Sum of Reciprocals of Primes in Imaginary Quadratic Field Diverges (2014 Miklós Schweitzer)

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.