# Verlagerung made “explicit”

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!

Summary:

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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## 1 Answer

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