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I am trying to prove that $\mathbb{Z}_p^*$ has a primitive $p - 1$-th root of unity. I already proved that $\mathbb{Z}_p^*$ has a $p - 1$-th root of unity using Hensel's lemma.

Here is my proof: if we consider the polynomial $x^{p - 1} - 1$ it is not hard to see that there exists $\alpha_1 \in \mathbb{Z}_p^*$ such that $p(\alpha_1) \equiv 0(p\mathbb{Z}_p)$. Moreover, $p'(\alpha_1) = (p - 1)\alpha_1^{p - 2} \equiv p - 1(\mod p)$. This implies that $p'(\alpha_1) \not\equiv 0(\mod p)$. In virtue of Hensel's lemma, there exists a $p - 1$-th root of unity in $\mathbb{Z}_p^*$, but I would like to have not any root but a primitive one. How can I do it?

Thank you for your help!

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  • $\begingroup$ Think about what happens to your $\alpha_1$ in the quotient $\mathbb{Z}_p/p$ $\endgroup$
    – Kevin Carlson
    May 7, 2017 at 17:38
  • $\begingroup$ @KevinCarlson according to Hensel's lemma, $\alpha_1$ goes to the $(p - 1)$-th root of unity, but I do not see how this is related with whether or not $\mathbb{Z}_p^*$ contains a primitive $(p - 1)$-th root. $\endgroup$
    – Robert
    May 7, 2017 at 18:44
  • $\begingroup$ Ah, let's say $\alpha$ is your actual $p-1$st root, rather than the approximation $\alpha_1$. Then if $\alpha$ represents a primitive root mod $p$, is must be primitive in the $p$-adics as well, no? $\endgroup$
    – Kevin Carlson
    May 7, 2017 at 19:05
  • $\begingroup$ That's true @KevinCarlson if $\alpha$ represents a primitive root $\mod p$, but as far as I know, $\alpha$ represents a regular root. $\endgroup$
    – Robert
    May 7, 2017 at 19:13
  • $\begingroup$ But you can just pick the mod $p$ residue however you like, to start of the Hensel application. $\endgroup$
    – Kevin Carlson
    May 7, 2017 at 20:00

1 Answer 1

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One way is to take the limit of $\alpha^{p^n}$ as $n$ goes to infinity where $\alpha$ is any primitive root in $\mathbb{Z}/p\mathbb{Z}$.

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