# Unique quadratic subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}(\sqrt{p})$ if $p \equiv 1$ $(4)$, and $\mathbb{Q}(\sqrt{-p})$ if $p \equiv 3$ $(4)$

I want to prove the assertion:

The unique quadratic subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}(\sqrt{p})$ when $p \equiv 1 \pmod{4}$, respectively $\mathbb{Q}(\sqrt{-p})$ when $p \equiv 3 \pmod{4}$.

My first attempt is this. In $\mathbb{Z}[\zeta_p]$, $1-\zeta_p$ is prime and $$p = \epsilon^{-1} (1-\zeta_p)^{p-1}$$ where $\epsilon$ is a unit. Since $p$ is an odd prime, $(p-1)/2$ is an integer and $$\sqrt{\epsilon p } = (1-\zeta_p)^{(p-1)/2}$$ makes sense and belongs to $\mathbb{Z}[\zeta_p]$.

How do I deal with the $\epsilon$ under the square root? I guess the condition on the congruence class of $p$ comes from that. Is this even the right way to proceed?

Uniqueness is not clear to me either. I thought about looking at the valuation $v_p$ on $\mathbb{Q}$, extending it to two possible quadratic extensions beneath $\mathbb{Q}(\zeta_p)$, then seeing how those have to extend to common valuations on $\mathbb{Q}(\zeta_p)$, but I didn't see how to make it work.

I would appreciate some help.

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LaTeX note: \pmod{p} will automatically produce "$\pmod{p}$" with appropriate spacing and all. – Arturo Magidin Mar 31 '11 at 13:14
the galois group is cyclic so there is a unique quadratic subfield – yoyo Mar 31 '11 at 16:09

To show that the quadratic subfield is $\mathbb{Q}(\sqrt{\pm p})$, remember that only $p$ can ramify in $\mathbb{Q}(\zeta_p)$.

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See this exercise sheet for a more or less guided solution. Re uniqueness: what is the Galois group of the cyclotomic field? What does the Galois correspondence tell you?

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My favorite way to prove this is to explicitly write down the quadratic Gauss sum: $$g_p = \sum_{a \in \mathbb F_p} \left( \frac{a}{p} \right) \zeta_p^{a}$$ Then you can show $g_p^2 = (-1)^{\frac{p-1}{2}} p$ by direct manipulation. This gives the result very explicitly!

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Recall that the discriminant of $x^n + b$ is $(-1)^{n(n-1)/2}n^n b^{n-1}$. You can calculate this using the Vandermonde determinant and some facts about power sums. If we substitute in $n = p$ and $b = -1$ we get the discriminant of $x^p - 1$ is $(-1)^{p(p-1)/2}p^p$. Now we already know that $\Bbb{Q}(\zeta_p)$ is a splitting field for $x^p - 1$. Now recall that $$\sqrt{(-1)^{p(p-1)/2}p^p} = (-1)^{p(p-1)/2} \prod_{i < j} (\alpha_i - \alpha_j)$$

so that the expression on the left hand must be in the splitting field $\Bbb{Q}(\zeta_p)$. It follows that $\Bbb{Q}(\sqrt{(-1)^{p(p-1)/2}p^p}) =\Bbb{Q}(p^{(p-1)/2}\sqrt{(-1)^{p(p-1)/2}p})$ is a subfield of $\Bbb{Q}(\zeta_p)$. Now it is easy to see that $$\Bbb{Q}(p^{(p-1)/2}\sqrt{(-1)^{p(p-1)/2}p}) = \Bbb{Q}(\sqrt{(-1)^{p(p-1)/2}p}).$$

Hence if $p \equiv 1 \mod 4$, $(-1)^{p(p-1)/2} =1$ so that $\Bbb{Q}(\sqrt{p})$ is a subfield of $\Bbb{Q}(\zeta_p)$ and if $p \equiv 3 \mod 4$ then $(-1)^{p(p-1)/2} =-1$ so that $\Bbb{Q}(\sqrt{-p})$ is a subfield of $\Bbb{Q}(\zeta_p).$

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