Verlagerung made "explicit" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:28:40Z http://mathoverflow.net/feeds/question/56823 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56823/verlagerung-made-explicit Verlagerung made "explicit" BY 2011-02-27T14:29:55Z 2011-03-01T14:59:28Z <p>Suppose given a finite extension $L/K$ of number fields. I would like to develop a better intuition for the Verlagerung giving an embedding $Ver : Gal(K^{ab}/K) \to Gal(L^{ab}/L)$. </p> <p>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 $Gal(K'/K) \cong Gal(K^{ab}/K)/Ker\phi$.</p> <p>Is it true that $K' \subset L'$ or even $K' = K^{ab}\cap L'$? (we think of every field being embedded into $L^{ab}$)</p> <p>Further I would be happy about literature dealing with the Verlagerung in this context.</p> <p>Thank you very much in advance!</p> <p>Summary:</p> <p>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 $K^{\widetilde{\mathfrak m}}$ $\subset L^{\widetilde{\mathfrak m}}$.</p> <p>Moreover, it seems that it is generally not true (except for trivial cases) that $K^{\widetilde{\mathfrak m}} = L^{\widetilde{\mathfrak m}} \cap K^{ab}$.</p> <p>But, nevertheless, the $K^{\mathfrak m}$ exhaust the maximal abelian extension $K^{ab}$ of $K$, i.e. $$\bigcup_{\mathfrak m}K^{\mathfrak m} = \ \ \bigcup_{\mathfrak m}K^{\widetilde{\mathfrak m}} \ = \ K^{ab}$$</p> http://mathoverflow.net/questions/56823/verlagerung-made-explicit/56866#56866 Answer by GH for Verlagerung made "explicit" GH 2011-02-28T01:27:27Z 2011-02-28T20:31:00Z <p>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.</p> <p><strong>EDIT:</strong> 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.</p>