4
$\begingroup$

While reading a paper about the Kronecker-Weber Theorem, I noticed a theorem saying that for a Galois extension $K/\mathbb{Q}$, its Galois group is generated by $I_p$s, being the inertia groups of primes $p$ that ramify in $K$.

In the same paper however, they define the inertia group $I_P$ as depending on the prime $P$ that lies over $p$, so choosing a different $P$ gives another inertia group.

How should I interpret this ? That it doesn't matter which $P$ you choose, each $I_P$ will do ?

Any help would be appreciated.

1 Answers 1

4

If the extension is Galois, the action is transitive on the set of prime lying over $p$. Therefore, the inertia group depends only on $p$ up to conjugation: if $P$ and $Q$ are two primes lying over $p$, then there is a $\sigma$ in the Galois group with $\sigma(P)=Q$. Then $I_P = \sigma I_Q\sigma^{-1}$.

If the extension is abelian (as in the Kronecker-Weber case), this means that all decomposition groups are equal.

  • 0
    @KevinDL: Careful, I messed up the last one: the ramification index of $P\cap L^{I_P}$ over $p$ is $1$, not that of $P$ over $P\cap L^{I_P}$.2011-11-29