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I'm trying to understand the concept of maximal order and unit group of an algebraic number field. By definition, the maximal order of an algebraic number field $F$ is the set of algebraic integers in F, i.e.

$O(F)=\{a \in F \mid \text{there exists a monic} \; f_a(x) \in \mathbb{Z}[x] \; \text{with} \; f_a(a)=0\}$

and the unit group of $F$ is

$U(F)=\{ a \in O(F) \mid a \neq 0 \; \text{and} \; a^{-1} \in O(F)\}$

To play with this, I'm looking at the splitting field $K$ of $f(x)=x^3-2$ over $\mathbb{Q}$. The roots of this polynomial are $\sqrt[3]{2}, \sqrt[3]{2} \left( \frac{-1+\sqrt{-3}}{2} \right), \sqrt[3]{2} \left( \frac{-1-\sqrt{-3}}{2} \right)$. So $K=\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})$ and $[K:\mathbb{Q}]=6$.

I think the maximal order of $K$ is $O(K)=\{a_1+a_2\sqrt[3]{2}+a_3\sqrt[3]{4}+a_4\sqrt{-3}+a_5\sqrt[3]{2}\sqrt{-3}+a_6\sqrt[3]{4}\sqrt{-3} \mid a_i \in \mathbb{Q} \}$. Is this correct?

Slightly unrelated question: I know a bit of GAP (the algebraic computation system). Can GAP tell me about maximal order of algebraic number field?

Edit: Define $a_i$ from $O(K)$ more carefully to be in $\mathbb{Q}$

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Hint: $\frac{1+\sqrt{-3}}2\in O(K)$ –  Thomas Andrews Nov 2 '11 at 17:45
    
I forgot to put in $a_i$ from $O(K)$ is in $\mathbb{Q}$. Then $\frac{1+\sqrt{-3}}{2} \in O(K)$ –  nullgraph Nov 2 '11 at 18:16
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But then $O(K)$ contains lots of non-integer rationals. –  Thomas Andrews Nov 2 '11 at 18:29
    
That's fine. $O(K)$ is the set of algebraic integers. An element $\alpha \in \mathbb{C}$ is an algebraic integer if there exists a monic irreducible polynomial $f(x) \in \mathbb{Z}[x]$ s.t. $f(\alpha)=0$. Btw, my definitions are from D. Holt "Handbook of Computational Group Theory", maybe you're using a different definition? –  nullgraph Nov 2 '11 at 19:02
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No, when you write: $$O(K)=\{a_1+a_2\sqrt[3]{2}+a_3\sqrt[3]{4}+a_4\sqrt{-3}+a_5\sqrt[3]{2}\sqrt{-3}+‌​a_6\sqrt[3]{4}\sqrt{-3} \mid a_i \in \mathbb{Q} \}$$ that means that every linear combination with all $a_i$ is in $O(K)$. It is true that every element of $O(K)$ can be written this way, but that's not useful, since every element of $K$ can be written that way. The right hand side of the equation is $K$, so you have written $O(K)=K$. If you meant $O(K)\subset K$, that would be true, but hardly useful –  Thomas Andrews Nov 2 '11 at 19:14

1 Answer 1

up vote 4 down vote accepted

This should be doable fairly easily by hand, and you'll learn much more that way. I recommend doing the problem first for the real cubic field ${\mathbb Q}(2^{1/3})$, where you'll find that an integral basis of the ring of algebraic integers is the obvious one, $\lbrace 1, 2^{1/3}, 2^{2/3} \rbrace $. A bit more fussing will persuade you that you probably have found a fundamental unit in $2^{1/3}-1$; general theory says that for this field, the complete group of units is $\lbrace \pm U^n\rbrace$, where ordinarily it's not easy to find $U$. The way things turn out is that it's easy to know the shape of the unit group, but hard to compute it (except in fields quadratic over $\mathbb Q$), while it's hard to predict the "class number" of the field without doing a specific computation, but relatively easy to find it computationally.

I confess I'm too lazy to look up the facts about the normal closure, ${\mathbb Q(2^{1/3},\omega})$, where $\omega$ is a primitive cube root of unity. My guess is that one possible integral basis is made up of the three elements mentioned above together with what you get by multiplying them by $\omega$. Again, general theory says that now there are two independent fundamental units, and I'll bet a nickel that you can take them to be the one I mentioned before and $\omega 2^{1/3}-1$

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