1
$\begingroup$

Let $\zeta$ be a primitive $m$th root of unity, and $L = \mathbb{Q}(\zeta)$. Then $B = \mathbb{Z}[\zeta]$ is the integral closure of $\mathbb{Z}$ in $L$. If $P$ is a prime ideal of $B$ and $\mathbb{Z}p = P \cap \mathbb{Z}$, I know that for $p$ odd, $P$ is ramified if and only if $p$ divides $m$, and for $p = 2$, $P$ is ramified if and only if $m \equiv 4 \pmod 4$.

In the special case $m = p$, I know that the ramification index of $P/\mathbb{Z}p$ is exactly $[L : \mathbb{Q}] = p - 1$. Is there a general formula for a ramification index of a prime ideal of $B$ when $m$ is arbitrary?

$\endgroup$
0

1 Answer 1

3
$\begingroup$

Yes, there is a general formula. If $p$ ramifies in $\mathbb{Q}(\zeta_m)$, then $p\mid m$. Then the ramification index $e=e(P;p)$ equals $\phi(p^r)=p^{r-1}(p-1)$, if $p^r\mid\mid m$, i.e., if $p^r$ is the exact power of $p$ dividing $m$. If $p\nmid m$, then $e(P;p)=1$, and $p$ is unramified.

$\endgroup$
1
  • $\begingroup$ Thanks, I was able to prove this using the tower formula for discriminants and the explicit formula for $\mathfrak d(\mathbb{Q}(\zeta_m)/\mathbb{Z})$. Is there also anything that can be said about the inertia constant? $\endgroup$
    – D_S
    Commented May 3, 2014 at 2:24

You must log in to answer this question.

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