If $K=\mathbb Q(\sqrt{15})$ then $\mathcal O_K=\mathbb Z[\sqrt{15}]$. $13$ is prime in $\mathcal O_K$.
Question: Does there exist and ideal $\mathfrak {a}$ of $\mathcal{O}_K$ such that $N (\mathfrak a)=|\mathcal{O}_K/\mathfrak {a}|=13$?
Thank you all.
Because $N(IJ) = N(I)N(J)$, we know that this ideal should be prime. Indeed, by factorizing the ideal into prime ideals, we will get that the norm of every ideal is $1$ except for one prime ideal. Suppose the prime ideal $P$ satisfies $N(P) = 13$. Then $P \mid (\,p)$ for some prime $p$, where $(p) = p\mathcal{O}_K$. But then $$13 = N(P) \mid N(p\mathcal{O}_K)= N_{\mathbb{Q}(\sqrt{15})/\mathbb{Q}}(p) = p^2$$ Which implies that $p = 13$.
But as you said, $13$ is prime in $\mathcal{O}_K$, so $p\mathcal{O}_K$ is a prime ideal. This would imply that $P = p\mathcal{O}_K$, clearly a contradiction.