Prime decomposition in algebraic integer rings Let $\mathcal{O}_K \subset \mathcal{O}_L$ be algebraic integer rings and $\theta \in \mathcal{O}_L$ such that $L=K( \theta )$. Let $\mathfrak{F}_\theta$ be the unique maximal ideal of $\mathcal{O}_L$ contained in $\mathcal{O}_K[\theta]$. We know that if a prime $\mathfrak{p}$ of $\mathcal{O}_K$ is prime to $\mathfrak{F}_\theta$, then we can get the prime decomposition of $\mathfrak{p}$ in $\mathcal{O}_L$ using the minimal polynomial of $\theta$ over $K$.
My question is that is there any conclusion saying that we can get the prime decomposition of all primes of $\mathcal{O}_K$ using this method by changing the $\theta$.
 A: First let me clarify what the $\mathfrak{F}_\theta$ is. $\mathfrak{F}_\theta=\{ \alpha \in \mathcal{O}_K[\theta] \mid \alpha \mathcal{O}_L \subset \mathcal{O}_K[\theta] \}$ is the largest ideal of $\mathcal{O}_L$ contained in $\mathcal{O}_K[\theta]$.
Then, if $K=\mathbb{Q}$ and $\mathcal{O}_K=\mathbb{Z}$, the condition $p$ prime to $\mathfrak{F}_\theta$ is equivalent to $p \nmid |\mathcal{O}_L/\mathbb{Z}[\theta]|$. For the necessity, if $p$ is prime to $\mathfrak{F}_\theta$, then $p \nmid |\mathcal{O}_L/\mathfrak{F}_\theta |$. Hence $p \nmid |\mathcal{O}_L/\mathbb{Z}[\theta]|$ since $|\mathcal{O}_L/\mathbb{Z}[\theta]| \mid |\mathcal{O}_L/\mathfrak{F}_\theta |$. For the sufficiency, if  $p \nmid |\mathcal{O}_L/\mathbb{Z}[\theta]|$, then the map from $\mathcal{O}_L/\mathbb{Z}[\theta]$ to itself which sends every element to its multiple by $p$ is an isomorphism of the additive group. Then $p \mathcal{O}_K + \mathbb{Z}[\theta]=\mathcal{O}_K$. Therefore, $p \mathcal{O}_K + \mathfrak{F}_\theta=\mathcal{O}_K$, that is, $p$ prime to $\mathfrak{F}_\theta$.
Next is our counterexample. Let $L=\mathbb{Q}(\theta)$ where $\theta^3+\theta^2-2\theta+8=0$. As in the handout mentioned above, $2 \mid |\mathcal{O}_L/\mathbb{Z}[\alpha] |$ for every integral primitive element $\alpha$. Hence $2$ cannot be decomposed using this method in $\mathcal{O}_L$.
This shows that the general answer is NO. As mentioned in the handout, the prime decomposition lies somewhat deeper than factoring in $\mathbb{F}_q[X]$.
