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Let $\zeta$ be the 15th primitive root of unity in $\mathbb{C}$, show that $\mathbb{Q}(\zeta)$ contains one of the two numbers $\sqrt{\pm5}$ and decide which one is contained in $\mathbb{Q}(\zeta)$.

First consider the element $\zeta^{3}\in\mathbb{Q}(\zeta)$; it is a primitive 5th root of unity since $(\zeta^{3})^{5}=\zeta^{15}=1$.

We are allowed to use the following theorem; let $p$ be an odd prime and let $\zeta$ be a primitive $p$th root of unity in $\mathbb{C}$ then $S^{2}=(\frac{-1}{p})p$. In particular, the cyclotomic field $\mathbb{Q}(\zeta)$ contains at least one of the quadratic fields $\mathbb{Q}(\sqrt{p})$ or $\mathbb{Q}(\sqrt{-p})$.

The Legendre symbol $(\frac{-1}{5})=1$ since $5\equiv 1(mod4)$ so we get the following: $$S^{2}=(\frac{-1}{5})5=5$$ Therefore $\mathbb{Q}(\zeta)$ contains $\mathbb{Q}(\sqrt{5})$ by the theorem above.

So I can prove that $\mathbb{Q}(\zeta^{3})$ contains $\mathbb{Q}(\sqrt{5})$ therefore $\mathbb{Q}(\zeta)$ contains $\mathbb(\sqrt{5})$ but I don't know how to show that $\sqrt{-5}$ is not contained in $\mathbb{Q}(\zeta)$.

I was thinking that it might involve showing that $i\notin\mathbb{Q}(\zeta)$ but I'm not sure how to go about this neither.

I also looked into showing that if $\mathbb{Q}(\sqrt{-5})\subseteq\mathbb{Q}(\zeta)$ then $[\mathbb{Q}(\zeta)\colon\mathbb{Q}]$ is divisible by $$[\mathbb{Q}(\sqrt{5},\sqrt{-5})\colon\mathbb{Q}]=4$$ But I cannot use this to disprove anything because the minimal polynomial of $\zeta$ is $x^8-x^7+x^5-x^4+x^3-x+1$ and clearly 4 does divide 8.

Any help would be appreciated, thank you.

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If $\sqrt 5$ and $\sqrt{-5}$ were both contained in $\mathbb Q(\zeta)$, then $\mathbb Q(\zeta)$ would also contain their quotient $\sqrt{-5}/\sqrt 5 = i$. But the only roots of unity in $\mathbb Q(\zeta)$ are of the form $\pm \zeta^k$ (the $30$th roots of unity). Since 4 does not divide 30, $i \not\in \mathbb Q(\zeta)$.

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