The final task is to show that the integral closure of $\mathbb Q[\theta]$ over $\mathbb Q$ is $\mathbb Z[\theta]$, where $\theta = \sqrt[3]3$.
I was given a list of steps, and I would like to check that I have done them right. First, denote the integral closure (which we must show is equal to $\mathbb Z[\theta]$) by $O$.
1 . Show that $Z[\theta] \subset O$.
If $x = a + b \theta + c \theta^2$ then $(T- x)(T - x_1)(T-x_2) = 0$ is a monic polynomial in $T$ with integer coefficients which $x$ satisfies, where $x_1$ and $x_2$ are the expression for $x$ with $\theta$ replaced by $\omega \theta$ and $\omega^2 \theta$ where $\omega^3 = 1$ is an imaginary cube root of unity.
2 . Use the dual basis to show that $3O \subset \mathbb Z[\theta]$.
The dual basis(under the trace) for the (ordered) standard basis $\{1,\theta,\theta^2\}$ for $\mathbb Q[\theta]$ over $\mathbb Q$ can be found to be $\{\frac 13, \frac{\theta^2}{9},\frac{\theta}{9}\}$. From here, we conclude that $\mathbf 9O \subset \mathbb Z[\theta]$. I do not see how to get from $9$ to $3$ using just the dual basis argument alone.
3 . Show that $O / (\theta) \cong \frac {\mathbb Z}{3\mathbb Z}$.
To see this, we note that $O/(3) \cong (\frac{\mathbb Z}{3 \mathbb Z})^3$ (I will not elaborate here : this is a standard result I can use). Now, define the maps $O/(3) \to O/(\theta^2) \to O/(\theta)$ by $a + (3) \to a + (\theta^2) \to a + (\theta)$.
Clearly, these are well defined as $(3) \subset (\theta^2) \subset (\theta)$. Furthermore, they are surjective maps, and are not isomorphisms, since for example, $\theta^2 + (3) \to 0$ and $\theta + (\theta^2) \to 0$. (Since $\theta^{-1} \notin O$, which is easy to check, the above cosets are non-zero cosets).
Consequently, $O/(\theta)$ is isomorphic to a strict subgroup of a strict subgroup of $(\frac{\mathbb Z}{3\mathbb Z})^3$, which leaves only two options : either it is trivial, or it is equal to $\frac {\mathbb Z}{3\mathbb Z}$. But of course it cannot be trivial : $1 + (\theta) \neq 2 + (\theta)$ are two distinct cosets, for example, as $1 \notin (\theta)$. So the result follows.
4 . Show that $(\theta)^rO + \mathbb Z[\theta] = O$ for all $r \geq 1$.
I am slightly unsure on how to do this part. In particular, I would like to quotient out $(\theta)$ and claim that the left hand side is just $\frac{\mathbb Z[\theta]}{(\theta)}$, since the other subgroup consists of multiples of $\theta$. This quotient has exactly three elements, so the isomorphism should follow.
5 . Deduce the conclusion.
We have for $r = 6$, that $(\theta^6)O + \mathbb Z[\theta] = 9O + \mathbb Z[\theta] = \mathbb Z[\theta]$ by the dual basis argument, but then from 4. we are done.
In particular, point 4 is my only hindrance, but I have put out the entire exercise so that somebody can point out other mistakes or places to be careful if required.