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Suppose we are considering the field $Q(m^{1/3})$. According to Marcus, Pg-18, if $m$ is a square free integer, and $m\not\equiv 1\text{ or }-1\pmod9$, then $1,m^{1/3},m^{2/3}$ forms an integral basis. Can anyone help me to reach this conclusion? I am confused as the discriminant of this basis comes out to be $-27m^2$, which is not square free, hence, this should not be the basis. Am I wrong somewhere?

Lios
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roydiptajit
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  • There are fields with non-square free discriminants. The usefulness of squares in determining the ring of integers is that it limits the type of bigger rings that could possibly exist. IIRC in your example case you can use the properties of trace to show that any denominators (in the coefficients of an algebraic integer w.r.t. this basis) must be divisors of three. After that it becomes more complicated. – Jyrki Lahtonen Oct 27 '20 at 07:26
  • See this thread (and the other linked to it) for a discussion. You see that the logic flows along a one-way street here. – Jyrki Lahtonen Oct 27 '20 at 07:28
  • Okay, so the other direction does not work I see.. Then how to prove that $1,m^{1/3},m^{2/3}$ generates the ring? Are there any thread that discusses about this? – roydiptajit Oct 27 '20 at 08:00
  • To what extent can you generalize this approach? Veteran tip: look at the right margin for Related threads. When your question is as high quality as this (a descriptive title etc), then the heuristics producing that list works remarkably well :-) – Jyrki Lahtonen Oct 27 '20 at 08:09
  • This prove introduces a lemma which I am not familiar with. I tried to find a more easy explanation, but unable to find it. In Ram Murthy a proof is given, but it is taking me a lot time to understand. Maybe I need to read a couple times more.. – roydiptajit Oct 27 '20 at 19:30
  • I'm fairly sure that the collected wisdom/knowledge in those answers is pretty much everything that can be explained in a small space. – Jyrki Lahtonen Oct 28 '20 at 11:20
  • Do check, if new answers come to this question. – Jyrki Lahtonen Oct 28 '20 at 16:18
  • Okay, thanks for your help – roydiptajit Oct 28 '20 at 16:31

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