# Equivalence of different prime factorizations in $\mathbb{Z}[\zeta_3]$

I'm reading that in the ring $\mathbb{Z}[\zeta_3]$, where $\zeta_3$ is the cubic root of unity, two prime factorizations of $4 = 2 \times 2 = (1 + \sqrt{-3})(1 - \sqrt{-3})$ are equivalent, because up to unit factors $\frac{1\pm\sqrt{-3}}{2}$ they are: $4 = 2 \times 2 = 2\left(\frac{1+\sqrt{-3}}{2}\right)\times 2\left(\frac{1-\sqrt{-3}}{2}\right) = (1 + \sqrt{-3})(1 - \sqrt{-3})$.

Here factorization of $4 = 2 \times 2$ had to be multiplied by both units $\frac{1\pm\sqrt{-3}}{2}$ to get its equivalent factorization. But what if I multiply by only one of them? Will I still get equivalent factorization of $4$? For example, are these factorizations $4 = 2 \times 2 = 2\left(\frac{1+\sqrt{-3}}{2}\right)\times 2 = (1 + \sqrt{-3}) \times 2$ equivalent?

Intuitively, I understand that they are not, but I'm not quite content with an explanation for why they are not that I'm being able to come up with. My explanation is that just like factorizations $3 \times 7$ and $-3 \times 7$ of $21$ are not equivalent simply because one was multiplied by unit $-1$, factorizations of $2 \times 2$ and $(1 + \sqrt{-3}) \times 2$ are not equivalent just because one can be obtained from the other by multiplication by some unit - you have to make sure that the unit(s) used is the right one.

Is that it? Thanks.

• Equality $2\times 2=2(\frac{1+\sqrt{-3}}{2})\times 2$ is wrong, because $\frac{1+\sqrt{-3}}{2}\neq 1$. Commented Jul 18, 2015 at 9:28
• Factorizations as a product of non-primes can be varied, in the natural numbers too. Commented Jul 18, 2015 at 9:38
• $\frac{1+\sqrt{-3}}{2}\ne1$, but equivalence of factorizations is established up to unit factors, not necessarily 1, and $\mathbb{Z}[\zeta_3]$ includes $\frac{1+\sqrt{-3}}{2}$ as a unit factor. Also both $2$ and $\frac{1+\sqrt{-3}}{2}$ are prime factors in $\mathbb{Z}[\zeta_3]$, so it's not about factorizations into non-primes either. Commented Jul 18, 2015 at 9:52

The point is one wants to understand the analogue of unique factorization of natural numbers in more general classes of integers. For a natural number, the factorization into natural number (i.e., positive) primes is unique, but if one works with larger sets of numbers, such as the ring $\mathbb Z$, this is not true for the simple reason that one has negative prime, and can stick in an even number of $(-1)$'s into a factorization $$n = p_1 \cdots p_r$$ which only changes the factorization in a trivial way. (One can also reorder the factorization, but this is also a trivial change.)
For general integer rings, there is no nice notion of "natural numbers" so one looks at what unique factorization means in rings. Namely, if in a (commutative) ring one has a factorization $$x = \alpha_1 \cdots \alpha_r$$ of $x$ into irreducible elements, one can trivially get other factorzations in an analogous way--we also have $$x = (u_1 \alpha_1) \cdots (u_r \alpha_r)$$ where $u_1, \ldots, u_r$ are units such that $u_1 \cdots u_r = 1$. (Note multiplying an irreducible by a unit results in another irreducible, just by definition of these terms.)