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.

  • $\begingroup$ 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 $2$ ? $\endgroup$ – mercio Dec 8 '12 at 15:10
  • $\begingroup$ 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 $E^G = \mathbb{Q}$. $\endgroup$ – LinAlgMan Dec 8 '12 at 15:44
  • $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.