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Probably the following proposition can be proved using class field theory. But I don't know how.

My question: Is the following proposition true? If yes, how would you prove this?

Proposition Let $K$ be an algebraic number field. Let $L$ be a finite abelian extension of $K$. Let $\mathcal{I}$ be the group of fractional ideals of $K$. Let $\mathcal{P}$ be the group of principal ideals of $K$. Let $\mathcal{H}$ be a subgroup of $\mathcal{I}$ such that $\mathcal{I} \supset \mathcal{H} \supset \mathcal{P}$. Suppose that every prime ideal $P$ of $K$ of absolute degree 1 in $\mathcal{H}$ splits completely in $L$. Then [$L : K$] = [$\mathcal{I} : \mathcal{H}$] and $L/K$ is unramified at every prime ideal of $K$.

EDIT The above proposition is false as David Loeffler pointed out. So I change the assertion: Then [$L : K$] | [$\mathcal{I} : \mathcal{P}$] and $L/K$ is unramified at every prime ideal of $K$.

Motivation

Let me explain that we can get a useful information about the class number of a cyclotomic number field by the above proposition.

Let $k$ be an algebraic number field. Let $K$ be the Hilbert class field over $k$. Let $L$ be a finite extension of $k$. Let $E = KL$. Let $\mathcal{I}$ be the group of fractional ideals of $L$. Let $\mathcal{P}$ be the group of principal ideals of $L$. Let $\mathcal{H}$ = {$I \in \mathcal{I}$; $N_{L/k}(I)$ is principal}. Note that $\mathcal{H} \supset \mathcal{P}$. Let $\mathfrak{P}$ be a prime ideal of absolute degree 1 in $\mathcal{H}$. Then by this result, $\mathfrak{P}$ splits completely in $E$. Hence, by the above proposition, [$E : L$] | [$\mathcal{I} : \mathcal{P}$] and $E/L$ is unramified at every prime ideal of $L$. Suppose $L$ is a cyclotomic number field of an odd prime order and $k$ is the unique quadratic subfield of $L$. Let $h$ be the class number of $k$. Let $h'$ be the class number of $L$. Since $K/k$ is unramified, $K \cap L = k$. Hence [$E : L$] = [$K : k$] = $h$. Hence $h | h'$. Note that $h$ can be computed relatively easily when the disciminant of $k$ is small.

Effort

Let $\mathcal{I}_L$ be the group of fractional ideals of $L$. Perhaps, by the assumption, $\mathcal{H} \subset N_{L/K}(\mathcal{I}_L)\mathcal{P}$. By class field theory, $[L : K] = [\mathcal{I} : N_{L/K}(\mathcal{I}_L)\mathcal{P}]$. Hence, $[L : K] | [\mathcal{I} : \mathcal{H}]$.

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    I'd strongly advice to stop asking all these questions the way you've been doing the last two days or so. Not only you've received galore of downvotes (and I'm afraid this will continue), but way more important: people is going to stop taking your questions & other stuff seriously. For example, this one: this is advanced stuff in one of the most beautiful and active of all subjects in modern mathematics. Don't you think it is worthwhile to add your insights into this problem, what've you done so far, examples you've worked on, etc.? As it stands it looks as if you're bored...2012-07-26
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    @DonAntonio Could you explain what's wrong with my question?2012-07-26
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    @Makotokato He literally just did. That would be "add your insights into this problem, what've you done so far, examples you've worked on, etc." Do we really have to go through this again? If you can write well enough to form this question, you can read well enough to understand advice.2012-07-26
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    The proposition is not false *thanks to* David: it is just false.2012-07-26
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    @RobertMastragostino I don't think a question should be banned just because it lacks motivation, background explanation, etc..2012-07-26
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    @Mariano I edited it. Thanks.2012-07-26
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    @MakotoKato, no one wrote above that your question should be banned, only that if you keep this on then lots of people will eventually get all bored up with your questions and just ignore them: and that is something you probably do not want.2012-07-26
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    @Makoto: Put the mouse cursor over the downvote arrow. "This question does not show any research effort", it seems that people see your questions as such, and downvote, in addition to all that happened before, for this reason, which seems to be a legitimate reason for downvoting.2012-07-26
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    @MakotoKato, I explained both in the very first comment of this question and also in other of your questions what I think is lacking. I can't do more than that.2012-07-26
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    @DonAntonio I just don't think that motivations, etc. are absolutely necessary.2012-07-27
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    Well @MakotoKato, then perhaps you haven't yet understood well enough what's this forum's purpose, requirements and etc., otherwise I can't understand how you don't take heed of all the downvotes you've received in the last days.2012-07-27
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    @DonAntonio May I ask What do you think the purpose of this forum is?2012-07-27
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    By the way, I added my effort. I hope all the downvoters change their mind.2012-07-27
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    @MakotoKato, no: you may not. Here http://math.stackexchange.com/faq in the site's FAQ you can read about general norms.2012-07-27
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    @DonAntonio I knew that. I think they don't mention about motivation, efforts, background, etc..2012-07-27
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    @MakotoKato. in the right side of that link, in "How do I ask?", we can read: "If other users ask you for more information in the comments, edit your question using the edit link just below your original question. "2012-07-27
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    @DonAntonio I think I answered most of comments concerning clarification.2012-07-27
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    @AsafKaragila "This question does not show any research effort" Come on. I invested a lot of time studying the subjects of my questions. Who can ask those questions without effort?2012-07-27
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    @DonAntonio I rewrote motivation. I hope you understand where the above proposition came from and its importance. Regards,2012-07-29
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    What's the reason for the downvotes and the close flag? Unless you make it clear, I can't improve it.2012-07-29
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    To someone who downvoted after I rewrote the motivation. What's the problem with my question? Unless you make it clear, I can't correct it as you wish.2012-07-29
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    @MakotoKato, perhaps now it's too late. Kudos for adding info but maybe the best thing is to apply all this in the following questions.2012-07-29
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    @DonAntonio What do you mean by it's too late?2012-07-29
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    @DonAntonio Well, apparently Makoto's techniques worked. All his questions got him tons of downvotes at first, but by now he has 18.5K reputation. Too bad I myself don't care enough about rep to try what he did, and see if it yields the same results...2016-03-03

2 Answers 2

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In particular, all principal prime ideals of absolute degree $1$ split in $L$, and so $L$ is contained in the Hilbert class field of $K$. The conclusion of the proposition follows by class field theory.

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This is clearly false: if the condition that every prime in $\mathcal{H}$ is split in $L$ is satisfied, then it is also satisfied if we replace $\mathcal{H}$ with any smaller subgroup $\mathcal{H}' $ contained in $\mathcal{H}$, but $[\mathcal{I} : \mathcal{H}']$ won't be the same as $[\mathcal{I} : \mathcal{H}]$.

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    That's right. Thanks.2012-07-26