Is there an algebraic non-rational extension of the integers, whose set of prime elements contains the prime integers?

Let the ring $\mathbb{Z}[\alpha]$ with $\alpha$ an algebraic number.
Let $P(\mathbb{Z}[\alpha])$ be the set of all the prime elements of $\mathbb{Z}[\alpha]$.

Question: Is there $\alpha$ algebraic and non-rational such that $P(\mathbb{Z}) \subset P(\mathbb{Z}[\alpha])$?

Solution 1:

Let $f \in \mathbb Z[X]$ be the (maybe non-monic, but definitely primitive) minimal polynomial of $\alpha$. We can assume the leading coefficient to be positive.

Consider the homomorphism $\mathbb Z[X] \to \mathbb Z[\alpha], X \mapsto \alpha$. The kernel is a non-maximal prime ideal containing $f$, hence it is equal to $(f)$, since $(f)$ is a prime-ideal (f is irreducible since it is primitive and irreducible over $\mathbb Q$) and any prime ideal properly containing $(f)$ is maximal. We deduce $\mathbb Z[\alpha] \cong \mathbb Z[X]/(f)$.

Pick $x \in \mathbb Z$ suitable large, such that $f(x) > 1$. In particular there is a prime $p$ with $p|f(x)$. We compute

$$\mathbb Z[\alpha]/(p) \cong \mathbb Z[X]/(f,p) \cong \mathbb F_p[X]/(f)$$

The latter is no integral domain, since $f(x) = 0$ in $\mathbb F_p$, in particular $f$ is not irreducible.