Let $\mathcal{O}$ be an order inside a quadratic number field (not necessarily maximal). I want to show that if $I$ is an invertible $\mathcal{O}$-ideal, then there is $\alpha \in I$ such that $N(\alpha I^{-1})$ is comprime with $N(I)$.
I know how to prove it in the case $\mathcal{O}$ is Dedekind. But, in general, it’s not necessarily true that every prime ideal is invertible, so $I$ might not factor into prime ideals. I tried arguing like this: let $P_1,…,P_n$ be the prime ideals that lie above the prime numbers that divide $N(I)$. Then, I want to show that there is $\alpha$ such that $\alpha \notin P_i I$ for every $i$. In other words, I want to show that it is not true that $$I=\bigcup P_i I.$$ I feel like I somehow need to use the invertibility of $I$ and prime avoidance to arrive to a contradiction. But I haven’t been able to do it.
This problem comes from a more general statement where $\mathcal{O}$ is an order inside a quaternion algebra, and $I$ is an invertible fractional right $\mathcal{O}$-ideal. The hint says to look for $\alpha$ such that $I^{-1}\alpha \subseteq \mathcal{O}$ and then look locally. But I haven’t found that useful.
Do you have any suggestions on how to proceed, or an other idea for a proof?
Thanks!
Let $\mathfrak{p}_1,\ldots,\mathfrak{p}_N$ be the primes above $(N(I))$.
Find $r$ such that for each $n$ $$\exists \beta_n\in\mathcal{O},\qquad \beta_n \in (I,\mathfrak{p}_n^r),\not \in (I\mathfrak{p}_n,\mathfrak{p}_n^r)$$
The $(I\mathfrak{p}_n,\mathfrak{p}_n^r)$ being pairwise comaximal we can find $\alpha\in\mathcal{O}$ such that for each $n$ $$\alpha =\beta_n\bmod(I\mathfrak{p}_n,\mathfrak{p}_n^r)$$ We get that $(\alpha)=IJ$ where $J$ is coprime with all the $\mathfrak{p}_n$ which means that $\gcd(N(J),N(I))=1$.