<s>[INST] Showing $(3 + \sqrt{3})$ is not a prime ideal in $\mathbb{Z}[\sqrt{3}]$ Let $I = (3+\sqrt{3})$
Looking at the field norm we note that $N(3 + \sqrt{3}) = 6$. We also know that $\mathbb{Z}[\sqrt{3}]$ is a Euclidean Domain.
We want to find some $\alpha, \beta \in \mathbb{Z}[\sqrt{3}]$ s.t. $\alpha \cdot \beta = 3 + \sqrt{3}$. This requires $N(\alpha)\cdot N(\beta) = 6$. 
So $N(\alpha) \in \{\pm 2\}$ and $N(\beta) \in \{\pm 3\}$. 
$N(a+b\sqrt{3}) = a^2 - 3b^2 = -2$ when $\alpha = 1 - \sqrt{3}$ which is a non-unit since $N(\alpha) \neq 1$. 
Then we have $N(c + d\sqrt{3}) = c^2 - 3d^2 = -3$ when $\beta = -3 - 2\sqrt{3}$ which is also a non-unit.
Then we see $\alpha\cdot\beta=(1-\sqrt{3})(-3-2\sqrt{3}) = 3 + \sqrt{3}$ 
Since this is a non trivial factorization of $3+\sqrt{3}$, then we see that $(1-\sqrt{3})\cdot(-3-2\sqrt{3})\in I$.
It remains to show that neither $\alpha$ or $\beta$ are in $I$.
Taking $\frac{1-\sqrt{3}}{3+\sqrt{3}} = \frac{(1-\sqrt{3})(3-\sqrt{3})}{6} = \frac{(3-\sqrt{3} - 3\sqrt{3} +3)}{6} = \frac{-4\sqrt{3}}{6}$ which is not in $\mathbb{Z}[\sqrt{3}]$. So there will be some remainder implying that $\alpha$ is not in $I$.
Doing the same thing we calculate $\frac{-3-2\sqrt{3}}{3+\sqrt{3}} = \frac{-3-\sqrt(3)}{6}$. Again, it will yield a remainder so we conclude that both $\alpha$ and $\beta$ are not in $I$, yet $\alpha \cdot \beta \in I$. 
Thus $(3+\sqrt{3})$ is not prime.
Is this attempt correct? Is there a shorter way to go about this? 
 [/INST] Why not write $3+\sqrt3=\sqrt3(1+\sqrt3)$?
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