(A post about Guillermo Mantilla-Soler’s paper posted on the arXiv yesterday.)

The most natural arithmetic invariant of a number field K is its *discriminant* D_K, an integer congruent to either 0 or 1 mod 4 whose prime factors are precisely the primes where K/Q is ramified. Oftentimes D_K is a squarefree, in which case it’s just the product of the ramified primes; even if not, the multiplicity of a prime factor of D_K can be described quite cleanly in terms of the p-inertia subgroup of Gal(K/Q).

The situation is especially handsome for quadratic field. The discriminants of quadratic fields are just those integers congruent to 0 or 1 mod 4 which have no square factor larger than 4. Better still, the discriminant specifies the field uniquely! So in order to describe a quadratic field it suffices to write down a single integer.

Life gets worse in higher degree. There can be lots of number fields with the same discriminant D. For example: if D is squarefree and K is a cubic field with discriminant D, then the Galois closure L of K is an unramified (Z/3Z)-extension of the quadratic field M with discriminant D. So if the ideal class group of M has a lot of (Z/3Z) in it, there are going to be a lot of cubic fields with discriminant D!

Just how bad is this multiplicity? It’s widely believed that, for every n, there are at most D^eps number fields of discriminant D. But I think nobody has a good idea about how to prove this, even for n=3.

So it’s naturally interesting to ask whether there are other invariants which might uniquely specify a number field. Here’s one natural candidate. The ring of integers O_K is a free rank n Z-module, endowed with a natural quadratic form q(x,y) = Tr_{K/Q}(xy), called the *trace form.* The discriminant of this trace form is, up to known factors, the discriminant of K. So you can think of the isomorphism class of the trace form as a refinement of the discriminant. The question is: is it such a good refinement that it actually specifies the field?

My former student Guillermo Mantilla-Soler, now at UBC, just posted a preprint offering the first real insight into this question, which he colorfully phrases “Can the trace hear the shape of its field?” He shows that the answer to the original question is *no: *for instance, he displays two non-isomorphic quintic fields of discriminant 34129 which have isomorphic trace form. More generally, he gives a necessary condition which I would expect is satisfied by examples in every degree (though it might be hard to prove this.)

But in some sense this is a local issue; the fields in these examples are not totally real, so the trace forms aren’t definite, and so, as Guillermo observes, it suffices to show the forms lie in the same spinor genus. In the definite case, it’s “harder” for quadratic forms to be isomorphic. Are there two non-isomorphic *totally real* number fields with isomorphic trace forms? Guillermo includes the results of a fairly large computer search which finds no examples in degree < 10 and discriminant < 10^9.

“Are there two non-isomorphic totally real number fields with isomorphic trace forms? ”

Your faithful reader conjectures anonymously that there are infinitely many in degrees 3 and 4, not so above.

Dear Anonymous: I think your conjecture seems very believable in degree 3. Here is a possible argument. Any two Galois cubic fields of the same discriminant have the same trace and trace zero forms ( see for example Thm 3.1 in http://arxiv.org/abs/1104.4598v1) so one can reduce the problem to find infinitely many pairs of Galois cubic fields of the same discriminant. I think finding cyclotomic extensions with Galois groups with 3-rank at least 2 should do the trick (maybe with some conditions on the discriminant.) For degree 4 I have no idea. I’d be curious to see the heuristics leading for your conjecture in degree 4. I should mention that the above is bit far from what is discussed in the paper. The big difference is that in the paper the main characters are fields of fundamental discriminant. So for example if one does not put any restriction on the discriminant it is possible to find several examples of totally real number fields with the same trace form. However, if one restricts this to fundamental discriminants there are no such a examples with bounded discriminants as described in the post. In fact if one works over totally real cubic fields the trace zero form totally characterizes the field, and I have conjectured the same is true in degree 4.

PS: I’m in the middle of a 20 hours trip back home so I hope the above is not to incoherent.

One can find arbitrarily many cyclic cubic fields of the same discriminant.

This is immediate from a theorem of Cohn, which is restated as (1.1) here: