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Let $\vartheta$ be an algebraic integer of degree $n$. The algebraic number field $\mathbb{Q}(\vartheta)$ has always an integral basis of the form

$\displaystyle\omega_{1}=1,$

$\displaystyle\omega_{2}=\frac{a_{{21}}\!+\!\vartheta}{d_{2}},$

$\displaystyle\omega_{3}=\frac{a_{{31}}\!+\!a_{{32}}\vartheta\!+\!\vartheta^{2}%
}{d_{3}},$

$\vdots\,\qquad\vdots\,\qquad\vdots$

$\displaystyle\omega_{n}=\frac{a_{{n1}}\!+\!a_{{n2}}\vartheta\!+\ldots+\!a_{{n,%
n-1}}\vartheta^{{n-2}}\!+\!\vartheta^{{n-1}}}{d_{n}}$,

where the $a_{{ij}}$’s and $d_{i}$’s are rational integers such that

$d_{2}\mid d_{3}\mid d_{4}\mid\ldots\mid d_{n},$ |

i.e.

$d_{i}\mid d_{{i+1}}\quad\forall\,i=2,\,3,\,\ldots,\,n\!-\!1.$ |

The integral basis $\omega_{1},\,\omega_{2},\,\ldots,\,\omega_{n}$ is called a canonical basis of the number field.

Remark. The integers $a_{{ij}}$ can be reduced so that for all $i$ and $j$,

$-\frac{d_{i}}{2}<a_{{ij}}\leqq\frac{d_{i}}{2}.$ |

Then one may speak of an adjusted canonical basis. In the case of a quadratic number field $\mathbb{Q}(\sqrt{d})$ with $d\equiv 1\,(\mbox{mod}\,4)$ we have (see the examples of ring of integers of a number field)

$\omega_{1}=1,\quad\omega_{2}=\frac{1\!+\!\sqrt{d}}{2}.$ |

The discriminant^{} of this basis is $d$.

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## Comments

## Canonical basis

If \theta^3 = 2, what is like a canonical basis of Q(\theta)?

## Re: Canonical basis

I noticed that no one has replied to this for over a year!

A canonical basis for Q(\theta) where \theta^3=2 is {1, \theta, \theta^2}. Using the notation of the entry to which the original post is attached, all of the a_{ij}'s for i>1 are 0, and all of the d_i's are 1.

## Re: Canonical basis

What is not said in the theorem the way it's written here is that the $d_i$ are unique... That is, if you find a basis that works, then you've found The basis (well, you still have some freedom as how to choose the a_{i,j}).

The real problem is not so much to find a canonical basis, it's to find a basis. Characterizing the integral closure of Z in a weird ring is the difficult part (once you have a basis you'll easily find a way to torture it to make it canonical, this is classical Z-module algebra).

What you know for sure is that $1,\theta, \tetha^2$ are algebraic integers. The question is, did you forget some of them. Q(\theta) can be represented as a 3 dimensional vector space over Q and theta is an endomorphism with matrix

0 0 2

1 0 0

0 1 0

any element of Q(\theta) is therefore a matrix

a 2c 2b

b a 2c = a+b\theta+c\theta^2

c b a

And its characteristic polynomial is (I'll spare you the calculation)

-x^3+3ax^2+(-3a^2+6bc)x+a^3+2b^3+4c^3-6abc

The fact that the elements are in the integral closure of Z means that

3a, (-3a^2+6bc) and a^3+2b^3+4c^3-6abc are all integers (but a,b and c are only rational numbers !)

As I said before you may be missing generators of the integral closure but you have already 1,\theta and \theta^2, so you may restrict your search to new generators that lie in [0,1]^3, that is, you can suppose 0<=a<1, 0<=b<1 and 0<=c<1. This makes the determination of the factors possible (though tedious, as was computing the characteristic polynomial...)

Ok, let's see what happens in this case, I'll just use my computer because it is very boring and clearly error prone to do it by hand:

3a is an integer (first coefficient), therefore a=0,1/3 or 2/3

Lets look at the second equation:

if a=0 6bc integer, b=1/2,c=1/3 or b=1/3, c=1/2

if a=1 6bc-1/3 integer, hence 3|(18bc-1) and in particular 18bc integer

if a=2 6bc-4/3 integer and in particular 18bc integer.

therefore 18b and 18c are always integers, and we have finitely many possibilities to test.

The simplest way to do it is to try everything with a computer. Everything being try every combination of b and c rational <1 such that 18bc integer. b=0,1/18,1/9,1/6,1/3,1/2 same for c, and a=0,1/3,2/3

I used the 3 maple lines :

g:=a^3+2*b^3+4*c^3-6*a*b*c;

L:=[0,1/18,1/9,1/6,1/3,1/2];

simplify(seq(seq(seq(eval(g,{a=i/3,b=L[j],c=L[k]}),i=0..2),j=1..6),k=1..6));

The output contains no integer (except for a=b=c=0...), therefore there is not elements of the algebraic integer ring that we have to add to 1,\theta,\theta^2.

In other words, 1,\theta,\theta^2 is already a set of generators (hence basis, they are linearly independent!), and it is obviously canonical.

To consider an example where the integer ring is not as trivial as here, you can do the same for Q(sqrt(d)) (assume that d has no square factor), you get the characteristic polynomial

x^2-2ax+a^2-db^2

2a is an integer, a=0 or 1/2

if a=0, -d b^2 integer, d has no square factor therefore b has to be an integer itself and <1, therefore b=0, this does not add anything

now if a=1/2, 1/4-db^2 integer, 4|(4db^2-1), in particular 4db^2 is an integer, therefore b can be 0 or 1/2.

For b=0, 1/4-db^2=1/4, that doesn't work.

For b=1/2, 1/4-d/4 is an integer if and only if d mod 4=1

Therefore if d mod 4=1, we have to add the extra element a=1/2,b=1/2 ie. (1+sqrt(d))/2 to the ring, and we have a set of generators with (1, sqrt(d),(1+sqrt(d))/2) but as I mentioned at the very beginning, you can easily modify that set of generators to find a basis (canonical), here you see that it's 1,(1+sqrt(d))/2

If d mod 4 \neq 1, then there is no extra element, and, as in the case of Q(2^(1/3)), (1,sqrt(d)) is already a set of generators, and that is obviously in canonical form already.

Final note, the structure of those rings is complicated and there is no general way to deal with them. Although some subtle theorems can help you find a basis in some special cases (for your example there are ways to prove that there is no element to add without doing explicit calculations), the only way to treat all the cases in the one I gave you, that is, reduce to a finite number of possibilities and try them all. This process can be completely automated.

Good luck now :)

## Re: Canonical basis

Thanks for your interesting answer!

## Re: Canonical basis

You should also check out

http://www.math.uconn.edu/~kconrad/blurbs/Qw2.pdf (as well as Qw3,5,6,7) and

http://www.math.uconn.edu/~kconrad/blurbs/powerbasis.pdf

## Re: Canonical basis

Thanks, Roger, for very interesting references!

Jussi