Unique factorization theorem in algebraic number theory Consider the set $S = a + b \sqrt {-6}$, where $a$ and $b$ are integers. Now, to prove that unique factorization theorem does not hold in set $S$, we can take the example as follows:
$$
10 = 2 \cdot 5 = (2+\sqrt {-6}) (2-\sqrt {-6})
$$
"Thus we can conclude  that there is not unique factorization of 10 in set $S$. Note that this conclusion does not depend on our knowing that $2+\sqrt {-6}$ and $2-\sqrt {-6}$ are primes; they actually are, but it is unimportant in our discussion. "
Can someone explain why the conclusion is independent of nature of $2+\sqrt {-6}$ and $2-\sqrt {-6}$. Basically, unique factorization theorem is based on the fact that factors are primes. So, why is it independent?
Note: This is from the book An Introduction to the Theory of Numbers, 5th Edition by Ivan Niven, Herbert S. Zuckerman, and Hugh L. Montgomery.

$2$ is Irreducible but not Prime.
In fact if $2=cd$, then $N(c)=2$ but there is no solution to $a^2 + 6b^2 = 2$ reducing modulo 6. Thus $2$ is Irreducible.
$2  |10 = (2+\sqrt {-6}) (2-\sqrt {-6})$, but if $2 |(2+\sqrt {-6})$ then $2 |\sqrt{-6}$ which is impossible since $2(a+b\sqrt{-6})=\sqrt{-6}$ has no solutions. Same for the minus. Thus $2$ is not Prime.
