Let $R=\mathbb Z[\sqrt{-5}]$ and $I=(3,2+\sqrt {-5})$ be the ideal generated by $3$ and $2+\sqrt{-5}$. I'm trying to prove that $I$ is a projective $R$-module.
I'm using the lifting property which is the most used definition of projective modules. I'm having troubles to prove that $I$ is projective using this definition, maybe it's easy and I'm forgetting something or there is another definition which could help more?
I need help.
Thanks
The general strategy here is to take the exact sequence $$ 0\to \ker \pi\to R^2\overset{\pi}\to I\to 0 $$ and find a splitting for $\pi$. Then we'd have $R^2\cong \ker\pi\oplus I$, so $I$ is projective as a direct summand of a free module. You have to implicitly use fractional ideals of $R$, so if you know what that is do some Googling, but if you don't then just try to split the sequence.