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Let $K/\mathbb{Q}$ be a cubic extension of the rationals. Skimming the internet and Alan Baker's A Comprehensive Course in Number Theory I get the impression that, unlike the quadratic case, there is no explicit description of $K$'s ring of integers $\mathcal{O}_K$ except for special cases.

  • The case $K=\Bbb{Q}(\sqrt[3]{d})$ with $d$ squarefree is described here
  • Baker mentions (Example 10.1) that if $\alpha$ is a root of $x^3+px+q$ with $p,q$ integers such that the discriminant $-(27q^2 + 4p^3)$ is squarefree, then $(1,\alpha,\alpha^2)$ is an integral basis of ${\cal O}_{\Bbb{Q}(\alpha)}$.

Is this impression correct? If so, what makes the cubic case hard? If not, where can I read about the cubic case?


The paper The integral closure of cubic extensions by Keith Dunbar McLean contains a discussion of the topic. Paragraph 2 of the first chapter of the second chapter says "Few, if any, elementary texts discuss cubic extensions, possibly because of the unmanageable computations involved [...]"

It contains a discussion of the "pure cubic" case where $K=\Bbb{Q}(\sqrt[3]{d})$ with $d$ cube free (pages 36 and onwards).


The pure cubic case is described in @nguyenquangdo's answer here.


There is also the paper The simplest cubic fields by Daniel Shanks that seems far more advanced and on cursory inspection does not provide an answer to my question.


EDIT. There are explicit descriptions of integral bases for the ring of integers of cubic extensions, see On Voronoi's Method for Finding an Integral Basis of a Cubic Field and An Explicit Integral Basis for a Pure Cubic Field

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    $\begingroup$ Vague but possibly useful comment: I would claim that the cubic case isn't hard—it's the problem of finding rings of integers that is in general hard, and the quadratic case happens to be much easier than typical. $\endgroup$ – Greg Martin Jan 23 at 23:43
  • $\begingroup$ @GregMartin Yes, indeed. I thought of including the question "alternatively, what makes the quadratic case easier" but decided against it as I thought that it might be "vague" as you put it. $\endgroup$ – Olivier Bégassat Jan 23 at 23:54
  • $\begingroup$ For a unified solution for pure cubic fields, see ethe ref. given in math.stackexchange.com/a/3508522/300700 $\endgroup$ – nguyen quang do Jan 24 at 7:42
  • $\begingroup$ Thanks for the link @nguyenquangdo! $\endgroup$ – Olivier Bégassat Jan 24 at 9:19
  • $\begingroup$ Related but different question: math.stackexchange.com/questions/2197359/… $\endgroup$ – Mr. Brooks Jan 25 at 22:33

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