Infinitely many primes in the ring of integers

Let $K = \Bbb{Q}(\alpha)$; write $f(x)$ for the minimal polynomial of $\alpha$ over $\Bbb{Q}$. Our task now is to create infinitely many prime ideals $P$ in $\mathcal{O}_K$ such that $f(P|p) = 1$ where $p$ is the prime in $\Bbb{Z}$ lying under $P$. Now we know that there are infinitely many primes $p$ for which $f$ has a root mod $p$ . Thus for each such prime $p$ is an integer $n_p$ such that

$$g(x)(x - n_p) = f(x) \pmod{p}.$$

Then we claim that for all but finitely many primes $p$ the ideal $P = (p, \alpha- n_p)$ is a prime ideal. When $p \nmid n$ where $n = \left[\mathcal{O}_K : \Bbb{Z}[\alpha]\right]$, it is shown in pg.3 of the document "Factoring After Dedekind" that the map $$\overline{\iota} : \Bbb{Z}[\alpha]/(p) \to \mathcal{O}_K/(p)$$ induced by inclusion $\iota: \Bbb{Z}[\alpha] \to \mathcal{O}_K$ is an isomorphism. Now consider the diagram

where the bottom row is an isomorphism by the first isomorphism theorem. The bottom left and right rings are each isomorphic respectively to $\Bbb{Z}[\alpha]/(p,\alpha - n_p)$ and $\mathcal{O}_K/(p,\alpha - n_p)$ by the third isomorphism theorem. Thus the primality of the ideal $(p,\alpha - n_p)$ in $\mathcal{O}_K$ will follow immediately from the fact that \begin{eqnarray*} \Bbb{Z}[\alpha]/(p,\alpha - n_p) &\cong& \bigg(\Bbb{Z}[x]/(f(x)) \bigg) \bigg/ \bigg((p,x - n_p)/(f(x))\bigg)\\ &\cong& \Bbb{Z}[x] / (p, x -n_p) \\ &\cong& \Bbb{Z}/p\Bbb{Z}[x]/ (x- \overline{n_p}) \\ &\cong& \Bbb{Z}/p\Bbb{Z}.\end{eqnarray*}

Note that my answer in some ways is a generalisation of Rankeya's because we don't assume that $\mathcal{O}_K = \Bbb{Z}[\alpha]$.

Hint: The inertial degree is the degree of the field extension $\mathbb{Z}/(p) = \mathbb{F}_p \subset \mathbb{Z}[\alpha]/\mathcal{P}$.

Given what you already know, the only sensible choice of the polynomial $f$ is to take $f$ to be the minimal polynomial of $\alpha$ over $\mathbb{Q}$, which since $\alpha$ is integral, will actually have coefficients in $\mathbb{Z}$.

Now there are infinitely map primes $p$ for which $f(x) \equiv 0 (mod p)$ has a solution $r_p$ in $\mathbb{Z}$. Consider the ideal $(\alpha - r_p, p) \subset \mathbb{Z}[\alpha]$.

We would like $(\alpha - r_p, p)$ to be a prime ideal. To show this, observe that we have $\mathbb{Z}[\alpha]/(\alpha - r_p, p) \cong \mathbb{Z}[x]/(f(x), x - r_p , p) = \mathbb{Z}[x]/(x-r_p,p)$.

Why does the last equality hold (hint divide $f(x)$ by $x - r_p$ and deduce that the constant remainder is a multiple of $p$ from the fact that $f(r_p) \equiv 0 (mod \hspace{1mm} p)$)?

What is the ring $\mathbb{Z}[x]/(x - r_p , p) $ isomorphic to?

Now can you conclude that $(\alpha - r_p, p) \cap \mathbb{Z} = (p)$.

I think my solution is okay (can someone confirm this?) given that in your question $\mathcal{O}_K = \mathbb{Z}[\alpha]$. Things would have been slightly more complicated if $\mathcal{O}_K$ was just the ring of integers of $K = \mathbb{Q}(\alpha)$ for an algebraic interger $\alpha$.