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 $\mathbb{Q}$