Is $p=257$ prime in $\mathbb Z[\sqrt{ -92}]$?
My thoughts: We have $\mathbb Z[\sqrt{ -92}]\cong\mathbb Z[X]/(X^2+92)$, so $\mathbb Z[\sqrt{ -92}]/(p)\cong\mathbb Z_{257}[X]/(X^2+92)$. Then $p$ is prime if $(p)$ is prime and $(p)$ is prime iff the ring above is an integral domain. But $X^2+92$ is not irreducible over $\mathbb Z_{257}$ since the equation $a^2\equiv 165\bmod 257$ has a solution (the Jacoby-Symbol is $\left(\frac{165}{257}\right)=1$). So the ring above is not an integral domain and $p$ is not prime.