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 $p$.
Roots of 1 in $\mathbb Q_p$
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0Iām sure you meant to specify that the claim was to be proven in case p>2. ā 2012-12-15
2 Answers
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 $x^{\text{lcm}(p-1, 2)} - 1$.
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0@Hurkyl: What you say about the $2$-adic logarithm is correct: c.f. Exercise 5.3 on p. 70 of http://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.) ā 2012-12-15
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 $p-1$.