Is $1+\sqrt{5}$ a prime under the $\mathbb{Z}[{\sqrt{5}}]$ domain? The title is self-explanatory. I know it's irreducible but is it a prime? How to prove these primality and/or irreducibility of $1+\sqrt{5}$.
Can you just briefly state how a prime is defined under $\mathbb{Z}[{\sqrt{5}}]$? I know that it will only be divisible by its associates and the unity. But please tell about the norm conditions and other properties or the complete rigorous definition of primes.

One would usually define a non-zero element $a$ of a ring $R$ to be prime if the ideal it generates is a prime ideal.
Under this definition, $1+\sqrt 5$ is not prime, since $4 =2\times2= (1+\sqrt5)(\sqrt5-1) \in \langle1+\sqrt 5\rangle$, but $2\notin\langle1+\sqrt 5\rangle$.
You should note that $\mathbb Z[\sqrt 5]$ is not the ring of integers of $\mathbb Q(\sqrt 5)$. Rather $\mathbb Z[\frac{1+\sqrt 5}2]$ is, and $1+\sqrt 5$ is prime in this ring.
