If almost all primes of $K$ that are totally split in $M$ are totally split in $L$, then $L \subset M$.

Question

In Cox's book Primes of the form $x^2+ny^2$, we have the following proposition 8.20.

Let $K$ be a number field, $L,M$ be finite extensions of $K$ such that $M/K$ is Galois. Then $$L \subset M \\\iff\\ \text{all but finitely many of the primes of $K$ that are totally split in $M$ are also totally split in $L$.}$$

The direction $\implies$ is not difficult. However, the direction $\Longleftarrow$ is proved using Čebotarev theorem. Is there a way to prove this proposition without using this theorem (nor class field theory)?

Let's assume first that $L/K$ is Galois. I tried to show that $LM=M$ by looking at the degree $[LM:M]=efr$ where I take a prime of $LM$ above a prime of $M$... I know that "totally split in $L$ and in $M$" implies "totally split in $LM$". But I'm stuck there.

Thank you!

Answer 1

Thanks to Franz Lemmermeyer's comments above, I am able to provide an answer. I will however assume that $L/K$ is Galois. The direction $\implies$ is not really difficult. The converse is more interesting.

Let's introduce the following definition:

Definition. If $T$ is a set of prime ideals of a number field $K$ and if $\zeta_{T,K}^n$ has a pole of order $m$ at $s=1$ where $$\zeta_{K,T}(s):=\prod\limits_{p \in T} \dfrac{1}{1-N(p)^{-s}},$$ then $$d(T)=m/n$$ is called the polar density of $T$.

Here $N(p) = |O_K/p|$ is the absolute norm of $p$.

We can check the following properties:

— The polar density is additive : if $T$ and $T'$ are two disjoint sets of primes of $K$, then $$d(T \sqcup T') = d(T)+d(T')$$

– The polar density is monotonic : if $T \subset T'$ then $$d(T) \leq d(T')$$

– A finite set of primes of $K$ has density $0$, and the set of all prime ideals of $K$ has density $1$.

$\newcommand\Spl{\mathrm{Spl}}$ The main claim is the following:

Theorem: The set $\Spl(L/K)$ of prime ideals of $K$ that are totally split in $L$ has polar density $1/[L:K]$.

Then the inclusion $L \subset M$ follows from $\Spl(M/K) \setminus T \subset \Spl(L/K)$, where $T$ is a finite set of primes of $K$. Indeed, we want to show $LM=M$, i.e. $[M:K]=[LM:K]$. Since $\Spl(M/K) \cap \Spl(L/K) = \Spl(LM/K)$, we get $$\Spl(M/K) \setminus T \subset \Spl(LM/K) \subset \Spl(M/K)$$ Then $d(\Spl(M/K))=d(\Spl(M/K) \setminus T ) + d(T) = d(\Spl(M/K) \setminus T )$, since $T$ is finite. Therefore, we obtain $$d(\Spl(M/K))=d(\Spl(M/K) \setminus T ) \leq d(\Spl(LM/K)) \leq d(\Spl(M/K))$$ whence $$\dfrac{1}{[LM:K]} = d(\Spl(LM/K)) = d(\Spl(M/K)) = \dfrac{1}{[M:K]},$$ which shows $[M:K]=[LM:K]$ as desired.

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Let's prove the theorem.

Proof. $\newcommand\Spec{\mathrm{Spec}}$

Step 1. Let $S=\Spl(L/K)$ and $T = \{q \in \Spec(\mathcal{O}_L) \mid q \cap K \in S\}$. We claim that $$\zeta_{L,T}(s) = \zeta_{K,S}(s)^{[L:K]}.$$ (in the next steps, we will show that $\zeta_{L,T}$ has a pole of order $1$ at $s=1$).

