Let $R$ be an integral domain and $I$ a projective ideal. Then $I$ is finitely generated as an $R$-module.
This seems like it should be easy but I don't know what to argue, or where to bring in the integral domain condition.
I tried localising at a prime $P$ containing $I$, and we get that $I_P\subseteq PR_P\subseteq R_P$ is a projective $R_P$ module. Then if $0\to I_P\to R_P\to R_P/I_P\to 0$ is short exact, since projective modules are flat, we can tensor with $I_P$, and the last term is $0$. But there is a result that if $M$ is a flat module over a ring then if $J$ is an ideal, then $J\otimes M\cong JM$. Hence, $I_P^2\cong I_P\otimes I_P\cong R_P\otimes I_P\cong I_P$.
But if $I$, hence $I_P$ were finitely generated, then wouldn't that mean $I\subseteq I_P=0$ by Nakayama?