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Apparently this should be a straightforward / standard homework problem, but I'm having trouble figuring it out.

Let $D$ be a square-free integer not divisible by $3$. Let $\theta = \sqrt[3]{D}$, $K = \mathbb{Q}(\theta)$. Let $\mathcal{O}_K$ be the ring of algebraic integers inside $K$. I need to find explicitly elements generating $\mathcal{O}_K$ as a $\mathbb{Z}$-module.

It is reasonably clear that $\theta$ is itself an algebraic integer and that $\mathbb{Z}[\theta] \le \mathcal{O}_K$, but I strongly suspect it isn't the whole ring. I'm not sure where the hypotheses on $D$ come in at all... any hints would be much appreciated.

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    $\begingroup$ A general element of $\mathbb{Q}(\theta)$ has the form $a+b\theta+c\theta^2$ for some $a,b,c \in \mathbb{Q}$. Work out the minimal polynomial of such an element, and check when it's monic. $\endgroup$ Feb 12 '11 at 15:49
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    $\begingroup$ I don't think there is any really clean way to do this computation. If you look at math.uconn.edu/~kconrad/blurbs , at the files entitled "Invariants of the Splitting Field of a Cubic n", for $1 \leq n \leq 5$, you'll find many worked examples. $\endgroup$ Feb 12 '11 at 17:01
  • $\begingroup$ @David Speyer: Those look nice, but unfortunately I know nothing about local fields or ramification of primes. Yet. Perhaps I'll come back to them when I know more. $\endgroup$
    – Zhen Lin
    Feb 14 '11 at 18:45
  • $\begingroup$ When $D = 2$, for example, $\mathbf Z[\theta]$ is the whole thing. $\endgroup$ Jan 23 '12 at 16:10
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A very belated answer: This is (part) of the content of Exercise 1.41 of Marcus' Number Fields (a great source of exercises in basic number theory). In it, one proves that, for $K = \mathbf{Q}(m^{1/3})$, $m$ squarefree, an integral basis is given by

$\begin{cases} 1, m^{1/3}, m^{2/3},& m \not \equiv \pm 1 \pmod 9 \\ 1, m^{1/3}, \frac{m^{2/3} \pm m^{1/3} + 1}{3},& m \equiv \pm 1 \pmod 9 \end{cases}$

This is leveraged out of his Theorem 13 (among other things).

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  • $\begingroup$ It's actually exercise 2.41 of Marcus' Number Fields (in the 1987 edition which is the only one to my knowledge), sorry to resurrect the post. $\endgroup$
    – TrostAft
    Nov 14 '18 at 1:48
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I thought this was a fun problem. Let $z = a + b \theta + c \theta^2$ be an algebraic integer, $a, b, c \in \mathbb{Q}$. Then the coefficients of the characteristic polynomial of $z$ must be integers. This tells you, for example, that $\text{tr}(z) = 3a$ is an integer. The other two coefficients are slightly harder to work with, but here's a start: since $\theta z$ and $\theta^2 z$ are algebraic integers, their traces are also integers, so...? And then you work with the other coefficients of the minimal polynomial.

The hypotheses on $D$ come from some divisibility arguments you will need to make later; without them the problem is harder. Note that the discriminant of $\mathbb{Z}[\theta]$ has absolute value $27D^2$, which means that the index of $\mathbb{Z}[\theta]$ in $\mathcal{O}_K$ divides $3D$. (Actually it will turn out to divide $3$.)

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  • $\begingroup$ Ick. I was hoping not to have to directly do such case-bashing. But if you say so... $\endgroup$
    – Zhen Lin
    Feb 12 '11 at 15:49
  • $\begingroup$ Another hint that might be worth giving: you do not need to solve a nontrivial-looking Diophantine equation to solve this problem. (And I haven't indicated the need for any case-bashing so far! If you are careful with your casework you will only need two cases corresponding to the value of D mod 3.) $\endgroup$ Feb 12 '11 at 15:54
  • $\begingroup$ @Qiaochu Yuan, how do you know the trace is in $\mathbb{Z}$? $\endgroup$ Nov 22 '17 at 13:48
  • $\begingroup$ @Koenraad: let $K$ be a number field and $a \in K$ be an element. $a$ has a characteristic polynomial given by the characteristic polynomial of the linear operator on $K$ given by multiplication by $a$, and the trace is one of its coefficients. It's an exercise to show that $a \in \mathcal{O}_K$ iff the coefficients of the characteristic polynomial are integers. (I misspoke slightly above; the characteristic and minimal polynomial need not agree, e.g. if $a \in \mathbb{Q}$.) $\endgroup$ Nov 22 '17 at 18:03

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