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

Proposition Let $k$ be an algebraic number field. Let $K$ be a finite abelian extension of $k$. Suppose every principal prime ideal of $k$ splits completely in $K$. Let $L$ be a finite extension of $k$. Let $E = KL$. Let $\mathfrak{P}$ be a prime ideal of $L$ such that the relative degree of $\mathfrak{P}$ in $L/k$ is 1 and $N_{L/k}(\mathfrak{P})$ is principal. Then $\mathfrak{P}$ splits completely in $E$.


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}. Then $\mathcal{H}$ is a subgroup of $\mathcal{I}$ such that $\mathcal{I} \supset \mathcal{H} \supset \mathcal{P}$. If the proposition is true, it satisfies the assumption of this proposition.


Let $N_{L/k}(\mathfrak{P}) = \mathfrak{p}$. Let $M = K \cap L$. Let $N_{L/M}(\mathfrak{P}) = \mathfrak{p}_M$. Then $\mathfrak{p}_M$ splits completely in $K$. Let $\mathfrak{p}_M = \mathfrak{Q_1}\cdots\mathfrak{Q_m}$, where $m = [K : M]$. Let $\mathfrak{Q_i'}$ be the ideal in $E$ generated by $\mathfrak{P}$ and $\mathfrak{Q_i}$. Then perhaps we can prove that $\mathfrak{P} = \mathfrak{Q_1'}\cdots\mathfrak{Q_m'}$ and $\mathfrak{Q_i'}$s are distinct primes. Hence $\mathfrak{P}$ splits completely in $E$.

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    $\begingroup$ Please let me know the reason for the downvotes. Unless you make it clear, it is hard to improve my question. $\endgroup$ Jul 28, 2012 at 6:46

1 Answer 1


The assumption on $K$ is equivalent to $K$ being contained in the Hilbert class field $H$ of $k$. Thus $E$ is contained in the Hilbert class field of $L$. Thus $Gal(K/k)$ is naturally a quotient of $C_k$ (the class group of $k$), while $Gal(E/L)$ is naturally a quotient of $C_L$ (the class group of $L$).

There is an embedding $Gal(E/L) \hookrightarrow Gal(K/k)$ (given by restriction to $K$), which in terms of (quotients of) ideal class groups is induced by the norm.

Now $\mathfrak P$ splits completely in $E$ if and only if its associated Frobenius element is trivial. This can be tested by restricting to $k$ (since this restriction map is injective, as noted above).

Now $Frob_{\mathfrak P}$ corresponds precisely to the image of $\mathfrak B$ in (the relevant quotient of) $C_L$. Restricting to $k$ is given by applying the norm map, and by assumption $N_{L/k}(\mathfrak P)$ is trivial in $C_k$. Thus the restriction of $Frob_{\mathfrak P}$ is trivial, and so $Frob_{\mathfrak P}$ is trivial itself.

(Unless I'm blundering, the assumption that $\mathfrak P$ be of relative degree $1$ over $k$ is not needed.)


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