8

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?

Good Morning Captain
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  • If $\alpha\in I$, then $3+\sqrt{3}$ and $\alpha$ would be associates, but then $\beta$ would be a unit, which you have already proved is not true. – vadim123 Mar 28 '18 at 04:25
  • @vadim123 Ah, so is that a quicker way to show that neither of them separately are in $I$? – Good Morning Captain Mar 28 '18 at 04:27
  • Note that $u$ is a unit if and only if $N(u)=\pm 1$. Once you showed that your element is a product of elements which are not units, it is reducible, so cannot be prime. – orangeskid Mar 28 '18 at 07:41

2 Answers2

6

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...)

user103697
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4

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

Lubin
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