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Here is a question that I am trying to answer:

Let $p$ be a prime greater than $2$. For which $d \in \mathbb{Z}$ contains $\mathbb{Q}(\sqrt{d})$ a primitive root of power $p$?


What I did

If $\mathbb{Q}(\sqrt{d})$ contains a primitive root of power $p$, lets say $\zeta$, then we also know, be rewriting, that $\sqrt{d} \in \mathbb{Q}(\zeta)$. (I don't think that this statement doesn't hold in the opposite way.) So then there are $a_0, \cdots , a_{p-1}$ such that $$ d \quad = \quad (a_0 + a_1 \zeta + a_2 \zeta^2 + \cdots + a_{p-2}\zeta^{p-2})^2 $$ I know how I can rewrite it but I don't know if it would be useful to do so. Could you give me some advice. (I need rather advice than a whole answer)

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  • $\begingroup$ If you know about degrees of field extensions, it may help. Also, and related, have you done Eisenstein's criterion yet? $\endgroup$ – Geoff Robinson Feb 26 '15 at 20:59
  • $\begingroup$ yes I know about those things. $\endgroup$ – Koenraad van Duin Feb 26 '15 at 21:23
  • $\begingroup$ Well, then (as is made expicit in the answer below), you know that $\mathbb{Q}[\sqrt{d}]$ is an extension of $\mathbb{Q}$ of degree $2$, while if $\zeta$ is even a $p$-th root of unity, $\mathbb{Q}[\zeta]$ is an extension of degree $p-1$ of $\mathbb{Q}$. When $p =2$ or $3$ you need to work a little more. $\endgroup$ – Geoff Robinson Feb 26 '15 at 21:31
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$\mathbb{Q}(\sqrt{d})$ has degree two over $\mathbb{Q}$, while $\zeta_p$ is an algebraic number with degree $\varphi(p)=p-1$.

So $\zeta_p$ belongs to a quadratic extension of $\mathbb{Q}$ iff $p=3$, and in such a case $d$ is just the discriminant of the minimal polynomial of $\zeta_p$ over $\mathbb{Q}$, i.e. $d=-3$.

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