Show that $\Bbb Z_p$ contains all the $(p-1)$th roots of unity. For which primes $p$ does $\Bbb Z_p$ contains primitive fourth roots of unity? Here $\Bbb Z_p$ is the ring of $p$-adic integers. Proving that it has a $(p-1)$th root of unity is easy, but all roots is another matter. Please help me with these questions.

I think for the second question, $p$ has to be $5$, but maybe there are other answers that I didn't think of.

Edit: I solved the first question, only need the second one. From what I can see, to be the primitive $4$th root, $4\mid(p-1)$, is that right?


Hint: use Hensel's lemma. $\text{ }$

Added because you made an effort:

Second hint: what are the roots of $x^{p-1}-1$ in $\mathbf F_p$?

Third hint: for which $p$ does $x^2+1$ have a root in $\mathbf F_p$?

  • $\begingroup$ I tried to apply hensel's lemma but failed miserably, can you explain it to me a bit more?. Thanks $\endgroup$ – user108680 Nov 22 '13 at 3:47
  • 1
    $\begingroup$ @user108680 Please improve your answer to include what you have, so that I can see where you are stuck. Thanks. $\endgroup$ – Bruno Joyal Nov 22 '13 at 3:48
  • $\begingroup$ This is what i tried.Let f(x)=x^(n-1)-1,there exist x such that gcd(x,p)=1, just let x=1 and x^(n-1)=1 mod p. f'(1)=/=0 since it's still 1. so there exist a solution in Zp that f(x)=0 $\endgroup$ – user108680 Nov 22 '13 at 4:17
  • $\begingroup$ ,And part 2, since with part 1, Zp has all (p-1)th root, it's obvious that if p=5, Zp had all 4th roots, one of them is a primitive root. But that answer is too stupid $\endgroup$ – user108680 Nov 22 '13 at 4:19
  • $\begingroup$ In the previous comment,i meant f(x)=x^(p-1)-1 $\endgroup$ – user108680 Nov 22 '13 at 4:28

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.