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I want to prove that $a+b\sqrt{-3}$ and $a-b\sqrt{-3}$ are coprime in $\mathbb{Z}+ \omega \mathbb{Z}$ with $\omega$ a primitive third root of unity and $\gcd(a,b)=1$.

Approach: Assume they are not, let $s$ be an irreducible factor of $a+b\sqrt{-3}$ and $a-b\sqrt{-3}$, then $s|2b\sqrt{-3}$. Then $N(s)|6b^2$ and $N(s)|(a^2+3b^2)$….

Can someone help me how to approach a contradiction from this? Thanks in advance!!

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Actually, there’s no contradiction.

Even if $a,b$ are relatively prime, the claim is false.

For example, let

$$y = 1 + \sqrt{-3}$$

$$z = 1 – \sqrt{-3}$$

Then $y,z$ are non-units, but each divides the other, hence they have a common non-unit factor.

So many things that need clarification here. What are $a$ and $b$? Are they integers, that is $a, b \in \mathbb Z$? And are you sure you’re only looking at numbers of the form $a + b \sqrt{-3}$, and not also numbers of the form $a + b \omega$?

It needs to be said explicitly that $$\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2}.$$ With that in mind, we can rewrite $1 – \sqrt{-3}$ and $1 + \sqrt{-3}$ as $-2 \omega$ and $2 + 2 \omega$ respectively. We have changed $a$ and $b$, of course, but it should still be clear that $2 \omega$ and $-2 \omega$ have at least one prime factor in common, as do $2 – 2 \omega$ and $2 + 2 \omega$.

Also note that $N(a + b \omega) = a^2 – ab + b^2$.

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