October 27th
Today I learned a proof (after having skimmed one a few days ago) that a Galois extension is completely determined by its completely split primes. For a Galois extension $K/\QQ$ (or $L/K$; the proof extends), let $\op{Spl}(K/\QQ)$ be the set of primes which split completely.
Now, the claim we show is that for two Galois extensions $K/\QQ$ and $L/\QQ,$ $K$ is a subfield of $L$ if and only if more primes completely split in $K$ than in $L,$ in the sense that $\op{Spl}(L/\QQ)\subseteq\op{Spl}(K/\QQ).$ Indeed, if $K$ is a subfield of $L,$ then a prime splitting completely in $L$ will split completely in $K$ for free because the ramification and inertial information are multiplicative in towers and so upper-bounded by $1.$
The converse is harder. The setup is to consider the field composite $LK,$ and the main idea is to use our connection between completely splitting primes and field invariants—namely, the Dirichlet density of completely split primes is the reciprocal of the degree. In symbols,\[\frac1{[LK:\QQ]}=\lim_{s\to1^+}\dfrac{\displaystyle\sum_{p\in\op{Spl}(LK/\QQ)}\frac1{p^s}}{\displaystyle\sum_p\frac1{p^s}}\]from our discussion yesterday. However, a prime splits completely in $LK$ if and only if it splits completely in both $L$ and $K$; we show the converse later, but the forward direction is discussed above. So the fact that the primes which split completely in $L$ also split completely in $K$ tells us that we only care about the primes which split completely in $L.$ Namely, $\op{Spl}(LK/\QQ)=\op{Spl}(L/\QQ),$ which implies\[\frac1{[LK:\QQ]}=\lim_{s\to1^+}\dfrac{\displaystyle\sum_{p\in\op{Spl}(LK/\QQ)}\frac1{p^s}}{\displaystyle\sum_p\frac1{p^s}}=\lim_{s\to1^+}\dfrac{\displaystyle\sum_{p\in\op{Spl}(L/\QQ)}\frac1{p^s}}{\displaystyle\sum_p\frac1{p^s}}=\frac1{[L:\QQ]}.\]To finish, we write\[[LK:L]=\frac{[LK:\QQ]}{[L:\QQ]}=1,\]so $L=LK,$ and $K$ is indeed a subfield of $L.$
To get the result we want, it follows that if $\op{Spl}(K/\QQ)=\op{Spl}(L/\QQ),$ then each is a subset of the other, so each of $K$ and $L$ are a subfield of the other, so $L=K.$ This is the statement that the set of completely split primes entirely determines the field extension, which is equivalent to the statement $K\mapsto\op{Spl}(K/\QQ)$ is an injective function, as shown.
I guess I should show that if a prime splits completely in $L$ and $K,$ then it also splits completely in the composite $LK.$ The trick is to take $M$ a normal extension containing $LK$ and then note that the decomposition field of a certain prime contains $L$ and $K$ (because the prime splits completely) and so contains $LK,$ implying that the prime splits completely. In particular, we are using the characterization that the decomposition field is the largest intermediate field such that the prime $\mf p/p$ has $e(\mf p/p)=f(\mf p/p)=1.$