Why is $\mathbb{Z}[i \sqrt{2}]$ a Unique Factorization Domain?
We know that $\mathbb{Z}[i \sqrt{5}]$ is not a UFD as $$(1 + i \sqrt{5})(1 - i \sqrt{5}) = 6$$ and $6$ is also equal to $2 \times 3$.
Now $\mathbb{Z}[i \sqrt{2}]$ is a UFD since $2$ is a Heegner number, however the simple factorization
$$(2 + i \sqrt{2})(2 - i \sqrt{2}) = 4 + 2 = 6 $$
which is also equal to $2 \times 3$, and therefore not unique.
What is wrong here?
The fallacy is that none of the four numbers \begin{align*} 2+i\sqrt2 &= i\sqrt2 (1-i\sqrt2) \\ 2-i\sqrt2 &= -i\sqrt2 (1+i\sqrt2) \\ 2 &= -i\sqrt 2 (i\sqrt 2) \\ 3 &= (1+i\sqrt2)(1-i\sqrt2) \end{align*} is irreducible in $\Bbb Z[i\sqrt2]$. Once we observe this, the error is the same as if we asserted that $14\times15=210=10\times21$ shows that $\Bbb Z$ is not a unique factorization domain.