MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to show that all possible splitting fields occur for a class of cubic polynomials, so have started by looking at the discriminants.

Clearly, given that they are products of squares, all discriminants are congruent to 0 or 1 mod 4. And it is easy to show that all numbers of this form appear as discriminants of quadratics. But this does not seem to be the case for discriminants of cubics.

One would think that this would have been determined, but I can't seem to find any mention of it in any literature. After a lengthy search, I can only produce about a third of all numbers 0 or 1 mod 4 lying between -100 and 100.

Does anyone know of any other restrictions? Am I missing something obvious?!

share|cite|improve this question
up vote 18 down vote accepted

There are two different types of cubic fields: cyclic cubics and cubics with Galois group $S_3$.

Cyclic cubic fields are, by Kronecker Weber, subfields of cyclotomic fields. There is no cyclic cubic field with discriminant $5^2$ since the field of $5$th roots of unity does not have a cubic subfield; the cyclic cubic field with discriminant $7^2$ is contained in ${\mathbb Q}(\zeta_7)$.

Nonnormal Cubic fields can be explained by class field theory: the existence of the field with discriminant $-23$ comes from the fact that the quadratic number field with discriminant $-23$ has class number divisible by $3$. More generally, there is a cubic number field with discriminant $-4f^2$ if the order in ${\mathbb Z}[i]$ with conductor $f$ has class number divisible by $3$.

All of this was worked out by Hasse (Arithmetische Theorie der kubischen Zahlkörper auf klassenkörpertheoretischer Grundlage, Math. Z. 31 (1930), 565-582). For related material in English see Daniel Mayer, Multiplicities of dihedral discriminants, Math. Comput. 58 (1992), 831--848.

share|cite|improve this answer

Do you need more than is in ? The discriminants of cubic polynomials are there explained in terms of discriminants of cubic fields, and asymptotics given for the latter.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.