$\mathbb{Q}(\sqrt{-23}) \subseteq \mathbb{Q}(\zeta_{23})$ I want to prove $\mathbb{Q}(\sqrt{-23}) \subseteq \mathbb{Q}(\zeta_{23})$ using the hint from "Introduction to Cyclotomic Fields" Exercise 2.1. $\zeta_{23}$ is the primitive 23-th root of unity.
I know that $p=23$ is the only prime ramified in $\mathbb{Q}(\zeta_{23})$ hence in $\mathbb{Q}(\sqrt{d})$, thus we have $d=\pm 23$. And I'm wondering how to show $d=-23$ instead of $d=23$. The book says when $p \equiv 1 \pmod 4$ we choose $+$, other we choose $-$. I don't know how to prove it.
Thanks for helps!
 A: I'll just finish @Chris_H's comment, which requies only Galois theory, not ramification theory.
Since the Galois group is isomorphic to $\mathbb F_p^{\times}$ which is cyclic, we have there is a unique subgroup of order $\frac{p-1}{2}$, or in other words, there is a unique subgroup of index $2$, hence there is a unique subextension of degree $2$.
Similarly, there is a unique subgroup of order $2$, or equivalently there is a unique element of order $2$ in $\mathbb F_p^{\times}$. This element must be the complex conjugate from the Galois group. The question about whether $\sqrt{p}$ or $\sqrt{-p}$ is in $\mathbb Q(\zeta_p)$ (assuming one of the two holds, which can be established through ramification theory) reduces to the problem of whether the complex conjugate lies in the unique subgroup of order $\frac{p-1}{2}$, which fixes the unique subextension of order $2$ through the Galois correspondence.
If $p\equiv 1\mod 4$, then $\frac{p-1}{2}$ is even, the subgroup contains an element of order $2$ by Cauchy. While when $p\equiv 3\mod 4$, $\frac{p-1}{2}$ is odd, and the subgroup of such order cannot contain an element of order $2$.
A: For an odd prime $p$, the only prime that ramifies in $\mathbf Q(\zeta_p)$ is $p$, so in a quadratic subfield $K$ the only prime that could ramify is $p$.  Some prime ramifies in $K$, so $p$ is the unique prime ramifying in $K$.
In $\mathbf Q(\sqrt{d})$ for squarefree $d$, each prime factor of $d$ ramifies in this field, so $d = \pm p$. If $p \equiv 1 \bmod 4$ then $\mathbf Q(\sqrt{p})$ has discriminant $p$ and $\mathbf Q(\sqrt{-p})$ has discriminant $4p$. If $p \equiv 3 \bmod 4$ then $\mathbf Q(\sqrt{p})$ has discriminant $4p$ and $\mathbf Q(\sqrt{-p})$ has discriminant $p$.  Thus a quadratic field whose only ramified prime is $p$ is $\mathbf Q(\sqrt{p})$ when $p \equiv 1 \bmod 4$ and is $\mathbf Q(\sqrt{-p})$ when $p \equiv 3 \bmod 4$.
Thus $\mathbf Q(\sqrt{p}) \subset \mathbf Q(\zeta_p)$ when $p \equiv 1 \bmod 4$ and $\mathbf Q(\sqrt{-p}) \subset \mathbf Q(\zeta_p)$  when $p \equiv 3 \bmod 4$, so $\sqrt{p} \in \mathbf Q(\zeta_p)$ when $p \equiv 1 \bmod 4$ and $\sqrt{-p} \in \mathbf Q(\zeta_p)$ when $p \equiv 3 \bmod 4$. In particular, $\sqrt{-23} \in \mathbf Q(\zeta_{23})$.
A more explicit argument for $\sqrt{p}$ or $\sqrt{-p}$ lying in $\mathbf Q(\zeta_p)$ uses Gauss sums: set
$$
G_p = \sum_{a = 1}^{p-1} \left(\frac{a}{p}\right)\zeta_p^a
$$
where $(\frac{a}{p})$ is a Legendre symbol. This lies in $\mathbf Q(\zeta_p)$, and Gauss showed
$$
G_p^2 = \left(\frac{-1}{p}\right)p, 
$$
so $G_p^2 = p$ when $p \equiv 1 \bmod 4$ and $G_p^2 = -p$ when $p \equiv 3 \bmod 4$. Thus $G_p = \pm\sqrt{p}$  when $p \equiv 1 \bmod 4$ and $G_p = \pm\sqrt{-p}$ when $p \equiv 3 \bmod 4$.
Gauss also proved a much more delicate result: when $\zeta_p$ is the specific complex $p$th root of unity $e^{2\pi i/p}$, $G_p = \sqrt{p}$ (positive square root) when $p \equiv 1 \bmod 4$ and $G_p = i\sqrt{p}$ (a specific choice of square root of $-p$) when $p \equiv 3 \bmod 4$, but this is not needed in order to show $\sqrt{p}$ or $\sqrt{-p}$ is in $\mathbf Q(\zeta_p)$ depending on $p \bmod 4$.
A: Otherwise simply check that $$(-1)^{23\cdot 22/2}\prod_{0\le i< j<23} (\zeta_{23}^i- \zeta_{23}^j)^2=
\prod_{0\le i< j<23} (\zeta_{23}^i- \zeta_{23}^j)(\zeta_{23}^j- \zeta_{23}^i)$$ $$=
\prod_{i\ne j} (\zeta_{23}^i- \zeta_{23}^j)
 = \prod_{i=0}^{22} \zeta_{23}^i \prod_{n=1}^{22}(1-\zeta_{23}^n)
 = 23^{23}$$
A: also, if $x^{23} = 1$  but $x \neq 1,$ take
$$z = x + x^2 + x^3  + x^4 + x^6  + x^8 +x^9 + x^{12} + x^{13} + x^{16} + x^{18} $$
Then
$$   z^2 + z + 6  = 0  $$
because gathering the same exponents mod 23 comes out to
$$z =6 \left( 1 + x + x^2 + x^3  + x^4 + x^5 + x^6 +x^7  + x^8 +x^9 + x^{10} + x^{11} +x^{12} + x^{13} + x^{14} + x^{15} + x^{16} + x^{17} +  x^{18} + x^{19} + x^{20} + x^{21} +x^{22}
 \right) $$
A: The only quadratic field where 23 is the only ramified prime is $\mathbb{Q}(\sqrt{-23})$, since it's discriminant is 23. You cannot take $\mathbb{Q}(\sqrt{23})$, because 2 is also ramified in there since its discriminant is $4\times 23$.
