factor $3+\sqrt{3}$ into irreducibles in $\mathbb{Z}[\sqrt{3}]$

Question

Factor $3+\sqrt{3}$ into irreducibles in $\mathbb{Z}[\sqrt{3}]$.

I have done the norm which is $6= 2 \cdot 3$. And i have tried two find $a$ and $b$ such that the norm of $a+b\sqrt{3}$ is $2$ and similarly with $3$, but then i get that there are no solutions (and that means that it is irreducible?) so I think I may be doing something wrong. I am not quite sure how to solve this.

Thank you Sorry for the format i am not used to it-

Answer 1

$$3 + \sqrt{3} = (1 - \sqrt{3})(-3 - 2\sqrt{3})$$

This is one possible factorization of $3 + \sqrt{3}$ into irreducibles. Noting that the first factor has norm $2$ and the second one $3$, proves that the factors on the RHS are indeed irreducibles.

Answer 2

You need to consider factors with norm $\pm 2, \pm 3$.

Anyway, it is easy to find a factorization of $3+\sqrt{3}$ by inspection: $$ 3+\sqrt{3} = \sqrt{3}(\sqrt{3}+1)=(0+\sqrt{3})(1+\sqrt{3}) $$ The first factor has norm $-3$ and the second factor has norm $-2$. Since the norms are prime, the factors are irreducible.