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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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    @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. Than$k$s.2012-07-26