Irreducible but not prime：Irreducible but not prime in $mathbbZ}[sqrt

The proof that $1-\sqrt{-5}$ is irreducible is incorrect. In addition to a minor (and alas, common) mistake in writing that makes what you write not what you intended to write, there is an assertion which is just plain false.

Explicilty, you write:

Assume $1-\sqrt{-5}$ is reducible, then there must exist $a,b \in \mathbb{Z}[\sqrt{-5}]$ so that $N(1-\sqrt{-5})=N(a)N(b) \Rightarrow N(a)=N(b)= \pm (1-\sqrt{-5})$

First: the use of $\Rightarrow$ is incorrect. What you have written is that there exist $a$ and $b$ in $\mathbb{Z}[\sqrt{-5}]$ for which the following statement holds:

If $N(1-\sqrt{-5}) = N(a)N(b)$, then $N(a)=N(b) = \pm (1-\sqrt{-5})$.

What you actually wanted to write was that

If there exist $a,b\in\mathbb{Z}[\sqrt{-5}]$ such that $N(1-\sqrt{-5}) = N(a)N(b)$, then $N(a)=N(b)=\pm(1-\sqrt{-5})$.

Whats the difference? The first statement will be true if you can find an $a$ and a $b$ for which $N(1-\sqrt{-5})...