Let $K/\mathbb{Q}$ be a number field. We say that a rational prime $p$ splits in $K$ if there exists a prime $\mathfrak{p}$ of $K$ above $p$ of interia degree $1$.

Is a number field $K$ uniquely determined by the set of primes which split in $K$?

A well-known application of the Chebotarev density theorem (Neukirch Cor. 13.10) says that this is true when $K/\mathbb{Q}$ is Galois (note that here a prime splits if and only if it splits completely). So I am really interested in what happens in the non-Galois case.

Note also that the answer to the analogous question for completely split primes is no; indeed a prime is completely split in $K$ if and only if it is completely split in the Galois closure of $K$.


1 Answer 1


See exercises (6.3) and (6.4) of Cassels-Frohlich book on algebraic number theory. In these exercises, an example of two number fields $E,E'$ with the same zeta function is given; therefore, the set of primes which split in $E,E'$ are the same.

This amounts to constructing two subgroups $H,H'$ in a finite group $G$, which are not conjugate but meet every conjugacy class of $G$ in the same number of elements.

Since the OP has asked for split (but not necessarily completely split), the fact that equality of zeta functions implies that the split primes are the same needs a small argument. Let $p$ be a prime and $f_1,\cdots, f_r$ be the residue class degrees in $E$ over $p$ (in decreasing order); simlarly $g_1,\cdots, g_s$ the residue class degrees in $E'$ over $p$ (in decreasing order). Assume $f_1\geq g_1$. The local zeta functions at $p$ are

$$(1-X^{f_1})\cdots (1-X^{f_r})=(1-X^{g_1})\cdots(1-X^{g_s})$$where $X=p^{-s}$. Since the multiplicity of the root $X=1$ is the same, we get $r=s$. The cyclotomic polynomial $\Phi _{f_1}(X)$ divides the left hand side and hence the right hand side. Therefore, by the uniqueness of irreducible factors on the LHS-RHS, $g_1=f_1$. We can thus cancel the factor $1-X^{f_1}$ on both sides and use induction on $r$ to conclude that $f_i=g_i$ for all $i$.

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    $\begingroup$ I just thought that it was worth pointing out that number fields with the same Dedekind zeta functions are called "arithmetically equivalent" and were studied by Robert Perlis (sciencedirect.com/science/article/pii/0022314X77900701#) where he showed that they can only arise from the the group theoretic condition mentioned by Venkataramana. (Number theorists tend to call these groups "Gassmann equivalent", whereas Riemannian geometers call them "almost conjugate". They are closely related to the construction of isospectral non-isometric Riemannian manifolds.) $\endgroup$
    – user1073
    Nov 5, 2014 at 14:29
  • $\begingroup$ The following answer to a MO question also addresses this question. mathoverflow.net/q/3450 $\endgroup$ Oct 24, 2016 at 12:53

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