How to prove that: Automorphism $\zeta_n\to\zeta_n^p$ leaves prime ideal factor of *p* unchanged.

Question

Let's

(n,p)=1

$\zeta_n$ - n-root of unity

$\mathfrak p$ - prime ideal factor of prime number p

How to prove that automorphism $\zeta_n\to\zeta_n^p$ leaves $\mathfrak p$ unchanged.

Answer 1

The answer to your question comes from (i) the definition of the Frobenius automorphism at a prime ideal $\mathfrak p$ and (ii) the uniqueness of the Frobenius at $\mathfrak p$ when $p$ is unramified. So it is not a direct calculation, but an indirect one.

Let $K$ be a finite Galois extension of $\mathbf Q$ and $G = {\rm Gal}(K/\mathbf Q)$. (If $K = \mathbf Q(\zeta_n)$, then $G \cong (\mathbf Z/n\mathbf Z)^\times$.) For a prime number $p$ and a prime ideal $\mathfrak p$ in $\mathcal O_K$ that divides $p\mathcal O_K$, we define a decomposition group at $\mathfrak p$ inside $G$: $$ D(\mathfrak p|p) = \{\sigma \in G : \sigma(\mathfrak p) = \mathfrak p\}. $$ (In older literature this subgroup is written as $Z(\mathfrak p|p)$ from the German word for "decomposition".) To each $\sigma$ in $D(\mathfrak p|p)$, it makes sense to apply $\sigma$ to the residue field $\mathcal O_K/\mathfrak p$: if $\alpha \equiv \beta \bmod \mathfrak p$ then $\sigma(\alpha) \equiv \sigma(\beta) \bmod \mathfrak p$ since $\sigma(\mathfrak p) = \mathfrak p$. So there's an induced automorphism $\overline{\sigma}$ of the residue field $\mathcal O_K/\mathfrak p$, and thinking of this on all such $\sigma$ gives us a reduction homomorphism $D(\mathfrak p|p) \to {\rm Gal}((\mathcal O_K)/\mathfrak p)/\mathbf F_p)$ where $\sigma \mapsto \overline{\sigma}$.

This reduction homomorphism has three fundamental properties: (i) it is surjective, (ii) its kernel has order equal to the ramification index $e(\mathfrak p|p)$, and (iii) the Galois group ${\rm Gal}((\mathcal O_K)/\mathfrak p)/\mathbf F_p)$ contains the $p$-th power map on the residue field $\mathcal O_K/\mathfrak p$. Because of (i) and (ii), when $p$ is unramified in $K$, which makes $e(\mathfrak p|p) = 1$, the reduction mapping $\sigma \mapsto \overline{\sigma}$ is a bijection, so by (iii) there is a unique $\sigma \in D(\mathfrak p|p)$ such that $\overline{\sigma}$ is the $p$th power map on $\mathcal O_K/\mathfrak p$. That means there is a unique $\sigma \in D(\mathfrak p|p)$ such that $$ \sigma(\alpha) \equiv \alpha^p \bmod \mathfrak p $$ for all $\alpha \in \mathcal O_K$. This $\sigma$ is called the Frobenius automorphism at $\mathfrak p$. If $p$ were ramified in $K$, then the reduction mapping from $D(\mathfrak p|p)$ to the Galois group of the residue field extension would not be injective, so the Frobenius automorphism at $\mathfrak p$ would not be unique.

In fact there is a unique $\sigma$ in $G$ that satisfies the above congruence, not just in $D(\mathfrak p|p)$, because when the above congruence holds for all $\alpha$ and we pick $\alpha \in \mathfrak p$ (so $\alpha^p \equiv 0 \bmod \mathfrak p$), we get $\sigma(\mathfrak p) \subset \mathfrak p$, so $\sigma(\mathfrak p) = \mathfrak p$ because $\sigma(\mathfrak p)$ is a maximal ideal. So an automorphism satisfying the above displayed congruence for all $\alpha$ in $\mathcal O_K$ must belong to $D(\mathfrak p|p)$: it must preserve the ideal $\mathfrak p$. Therefore when $p$ is unramified in a finite Galois extension $K$ of $\mathbf Q$, there is a unique $\sigma \in {\rm Gal}(K/\mathbf Q)$ such that the above congruence mod $\mathfrak p$ is satisfied for all $\alpha$ in $\mathcal O_K$.

Now let's specialize all of this to your situation: $K = \mathbf Q(\zeta_n)$ and $p$ is a prime such that $(p,n) = 1$. The elements of $G = {\rm Gal}(\mathbf Q(\zeta_n)/\mathbf Q)$ are determined by their effects on $\zeta_n$, which must look like $\zeta_n \mapsto \zeta_n^a$ with $(a,n) = 1$. Let $\mathfrak p$ be a prime ideal factor of $p$ in $\mathcal O_K$.

Claim: The condition $(p,n) = 1$ implies that the different $n$th roots of unity in $K$ are distinct in $\mathcal O_K/\mathfrak p$.

Proof of claim: this comes from the condition $(p,n) = 1$. When $(p,n) = 1$, $x^n - 1$ is separable in characteristic $p$, so it has no repeated roots in characteristic $p$. Since $x^n - 1 = \prod_{j=0}^{n-1} (x - \zeta_n^j)$ in $\mathcal O_K[x]$, reducing mod $\mathfrak p$ tells us $x^n - \overline{1} = \prod_{j=0}^{n-1} (x - \overline{\zeta_n}^j)$ in $(\mathcal O_K/\mathfrak p)[x]$. Thus the separability implies the different $n$th roots of unity $\overline{\zeta_n}^j$ for $j \in \mathbf Z/n\mathbf Z$ are distinct in the residue field $\mathcal O_K/\mathfrak p$, or in other words the different $n$th roots of unity $\zeta_n^j$ for $j \in \mathbf Z/n\mathbf Z$ are incongruent mod $\mathfrak p$: if $\zeta_n^j \equiv \zeta_n^k \bmod \mathfrak p$ then $\zeta_n^j = \zeta_n^k$, so $j \equiv k \bmod n$. That finishes the proof of the claim.

The condition $(p,n) = 1$ also implies $p$ is unramified in $\mathcal O_K$ (this is a very important property of cyclotomic fields), so there is a unique $\sigma \in G$ such that $$ \sigma(\alpha) \equiv \alpha^p \bmod \mathfrak p $$ for all $\alpha$ in $\mathcal O_K$, and in fact this $\sigma$ must preserve $\mathfrak p$ (it lies in $D(\mathfrak p|p)$). Taking $\alpha = \zeta_n$, we have $\sigma(\zeta_n) = \zeta_n^a$ where $(a,n) = 1$, so $$ \zeta_n^a \equiv \zeta_n^p \bmod \mathfrak p. $$
From the claim above, $\zeta_n^a = \zeta_n^p$, so $\sigma(\zeta_n) = \zeta_n^p$ and this value determines $\sigma$ on all of $K$ (since $\zeta_n$ is a primitive element for $K/\mathbf Q$). So we have shown for prime $p$ not dividing $n$ that the $\sigma$ in ${\rm Gal}(\mathbf Q(\zeta_n)/\mathbf Q)$ where $\sigma(\zeta_n) = \zeta_n^p$ is the Frobenius automorphism at each $\mathfrak p$ dividing $p$, so this $\sigma$ lies in $D(\mathfrak p|p)$: it preserves $\mathfrak p$.