Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to prove, that all roots of 1 in $\mathbb Q_p$ are roots of $x^{p-1}-1$? If we consider the ring homomorphism $$ \mathbb Z_p \to \mathbb F_p^*, $$ then we see, that all the roots in power $p-1$ are equal to 0 modulo $p$. Using Hensel's lemma we can construct a solusion to $x^{p-1}-1=0$, which is the same as the first modulo $p$.

share|improve this question
    
Who's he? ${}{}$ –  joriki Dec 15 '12 at 13:16
    
@joriki Hensel? I can't recall his lemma but I have "learned" it before in a number theory course. It's standard. en.wikipedia.org/wiki/Hensel%27s_lemma –  Graphth Dec 15 '12 at 15:35
2  
@Graphth: This is an instance of that frustratingly widespread practice of silently editing and deleting things in response to comments such that comments or answers no longer make sense, without marking it or leaving comments or otherwise interacting. There was a comment by the OP above mine that referred to a "he" without any "he" having been introduced -- that's what my comment referred to. I had heard of Hensel's lemma before ;-) –  joriki Dec 15 '12 at 18:29
    
I’m sure you meant to specify that the claim was to be proven in case $p>2$. –  Lubin Dec 15 '12 at 20:55
add comment

2 Answers

up vote 5 down vote accepted

Let $\pi\colon \mathbb{Z}_p\to \mathbb{F}_p$ be the reduction homomorphism. Suppose $\zeta\in \mathbb{Z}_p$ is an $n$th root of unity.

Assume that $p\nmid n$. Let $m$ be the order of $\pi(\zeta)$ as an element of the multiplicative group $\mathbb{F}_p^*$. Since $|\mathbb{F}_p^*| = p-1$, it follows that $m\mid p-1$. Moreover, since $m$ is the order of $\pi(\zeta)$, we have that $\pi(\zeta) = \pi(\zeta^{m+1})$. But then $\zeta$ and $\zeta^{m+1}$ are two $n$th roots of unity that are equivalent mod $p$. The uniqueness part of Hensel's lemma (which we can apply to the polynomial $x^n-1$ since $p\nmid n$) says that $\zeta = \zeta^{m+1}$, and hence $\zeta$ is an $m$th root of unity. Since $m\mid p-1$, we conclude that $\zeta$ is also a $(p-1)$st root of unity.

To complete the proof, one has to compute the $p$-th roots of unity. Any such root $\zeta$ must satisfy $\pi(\zeta) = 1$.

If $p > 2$, then the $p$-adic logarithm gives an isomorphism from the multiplcative group $1 + p \mathbb{Z}_p$ to the additive group $p \mathbb{Z}_p$, which is torsion free. Therefore, $\mathbb{Z}_p$ doesn't have any $p$-th roots of unity.

In the $p=2$ case, $-1$ is a root of unity. There are no others, because $1 + 2 \mathbb{Z}_2 \cong (1 + 4 \mathbb{Z}_2) \times \{ \pm 1 \}$, and the logarithm is an isomorphism $1 + 4 \mathbb{Z}_2 \to 4 \mathbb{Z}_2$ (The difference from the odd case is the domain of convergence of the exponential).

A corrected version of your question would be that all roots of unity are roots of $x^{\text{lcm}(p-1, 2)} - 1$.

share|improve this answer
    
Every $p$-th root of unity is in $1 + p \mathbb{Z}_p$. However, by taking logarithms, this multiplicative group is isomorphic to the additive group $\mathbb{Z}_p$. –  Hurkyl Dec 15 '12 at 14:58
    
@Hurkyl: Nice! This is a community wiki question, so feel free to add this to the answer! –  froggie Dec 15 '12 at 15:16
    
Unfortunately, I don't recall exactly what is isomorphic in the $2$-adic case, can't find a reference. (and am dissatisfied with the likelihood of making an error in the derivation) –  Hurkyl Dec 15 '12 at 15:34
    
I think I have it right. I wouldn't mind my facts being double-checked for the $p=2$ case. :) –  Hurkyl Dec 15 '12 at 20:29
    
@Hurkyl: What you say about the $2$-adic logarithm is correct: c.f. Exercise 5.3 on p. 70 of math.uga.edu/~pete/8410FULL.pdf. To my taste you could be a bit more explicit about why this finishes the computation. In particular, the reason that it suffices to compute the $p$th roots of unity in the $p > 2$ case is that there are none, hence there are no $p^k$th roots of unity for any $k$. When $p = 2$ there are $p$th roots of unity, so I think you should be explaining why there are no further $p^k$th roots of unity. (This does follow easily from what you've done.) –  Pete L. Clark Dec 15 '12 at 20:54
add comment

Chapter 2, $\S 1.4$ of my notes on local fields contains a discussion of roots of unity in local fields. In particular there is a complete proof that the group of roots of unity in $\mathbb{Q}_p$ is cyclic of order $p-1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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