# On a certain criterion for unramification of an abelian extension of an algebraic number field

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... –  DonAntonio Jul 26 '12 at 20:13
@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. –  Robert Mastragostino Jul 26 '12 at 21:38
The proposition is not false thanks to David: it is just false. –  Mariano Suárez-Alvarez Jul 26 '12 at 21:46
@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. –  Mariano Suárez-Alvarez Jul 26 '12 at 21:57
@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. –  DonAntonio Jul 26 '12 at 23:54

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. –  Makoto Kato Jul 26 '12 at 20:19