Quadratic integer $\alpha$ is coprime to its conjugate if its norm and trace are coprime

Question

in a certain proof I'm working with the ring of integers of $\mathbb{Q}(\sqrt{-d})$ where $d=n^g-1$ is squarefree, $n>1$ is odd and $g>1$.

Then, as $d \equiv 2 \hspace{1mm} (\operatorname{mod} 4)$ I know that this ring is $\mathbb{Z}[\sqrt{-d}]$, and in this ring I have the following factorization of the ideal generated by $n$ $$(n)^g=(n^g)=(d+1)=(1+\sqrt{-d})(1-\sqrt{-d})$$ Know, I need to show that this two ideal factors are coprime and the text I'm using states that this fact derives from $n$ being odd but I don't see why.

I know that if a prime ideal $P$ divides both $(1+\sqrt{-d})$ and $(1-\sqrt{-d})$ then it divides $(n)$ and as the norm of $P$ is a power of a prime and the norm of $(n)$ is $n^2$ there is an odd prime dividing the norms of the two factors, which are both $1+d=n^g$, but I don't see any contradiction following this reasoning.

Any ideas?

Answer 1

Hint: $\ (a,b)\supset (ab,a\!+\!b).\,$ You have $\,ab\,$ odd, $\,a+b\,$ even so $\ldots$

When $\,b = \bar a\,$ it says: $ $ if $\,a\,$ has coprime norm and trace then $\,a,\bar a\,$ are coprime.

See here for another example and further intuition (e.g. method of simpler multiples).

Generally, because the norm map is multiplicative it preserves many factorization properties. For example, in many favorable contexts (e.g. Galois) a number ring enjoys unique factorization iff its monoid of norms does. For further interesting examples see the papers of Bumby, Dade, Lettl, Coykendall cited in this answer.