Prove that $\mathbb{Z}$ is a UFD while $\mathbb{Z}[\sqrt{-5}]$ is not.

Question

An integral domain $R$ is called a unique factorization domain (UFD) if every nonzero, nonunit element of $R$ can be uniquely written as a product of irreducible elements, up to reordering the factorization and taking associates of the irreducible factors (e.g. $10 = (2)(5) = (-5)(-2)\in\mathbb{Z}$).

$1.$ Prove that $\mathbb{Z}$ is a UFD.

$2.$ Prove that $\mathbb{Z}[\sqrt{-5}]$ is not a UFD.

I think $1$ is equivalent to the proving the uniqueness part of the Fundamental Theorem of Arithmetic.

As for $2$, $5 = 1\cdot 5$, where $1$ and $5$ are irreducible, and $5 = (-1)\cdot (\sqrt{-5})^2$, where $(-1)$ and $(\sqrt{-5})$ are also irreducible, so it has two distinct factorizations in $\mathbb{Z}[\sqrt{-5}].$ Thus, $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. Do I need to prove that $-1,1,5,\sqrt{-5}$ are irreducible? And if not, is this proof still correct?

Answer 1

No!

$5$ and $-5=(\sqrt{-5})^2$ are associates to each other. So your factorization not works!

Instead, note that $$6= 2\times 3=(1+\sqrt{-5}) \times (1-\sqrt{-5})$$ and all are irreducible and neither of them is associate to each other!

Answer 2

For 1: the definition says "can be uniquely written", so you essentially have to prove the Fundamental Theorem of Artithmetic (not just the "uniqueness part).

For 2: are really 1,-1 and 5 irreducible? Instead, note that $2\cdot 3=6=(1+\sqrt{-5})\cdot(1-\sqrt{-5})$

PS: Remember that irreducible elements are not units by definition