TITLE: Prove that if $w \in \mathbb{Z}[\sqrt{3}]$ and $N(w)$ is a prime, then $w$ is prime also
QUESTION [11 upvotes]: Prove that if $w$ is an extended integer, $w \in\mathbb{Z}[\sqrt{3}]$, and $N(w)$ is a prime in the ordinary integers, then $w$ is a prime. From this, conclude that $7+2\sqrt{3}$ is a prime in $\mathbb{Z}[\sqrt{3}]$.
I'm kinda having a tough time dealing with primes so I'm wondering if this proof makes sense:
We say that $w = a+b\sqrt{3}$ and $N(w)$ is prime. Let $w = a+b\sqrt{3} = k\cdot z$ where $k$ and $z$ are integers. We now have $N(w) = N(k)\cdot N(z)$, but we know that $N(w)$ is prime which implies that either $N(k)$ is unit or $N(z)$ is unit. If $N(k)$ is unit, then $k$ is also unit. Similarly, if $N(z)$ is unit, then $z$ is also unit. Since $w = k\cdot z$ and either $k$ or $z$ is unit, we can conclude that $w$ is prime hence the proof is complete.
We have $w = a+b\sqrt{3} = 7+2\sqrt{3}$, so $N(w) = a^2-3b^2 = (7)^2-3(2)^2 = 49-3(4) = 37$. Since $37$ is a positive integer prime, this implies that $N(w)$ is prime. From the proof above, we can conclude that if $N(w)$ is prime then $w$ is also prime, hence $7+2\sqrt{3}$ is a prime in $\mathbb{Z}[\sqrt{3}]$.
Thank you for any help you may be able to offer.

REPLY [5 votes]: I will first prove
(1) An element $p \in R$ is prime id and only if the residue class ring $R/pR$ of the residue classes modulo $\pi$ is a domain.
In fact, the residue class ring oes not have a zero 
divisor if $ab \equiv 0 \bmod p$ implies that $a \equiv 0 \bmod p$ or $b \equiv 0 \bmod p$. But this is just a version of the definition of a prime element, which states that an element ist prime if $p \mid ab$ implies that $p \mid a$ or $p \mid b$.
Now we claim
(2) If $k$ is a quadratic number field with ring of integers $R ? {\mathcal O}_k$, then each $\pi \in {\mathcal O}_k$ with prime norm is prime.
We will show that the residue class ring $R/\pi R$ is a domain by showing that it is isomorphic to the field with $p$
elements.
To this end let $\{1, \omega\}$ be an integral basis of ${\mathcal O}_k$; then $\pi = a+b\omega$ for integers $a, b \in {\mathbb Z}$. We claim that $b$ is not divisible by $\pi$ (and thus not divisible by $p = |\pi \pi'|$). In fact, $\pi \mid b$ implies $\pi \mid a$ since $a = \pi - b\omega$, and
taking norms we find $p \mid a^2$ and $p \mid b^2$. Since $p$ is prime, this implies that $p \mid a$ and $p \mid b$. But then $\pi = a+b\omega$ would be divisible by $p$, hence $\pi'$ would be a unit: contradiction.
Thus there exists an integer $c \in {\mathbb Z}$ with $bc \equiv 1 \bmod p$, and in particular we have $bc \equiv 1 \bmod \pi {\mathcal O}_k$. We find $b\omega \equiv -a \bmod \pi$, after multiplying through by $c$ thus $\omega \equiv -ac \bmod \pi {\mathcal O}_k$. If any $\gamma = r+s\omega \in {\mathcal O}_k$ is given, then we find $\gamma \equiv r - sac \bmod \pi {\mathcal O}_k$, thus modulo $\pi$ every element is congruent to an ordinary integer. Reducing this number modulo $p$ (and $p$ is a multiple of $\pi$) we find that $\gamma$ is congruent to one of the numbers $0, 1, 2, \ldots, p-1$ modulo $\pi$.
Now it is easy to show that there are no zero divisors in the ring of residue classes: If we had $\alpha \beta \equiv 0 \bmod \pi$ and if $A, B \in \{0, 1, \ldots, p-1\}$ are integers with  $\alpha \equiv A \bmod \pi {\mathcal O}_k$ and $\beta \equiv B \bmod \pi {\mathcal O}_k$, then $\pi \mid AB$; taking norms yields $p \mid A^2B^2$, hence $p \mid A$ or $p \mid B$. Thus $A = 0$ or $B = 0$, and therefore $\alpha \equiv A = 0 \bmod \pi$ or $\beta \equiv B = 0 \bmod \pi$.