Prime elements in $\mathbb{Z}[\sqrt{5}]$ I need to determine which of the elements $3+2\sqrt{5}$, $9+4\sqrt{5}$ and $4-\sqrt{5}$ are prime elements in $\mathbb{Z}[\sqrt{5}]$, respectively which are associated.
My ansatz is as follows:
So let $x=3+2\sqrt{5}$ divide $ab$ for $a,b \in R$. Thus, there are $u,v \in \mathbb{Z}$, such that
$$ab=(a_1+b_1\sqrt{5})(a_2+b_2\sqrt{5})=(a_1a_2+5b_1b_2)+(a_1b_2+a_2b_1\sqrt{5})=(u+v\sqrt{5})(3+2\sqrt{5})=(3u+10v)+(3v+2u)\sqrt{5}.$$
How do I go from here in order to check whether $x|a$ or $x|b$?

$3+2\sqrt{5}$ has norm $-11$ and $4-\sqrt{5}$ has norm $11$. Perhaps they are associates. Indeed
$$
\frac{3+2\sqrt{5}}{4-\sqrt{5}} = 2+\sqrt{5}
$$
and $2+\sqrt{5}$ is a unit because it has norm $-1$. In fact, $(2+\sqrt{5})(-2+\sqrt{5})=1$.
$9+4\sqrt{5}$ is a unit because it has norm $1$ and so $(9+4\sqrt{5})(9-4\sqrt{5})=1$.
It remains to decide whether $3+2\sqrt{5}$ is prime. Consider
$$
\frac{\mathbb Z[\sqrt{5}]}{\langle 3+2\sqrt{5} \rangle}
=
\frac{\mathbb Z[\sqrt{5}]}{\langle 3+2\sqrt{5},11 \rangle}
=
\frac{\mathbb Z[X]}{\langle X^2-5,3+2X,11 \rangle}
\cong
\frac{\mathbb Z_{11}[X]}{\langle X^2-5,3+2X \rangle}
=
\frac{\mathbb Z_{11}[X]}{\langle X^2-5,2(7+X) \rangle}
=
\frac{\mathbb Z_{11}[X]}{\langle X^2-5,X+7 \rangle}
=
\frac{\mathbb Z_{11}[X]}{\langle X+7 \rangle}
\cong
\mathbb Z_{11}
$$
which is a domain. Therefore, $3+2\sqrt{5}$ is prime.
