Find prime factorization in the ring $\mathbb{Z}\left[\frac{-1 + \sqrt{3}}{2}\right]$ Find prime factorization of the number $13$ in the ring $\mathbb{Z}\left[\frac{-1  + \sqrt{3}}{2}\right]$
My progress:
Let $w = \frac{-1 + \sqrt{3}}{2}$ and let $N(z) = z \bar z$ be the norm function.
$N(a + bw) = a^2 - ab + b^2$
$13 = (a + bw)\overline{(a + bw)}  =  a^2 - ab + b^2$. Let's divide the equation by $4$:
$13 = \frac{(2a-b)^2 + 3b^2}{4} \Rightarrow (2a-b)^2 + 3b^2 = 52 \Rightarrow$
$2a-b = 5 , b = 3 \Rightarrow a = 4, b = 3$
Thus $13 = (4 + 3w)\overline{(4 + 3w)}$
My questions:


*

*Is it correct?

*How can one show that $4 + 3w$ is prime in $\mathbb{Z}[w]$?

 A: There are several problems here. First of all, $\frac{-1 + \sqrt{3}}{2}$ is not an algebraic integer (its minimal polynomial is $2x^2 - 2x - 1$). Presumably you meant $\frac{-1 + \sqrt{-3}}{2}$. Your usage of $w$ instead of $\omega$ would normally not be worth mentioning, except that here it masks a serious mistake. So as to be on the same page, I will assume you mean $\mathbb{Z}\left[\frac{-1 + \sqrt{-3}}{2}\right]$. So as to stay on the same page, let's set $\omega = \frac{-1 + \sqrt{-3}}{2}$.
With that cleared up, everything else falls into place. Indeed $(4 + 3\omega)(4 + 3\omega^2) = 13$. To be sure that $4 + 3\omega$ is indeed prime, I could embark on an argument by contradiction. But today I'll just invoke Theorem 9.24 in Ivan Niven & Herbert S. Zuckerman, An Introduction to the Theory of Numbers 4th Ed. New York: John Wiley & Sons (1980), which asserts that in a quadratic integer ring which is a unique factorization domain (which $\mathbb{Z}[\omega]$ is), if the norm of an algebraic integer in that ring is a prime in $\mathbb{Z}$, then that algebraic integer is a prime in that ring.
A: It's very obvious to me that you're talking about the domain of Eisenstein integers, despite your consistent omission of an "$i$" or a "$-$" in three instances above. It does my ego no good to dump on you over such a little mistake.
Nevertheless, I strongly recommend you use $\omega$ rather than $w$ for that complex cubic root of 1. You might think you're saving five bytes, but it has to be rendered regardless. Besides, it leaves $w$ available for use as an arbitrary variable.
Anyway, on to your questions:


*

*Yes, it's correct. Review your book and/or class notes to reassure yourself that $$\overline{a + b \omega} = a + b \omega^2.$$ Then define omega in your favorite computer algebra system and have it compute (4 + 3omega)(4 + 3omega^2). The answer should be 13.

*You have to show that $4 + 3 \omega$ is irreducible and you have to show that $\mathbb{Z}[\omega]$ is a unique factorization domain. (I don't know if you've been taught the distinction between "irreducible" and "prime," a distinction that is moot in a UFD). To show that $4 + 3 \omega$ is irreducible, assume that it isn't. This means that there are numbers $\alpha, \beta \in \mathbb{Z}[\omega]$ such that $N(\alpha) N(\beta) = N(4 + 3 \omega)$, yet $N(\alpha) \neq 1$ and likewise $N(\beta) \neq 1$. The norm is always an integer, and, what's more, in an imaginary ring (like this $\mathbb{Z}[\omega]$) the norm is never negative. There are only two possibilities, one of which is $N(\alpha) = 1, N(\beta) = 13$, directly contradicting the assertion earlier. Maybe you doubt what I'm saying, even though it might as well be plagiarized from any standard algebraic number theory text. If you doubt this, try this for yourself: show that $8 + 6 \omega$ is composite.

