4
$\begingroup$

Let $\zeta$ be a primitive $l$th root of unity, where $l$ is prime. If $p$ is another prime number, let $f$ be the order of $p$ in $U(\mathbb{Z}/l \mathbb{Z})$. Then in $\mathbb{Z}[\zeta]$, $p$ factors as a product of $\frac{l-1}{f}$ unramified primes, each with inertia degree $f$.

What about in an arbitrary cyclotomic extension though? If $\zeta$ is a primitive $n$th root of unity and $p$ does not divide $n$ (hence $p$ is unramified in $\mathbb{Z}[\zeta]$), how does $p$ factor? I'm aware this is equivalent to the factorization of the $n$th cyclotomic polynomial modulo $p$, but I was wondering if any general result on this was known.

$\endgroup$

1 Answer 1

4
$\begingroup$

If you identify $G = \text{Gal}(\mathbb Q(\zeta_n)/\mathbb Q)$ with $(\mathbb Z/n\mathbb Z)^\times$ in the usual way, and $(n,p)=1$, then the Frobenius $\sigma_p \in G$ at $p$ is precisely the class of $p$ in $(\mathbb Z/n\mathbb Z)^\times$. The order of $\sigma_p$ is therefore equal to the multiplicative order $f$ of $p$ modulo $n$. Since $p$ is unramified, it follows that the decomposition group at $p$ has order $f$, and that $p$ factors as a product of $|G|/f = \varphi(n)/f$ distinct primes in $\mathbb Z [\zeta_n]$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .