Showing $(3 + \sqrt{3})$ is not a prime ideal in $\mathbb{Z}[\sqrt{3}]$

Question

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?

Answer 1

In addition to what was already said in the comments, I think there is another possibly faster way worth mentioning, which is using the following criterion for some Ideal $I$ being prime in a ring $R$ (which results immediately from the definitions):

$I\subset R$ is prime if and only if $R/I$ is an integral domain.

Using this, you would get

$$\mathbb{Z}[\sqrt{3}]/(3+\sqrt{3})\simeq(\mathbb{Z}[X]/(X^2-3))/(3+[X])\simeq \mathbb{Z}[X]/(X+3,X^2-3)\simeq\mathbb{Z}/(6),$$

which is certainly no integral domain, so $(3+\sqrt{3})\subset\mathbb{Z}[\sqrt{3}]$ is not prime.

(Whether or not this version is really shorter than the one you presented does however certainly depends on the amount of detail you would like to add to the isomorphisms used above...)

Answer 2

Why not write $3+\sqrt3=\sqrt3(1+\sqrt3)$?