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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 $I_P = \sigma I_Q\sigma^{-1}$.

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

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    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 $p$ ?2011-11-29
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    @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).2011-11-29
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    Yes, the theorem says it holds for a general Galois extension (otherwise it wouldn't be a problem). The exact paper can be found here: http://goo.gl/6nQV12011-11-29
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    @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 $p$.2011-11-29
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    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 $L^{I_P} \neq L^{I_Q}$ ?2011-11-29
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    @KevinDL: Mind you, I *think* it's sloppy notation, but I could be wrong. Careful with your nomenclature: $\mathbb{Q}$ has no unramified extensions. $L^{I_P}$ is the inertia field of $P$, so $P\cap L^{I_P}$ is unramified in $L$; similarly, $Q\cap L^{I_Q}$ is unramified in $L$. Whether or not $L^{I_P}\neq L^{I_Q}$ depends on the normalizer of $I_P$ in $G$: if the normalizer includes the elements $\sigma$ such that $\sigma(P)=Q$, then the fields are the same; if not, then the fields are different.2011-11-29
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    @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