt{p^*})$, as desired.

If you don't like ramification technology--which is by far the cleanest and most beautiful way to see the result--you can rely on fossils like Gauß sums, which is an oldie, but a goodie, but other than that there's not a really nice, qualitative reason.

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