Let $R$ be a Noetherian ring and $I$ be a proper ideal of $R$ and let $b\notin I$ be such that $I+Rb$ and $(I:b)=\{x\in R: xb\in I\}$ are projective as $R$-modules.
Then is the ideal $I$ projective as an $R$-module?
NOTE: If this is true, then it would imply that in a Noetherian ring, if every prime ideal is projective, then every ideal is projective.
Here goes. To say $I$ is projective is a local condition, so we localize at a prime ideal and call this $R$ etc. Now, one possibility is, when you localize, $b\in I$. Then $I+Rb=I$ is projective since $I+Rb$ is. The other possibility is $b\notin I$. Then, as I explained earlier, $I+Rb=Rb$ and so $I\subset Rb$ and $b$ is not a zero divisor in $R$, since $Rb$ is free. Thus $I=Jb$ where $J=(I:b)$, which is projective. So, $I\cong J$ and thus $I$ is projective. Since at every localization $I$ is projective, so is $I$ without localization.