Let $R= \mathbb{Z}[\sqrt{-5}]$ and $I= (2, 1+\sqrt{-5})$. I'm trying to show $\tilde{I}$ is an invertible sheaf on $\mathrm{Spec}R$. The spectrum is covered by $D(2)$ and $D(3)$ so I claim the restriction of $\tilde{I}$ on each open affine could be trivial.
On $D(3)$, I get $(2, 1+\sqrt{-5})_3= (\frac{1-\sqrt{-5}}{3})$ as $R_3$-modules. How about $D(2)$?
My teacher explained $(2, 1+\sqrt{-5})_2= (\frac{1+\sqrt{-5}}{2})$ as $R_2$-modules, but I found it wrong because $2$ cannot be generated by $\frac{1+\sqrt{-5}}{2}$ as a $R_2$-module. I'm stuck here and I can't find a suitable generator. What should I do on $D(2)$? Or I doubt $\tilde{I}$ is trivial on $D(2)$. Thanks to your help!