Let $\theta = \sqrt[3]2$ be the real cube root of $2$. While proving that the integral closure $O$ of $\mathbb Q[\theta]$ over $\mathbb Q$ is $\mathbb Z[\theta]$, I was doing some computations, which I would like to verify.
It is easy to see that $\mathbb Z[\theta] \subset O$. We have a lemma which states that if $\{x_i\}$ is a basis of $\mathbb Z[\theta]$ as a module over $\mathbb Q$, then $O$ is contained in $\mathbb Z[y_i]$, where $y_i$ is a dual basis to $x_i$, where $y_i \in \mathbb Q[\theta]$.
Now, I chose the basis $x_i = \{1,\theta,\theta^2\}$. With respect to this, the trace matrix $T_{ij} = \mbox{tr}(x_ix_j)$ should look like : $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \\ \end{pmatrix} $$
Where $\theta,\theta^2$ have trace zero since their minimal polynomials are $x^3 - 2$ and $x^3 - 4$ respectively, whose next-to-leading coefficient is zero. Now, computing the dual basis is tantamount to finding the inverse of this matrix and taking the columns. Turns out the inverse of $T$ is just $\frac 14 T$.
Consequently, the dual basis is just the original basis scaled down by $4$.
This allows us to conclude that $4O \subset \mathbb Z[\theta]$.
All I would like to ask is : is the computation of the dual basis, and the resulting conclusion correct? The problem is, my printing material says $3 O \subset \mathbb Z[\theta]$(which is eventually correct) but this seems incorrect for now, in the light of my computation, since I was asked to conclude this using the dual basis argument.