Questions about $\mathbb{Z}[\sqrt{-3}]$ I have some questions about $R=\mathbb{Z}[\sqrt{-3}]=\lbrace a+b\sqrt{-3}| a,b \in \mathbb{Z} \rbrace$.  I am working on a problem for comprehensive exam review and think I am making it harder than I need to or missing something obvious.  For the first and fourth I think I have the answers but want to make sure you agree but for the second and third I am having issues and need help.  
1)Is $R$ an integral domain?  Yes!  Let $r_1r_2=(a+b\sqrt{-3})(c+d\sqrt{-3})=0$ then $ac-3bd=0, ad+bc=0$.  I can then manipulate these equation to show either $r_1$ or $r_2$ must be zero.
2)What are the units in R?  I tried following a similar method as number $1$ where this time $ac-3bd=1, ad+bc=0$.  However, when I try to manipulate things in the same manner I run in circles.
3)Is 2 irreducible in R?  Again, I tried following a similar method as number $1$ where this time $ac-3bd=2, ad+bc=0$.  However, when I try to manipulate things in the same manner I get a bunch of junk.
4)If $x,y \in R$ and $2|xy$ does it follow that $2$ divides either $x$ or $y$?  No! $4 = (1+\sqrt{-3})(1-\sqrt{-3})$ but both are irreducible so $2$ divides neither.
Thanks!
 A: 1) Or, you could note that $\mathbb{Z}[\sqrt{-3}]$ has an embedding into $\mathbb{C}$ and so is, of course, an integral domain.
2) Since $\mathbb{Z}[\sqrt{-3}]$ is contained in the integers of $\mathbb{Q}(\sqrt{-3})$ you know that $a+b\sqrt{-3}$ is a unit if and only if $N(a+b\sqrt{-3})=\pm 1$, where $N=N_{\mathbb{Q}(\sqrt{-3})/\mathbb{Q}}$. So the units are the $a+b\sqrt{-3}$ solving $a^2+3b^2=\pm 1$. That shouldn't be too bad.
Edit: If you want to do number theory without admitting it. Just note that in this case the extension is Galois, and $N(\alpha)=\alpha\overline{\alpha}$ (where $\overline{\alpha}$ is the Galois conjugate). From this it's pretty easy to see that $\alpha$ is a unit if and only if $N(\alpha)$ is a unit, but a unit in $\mathbb{Z}$, so $\pm 1$.
3) Note that $N(2)=4$. If $2=\alpha\beta$ then this implies that $4=N(\alpha)N(\beta)$. If neither of $\alpha$ or $\beta$ is a unit (so their norm is not one) then this implies that $2=N(\alpha)$ has a solution. This says that $2=a^2+3b^2$ for some $a,b\in\mathbb{Z}$. Is this possible?
4) Correct! 
