Let $x = a + b\sqrt{-7} \in R$. You can define the norm of $x$ by $N(x) = x \cdot \bar x = a^2 + 7b^2$. You can prove that the norm is multiplicative.
This means that if $x$ is invertible in $R$, then $N(x)N(x^{-1}) = 1$, so $N(x) = 1$. What are the possible $x$'s?
For your gcd problem, if ther...
Yes. We still can say $c \ | \ a$ if $a = qc$ for some $q$. And a gcd of $a$ and $b$ is an element $d$ such that (i) $d \ | \ a$ and $d \ | \ b$; and (ii) for any $d'$ such that $d' \ | \ a$ and $d' \ | \ b$, $d \ | \ d'$. So it need not be unique (can be up to a unit), but is still defined.
Because the norm of the gcd can actually be smaller. For instance, the gcd of $1 + \sqrt{-7}$ and $1 - \sqrt{-7}$ is 1 since $1 + \sqrt{-7}$ is not divisible by $1 - \sqrt{-7}$ or 2 (just divide them in $\mathbb Q[\sqrt{-7}]$ and verify that the quotient is not in $R$).
The elements in $R$ with norm dividing 8 are, $\pm(1 \pm \sqrt{-7})$, $\pm 2$ and $\pm 1$. You need to check divisibility for each. After calculation you shall see that the gcd is something of norm 8 in this case.
Ok so let's say if $R = \mathbb{Z}[i\sqrt{3}]$ and I need to find gcd of $(3+i\sqrt{3},2-i\sqrt{3})$. I found its norm of gcd must divide 1. So the only case is + and - 1, am I right?
Let $R = \mathbb Z[\sqrt{-3}]$, $a = 4 = 2 \times 2 = (1+\sqrt{-3})(1-\sqrt{-3})$ and $b = 2(1+\sqrt{-3})$. Both 2 and $1+\sqrt{-3}$ are common divisors, but neither is "greater" than the other.
So in the same $R=\mathbb{Z}[\sqrt{-3}]$ the gcd of $a = 6+2\sqrt{-3}, b = 4-2\sqrt{-3}$ I can't find it because both $2+2\sqrt{-3}$ and $-2-\sqrt{-3}$ are common divizors but neither is greater then the other, right?
I am unsure whether you are just having a typo somewhere, but actually $2+2\sqrt{-3}$ does not divide $4-2\sqrt{-3}$. If you divide them the quotient is not in $R$.
Discussion between Hw Chu and Raducu …
Discussion between Hw Chu and Raducu …