Are the sets of valuations uniquely determined

Question

Let $\mathcal{V_K}$ be the set of valuations of a number field $K$.

Can it be that $\mathcal{V_L}=\mathcal{V_K}$, for the set of valuations of another number field $L$ non-isomorphic to $K$?

Answer 1

Your question is rather vague, because of the term « valuations ». Speaking of number fields, perhaps do you mean « places » or « primes » (finite or infinite) in the usual language of ANT. Then :

1). If $K$ and $L$ have the same set of infinite primes, i.e. of archimedean absolute values induced by embeddings into an algebraic closure $\bar {\mathbf Q} $, they must have the same number of real (resp. complex) embeddings, hence in particular the same degree over $\mathbf Q$.

2). What does it mean that $K$ and $L$ have the same set of finite primes? The only definition which makes sense is that the rational primes in $\mathbf Q$ share the same ramification-decomposition properties in $K / \mathbf Q$ and in $L / \mathbf Q$. In particular, as pointed out by Lubin, $K$ and $L$ have the same set of rational primes which split completely . It then follows from the Tchebatorev density theorem that $K$ and $L$ are isomorphic (see e.g. exercise 6 at the end of Cassels-Fröhlich’s book "ANT"). Note that this remains true when $\mathbf Q$ is replaced by any common subfield of $K$ and $L$ .