Let $K=\mathbb{Q}(\theta)$ where $\theta$ is a root of $X^3-2X-2$. Integral basis for $\mathcal{O}_{K}$? 
Let $K=\mathbb{Q}(\theta)$ where $\theta$ is a root of $X^{3}-2X-2$. Compute an integral basis for $\mathcal{O}_{K}$.

I have computed the discriminant as $\Delta_{K}=-2^2.19$. I want to apply the algorithm in the book of Frazer Jarvis Algebraic Number Theory. We know that if $\beta=\frac{u_0+u_1\theta+u_2\theta^2}{2}$ is an integral element then its trace and norm is an integer. $Trace(\beta)=\frac{3u_0+4u_2}{2}$. Since $0 \leq u_i<2$ for any i, $u_0=0.$ Then the only possible $\beta$ are : $\frac{\theta}{2},\frac{\theta^2}{2},\frac{\theta+\theta^2}{2}$. But when I computed the norm of all possiblities, Norms are 1/4, 1/2, 1/4 respectively, then they are not integer.
Is there any computational mistakes or what is wrong?
 A: The reason why neither of the norms is an integer is because the elements listed are not algebraic integers. Indeed, for example $\frac{\theta + \theta^2}2$ has $4X^3 - 8X^2 - 4X - 1$ as a minimal polynomial. In fact, it turns out that the ring of integers $\mathcal O_K$ is equal to $\mathbb Z[\theta]$, which can be deduced by eliminating all possibilites that come from applying the algorithm. So an integral basis of it is $\{1,\theta,\theta^2\}$.
There is another way to compute the disciminant $d_K$ of the extension $K/\mathbb Q$, which is much more theory dependent, but doesn't involve long computations. In shortly, the only unknown prime power of $d_K$ is the power of $2$. To find out, you make use of the fact that $X^3 - 2X - 2$ is an Eisenstein polynomial with respect to $2$, compute the $2$-adic different of $K/\mathbb Q$ and deduce that $4$ is the highest power of $2$ dividing $d_K$. So as $d_K = \text{disc} f$ we deduce that $\mathcal O_K = \mathbb Z[\theta]$.
However, I suspect that this method might be too advanced. But, if you want me to, I can include the computation steps, as well as giving you references for the theoretic stuffs.
