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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!

Bruno Joyal
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P. M. O.
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    The way this is phrased, it would seem there is something called the Frobenius map on $K/{\Bbb Q}$, and your task is to show that it is given by the unique field automorphism extending $\zeta_m\mapsto\zeta_m^p$. How is a Frobenius map in characteristic zero defined? – anon May 23 '13 at 01:39
  • @anon The roots are permuted, not necessarily by the power p – P. M. O. May 23 '13 at 01:50
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    How is a Frobenius map in characteristic zero defined? Does the original problem statement use the word "Frobenius"? Can you please state the original problem verbatim? Your comment "The roots are permuted, not necessarily by the power p" seems to be describing "Galois action" or "field automorphism," somewhat vaguely, in particular on cyclotomic number fields. – anon May 23 '13 at 01:52

1 Answers1

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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.

Bruno Joyal
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