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Franz Lemmermeyer
Franz Lemmermeyer
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Addition. It is of course also possible to define the Hilbert class field $H$ of a number field $K$ as the maximal abelian unramified extension of $K$.

A naive idea for showing that exactly the principal prime ideals split completely in $H/K$ would be trying to prove this in its cyclic subextensions; but this is bound to fail: if $K$ has class group of type $(2,2)$, then the prime ideals from a class of order $2$ remain inert in two of the three quadratic unramified extensions.

Thus for proving the decomposition law you will have to use the maximality condition, which brings me to the first question:

Is there a simple proof that the maximal abelian unramified extension of a number field is finite?

If one could generalize Hilbert's Satz 94 to noncyclic extensions I would guess that the answer to this question is yes. The main ingredients of such a proof certainly would be Dirichlet's unit theorem (finite generation of the unit group) and the finiteness of the class number. Or is there a different way of linking unramified abelian extensions to the class group?

The classical proof of the finiteness of the Hilbert class field runs as follows. Let $L/K$ be a cyclic unramified extension, and consider the index $$ h_{L/K} = (D_K : ND_L \cdot P_K). $$ Using elementary transformations and the ambiguous class number formula (essentially the Herbrand unit index calculation) this can be shown to equal $$ h_{L/K} = (L:K) (E_K : E_K \cap NL^\times) (Cl(L)[N] : Cl(L)^{1-\sigma}), $$ where $\sigma$ generates the Galois group of $L/K$.

By the first inequality (a consequence of the fact that the Dedekind zeta function has a pole of order $1$ at $s = 1$) we must have $h_{L/K} \le (L:K)$ for all cyclic unramified extensions, from which we conclude that $h_{L/K} = (L:K)$, and that $$(E_K : E_K \cap NL^\times) = (Cl(L)[N] : Cl(L)^{1-\sigma}) = 1$$ for all cyclic unramified extensions $L/K$. From here it is not difficult to prove that $h_{L/K} = (L:K)$ holds for all abelian unramified extensions. This in turn proves that $(L:K) = h_{L/K} \le h_K$ for any abelian unramified extension.

I do not think that this inequality is sufficient for getting hold of the decomposition law. If the maximal unramified abelian extension of $K$ has degree $< h_K$ then you are in trouble. If, for example, ${\mathbb Q}(\sqrt{-5})$ were its own Hilbert class field, all ideals would split in $H/K$, not just the principal ones. Thus I fear that for proving the decomposition law you actually need the existence of an unramified abelian extension of degree $h_K$. If this is true, then I guess that there really is no simple proof of the decomposition law: the existence theorem in class field theory is a technical tour de force even (and perhaps even more so) if you restrict your attention to unramified extensions.

Addition. It is of course also possible to define the Hilbert class field $H$ of a number field $K$ as the maximal abelian unramified extension of $K$.

A naive idea for showing that exactly the principal prime ideals split completely in $H/K$ would be trying to prove this in its cyclic subextensions; but this is bound to fail: if $K$ has class group of type $(2,2)$, then the prime ideals from a class of order $2$ remain inert in two of the three quadratic unramified extensions.

Thus for proving the decomposition law you will have to use the maximality condition, which brings me to the first question:

Is there a simple proof that the maximal abelian unramified extension of a number field is finite?

If one could generalize Hilbert's Satz 94 to noncyclic extensions I would guess that the answer to this question is yes. The main ingredients of such a proof certainly would be Dirichlet's unit theorem (finite generation of the unit group) and the finiteness of the class number. Or is there a different way of linking unramified abelian extensions to the class group?

The classical proof of the finiteness of the Hilbert class field runs as follows. Let $L/K$ be a cyclic unramified extension, and consider the index $$ h_{L/K} = (D_K : ND_L \cdot P_K). $$ Using elementary transformations and the ambiguous class number formula (essentially the Herbrand unit index calculation) this can be shown to equal $$ h_{L/K} = (L:K) (E_K : E_K \cap NL^\times) (Cl(L)[N] : Cl(L)^{1-\sigma}), $$ where $\sigma$ generates the Galois group of $L/K$.

By the first inequality (a consequence of the fact that the Dedekind zeta function has a pole of order $1$ at $s = 1$) we must have $h_{L/K} \le (L:K)$ for all cyclic unramified extensions, from which we conclude that $h_{L/K} = (L:K)$, and that $$(E_K : E_K \cap NL^\times) = (Cl(L)[N] : Cl(L)^{1-\sigma}) = 1$$ for all cyclic unramified extensions $L/K$. From here it is not difficult to prove that $h_{L/K} = (L:K)$ holds for all abelian unramified extensions. This in turn proves that $(L:K) = h_{L/K} \le h_K$ for any abelian unramified extension.

I do not think that this inequality is sufficient for getting hold of the decomposition law. If the maximal unramified abelian extension of $K$ has degree $< h_K$ then you are in trouble. If, for example, ${\mathbb Q}(\sqrt{-5})$ were its own Hilbert class field, all ideals would split in $H/K$, not just the principal ones. Thus I fear that for proving the decomposition law you actually need the existence of an unramified abelian extension of degree $h_K$. If this is true, then I guess that there really is no simple proof of the decomposition law: the existence theorem in class field theory is a technical tour de force even (and perhaps even more so) if you restrict your attention to unramified extensions.

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Franz Lemmermeyer
Franz Lemmermeyer
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Let $K/F$ be an extension of number fields. We say that $K/F$ is

  • a Weber-Hilbert class field if the prime ideals of $F$ that split completely in $K$ are exactly the principal prime ideals.

  • a Takagi-Hilbert class field if the norms of all ideals from $K$ are principal ideals in $F$: $N_{K/F} D_K \subseteq P_K$, where $D_K$ is the group of nonzero fractional ideals in $K$ and $P_K$ the group of nonzero fractional ideals in $F$.

  • an Artin-Hilbert class field if $K/F$ is abelian and if the Artin symbol induces an exact sequence $$ 1 \longrightarrow P_K \longrightarrow D_K \longrightarrow Gal(K/F) \longrightarrow 1 $$

Since these definitions involve ideals rather than ideles my guess is that the most simple proof of the equivalence of these definitions uses the classical approach. This approach in turn uses analytic tools, and for these tools sets of prime ideals with density $0$ are invisible. Thus in the classical theory, which is based on Takagi's definition, the equivalence with Weber's definition for all ideals (and not just all except for a set of density $0$) is proved after all the main theorems.

What is clear from the definitions is that if $K/F$ is a Takagi-Hilbert class field, then each prime ideal that splits completely in $K/F$ must be principal. The proof that in a Takagi-Hilbert class field exactly the princiapal primes split completely up to a set of exceptions with density $0$ may be accomplished using analytic tools and should not be too difficult.

It is also clear that an Artin-Hilbert class field is a Weber-Hilbert class fields, since prime ideals have trivial Artin symbol if and only if they split completely. For proving the converse you have to show that a Weber-Hilbert class field is abelian. Arnold Scholz ( Zur Klassenkörpertheorie auf Takagischer Grundlage, Math. Z. 30 (1929), 332--356) has shown that Takagi-Hilbert class fields must be abelian, and perhaps this proof may be transferred to Weber-Hilbert class fields.

You will get a proof that Takagi-Hilbert class fields are Weber-Hilbert class fields by taking the classical approach to class field theory based on Takagi's definition and following the proof up to the decomposition law, restricting everything to unramified extensions (which simplifies a few things, but not dramatically). If anyone has a better idea, I'm all ears.