can anyone give a proof by definition: $11$ is prime in $ \mathbb{Z}[\sqrt{-5}] $ what i did is:
assume $\alpha  \notin (11),\beta\notin (11), \alpha\beta \in (11)\Rightarrow\exists \gamma, s.t.$  $ \alpha\beta = 11 \gamma$, $\Rightarrow N(\alpha)N(\beta) = 11^2N(\gamma)  $
then there are 2 cases
one case is $11|N(\alpha)$ and  $11|N(\beta)$
the other is $11^2|N(\alpha)$ or $11^2|N(\beta)$
my prof said both cases will lead to a contradiction, but I cannot figure it out.
 A: We will show that if $11$ divides $\alpha\beta$ then $11$ divides $\alpha$ or $11$ divides $\beta$. Since $11$ is obviously not a unit  of $\mathbb{Z}[\sqrt{-5}]$,
by the definition of prime it follows that $11$ is prime in $\mathbb{Z}[\sqrt{-5}]$.
From the post, it is clear that you know that if $11$ divides $\alpha\beta$, then $11$ divides $N(\alpha)N(\beta)$ and therefore $11$ divides $N(\alpha)$ or $11$ divides $N(\beta)$.
Let $\alpha=s+t\sqrt{-5}$.  Then $N(\alpha)=s^2+5t^2$. Suppose that $11$ divides $s^2+5t^2$. We will show that $11$ divides $t$ (and therefore $11$ divides $s$).
Suppose that $s^2+5t^2=11k$, and $11$ does not divide $t$. Let $wt\equiv 1\pmod{11}$ (there is such a $w$ since $t$ has an inverse modulo $11$).
We get $(ws)^2\equiv -5\equiv 6\pmod{11}$. This is impossible, since $6$ is not a quadratic residue of $11$ (just try all squares up to $5^2$). 
Since $11$ divides $t$ and $s$, it follows that $11$ divides $\alpha$.
If $11$ does not divide $N(\alpha)$, it must divide $N(\beta)$, and then we conclude in the same way that $11$ divides $\beta$.
