I have seen complicated arguments to establish whether $\mathbb Z$ extended by this or that radical is a UFD. They usually suppose that $u$ is a unit and then compute the reciprocal and then make arguments about each component being an integer. Some of them go by defining a norm and proving things about that. I'm a little confused about why these arguments are so elaborate when it seems to me that there is a very straight-forward argument in many cases. My guess is that I'm overlooking some important caveat in one of my steps but I'm not sure where.
Take for example $\mathbb Z[\sqrt{7}]$. To prove it's not a UFD I look at two factorizations of 6:
$$ 6=3\cdot 2 $$
$$ 6 = (\sqrt 7 – 1)(\sqrt 7 + 1) $$
I then argue that there is no unit (in fact no element at all) such that $3u=\sqrt 7-1$. Since $u=a+b\sqrt 7$ then we would need
$$ 3a = -1 $$
but this is not true for any $a\in\mathbb Z$. Of course the same argument could be used on any of $\pm 3,\pm 2$ on the left and $\pm (\sqrt 7-1),\pm (\sqrt 7+1)$ on the right.
Is there some logical error in saying that if $w+x\sqrt 7 = y+z\sqrt 7$ for integers $w,x,y,z$ then we must have $w=y$ and $x=z$?
Best Answer
There is a counter example. Since $4=2 \cdot 2 = (\sqrt{13}+3)(\sqrt{13}-3)$, it suffices to show those factors are irreducible. Suppose to the contrary that $2$ is not irreducible. It follows that there exists $a,b \in \mathbb{Z}[\sqrt{13}]$ such that $2=ab$, $N(a) \ne 1$ and $N(b) \ne 1$. It boils down to show that there are no solutions to $u^2=13v^2 \pm 2$ in $\mathbb{Z}$. Observe that $2 \nmid \sqrt{13} \pm 3$ and we are done.
Someone asked the same about $\mathbb{Z}[\sqrt{8}]$. You might find Is $\mathbb{Z}[{ \sqrt 8 } ] $ a Euclidean domain? relevant.