Let $A$ be a Noetherian commutative ring and $I$ an ideal in $A$. It is pretty much trivial to see that every free $A/I$-Module is obtained from a free $A$-module by tensoring over $A$ with $A/I$: Just choose preimages for every generator and take the free module generated by them. I'm wondering whether the same remains true if I replace ‘free’ by ‘projective and finitely generated’.


One way to prove this would be the following: Let $X=\operatorname{Spec} A$, $Z=\operatorname{Spec}(A/I)$, and $U=Z^c$. If there were something like an exact sequence


$$K\_0(U) \rightarrow K\_0(X) \rightarrow K\_0(Z) \rightarrow 0$$


the claim would follow, right? Does such a sequence exist? If I'm not mistaken, such a sequence exists if $A$ is a Dedekind ring. Just to make it clear, here is the precise question:


Question: Given a Noetherian commutative ring $A$, an ideal $I$ and a projective, finitely generated $A/I$-module $M$, does there exist a projective, finitely generated $A$-module $N$ such that $N/IN=M$?



Does every projective A/I-module come from A?
Thomas already [gave](https://mathoverflow.net/a/19564) an example, but let me make a general point I wish he had said: lots of rings with complicated sets of projectives are quotients of rings with simple sets of projectives. For example, any finitely generated projective module over a polynomial ring in any field is free (this is actually a hard theorem; it was proved by Suslin and Quillen in 1976 ([Projective modules over polynomial rings](https://doi.org/10.1007/BF01390008))).


So, if the answer to your question were affirmative, every projective module over a finitely generated commutative $k$-algebra for any field $k$ would be free, which is extremely false. In particular, you would have proven that every vector bundle on an affine variety is trivial, which as Tom points out, is not the case.


