2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Think about what happens to your $\alpha_1$ in the quotient $\mathbb{Z}_p/p$ $\endgroup$ – Kevin Carlson May 7 '17 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 '17 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 '17 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 '17 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 '17 at 20:00
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.