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I am having trouble seeing the following: Let $\mathbb{Q}_p$ be $p$ adic numbers. And $\zeta$ be the third root of unity, so that $\zeta^3 = 1$. Does there exist $p$ for which $\zeta \in \mathbb{Q}_p$? Any comments are appreciated. Thank you.

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    $\begingroup$ See section $3$ in K. Conrad's notes here about the roots of unity in $\mathbb{Q}_p$ via Hensel's Lemma. $\endgroup$ – Dietrich Burde Jan 27 '18 at 20:48
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    $\begingroup$ Nitpick: You really shouldn't first describe an element, here $\zeta$, and then ask which fields it belongs to. By the act of calling it a third root of unity you already tacitly assumed at least a ring where this element resides in. For otherwise how could you compare $1$ and $\zeta^3$. Also, there is no obvious way to relate the third roots of unity in the field $\Bbb{Q}_7$ to those in $\Bbb{Q}_{13}$ and to those in $\Bbb{C}$ etc. I would phrase this as: which of the fields $\Bbb{Q}_p$ have third roots of unity? or some such. $\endgroup$ – Jyrki Lahtonen Jan 28 '18 at 5:24
  • $\begingroup$ @JyrkiLahtonen, could you please elaborate on the last two sentences? $\endgroup$ – 1717 Jan 28 '18 at 16:08
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If $3 \mid (p-1)$, then by the cyclicity of $\Bbb F_p^\times$, there's a third root of unity in $\Bbb F_p^\times$

As the third cyclotomic polynomial is separable mod $p$ for $p \neq 3$, we can lift this solution of $x^2+x+1$ by Hensel's lemma.

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For odd $p$, the roots of unity within $\Bbb Q_p$ are the $(p-1)$-th roots of unity (think about the Teichmuller map). So $\Bbb Q_p$ contains all $r$-th roots of unity iff $r\mid(p-1)$.

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In fact, if $p$ is an odd prime, $\mathbf{Q}_p$ contains exactly the $(p-1)$st roots of unity (this is sometimes used to show that for odd primes, $\mathbf{Q}_p$ and $\mathbf{Q}_q$ are not isomorphic). This statement is proved by considering the polynomial $x^{p}-x$ in $\mathbf{Z}_p[x]$ and noting that it splits in $\mathbf{F}_p$, and then applying Hensel's Lemma we get that it splits in $\mathbf{Z}_p$. So consider $p=13$. Then $\mathbf{Q}_{13}$ contains all the $12$th roots of unity, hence contains the cube roots of unity. It's probably not too hard to construct a cube root of unity in $\mathbf{Q}_{13}$ explicitly.

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  • $\begingroup$ By $\mathbf Z_p$ you mean $\mathbb Z/p\mathbb Z$, I guess. In a context where $\mathbf Q_p$ is also used that's weird :-) $\endgroup$ – Mariano Suárez-Álvarez Jan 28 '18 at 5:17
  • $\begingroup$ I meant that the polynomial splits in $\mathbf{Z}/p\mathbf{Z}$, and then applying Hensel's Lemma repeatedly we get that it splits in the actual $\mathbf{Z}_p$. $\endgroup$ – 1717 Jan 28 '18 at 5:22

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