Let $R=\mathbb{Z}[\sqrt{-3}]=\{a+b\sqrt{-3}\ |\ a,b\in \mathbb{Z}\}$ with the sum and the product of $\mathbb{C}$. Then, $R$ is a ring. I would like to prove that the elements $a=4$ and $b=2+2\sqrt{-3}$ have no g.c.d. in $R$. By that, we mean an element $z$ in $R$ such that if $z$ divides both $a$ and $b$, then any other common divisor of $a$ and $b$ divides $z$. I've been asked to prove a g.c.d does not exist, yet for me it is natural to think that $2$ would work. Any help?
Similarly, I would like to know if $c=2$ and $d=1+\sqrt{-3}$ have a l.c.m. element $w$ in $R$ in the sense that if $c$ and $d$ divide some element $t\in R$ then it must be the case that $w$ divides $t$. Thank you in advance.
$a$ divides $b$ in $R$ if there exists $c\in R$ such that $b=ac$.
For the first one, you first need to demonstrate two divisors that $a$ and $b$ have in common, such that neither divisor divides the other. They clearly have $2$ in common, so that's probably a good first divisor. What can the second divisor be? (What other divisors does $b$ have? Are any of those also divisors of $4$? Are any of those neither divisors of, nor divisible by, $2$?)
As Tobias Kildetoft points out below, that's not entirely enough. You also need to show that there are no divisors "between" the ones you have, and $a$ and $b$, which could be a greatest common divisor. Part of that is showing that $a\nmid b$ and $b\nmid a$, such that $a$ and $b$ aren't $\gcd$'s themselves. The other part is showing that the are no other possible divisors between, say, $2$ and $4$.
Once you've solved that problem, you can probably see a connection between the numbers given in the two problems. What if that connection goes all the way?