Does every projective A/I-module come from A? 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 $N/IN=M$?
 A: Here's an explicit counterexample:
Let $A = \mathbb{Z}$, and $I=6 \mathbb{Z}$. Then $A/I = \mathbb{Z}/6$ which is isomorphic to $\mathbb{Z}/2 \oplus \mathbb{Z}/3$ as module over itself. Since $\mathbb{Z}/2$ and $\mathbb{Z}/3$ are direct summands of a free module they are projective and they are clearly finitely generated. Every finitely generated projective $\mathbb{Z}$-module $P$ is free so $P \otimes \mathbb{Z}/6$ must be free.
This can also be stated with $K_0$-groups: There is an isomorphism $K_0(\mathbb{Z}/6) \cong K_0(\mathbb{Z}/2) \oplus K_0(\mathbb{Z}/3)$ and each summand here is isomorphic to $\mathbb{Z}$. The map $K_0(\mathbb{Z}) \rightarrow K_0(\mathbb{Z}/2) \oplus K_0(\mathbb{Z}/3)$ corresponds to the diagonal map $\mathbb{Z} \xrightarrow{\Delta} \mathbb{Z} \oplus \mathbb{Z}$ so $(1,0)$ and $(0,1)$ are not in the image.
In this example the OP's sequence of $K_0$-groups cannot be exact. In Rosenberg's book Introduction to algebraic $K$-theory that Mariano mentions there is no zero group at the end of the sequence.
Geometrically this discussion amounts to the fact that the vector bundles on a disconnected subspace, in this case {$ \{ (2),(3) \}$}, can't all be restrictions of vector bundles on the total space which is connected, since the rank of a bundle is locally constant.
A: There is such a sequence, of sorts. Check, for example, Rosenberg's Introduction to algebraic $K$-theory.
A: No.  Here's an "example" (it's not quite completely explicit, but with a little effort you can make it absolutely explicit).  Let $A = {\mathbb C}[x,y]$ and $I = (y^2- x(x+1)(x-1))$.  [If you are comfortable with algebraic geometry, $A$ is the coordinate ring of the affine plane and $I$
is the ideal of a nonsingular plane cubic curve $C$ whose equation is $y^2 = x(x+1)(x-1)$, i.e. an "affine elliptic curve."]  
There 
are plenty of rank $1$ projective $A/I$-modules that are not trivial: geometrically, these
correspond to spaces of sections of nontrivial line bundles on the curve $C$.  So, for 
example, you can get one by picking a point $p$ on $C$ "at random" and taking the module $M$
consisting of rational functions on $C$ that have at worst a first-order pole at $p$ and are regular elsewhere on $C$.  On the other hand, every rank $1$ projective module over ${\mathbb C}[x,y]$ is trivial...   
A: Thomas already gave 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)).
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.
