Showing these prime ideals are principal Let $K=\mathbb{Q}(\theta)$ be a number field where $\theta$ has minimal polynomial $x^3-9x-6$. I had to factorise the ideals $(2)$ and $(3)$ into prime ideals, for which I got $(2) = (2,\theta)(2,\theta +1)^2$ and $(3) = (3,\theta)^3$ by using Dedekind's criterion.
Then I am to show these prime ideals are principal. I'm a bit stuck at this point. The best I can think to do is to choose some element I might believe generates the whole ideal and show that we can obtain the generators from it. I've had a few guesses for only $(2,\theta)$ and I can't manage to get there.
I imagine there is a better way to do this than to simply strike it lucky with a good guess. Is there a better way to proceed in finding a generator, or maybe some other method of showing the ideals are principal?  
 A: $\require{cancel}$
As is usual, we let $q(x)=q_\theta(x)=x^3-9x-6$. As Jyrki notes, $q(x-1)$ is the minimal polynomial for $\theta+1$, but then $(\theta+1,2)=(\theta+1, (\theta+1)p(\theta))$ for some polynomial $p(x)\in\mathcal{O}_K$, so that $(\theta+1,2)=(\theta+1)$ is principal.
So let's do the others. Since $2\in (\theta+1)$ we have that there exists $p(x)\in\mathcal{O}_K[x]$ so that $2=(\theta+1)p(\theta)$. We note that $\theta^3-9\theta-6=0$ is equivalent to

$$\theta(\theta-3)(\theta+3)=2\cdot 3$$

Then writing this as $\theta(\theta+1-4)(\theta+1+2)=2\cdot 3$ we see that this makes $\theta\cancel{(\theta+1)}(1-(\theta+1)p(\theta)^2)(1+p(\theta))=3\cancel{(\theta+1)}p(\theta)$.
Since the other two factors on the LHS are congruent to $1$ modulo $p(\theta)$ we must have that $\theta\in (p(\theta))$ since reduction modulo $(p(\theta))$ on both sides yields $0$. So $\theta=p(\theta)r(\theta)$. But then if we consider how the definition of ideal addition gives $(2,\theta)=(2)+(\theta)$ then we see

$$(2,\theta)=(2)+(\theta)=(\theta+1)(p(\theta))+(p(\theta))(r(\theta))=(p(\theta))\big((\theta+1)+(r(\theta)).$$

Since we already know $(\theta+1)$ to be prime, it must be that $(\theta+1)\not\big|(r(\theta))$ or else $(\theta+1)|(2,\theta)$ a contradiction since $q(x)$ is not a cube modulo $2$. We conclude $(\theta+1)+(r(\theta))=\mathcal{O}_K$ so that $(2,\theta)=(p(\theta))$.
We turn our attentions to $(3,\theta)$. By using $\theta^3=3(3\theta+1)$ we wish to compute the norm of $\theta^3$. By looking at the $\Bbb Q$-basis for $K$ $\{1,\theta,\theta^2\}$ we can compute the norm of $\alpha=3\theta+2$ by looking at the determinant of the matrix of the linear transformation induced by multiplication by $\alpha$ relative to this basis which is

$$M_\alpha=
\begin{pmatrix} 2 & 27 & 18 \\
3 & 2 & 0 \\
0 & 3 & 2
\end{pmatrix}$$

which has determinant $8$. This is most easily done (by-hand) by row reducing once--the operation is just $R_1\to R_1-9R_3$--then expanding along the top row which now has only one non-zero entry.
We conclude $N(\theta)=3^3\cdot 2^3$. Since $(3\theta+2)$ factors as primes above $8$ and must be a perfect cube, we determine that $(3\theta+2) =(2(\theta+1)+\theta)$ must be equal to $(p(\theta))^3$, since we've already shown that $(2)$ factors as principal ideals and since clearly $3\theta+2\equiv \theta\equiv 1\mod{(\theta+1)}$. From this and our earlier deduction that $\theta=p(\theta)r(\theta)$, we have that $(r(\theta))^3=(3,\theta)^3$, and again since the RHS is the cube of a prime ideal, we have that $(r(\theta))=(3,\theta)$.
A: Denote $p(x)=x^3-9x-6$. We see that $p(-1)=-1+9-6=2$. Consequently the constant term of $p(x-1)$ is equal to two. But $p(x-1)$ is the minimal polynomial of $\theta+1$. Expanding $p(x-1)$ we get that $\theta+1$ is a factor of two in the ring $\Bbb{Z}[\theta]$. This gives you the answer for the ideal $(2,\theta+1)$.
The other two are trickier. 
A: A way to restrict your search is to note that if $(\alpha)$ is a principal ideal, $N$ is the ideal on norms, and $\text{Norm}_{K/\mathbb{Q}}$ is the norm on elements, then
$$
N((\alpha)) = \text{Norm}_{K/\mathbb{Q}}(\alpha).
$$
