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?
Best Answer
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 further intuition (method of simpler multiples) and generalizations (Bumby).