TITLE: Question about proving $\mathbb{Z}[i]$ is not UFD
QUESTION [5 upvotes]: I know is a silly question, but why $10=\left(3+i\right)\left(3-i\right)=2\cdot5$ is not enough to prove that $\mathbb{Z}[i]$ is not a UFD. Thanks in advance!

REPLY [3 votes]: Actually, there are a few more different-looking factorizations you could pile in there: $$(-1 + i)(1 + i)(-1 + 2i)(1 + 2i) = (1 - 3i)(1 + 3i) = 10.$$ However, that's no more distinct than saying $-2 \times -5$ is a factorization of $10$ in $\mathbb{Z}$ that's distinct from $5 \times 2$.
For unique factorization, we ignore:

ordering
multiplication by units

In $\mathbb{Z}[i]$, the units are, clockwise from "twelve": $i, 1, -i, -1$. So $3i, -3i, -3$ are all just as prime as $3$.
And of course you also have to check that you really have primes, and not composites with no obvious prime factors. For instance, in z, $1729 = 19 \times 91$, but one of those is actually composite, though it sure looks like it could be prime.