Attempt on proving elements of $\Bbb Z[\sqrt{-3}]$ are relatively prime Let $a = 1 + \sqrt{-3}$ and $b = 1-\sqrt{-3}$. Prove that $gcd(a,b) = 1$ in $\mathbb{Z}[\sqrt{-3}]$.
I first consider $e = gcd(a,b) = c + d\sqrt{-3}$ with the norm $N(e) = c^2 + 3d^2$. If e is supposed to divide a and b, then the norm of a and b should also be divisible by the norm of e. The norm of a and b are 4, so $(c^2+3d^2)|4$, which gives $e \in \{\pm 1, \pm 2, \pm 1\pm \sqrt{-3}\}$. The problem is that I don't know how you prove on an abstract level that the gcd now must be 1.
Maybe the fact that $\pm 2$ and $\pm 1 \pm \sqrt{-3}$ has norm equal to 4, such that a and b must be associates, which they are not, because $\pm 1$ are the only units in $\mathbb{Z}[\sqrt{-3}]$...? And should I just argue that since $-1 \leq 1$
, then gcd(a,b) = 1? The definition of gcd does seem to fit both 1 and -1.
I would be happy for some feedback, thanks :)
EDIT: I'm now aware of that 1 and -1 are considered equal since they are associates, but still.
 A: The ring $R'=\Bbb Z[\sqrt{-3}\,]$ is not a principal ideal, which makes the story much more complicated. But we can look at it as a subring of $R=\Bbb Z\bigl[(1+\sqrt{-3}\,)/2\bigr]=\Bbb Z[a/2]$ in your notation. This ring $R$ is a PID, and is the ring of integers of $\Bbb Q(\sqrt{-3}\,)$. Its arithmetic is this: it’s PID, $2$ is a prime element, and the units of $R$ are six in number, the powers of $a/2$.
I’ll use the (fairly) standard terminology that two numbers are associates of each other if one of them is a unit times the other. In $R'$, the only units are $\{\pm1\}$, since these are the only powers of $a/2$ that are in $R'$. In $R'$, $a$ and $b$ are not associate, since $a/b$ is equal to the $R$-unit $(a/2)^2$, which isn’t in $R'$.
Now all I need to do is show that $a$ is an irreducible element of $R'$; the same proof shows that $b$ also is irreducible. Since they’re not associate, we have to consider them relatively prime.
But looking at $a=2\cdot a/2$ as element of $R$, you see that it’s $2$ times a unit, so it’s irreducible in $R$, and thus certainly irreducible in $R'$. That should do it.
(There may well be easier arguments!)
