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I have a problem understanding the interpretation of the ideal class group in the case of restricted ramifiction. Let $k$ be a number field and $S$ a set of primes of $k$. Then $k_S$ denotes the maximal Galois extension which is unramified outside of $S$.

Now the $S$-ideal class group $Cl_S(k)$ is stated to be "naturally isomorphic to the Galois group of the maximal abelian extension of $k$ inside $k_S$ in which all primes of $S$ split completely". By class field theory I know that $Cl(k)$ is isomorphic to the galois group of the maximal abelian unramified extension of $k$. But how does the relation to the splitting of primes work? Why is it not isomorphic to the Galois group of $k_S|k$?.

Does somebody can help me or provide me with a reference?

Thank you a lot, Tom :-)

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Here are a few reasons why one shouldn't expect $Cl_S(k)$ to be related to $Gal(k_S / k)$, or even to $Gal(k_S^{\mathrm{ab}} / k)$ where $k_S^{\mathrm{ab}}$ is the maximal abelian extension unramified outside $S$.

Firstly, the Galois group of $k_S^{\mathrm{ab}} / k$ will generally be quite large; e.g. for $k = \mathbf{Q}$ the extension $k_S^{\mathrm{ab}} / k$ is infinite as soon as $S$ is non-empty. On the other hand $Cl_S(k)$ is always finite.

Moreover, the field $k_S^{\mathrm{ab}}$ gets bigger and bigger as you enlarge $S$, while the S-ideal class group gets smaller -- for $S$ sufficiently large it will be trivial.

If you are familiar with the isomorphism of global class field theory between $Gal(k^{\mathrm{ab}} / k)$ and the idele class group $\mathbf{A}_k / \overline{k^\times k_\infty^\circ}$, you can easily see what's going on. The group $Gal(k^{\mathrm{ab}}_S / k)$ corresponds to the quotient of the idele class group by the image of the subgroup of $\mathbf{A}_k^\times$ given by $$ \left( \prod_{v \notin S}\mathcal{O}_{k, v}^\times\right) \times \left(\prod_{v \in S} 1 \right).$$ The group $Cl_S(k)$ corresponds to the quotient by $$ \left( \prod_{v \notin S}\mathcal{O}_{k, v}^\times\right) \times \left(\prod_{v \in S} K_v^\times \right).$$ So these are very different beasts, unless $S$ is empty.

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  • $\begingroup$ Thanks for your answer! It helps me a lot. Could you elaborate more on the relationship between $Cl_S(k)$ and the splitting of primes, since this is still a bit obstructive to me. Thank you :-) $\endgroup$ – BIS HD Feb 24 '14 at 13:54
  • $\begingroup$ @David Loeffler: Is it really true that the idele class group is isomorphic to the Galois group of the maximal abelian extension of k? Don’t we need to quotient out by the connected component of the identity? $\endgroup$ – Dedalus Jan 5 at 14:16
  • $\begingroup$ Or maybe that is what you are saying with your notation? $\endgroup$ – Dedalus Jan 5 at 14:18
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Relationship between Cl_S(k) and the splitting of primes : Cl_S (k) is the quotient of Cl(k) by the subgroup generated by the classes of the prime ideals in S. If we identify by class field theory Cl(k) with the Galois group over k of the maximal abelian unramified extension of k (the so called Hilbert class field), then the aforementioned subgroup is isomorphic to the common decomposition subgroup of all the primes in S.

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