In showing that the elements $6$ and $2+2\sqrt{5}$ in $\mathbb{Z}[\sqrt{5}]$ have no gcd, I was thinking of trying the following method.
If the ideal $(6)$ + $(2+2\sqrt{5})$ is not principal in $\mathbb{Z}[\sqrt{5}]$ then there does not exist $d \in \mathbb{Z}[\sqrt{5}]$ such that $(6)$ + $(2+2\sqrt{5}) = (d)$ and therefore there is no gcd.
However this uses the lemma that if $\gcd(x,y) = d$ then $(x) + (y) = (d)$, is this true?
Best Answer
It is not true in general that if $\gcd(x,y)=d$ then $(x)+(y)=(d)$. In particular, if $R=\mathbb{Z}[x]$ (the ring of polynomials over $\mathbb{Z}$ in the indeterminate $x$), then $\gcd(2,x)=1$ (since any divisor of $2$ in $R$ must be an integer, and the only integers that divide $x$ in $R$ are $\pm 1$), however $(2,x)=(2)+(x)\subsetneq R$.
More generally, integral domains for which gcds can be written as a linear combination are known as Bézout domains--domains where every finitely generated ideal is principal.
As Bill mentions in his answer, GCD domains (ie integral domains in which every pair of nonzero elements has a gcd) satisfy the property that every irreducible is prime. In particular, if $R$ is an integral domain in which every nonzero nonunit can factor into irreducibles, then $R$ is a GCD domain if and only if $R$ is a UFD.
As for showing that $6$ and $2+2\sqrt{-5}$ have no gcd, this can also be done easily enough using the norm function $N:\mathbb{Z}[\sqrt{-5}]\rightarrow\mathbb{N}\cup\{0\}$ defined by $N(a+b\sqrt{-5})=a^2+5b^2$. In particular for all $\alpha,\beta\in R$, $N(\alpha\beta)=N(\alpha)N(\beta)$, $N(\alpha)=0$ if and only if $\alpha=0$, and $N(\alpha)=1$ if and only if $\alpha\in U(R)$ (proving this is a canonical abstract algebra exercise). So, if there is a gcd of 6 and $2+2\sqrt{-5}$, then there must be a common divisor of 3 and $1+\sqrt{-5}$ (since 2 divides both 6 and $2+2\sqrt{-5}$ in $R$).
Thus, there exists $\alpha\in R$ such that $\alpha \mid 3$ and $\alpha \mid 1+\sqrt{-5}$. Taking the norm (and using the norm properties above), we have $N(\alpha)\mid 9$ and $N(\alpha) \mid 6$. Therefore, it follows that either $N(\alpha)=1$ or $N(\alpha)=3$. However, the fact that there are no integer solutions to the equation $a^2+5b^2=3$ (why?) implies that there is no element in $R$ having a norm of 3. Therefore $N(\alpha)=1$ and $\gcd(3,1+\sqrt{-5})=1$.
Now suppose that $6$ and $2+2\sqrt{-5}$ have a gcd. Then we have $2=2\gcd(3,1+\sqrt{-5})=\gcd(6,2+2\sqrt{-5})$. However, we reach a contradiction as $1+\sqrt{-5}$ is a common divisor of both 6 and $2+2\sqrt{-5}$, but $1+\sqrt{-5}$ does not divide 2 (Hint: suppose it does and use the norm). Therefore $\gcd(6,2+2\sqrt{-5})$ does not exist.