Least rational prime which is composite in $\mathbb{Z}[\alpha]$? Sébastien Palcoux asked if there was some irrational algebraic $\alpha$ such that all rational primes are primes in $\mathbb{Z}[\alpha].$ MooS answered that there are no such $\alpha.$ This leads to a natural question: 

Given some irrational algebraic $\alpha,$ what is the least rational prime $p$ such that $p$ is composite in $\mathbb{Z}[\alpha]$?

I'm looking for an upper bound on $p$. (I wasn't able to extract one from MooS' proof -- it can be greater than $f(x)$.)
Probably this question would work better in $\mathcal{O}_{\mathbb{Q}(\alpha)}$ than $\mathbb{Z}[\alpha]$; feel free to use whichever is more convenient.
 A: A partial result is the following:
For any finite set $\{p_1, \dotsc, p_s\}$ of prime numbers, there exists an algebraic irrational number $\alpha$ such that all $p_i$ are prime in $\mathbb Z[\alpha]$.
To that account, it suffices to show: There is a polynomial in $\mathbb Z[x]$ of degree at least $2$, which is irreducible over $\mathbb F_{p_i}$ for all $i$ and irreducible over $\mathbb Z$.
We proceed by induction on $s$, the case $s=1$ being the equivalent to the statement, that $\mathbb F_p$ is not algebraically closed.
Let (exists by induction hypothesis) $f \in \mathbb Z[x]$ be an irreducible polynomial of degree $n$ which is irreducible modulo $p_i$ for $1\leq i \leq s-1$. Let $P = p_1 \dotsb p_{s-1}$. $P$ is coprime to $p_s$, hence there is $Q \in \mathbb Z$ with $PQ=1 \mod {p_s}$
Let $g \in \mathbb Z[x]$ be a monic polynomial of degree greater than $n$, which is irreducible modulo $p_s$. Consider $$h := f + P(g-Qf)$$
Modulo $p_i, 1\leq i \leq s-1$, we obtain $h = f$, hence irreducible.
Modulo $p_s$, we obtain $h = Pg$, hence irreducible.
Furthermore, the leading coefficient of $h$ is $P$, hence co-prime to $p_s$. So the irreducibility in $\mathbb Z[x]$ follows from the irreducibility in $\mathbb F_{p_s}$.

This result shows, that there is no global upper bound. So the upper bound depends on $\alpha$. If you ask about being prime, the answer is just the least prime, that occurs among the the prime factors of $f(x)$, while $x$ runs through $\mathbb Z$ and $f$ is the minimal polynomial.
If you ask about being irreducible (which I guess you do), the question might be better investigated in $\mathcal O_{\mathbb Q(\alpha)}$. It comes down to the question which is the least prime among the norms of elements of that ring. But I cannot think any further.
