# Inertia Groups Generate Galois Group

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.

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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 $I_P = \sigma I_Q\sigma^{-1}$.
Yes, I understand that all decomposition groups are isomorphic (and the inertia group because of it), so up to isomorphism, they are equal. But as subgroups of the Galois group, they are different in the general case, but the theorem still says that the $I_p$s generate the Galois group (with $p$ a rational prime). So my question is: does it matter which $P$ you choose lying above $p$ ? – KevinDL Nov 29 '11 at 16:37
@KevinDL: Is the theorem for a general Galois extension, or just for an abelian extension? Usually, $I_p$ is used only when the groups $I_{P/p}$ are equal for all $P$ lying over $p$. ("Conjugate" is stronger than "isomorphic", by the way). – Arturo Magidin Nov 29 '11 at 17:08
@KevinDL: I think it's sloppy notation, meaning all inertia groups $I_{P/p}$ as $p$ ranges over all rational primes and $P$ ranges over all primes $P$ lying over $p$. – Arturo Magidin Nov 29 '11 at 17:45
Ah, thanks a lot, that would explain a lot. One more question: taking $P$ and $Q$ primes lying over $p$ with inertia $I_P$ in a general Galois extension, are $L^{I_P}$ and $L^{I_Q}$ still maximal unramified extensions of $\mathbb{Q}$, with $P \neq Q$ implying $L^{I_P} \neq L^{I_Q}$ ? – KevinDL Nov 29 '11 at 17:50