Why are there only finitely many cubic fields of a given discriminant? Is this true for higher dimension too?
What other invariants are needed to classify cubic fields? number of real and complex embeddings, Galois closure?...
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ To classify cubic fields, look up the Davenport-Heilbronn theorem. $\endgroup$– S. Carnahan ♦Commented Mar 3, 2013 at 13:54
-
6$\begingroup$ Dear Solovei, In fact there are finitely many number fields (of arbitrary degree --- i.e. I am not fixing the degree a priori) of bounded discriminant, see e.g. math.wisc.edu/~ellenber/papers.html#BEL Regards, $\endgroup$– EmertonCommented Mar 4, 2013 at 3:36
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
Your first two questions have already received good answers, so let me offer a few references about "classifying cubic fields", and point to some of my colleagues' interesting work in the subject.
What information determines the field? This question turns out to be quite subtle. In general, even the Dedekind zeta function is not enough -- see this paper of Bosma and de Smit (which does not address cubic fields though).
Scott Carnahan pointed you to the Davenport-Heilbronn theorem, which gives an asymptotic for the counting function of cubic fields with bounded discriminant. This paper of Bhargava, Shankar, and Tsimerman provides, among much else, a beautifully written self-contained introduction to the proof of this theorem.
For classifying cubic fields, knowing the discriminant is not enough. In particular, cubic fields with discriminant $D$ are, by class field theory, in bijection with subgroups of the class group of $\mathbb{Q}(\sqrt{D})$ of index 3, and it is expected (google the "Cohen-Lenstra heuristics") that these 3-torsion groups can be arbitrarily large.
Along these lines, an interesting paper was written by Guillermo Mantilla-Soler, see here, where he studies the extent to which cubic fields are determined by their integral trace forms.
You can also study cubic fields by means of their quadratic resolvent, for example in this paper of Cohen and Morra. You can also look at the related 3-torsion problem in terms of Heegner points -- see this paper of Bob Hough, and there is forthcoming work of Yongqiang Zhao, who develops a powerful approach to counting cubic function fields in the language of algebraic geometry. Or, you can use zeta functions.
There are a lot of ways to understand cubic fields -- and yet plenty of open questions remain.
answered Mar 4, 2013 at 6:12
-
$\begingroup$ It is apparently known that the number of cubic fields with discriminant $D$ is unbounded, so by the bijection with the subgroups of the class group of $Q(\sqrt{D})$, that you mention, these subgroups are arbitrarily large. $\endgroup$ Commented Mar 4, 2013 at 11:55
-
$\begingroup$ @solovei: I am unfamiliar with the result you mention, and would be extremely interested to read a proof. If you know of a reference would you please share it with me? Thank you! $\endgroup$ Commented Mar 4, 2013 at 16:39
-
$\begingroup$ @Frank See page 77 of Elementary and Analytic Theory of Algebraic Numbers By Wladyslaw Narkiewicz. $\endgroup$ Commented Mar 4, 2013 at 17:09
$\begingroup$
$\endgroup$
1
This a theorem of Hermite (true for number fields of any degree). I don't know of a reference, but I think it is good to look at the theorems of Hunter and Martinet which give you a finite search region for the minimal polynomial of a generator of a number field of given discriminant. A great reference is
Henri Cohen, {\em Advanced topics in computational number theory}, Springer GTM 193, 2000.
-
7$\begingroup$ It is proved in Serge Lang's book on algebraic number theory that the number of number fields of a given degree and discriminant is finite, using the "Minkowski estimate". $\endgroup$ Commented Mar 3, 2013 at 13:29
$\begingroup$
$\endgroup$
A theorem of Birch and Merriman proves that over $R$ the ring of integers of a number field $K$, there are finitely many $\text{GL}_2(R)$-orbits of homogeneous binary forms $f(x,y)$ with $\text{deg}(f) = n$ and $\text{Disc}(f) = D_0$.
Birch and Merriman write: "...Hermite proved a theorem with deceptively similar enunciation, but with a different less natural 'determinant' in place of our discriminant; his determinant is skilfully devised so that the theorem is provable by reduction theory".
http://plms.oxfordjournals.org/content/s3-24/3/385.full.pdf
Hope that helps!
answered Mar 4, 2013 at 1:47