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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!

EDIT

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 $K^{\widetilde{\mathfrak m}}$ $\subset L^{\widetilde{\mathfrak m}}$.

On the other hand

Moreover, it seems that it is never generally not true (except in special for trivial casesmaybe) that $K^{\widetilde{\mathfrak m}} = L^{\widetilde{\mathfrak m}} \cap K^{ab}$.

A natural question is now if

But, nevertheless, the $K^{\mathfrak m}$ exhaust the maximal abelian extension $K^{ab}$ of $K$, i.e. $$\bigcup_{\mathfrak m}K^{\mathfrak m} = \ \ \bigcup_{\mathfrak m}K^{\widetilde{\mathfrak m}} \ \ \overset {?} = \ \ K^{ab}$$If this it not true, can one say something about the index (finite, or infinite)?

5 added 43 characters in body

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!

EDIT:

As shown below, in the answer of GH plus comments, it is not alway true that $K' \subset L'$is not always true. 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 $K^{\widetilde{\mathfrak m}}$ $\subset L^{\widetilde{\mathfrak m}}$.

On the other hand, it is never true (except in special cases maybe) that $K^{\widetilde{\mathfrak m}} = L^{\widetilde{\mathfrak m}} \cap K^{ab}$.

A natural question is now if nevertheless the $K^{\mathfrak m}$ exhaust the maximal abelian extension $K^{ab}$ of $K$, i.e. $$\bigcup_{\mathfrak m}K^{\mathfrak m} = \ \ \bigcup_{\mathfrak m}K^{\widetilde{\mathfrak m}} \ \ \overset {?} = \ \ K^{ab}$$ If this it not true, can one say something about the index (finite, or infinite)?

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!

EDIT:

As shown below $K' \subset L'$ is not always true. 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 $K^{\widetilde{\mathfrak m}}$ $\subset L^{\widetilde{\mathfrak m}}$.

On the other hand, it is never true (except in special cases maybe) that $K^{\widetilde{\mathfrak m}} = L^{\widetilde{\mathfrak m}} \cap K^{ab}$.

A natural question is now if nevertheless the $K^{\mathfrak m}$ exhaust the maximal abelian extension $K^{ab}$ of $K$, i.e. $$\bigcup_{\mathfrak m}K^{\mathfrak m} = \ \ \bigcup_{\mathfrak m}K^{\widetilde{\mathfrak m}} \ \ \overset {?} = \ \ K^{ab}$$ If this it not true, can one say something about the index (finite, or infinite)?

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