Could anyone help me prove that $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$?
As $6=2*3=(1+\sqrt{-5})(1-\sqrt{-5})$ so $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. Therefore is not a PID or euclidean domain
Could anyone help me prove that $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$?
As $6=2*3=(1+\sqrt{-5})(1-\sqrt{-5})$ so $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. Therefore is not a PID or euclidean domain
The standard method is:
Define a function $N\colon \mathbb{Z}[\sqrt{-5}]\to\mathbb{Z}$ by $N(a+b\sqrt{-5}) = (a+b\sqrt{-5})(a-b\sqrt{-5}) = a^2+5b^2$.
This is a common technique for dealing with rings of the form $\mathbb{Z}[\theta]$, where $\theta$ is an algebraic integer.
It can also be done directly, though it is a bit more laborious. Here's what I came up with on the fly:
Assume that $(a+b\sqrt{-5})(c+d\sqrt{-5}) = 2$. Then $ac+5bd = 2$ and $ad+bc=0$. If $a=0$, then we must have $c=0$ (since $bc=0$ but we cannot have $a=b=0$), but then $5bd=2$ is impossible. Thus, $a\neq0$ and $c\neq 0$. If $b=0$, then $d=0$, so the factorization occurs in $\mathbb{Z}$ and is trivial; symmetrically if $d=0$. So we may assume that all of $a,b,c,d$ are nonzero.
Then $ad=-bc$, so $2d=acd + 5bd^2 = -bc^2 + 5bd^2 = b(5d^2 - c^2)$. If $b$ is odd, then $b|d$, so writing $d=bk$ we get $abk + bc=0$, so $ak+c=0$, thus $c=-ak$. Hence $c+d\sqrt{-5} = -ak+bk\sqrt{-5} = k(-a+b\sqrt{-5})$. But this gives $2 = (a+b\sqrt{-5})(c+d\sqrt{-5}) = k(a+b\sqrt{-5})(-a+b\sqrt{-5}) = -k(a^2+5b^2).$ Since $a$ and $b$ are both nonzero, $a^2+5b^2$ is at least 6, which is impossible. So $b$ is even, b=2b'.
Then b'(5d^2-c^2) = d, so setting $5d^2-c^2 = k$ we have a+b\sqrt{-5} = a+2b'\sqrt{-5},\qquad c+d\sqrt{-5} = c+kb'\sqrt{-5}. From $ad=-bc$ we get $ak=-2c$. If $a$ is even, then we have a=2a', so 2 = (2a'+2b'\sqrt{-5})(c+d\sqrt{-5}) = 2(a'+b'\sqrt{-5})(c+d\sqrt{-5}), which yields that $c+d\sqrt{-5}$ is a unit (in fact, this is impossible with $c$ and $d$ both nonzero, but that doesn't matter). If $a$ is odd, then $k$ is even and c=ak', with 2k'=k. So now we have \begin{align*} 2 &= (a+b\sqrt{-5})(c+d\sqrt{-5})\\ &= (a + 2b'\sqrt{-5})(ak' + 2b'k'\sqrt{-5})\\ &= k'(a+2b'\sqrt{-5})(a+2b'\sqrt{-5})\\ &= k'(a^2 - 20b'^2) + 4ab'\sqrt{-5} \end{align*} which implies $a=0$ or b'=0 (hence $b=0$), contradicting our hypotheses.
Thus, the only factorizations of $2$ in $\mathbb{Z}[\sqrt{-5}]$ are trivial.
(And you can probably see why the "standard method" is so much better....)
Let $A = \mathbb{Z}[\sqrt{-5}]$. For every element $\alpha$ of $A$ there exist unique $a, b \in \mathbb{Z}$ such that $\alpha = a + b \sqrt{-5}$. Consider the function $N \colon A \to \mathbb{N}$ defined by $N(a + b \sqrt{-5}) = a^2 + 5 b^2$. (If you know some field theory, $N$ is the restriction to $A$ of the norm of the field $\mathbb{Q}(\sqrt{-5})$ over $\mathbb{Q}$.) Prove the following facts:
Do there exist element in $A$ with norm equal to $2$ or $3$? Now, $N(2) = 4$, $N(3) = 9$, $N(1 + \sqrt{-5}) = N(1- \sqrt{-5}) = 6$. As in Henning Makholm's comment, if they were irreducible then what could the norms of the factors be?