Prove $4+\sqrt{5}$ is prime in $\mathbb{Z}[\sqrt{5}]$ Prove $4+\sqrt{5}$ is prime in $\mathbb{Z}[\sqrt{5}]$. I already proved it's irreducible but have no idea what to do next since I dom't know much about $\mathbb{Z}[\sqrt{5}]$, if it is a UFD or not. Anyway, help me with this problem
 A: Extended hint: Show that $\mathfrak{p}=(4+\sqrt5)$ is a prime ideal by showing that
$$
\Bbb{Z}[\sqrt5]/\mathfrak{p}\simeq\Bbb{Z}_{11},
$$
which is an integral domain.
This is accomplished (do you see why?), if you prove that


*

*$11\in\mathfrak{p}$, and

*every coset of $\mathfrak{p}$ contains an integer.

A: Here is another proof. Althought @Jyrki Lahtonen 's in quicker, this uses more of the theory, and it generalizes.
The ring of integers $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$ is an integral domain and $4+\sqrt{5}$ is prime there since $N(4+\sqrt{5})=11$ is prime.
This means that if $a,b\in \mathbb{Z}[\sqrt{5}]$ then 
$(4+\sqrt{5})|a$ or $(4+\sqrt{5})|b$ in $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$.
Take the former, that means we have a relation of the form
$$a=(4+\sqrt{5})\frac{x+y\sqrt{5}}{2}$$
this implies that 
$$4x+5y$$ and 
$$x+4y$$ are even and since the determinant of this matrix is $11$, we have that $x$ and $y$ are even.
Thus we have a relation 
$$a=(4+\sqrt{5})(x_1+y_1\sqrt{5})$$
 and so $(4+\sqrt{5})| a$ in $\mathbb{Z}[\sqrt{5}]$.
A: $a+b\sqrt 5$ is a multiple of $4+\sqrt5$ if and only if $4a-5b$ and $-a+4b$ are multiples of $11$. It's easy to prove that $11\mid4a-5b\iff11\mid-a+4b$
Now, suppose that $(a+b\sqrt 5)(c+d\sqrt 5)=(ac+5bd)+(ad+bc)\sqrt 5$ is a multiple of $4+\sqrt 5$. Then
$$11\mid -ac-5bd+4ad+4bc$$
or
$$11\mid a(4d-c)+b(4c-5d)$$
But
$$a(4d-c)+b(4c-5d)\equiv a(4d-c)-4b(-12c+15d)\equiv (a-4b)(4d-c)\pmod{11}$$
This implies that $(a+b\sqrt5)(c+d\sqrt 5)$ is a multiple of $4+\sqrt 5$ only if one of the factors is. That is, $4+\sqrt 5$ is prime.
A: Below I explain how to prove that $\,w = 4+\sqrt5\,$ is prime using a direct proof analogous to the classical proof for integers. This will also reveal the relationship between some of the other answers. First let's recall the classical Euclidean proof that atoms (irreducibles) $p$ are prime in $\,\Bbb Z$
$$ p\nmid a,\ p\mid ab\,\Rightarrow\,p\mid ab,pb\,\Rightarrow\,p\mid (ab,pb)=\color{#0a0}{(a,p)}b = b$$
since, applying Bezout's gcd identity we deduce that $\,p\nmid a\,\Rightarrow\, \color{#0a0}{(a,p) = 1}\, $ since $\,p\,$ is an atom.
By a short simple Lemma, $\,ww'=11\,\Rightarrow (w) = (w,11) = [w, 11] := w\Bbb Z + 11\Bbb Z,\,$ i.e. the ideal generators $\,w,11\,$ are actually module generators. Now we can use division with remainder to reduce any $\, \alpha = c+d\sqrt 5\,$ to a unique integer remainder $\  \alpha \bmod w \equiv n\pmod{11} $ as follows
${\rm mod}\ (w) = [4\!+\!{\sqrt 5},\, 11]\!:\,\ \color{#c00}{\sqrt5\equiv -4}\ $ $\Rightarrow$ $\ c\!+\!d\color{#c00}{\sqrt 5}\,\equiv\, (c\!\color{#c00}{-\!4}d)\bmod 11,\ $  by $\ 11\equiv 0$
For uniqueness: $\,n'\equiv \alpha\bmod w\equiv n\,$ $\Rightarrow$ $\,k = n'\!-\!n\in (w)\Rightarrow n'\equiv n\pmod{\!11}\, $ by
$$\begin{align} {\rm integer}\ k \in (w) = w\Bbb Z + 11\Bbb Z &\iff\! k = w\,j + 11\,i\ \ \ {\rm for\ some}\ \ i,j\in\Bbb Z\\ &\iff\! k = 11\,i\ \ \ {\rm for\ some}\,\ i\in\Bbb Z,\, \ {\rm by}\ \ w\not\in\Bbb Q\,\Rightarrow j=0\end{align}$$
Thus if $\, w\nmid \alpha \,$ then $\, 11\nmid n:= \alpha \bmod w,\,$ so $\,(11,n)= \color{#0a0}{1\in (\alpha,w)}$
Hence $\,w\nmid \alpha,\ w\mid \alpha\beta\,\Rightarrow\,w\mid \alpha\beta,w\beta\,\Rightarrow\, w\mid (\alpha\beta,w\beta) = \color{#0a0}{(\alpha,w)}\beta = \beta,\,$ as in the classical proof.
Remark $\ $ This explains the calculations in ajotatxe's answer, i.e. we are essentially doing remainder calculations modulo a triangular basis (something that will be clarified when one studies Hermite / Smith normal forms). It also gives another view on the hint in Jyrki's answer. 