Indeed, if $p \in S$, then $pO_L = \mathfrak{q}_1 \cdots \mathfrak{q}_{[L:K]}$ with $N_{L/\mathbb{Q}}(\mathfrak{q}_i) = N_{K/\Bbb Q}(p)$ and $\mathfrak{q}_i \in T$. Therefore $$ \zeta_{L,T}(s) = \prod\limits_{\mathfrak{q} \in T} (1-N(\mathfrak{q})^{-s})^{-1} = \prod\limits_{p \in S} \prod\limits_{\mathfrak{q} \mid p} (1-N_{L/\mathbb{Q}}(\mathfrak{q})^{-s})^{-1} = \prod\limits_{p \in S} (1-N(p)^{-s})^{-[L:K]} = \zeta_{K,S}(s)^{[L:K]} $$

Step 2. We know that $T$ contains $$Q := \{\mathfrak{q} \in \Spec(\mathcal{O}_L) \mid N(\mathfrak{q}) \text{ is prime in } \mathbb{Z} \;\text{ and }\; \mathfrak{q} \cap K \text{ is unramified in } L\}$$ so that writing $\Spec(\mathcal{O}_L) = T \sqcup T^c$, where $T^c$ is the complement of $T$ in $\Spec(\mathcal{O}_L)$ yields $T^c \subset Q^c = P \cup R$ where $$P:=\{\mathfrak{q} \in \Spec(\mathcal{O}_L) \mid N(\mathfrak{q}) \text{ is not prime in } \mathbb{Z} \}$$ and $$R:=\{\mathfrak{q} \in \Spec(\mathcal{O}_L) \mid \mathfrak{q} \cap K \text{ is ramified in } L \}.$$ Notice that $R$ is finite, so that $d(R)=0$.

Step 3. We show that $d(P)=0$. If $\mathfrak{q} \in P$, then $N_{L/\mathbb{Q}}(\mathfrak{q}) = p^f$ for some prime $p \in \mathbb{Z}$ and $f \geq 2$.

Conversely, given a prime $p \in \mathbb{Z}$, there are at most $[L:\Bbb Q]$ primes of $L$ above $p$.

Then $$\zeta_{L,P}(s) = \prod\limits_{\mathfrak{q} \in P} (1-N(\mathfrak{q})^{-s})^{-1} = \prod\limits_{\substack{p \in \mathbb{Z} \\ p \text{ prime }}} \prod\limits_{i=1}^{[L:\mathbb{Q}]} g_i(p;s)= \prod\limits_{i=1}^{[L:\mathbb{Q}]} \underbrace{\prod\limits_{\substack{p \in \mathbb{Z} \\ p \text{ prime }}} g_i(p;s)}_{=: g_i(s)} $$ where $p\mathcal{O}_L = \mathfrak{q}_1^{e_1} \cdots \mathfrak{q}_{r(p)}^{e_{r(p)}}$ and$$g_i(p;s):= \begin{cases} (1-p^{-f(\mathfrak{q}_i/p)s})^{-1} & \text{if }\; i \leq r(p) \leq [L:\mathbb{Q}] \;\text{ and }\; f(\mathfrak{q}_i/p) \geq 2\\ 1 & \text{else.} \end{cases}$$

But for all $i$, we have $$g_i(1) = \prod\limits_{\substack{p \in \mathbb{Z} \\ p \text{ prime }}} g_i(p;1) \leq \prod\limits_{\substack{p \in \mathbb{Z} \\ p \text{ prime }}} (1-p^{-2})^{-1} = \zeta(2) < \infty$$

Therefore $g_i$ is holomorphic at $s=1$and so is $\zeta_{L,P} = \prod_{i=1}^{[L:\mathbb{Q}]} g_i$.

Therefore $d(P)=0/1=0$.

Step 4. Finally, $d(T^c) = 0$ and by monotonicity, $d(T)=1$, which means that $\zeta_{L,T}$ has a pole of order $1$ at $s=1$. Since $\zeta_{L,T}(s) = \zeta_{K,S}(s)^{[L:K]}$, we can write $d(S)=\dfrac{1}{[L:K]}$, as desired. $\hspace{17cm}\blacksquare$

All these ideas can be found in Milne, Class field theory, chap. VI, §3.