<s>[INST] $\mathbb{Q}(\sqrt{-23}) \subseteq \mathbb{Q}(\zeta_{23})$ 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!
 [/INST] 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
$$z =6 \left( 1 + x + x^2 + x^3  + x^4 + x^5 + x^6 +x^7  + x^8 +x^9 + x^{10} + x^{11} +x^{12} + x^{13} + x^{14} + x^{15} + x^{16} + x^{17} +  x^{18} + x^{19} + x^{20} + x^{21} +x^{22}
 \right) $$
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