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Let $L/K$ be a finite, cyclic extension of number fields, say with $\mathrm{Gal}(L/K)=G$. In my context $G$ is actually of order $p$, an odd prime number, but let me state my question for every cyclic $G$.
Hilbert theorem 94 says that if $L/K$ is everywhere unramified (hence contained in the Hilbert class field $H$ of $K$), then the ''capitulation kernel'', namely the kernel of the natural map $\iota:Cl_K\to Cl_L$, has order divisibile by $\vert G\vert$. Some (but, to my knowledge, not many) generalizations have been proven, mainly removing the cyclicity assumption although losing something.

My question goes into another direction: if $L/K$ is allowed to ramify and is contained in the ray class field $H(\mathfrak{f})$ modulo some conductor $\mathfrak{f}$, what can we say about the capitulation kernel $\iota:Cl_K(\mathfrak{f})\to Cl_L(\mathfrak{F})$, where these groups are now the ray class groups modulo the respective conductors (and $\mathfrak{F}=\mathfrak{f}\mathcal{O}_L)$)? Can it be trivial?

I must confess my ignorance with respect to an even more basic question, namely: what is the state of the art concerning the Principal Ideal Theorem for ray class fields? Is it known – or false, or trivial, or... – that all classes in $Cl_K(\mathfrak{f})$ become principal (perhaps, only $\mathfrak{f}$-principal?) in $H(\mathfrak{f})$?

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Your second question is indeed well-known: the principal ideal theorem remains true with appropriate modification ($\mathfrak f$ principality) in the ray class group. This is a classical theorem of Iyanaga whose proof is identical to the principal ideal theorem (both reduce to the computation of the transfer map towards the derived subgroup). Then again, Grothendieck was famously stumped by this question. See also mathoverflow.net/questions/63465/… –  Olivier Apr 19 '12 at 7:37
    
Thanks! I did not think, when asking, that the proof is only group-theoretic. Still, I would be interested in the H94 stuff... –  Filippo Alberto Edoardo Apr 19 '12 at 12:34
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Iyanaga's article is in German: Ueber den allgemeinen Hauptidealsatz, Japanese Journ. of Math. 7 (1934), 315-333 –  Franz Lemmermeyer Apr 19 '12 at 12:38
    
Kein Problem! Und jetzt bin ich in Japan... ;-) Danke, Filippo –  Filippo Alberto Edoardo Apr 19 '12 at 13:31
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1 Answer

up vote 4 down vote accepted

The answer to both my question is that "adding conductors does not change anything". Olivier has already discussed this for the Principal Ideal Theorem, and for Hilbert 94 this is proven by Suzuki in
http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.nmj/1118782786

The whole point is that Hilbert 94 follows from a generalization of Furtwängler theorem, saying that the kernel of the transfer map $$\mathrm{ver}:G^\text{ab}\to N^\text{ab}$$ has order divisible by $[G:N]$, and this for all normal subgroups $N$ containing the commutators $[G,G]$ (or, equivalently, such that the quotient by them is abelian). The case $N=[G,G]$ is Furtwangler's theorem and applying it in my setting with $N=\mathrm{Gal}(H_L(\mathfrak{F})/K)$ in the notations of the question (now $H_L(-)$ are ray class fields of $L$) shows that $\iota$ is never invective.

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