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I have the ring $R = \mathbb{Z[i\sqrt{3}]}$, and I need to find the gcd of $a = 6+2i\sqrt{3}, b = 4-2i\sqrt{3}$. This is the way I solved it:

Let $d = x+yi\sqrt{3}$ be $gcd(a,b)$ then $d |a$ and $d|b$ and also $N(d)|N(a) = 48$ and $N(d)|N(b) = 28$ so $N(d)| 4$. Therefore $d$ could be + or -$1,2$ or $1+i\sqrt3$ because these numbers have norm that divide 4. So I would say that gcd is $1+i\sqrt3$. But it turns out that this can't be gcd because it is not dividing a and b.

So my question is why can't $1+i\sqrt3$ be gcd?

Raducu Mihai
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    In general, if $d\nmid a$ or $d\nmid b$ then clearly $d$ cannot be $\gcd(a,b)$. Just because the norm of $d$ divides the norms of $a$ and $b$ does not mean that $d$ divides $a$ and $b$. The implication goes the other way and only the other way. – Dave Jan 15 '18 at 21:40
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    How is the answer to this not "all common divisors of $a$ and $b$ divide both $a$ and $b$"? What alternative are you proposing? – Eric Towers Jan 15 '18 at 21:40
  • @EricTowers I don't know exactly what are you asking. Anywas I think Dave 's comment cleared the things – Raducu Mihai Jan 15 '18 at 21:53
  • "$\gcd$" = "greatest common divisor"; it is the maximal common divisor. All common divisors of $a$ and $b$ divide $a$ and $b$. How could something be a $\gcd$ and not also be a common divisor? – Eric Towers Jan 15 '18 at 21:56
  • I know this. But I thought using the method that I used to solve that exercise would lead me to the gcd. More specifically my question is: I did I not get the gcd by using that method? – Raducu Mihai Jan 16 '18 at 13:22

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