Is $\sqrt{-5}$ a prime in $\mathbb{Z}{[\sqrt{-5}]}$? I know it is irreducible over $\mathbb{Z}{[\sqrt{-5}]}$ but since it is not even UFD, so we can't conclude primality from irreducibility.  
My guess is yes since $N(\sqrt{-5})=5$ is prime. I started with $ab=c\sqrt{-5}$ where a, b and c are in $\mathbb{Z}{[\sqrt{-5}]}$, now we have to show one of $a$ or $b$ is divisible by $\sqrt{-5}$.  
I got $N(a)N(b)=5N(c)$, hence $5$ divide either $N(a)$ or $N(b)$. 
What should be my next step ?

Say $5|N(a)$. Since $N(x+y\sqrt{-5}) = x^2+5y^2$, if $a=x+y\sqrt{-5}$ then $5|x$. So $a=5z+y\sqrt{-5} = \sqrt{-5}(y-z\sqrt{-5})$. 

You can do this without invoking the norm, just low-tech: Note that an element $x+y\sqrt{-5}$ of $\mathbb{Z}[\sqrt{-5}]$ is a multiple of $\sqrt{-5}$ if and only if $x$ is a multiple of $5$ in $\mathbb{Z}$. Indeed, $5a+y\sqrt{-5} = \sqrt{-5}(y-a\sqrt{-5})$, and $\sqrt{-5}(m+n\sqrt{-5}) = 5(-n) + m\sqrt{-5}$.

Now say $\sqrt{-5}$ divides $(a+b\sqrt{-5})(x+y\sqrt{-5})$. Then it divides
$$(a+b\sqrt{-5})(x+y\sqrt{-5}) = (ax-5by) + (ay+bx)\sqrt{-5}$$
and therefore $ax-5by$ is a multiple of $5$; hence $ax$ is a multiple of $5$, so $5$ divides either $a$ or $x$, hence $\sqrt{-5}$ divides either $a+b\sqrt{-5}$ or $x+y\sqrt{-5}$ as required. 
