Sum of Reciprocals of Primes in Imaginary Quadratic Field Diverges (2014 Miklós Schweitzer) Problem 5 of the 2014 Miklós Schweitzer states: Let $\alpha$ be a non-real algebraic integer of degree two, and let $P$ be the set of irreducible elements of the ring $\mathbb{Z}[\alpha]$. Prove that $$\displaystyle\sum_{p\in P}\frac{1}{|p|^2}=\infty.$$
I think I've figured this out for the cases in which $\mathbb{Z}[\alpha]$ has class number $1$, but have no idea how to extend this reasoning (if it is even correct): Let $\delta$ denote the discriminant of $\mathbb{Z}[\alpha]=\mathcal{O}(\mathbb{Q}(\sqrt{-d}))$. Then, a prime $p$ is expressible as a quadratic form (here, the norm of our quadratic field) with fundamental discriminant $\delta$ of class number $1 \iff (p/\delta)=1$. Rational primes $q$ are inert if $(q/\delta)=-1$. Thus, the sum in question should look something like $$\sum_{(p/\delta)=1}\frac{1}{p}+\sum_{(q/\delta)=-1}\frac{1}{q^2},$$ where $p$ and $q$ range over all rational primes. By Dirichlet's theorem, the sum on the left contains $\frac{1}{2}$ of all primes, and should thus diverge, causing the entire sum to diverge. Note that this chain of reasoning holds only in the very specific case that $\mathbb{Z}[\alpha]$ is a UFD, since we can write things like $(\delta/p)=(p/\delta)$ by Quadratic reciprocity.
I'm clueless to how to approach the general case when $\mathbb{Z}[\alpha]$ is not a UFD, and any help would be greatly appreciated. Perhaps since number fields are Dedekind we could look at the norms of prime ideals? I've scoured the internet but am unable to find any solution to this. Thanks in advance, and hopefully what I've written isn't completely bogus.
 A: You can of course use Cebotarev to answer this question, but there is a completely elementary argument.
By the usual equivalence between the (absolute) convergence of sums and products, we see that the
$$\sum_{p \in P} \frac{1}{|p|^2}$$
is divergent if and only if the following product is divergent:
$$P = \prod_{p \in P} \left(1 - \frac{1}{|p|^2} \right)^{-1} = \prod_{p \in P} \left(1 + \frac{1}{|p|^2} + \frac{1}{|p|^4} + \ldots \right)$$
Now consider the sum:
$$S =  \sum \frac{1}{|n|^2},$$
where the sum is over all the non-zero elements of $\mathbf{Z}[\alpha]$. An elementary estimate shows that $S = \infty$ (for example, using the integral test). Now every non-zero element $n \in \mathbf{Z}[\alpha]$ may be written as a product of irreducible elements. The product decomposition is not unique in general, but there is always at least one such decomposition, and that is all that will matter for this argument. That means that every term in $S$ occurs in the expansion of the product of $P$, and since all the terms in $P$ are positive, we have
$$P \ge S = \infty,$$
and we are done.
