What is the ring of integers for $\mathbb{Q}(\sqrt{23},\sqrt{3})$?

So, these are numbers of the form $a+b\sqrt{3}+c\sqrt{23}+d\sqrt{69}$ where $a,b,c,d\in\mathbb{Q}$, and we want to find ones whose minimum polynomial is monic. But I'm not sure how to find the minimum polynomial for a number of this form. Are there any theorems/methods to help?

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There are general algorithms to compute the integral basis of any number field. For example, see here.

That said, it's an exercise in most standard number theory courses to find integral bases for any biquadratic extension. There is an exercise (with hints) in Marcus's Number Fields which discusses this (it is exercise 42 on page 51)--you should do it.

It tells you, in his notation, that if we let $m=69$, $k=23$, and $n=3$, then your field has integral basis


Also, SAGE is your friend. It has the ability to compute the integral basis of any number field. For example, it spat out the following:

K.<a,b> = NumberField([x^2-23,x^22-3]);
[1, 5/2*a - 11/2*b, -1/2*b*a + 13/2, 4*a - 9*b] 
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  • $\begingroup$ Thanks for your answer -- would you mind posting the hints, since I don't have that book? $\endgroup$ – JJ Beck Feb 14 '14 at 7:29
  • $\begingroup$ @JJBeck What have you tried? $\endgroup$ – Alex Youcis Feb 14 '14 at 9:40
  • $\begingroup$ As I wrote in the post, I would like to find the minimum polynomial for $a+b\sqrt{3}+c\sqrt{23}+d\sqrt{69}$, but don't know how to do it. $\endgroup$ – JJ Beck Feb 14 '14 at 14:36
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    $\begingroup$ $1$, $\sqrt{3}$, $\sqrt{23}$ and $\sqrt{69}$ certainly generate an order $\cal O$ in the ring of integers ${\cal O}_K$. You should compute its discriminant and have a guess of what $[{\cal O}_K:{\cal O}]$ could be. Then you move from there. $\endgroup$ – Andrea Mori Feb 14 '14 at 23:32

There is a useful standard-classical theorem about rings of integers in a compositum of two linearly disjoint field extensions, when the "differents" are relatively prime, and the base ring of integers is a PID, the ring of integers is the tensor product (which, by the linear disjointness, naturally injects to the compositum).

This must appear in many places, but/and it is easily available on-line in my notes http://www.math.umn.edu/~garrett/m/number_theory/Notes_2011-12.pdf on page 101 and following, illustrated starting p. 103 on cyclotomic fields.

Perhaps it is worth noting that the "natural", but naive" approach of looking for the monic satisfied by a linear combination of the obvious algebraic numbers and "solving" for the integrality conditions is not viable. For quadratic extensions, it's fine, obviously, and classically known. For most other scenarios, even if/when it is possible to do-the-thing more-or-less directly (as Adrian Albert did in some papers from the 1940s or so), it is not illuminating, nor persuasive. That is, we have learned that this sort of question is not best answered just from "the definitions".

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  • $\begingroup$ +1 for mentioning the general viewpoint, and also for the link to your very nice notes on number theory. $\endgroup$ – Bill Dubuque Feb 15 '14 at 16:47

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