TITLE: Nonisomorphic number fields with the same Galois group and discriminant
QUESTION [5 upvotes]: It is possible for nonisomorphic fields to have the same discriminant, and even to be arithmetically equivalent, meaning with the same Dedekind zeta function. Previous questions which discussed this are "Number fields with same discriminant and regulator?" and "Are there two non-isomorphic number fields with the same degree, class number and discriminant?". The usual way this comes about involves fields with isomorphic subfields. I would find it useful if the discriminant sufficed to characterize number fields whose splitting field had a simple Galois group, so that there were no subfields. So the question is, can there be two nonisomorphic fields with the same degree, the same discriminant and the same simple Galois group for the splitting field?

REPLY [2 votes]: I just came upon this (old?) thread.  Here is a handy way to produce lots of non-Abelian, non-isomorphic extensions of Q with the same field discriminant:
Let K3/Q be a non-Abelian (resp Abelian) cubic field.  Let h3 be the size of the elementary 2-class group of K3.  Then Heilbronn (On the 2-classgroup of cubic fields, in: Studies in Pure Math}., 1971) showed that there is a (h3-1)  (resp (h3-1)/3) quadruplets of conjugate quartic fields K4 such that disc(K4)=disc(K3) and that the Galois closure of K4 contains K3.
So if you pick a K3 with large elementary 2-rank, then you will get lots of non-isomorphic K4 with Galois group S4 or A4, depending on whether K3 is non-Abelian or otherwise.
There is also a similar (and old) theorem of Hasse for 3-rank of quadratic fields.