Frobenius element in cyclotomic extension Let $K=\mathbb{Q}(\zeta_m)$. Then if $p\nmid m$ is any odd prime, how i can show that Frobenius map is
$(p,K/\mathbb{Q})(\zeta_m)=\zeta_m^p$.
We know, if $P$ is a prime above $p$ 
$$(p,K/\mathbb{Q})(x) \equiv x^p (\mbox{mod } P) \quad\forall x \in \mathbb{Z}_K$$
Thanks!
 A: In an arbitrary Galois extension $K/\mathbf Q$ with Galois group $G$, and an unramified prime $\mathfrak p$ of $K$ sitting over $p$, we know that the Galois group of the residual extension $\mathcal O_K/\mathfrak p$ over $\mathbf F_p$ identifies with the decomposition group $D_{\mathfrak p/p} = \{ \sigma \in G : \sigma(\mathfrak p) = \mathfrak p\} \subseteq G$. The Galois group of a finite extension of finite fields is cyclic, and comes equipped with a canonical generator : the $p$-th power map, also commonly known as the Frobenius. Under the identification of $D_{\mathfrak p/p}$ with $\text{Gal}((\mathcal O_k/\mathfrak p)/\mathbf F_p)$, the element of $D_{\mathfrak p/p}$ corresponding to the $p$-th power map is called the Frobenius at $\mathfrak p$ and is denoted $\sigma_{\mathfrak p/p}$.
If $\mathfrak p$ and $\mathfrak p'$ lie both above $p$, there is an element of the Galois group sending one to the other, and it follows that the decomposition groups $D_{\mathfrak p/p}$ and $D_{\mathfrak p'/p}$ are conjugate. Under this correspondence, the various Frobenius elements at $\mathfrak p$, when $\mathfrak p$ ranges over all the primes above $p$, determine no longer an element but a conjugacy class of $G$. This conjugacy class is written $\sigma_p$  and is called the Frobenius at $p$. 
When $G$ is abelian, conjugacy classes are just elements of $G$, and therefore the Frobenius $\sigma_p$ is actually a well-determined element of $G$. Thus, for each unramified prime of an abelian extension, we obtain a canonical element of the Galois group attached to it.
In your case, you are looking at the cyclotomic extension $K=\mathbf Q(\zeta_m)$. This is an abelian extension, unramified away from the primes dividing $m$. The Galois group identifies canonically with $(\mathbf Z/m\mathbf Z)^\times$, by $\zeta_m \mapsto \zeta_m^a$. Now, look at the automorphism $\alpha : \zeta_m \mapsto \zeta_m^p$. Claim: we have $\alpha(x) \equiv x^p \mod p$. Indeed, since $\mathcal O_K = \mathbf Z[\zeta_m]$, we can write an arbitrary element as $x= \sum_{i}a_i \zeta_m^i$; since the binomial coefficients $p \choose i$ vanish for $1<i<m$, we have 
$$x^p \equiv\sum_i a_i^p \zeta_m^{ip} \equiv \sum a_i\zeta_m^{ip} \equiv \alpha(x) \mod p$$
(using the fact that $a_i \equiv a_i^p$.) Therefore, $\alpha$ induces the $p$-power map on the residual extension so by definition of the Frobenius at $p$, $\alpha = \sigma_p$.
This fact is a primitive example of a "reciprocity law", which describes the splitting of primes in an extension in terms of arithmetic in the base.
