apologies if this is a naive question. Consider two Galois extensions, K and L, of the rational numbers. For each extension, consider the set of rational primes that split completely in the extensions, say Split(K) and Split(L).
If Split(K) = Split(L), then is it necessarily true that K and L are isomorphic as Galois extensions of the rationals?
If so, for a given set of rational primes, S, is there a way to construct the extension over which S is the set of completely split primes?
References welcomed! Thanks, Martin
Best Answer
This result is due to Bauer and dated to 1916:
$\textbf{Theorem}:$ Let $K$ be an algebraic number field, $L/K$ and $M/K$ finite Galois extensions, and $\text{Spl}(L/K)$, $\text{Spl}(M/K)$ the set of prime ideals of $K$ which split completely in $L$ and $M$, respectively. Then $L \subseteq M$ if and only if $\text{Spl}(M/K) \subseteq \text{Spl}(L/K)$.
The case of $K=\mathbb{Q}$ answers your first question in the affirmative.
As for the second question, not every subset of rational primes is $\text{Spl}(K/\mathbb{Q})$ for a number field $K$, simply by cardinality arguments. Class field theory gives a description of $\text{Spl}(L/K)$ when $\text{Gal}(L/K)$ is abelian, but the general problem is open and very hard.
An excellent source for this material is Keith Conrad's notes on the history of class field theory, which you can find under the Expository Notes section on his website here: http://www.math.uconn.edu/~kconrad