Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z}[\sqrt{5}]$?

Question

Consider the integral domain $\mathbb{Z}[\sqrt{5}]$. Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z}[\sqrt{5}]$?

I do not know the answer, so any help is welcome.

Note that $4+\sqrt{5}$ is an irreducible element of $\mathbb{Z}[\sqrt{5}]$, since its norm $N(4+\sqrt{5})=11$ is a prime number (here as usual $N(a+b\sqrt{5})=a^2-5b^2$ for every $a, b \in \mathbb{Z}$). Anyhow $\mathbb{Z}[\sqrt{5}]$ is not a unique factorization domain, as it can be easily seen from the following factorizations $4=2 \cdot 2 = (3+\sqrt{5})(3-\sqrt{5})$. So the question is not so trivial, at least for me!

Answer 1

Note that $11=(4+\sqrt{5})(4-\sqrt{5})\in\langle 4+\sqrt{5}\rangle$, and so we have the chain of isomorphisms $$\frac{\mathbb{Z}[\sqrt{5}]}{\langle4+\sqrt{5}\rangle}=\frac{\mathbb{Z}[\sqrt{5}]}{\langle4+\sqrt{5},11\rangle}\cong\frac{\mathbb{Z}[t]}{\langle t^2-5,4+t,11\rangle}.$$ Also, $t^2-5=11-(4-t)(4+t)$, whence $\langle t^2-5,4+t,11\rangle=\langle 4+t,11\rangle$, and thus the above ring $\mathbb{Z}[\sqrt{5}]\big/\langle 4+\sqrt{5}\rangle$ is in fact isomorphic to $$\frac{\mathbb{Z}[t]}{\langle 4+t,11\rangle}\cong\frac{(\mathbb{Z}\big/11)[t]}{\langle \bar{4}+t\rangle}.$$ Now, $\mathbb{Z}\big/11$ is a field, so $(\mathbb{Z}/11)[t]$ is a principal ideal domain, and clearly $\bar{4}+t$ is irreducible – thus prime – in $(\mathbb{Z}/11)[t]$. This means that the above ring is a domain, and so $\langle 4+\sqrt{5}\rangle$ is indeed a prime ideal of $\mathbb{Z}[\sqrt{5}]$.