Split Primes Recognize Galois

We give a non-standard proof of the fact that the Hilbert class field of a Galois number field is Galois over \(\mathbb Q\).

We are after the following theorem.

Notation 1. Fix an extension \(L/K\) of number fields. Then we let \(\operatorname{Spl}(L/K)\) denote the collection of prime ideals of \(\mathcal O_K\) which are totally split in \(L\).

Theorem 2. Fix a Galois extension \(L/K\) of number fields, and assume that \(K\) is Galois over \(\mathbb Q\). If the set \(\operatorname{Spl}(L/K)\) is invariant under \(\operatorname{Gal}(K/\mathbb Q)\) (possibly away from a finite set), then \(L\) is Galois over \(\mathbb Q\).

Corollary 3. Fix a Galois extension \(K\) of \(\mathbb Q\). Then the Hilbert class field \(H\) of \(K\) is Galois over \(\mathbb Q\).

Proof. A prime \(\mathfrak p\) of \(K\) splits completely in \(H\) if and only if it is principal. However, being principal in \(K\) is a property invariant under \(\operatorname{Gal}(K/\mathbb Q)\), so the corollary follows from the theorem. \(\blacksquare\)

We now turn to the proof of Theorem 2. Our main input will be the following recognition result.

Proposition 4. Fix Galois extensions \(L\) and \(L’\) of a number field \(K\). Then \(L\subseteq L’\) if and only if \(\operatorname{Spl}(L’/K)\subseteq\operatorname{Spl}(L/K)\) (possibly away from a finite set).

The inclusion in the proposition takes place in a fixed algebraic closure of \(\mathbb Q\).

Proof. We show the directions separately.

• Observe that an unramified prime \(\mathfrak p\) of \(K\) splits completely in \(L\) if and only if the conjugacy class \(\mathrm{Frob}_{\mathfrak p}|_L\) is the identity. Thus, \(L\subseteq L’\) implies \(\operatorname{Spl}(L’/K)\subseteq\operatorname{Spl}(L/K)\).
• For the converse, we use the Chebotarev density theorem. Consider the composite \(LL’\), which is Galois over \(K\); we would like to show that \(LL’=L’\), for which it is enough to show that \([LL’:K]=[L’:K]\). Well, \[\operatorname{Spl}(LL’/K)=\operatorname{Spl}(L/K)\cap\operatorname{Spl}(L’/K)\] is just \(\operatorname{Spl}(L’/K)\) by hypothesis. On the other hand, the Chebotarev density theorem (and the observation of the previous point) tells us that the density of totally split primes in \(LL’\) and \(L’\) are \(1/[LL’:K]\) and \(1/[L’:K]\), respectively. These densities must be equal, so the equality of degrees follows. \(\blacksquare\)

Remark 5. We do not need the full power of the Chebotarev density theorem to show that the density of split primes in an extension \(L/K\) is \(1/[L:K]\). In short, the set of primes \(\mathfrak q\) in \(L\) which lie over a totally split prime of \(K\) has Dirichlet density \(1\) for norm reasons: not splitting completely increases inertial degree, which causes the norm to explode (namely, note \(\sum_p1/p^2<\infty\)). But then the Dirichlet density of totally split primes in \(K\) is \[\lim_{s\to1^+}\frac{\sum_{\mathfrak p\in\operatorname{Spl}(L/K)}\operatorname N(\mathfrak p)^{-s}}{\sum_{\mathfrak p}\operatorname N(\mathfrak p)^{-s}}=\lim_{s\to1^+}\frac{\frac1{[L:K]}(-\log(s-1))}{(-\log(s-1))},\] which is \(\frac1{[L:K]}\).

Observe that the proof of Proposition 4 more or less upgrades the following special case.

Corollary 6. Fix Galois extensions \(L\) and \(L’\) of a number field \(K\). Then \(L=L’\) if and only if \(\operatorname{Spl}(L’/K)=\operatorname{Spl}(L/K)\) (possibly away from a finite set).

Proof. This is the equality case of Proposition 4. \(\blacksquare\)

Proof of Theorem 2. Embed all fields into \(\overline{\mathbb Q}\). We would like to show that \(\sigma(L)=L\) for any automorphism \(\sigma\) of \(\overline{\mathbb Q}\). By Corollary 6, it is enough to check that \[\operatorname{Spl}(L/K)\stackrel?=\operatorname{Spl}(\sigma(L)/K),\] possibly away from a finite set. (Note \(\sigma(K)=K\) because \(K\) is Galois over \(\mathbb Q\).) Well, a prime \(\mathfrak p\) of \(K\) splits completely in \(\sigma(L)\) if and only if \(\mathrm{Frob}_{\mathfrak p}|_{\sigma(L)}\) is trivial. However, hitting everything in sight with \(\sigma^{-1}\), it is equivalent to show that \[\sigma^{-1}\mathrm{Frob}_{\mathfrak p}\sigma|_L\] is trivial, which is equivalent to \(\mathrm{Frob}_{\sigma^{-1}\mathfrak p}|_L\) being trivial. This last condition is equivalent to \(\sigma^{-1}\mathfrak p\in\operatorname{Spl}(L/K)\), which is equivalent to \(\mathfrak p\in\operatorname{Spl}(L/K)\) (possibly away from a finite set) by the Galois-invariance hypothesis. This completes the proof. \(\blacksquare\)

Now that we are done proving our theorem, let’s give another application.

Example 7. Let \(K\) be an imaginary quadratic field, and choose \(\alpha\) so that \(\mathcal O_K=\mathbb Z+\mathbb Z\alpha\). For each positive integer \(n\), we define the order \(\mathcal O_n=\mathbb Z+\mathbb Zn\alpha\), and we define the ring class field \(H_n\) to correspond to the adelic quotient \[K^\times\backslash\mathbb A_{K,f}^\times/\widehat{\mathcal O}^\times_n\] via class field theory. As such, away from a finite set, a prime \(\mathfrak p\) of \(K\) splits completely in \(H_n\) if and only if a uniformizer of \(K_{\mathfrak p}^\times\hookrightarrow\mathbb A_{K,f}^\times\) lands in \(K^\times\widehat{\mathcal O}^\times_n\). This latter condition is conjugation-invariant (because \(K^\times\) and \(\widehat{\mathcal O}^\times_n\) are both preserved by conjugation), so Theorem 2 implies that the extension \(H_n/\mathbb Q\) is Galois.

Remark 8. Of course, Theorem 2 is not the only way to prove the claim of Example 7, but it is a quick way. Note that the “usual” argument which shows that the Hilbert class field is Galois does not really apply: one usually shows that the Hilbert class field is Galois by using the fact that it is the maximal unramified abelian extension, which is a property preserved by composites and embeddings. It seems trickier to give an analogous description of the ring class fields in Example 7.