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Suppose given a finite extension $L/K$ of number fields. I would like to develop a better intuition for the Verlagerung giving an embedding $Ver : Gal(K^{ab}/K) \to Gal(L^{ab}/L)$.

For example, let $L'/L$ be a finite abelian (Galois) extension. Define $\phi:Gal(K^{ab}/K)\to Gal(L'/L)$ to be the composition of the Verlagerung $Gal(K^{ab}/K) \to Gal(L^{ab}/L)$ followed by the projection $Gal(L^{ab}/L) \to Gal(L'/L)$. Then there is a finite, abelian (Galois) extension $K' / K$ such that $Gal(K'/K) \cong Gal(K^{ab}/K)/Ker\phi$.

Is it true that $K' \subset L'$ or even $K' = K^{ab}\cap L'$? (we think of every field being embedded into $L^{ab}$)

Further I would be happy about literature dealing with the Verlagerung in this context.

Thank you very much in advance!


As shown below, in the answer of GH plus comments, it is not alway true that $K' \subset L'$. But, if we restrict to strict ray class fields $L^{\mathfrak m}$ of $L$, for $\mathfrak m$ an ideal of $L$, one can show that for every $\mathfrak m$ there is an ideal $\widetilde {\mathfrak m}$ of $L$ with $\mathfrak m | \widetilde {\mathfrak m}$ such that $ K^{\widetilde{\mathfrak m}}$ $\subset L^{\widetilde{\mathfrak m}}$.

Moreover, it seems that it is generally not true (except for trivial cases) that $K^{\widetilde{\mathfrak m}} = L^{\widetilde{\mathfrak m}} \cap K^{ab}$.

But, nevertheless, the $K^{\mathfrak m}$ exhaust the maximal abelian extension $K^{ab}$ of $K$, i.e. $$\bigcup_{\mathfrak m}K^{\mathfrak m} = \ \ \bigcup_{\mathfrak m}K^{\widetilde{\mathfrak m}} \ = \ K^{ab}$$

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up vote 5 down vote accepted

I am no expert, but let me share an idea. For a number field $M$ let us denote by $C_M$ the idele class group of $M$. By class field theory, $K' \subset L'$ is the same as $N_{L'/K}(C_{L'})\subset N_{K'/K}(C_{K'})$, where $N$ stands for the norm map. Let us denote by $U$ the open subgroup $N_{L'/L}(C_{L'})$ of $C_L$, and by $j$ the natural injection of $C_K$ into $C_L$. Then, by the transfer theorem, the previous relation can be rewritten as $N_{L/K}(U)\subset j^{-1}(U)$, that is, $j(N_{L/K}(U))\subset U$. For certain open subgroups $U$ of $C_L$ this relation follows from ramification theory, and further local analysis might extend it to more general open subgroups.

EDIT: I think now that the conclusion $K' \subset L'$ fails when $L$ has two places $w$ and $w'$ above the same place $v$ of $K$ such that $L'/L$ is much more ramified at $w$ than at $w'$. Indeed, viewing $U$ as a subgroup of the ideles $J_L$, this assumption implies that $U_w:=U\cap L_w^\times$ is much smaller than $U_{w'}:=U\cap L_{w'}^\times$. However, $j(N_{L/K}(U))\subset U$ would imply that $N_{L_{w'}/K_v}(U_{w'})\subset U_w$ which is false when $U_w$ is sufficiently small in terms of $U_{w'}$. Perhaps this argument can be rewritten in terms of the transfer map and ramification theory only, i.e. avoiding class field theory as a whole.

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Dear GH, thank you very much for your helpful answer! I will see if I can solve the problem in the adelic formulation. – user5831 Feb 28 '11 at 12:31
Well, if you find it helpful, vote for it! Perhaps you voted but someone also downvoted? Thanks for the comment anyways, and best of luck! – GH from MO Feb 28 '11 at 17:31
Dear GH, I was concentrating on strict ray class groups (and assuming further that $L/K$ is Galois) and if I am not mistaken then $j(N_{L/K}(U^{\mathfrak m})) \subset U^{\mathfrak m}$, for $\mathfrak m$ an ideal of $\mathcal O _L$, if and only if for every pair of primes $\mathfrak P$, $\mathfrak P '$ lying over the same prime $\mathfrak p$ in $\mathcal O _K$ we have $\mathfrak P | \mathfrak m \Leftrightarrow \mathfrak P ' | \mathfrak m$. – user5831 Feb 28 '11 at 23:03
And if $L/K$ is not Galois, then one gets more obstructions on the dividing orders of the primes $\mathfrak P$ lying above a prime $\mathfrak p$ in $\mathcal O_K$. But, for $L/K$ arbitrary one can say, that for every abelian extension $L'/L$ one finds a strict ray class field $L^\mathfrak m$ containing $L'$ such that $j(N_{L/K}(U^{\mathfrak m})) \subset U^\mathfrak m$. – user5831 Feb 28 '11 at 23:11
Dear Bora, I agree with your comments. I am glad my initial idea was helpful in the end! – GH from MO Mar 1 '11 at 2:12

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