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 $X=Spec A$, $Z=Spec(A/I)$, and $U=Z^c$. If there were something like a 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 $N/IN=M$?

  • $\begingroup$ Notice that your exact sequence (some version of which does exist per Mariano's answer) is far from sufficient to establish your claim. Surjectivity of $K_0(X)\rightarrow K_0(Z)$ is far from sufficient to establish that every $Z$-projective lifts to $X$. $\endgroup$ – Steven Landsburg Aug 18 '15 at 16:24

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).

So, if the answer to your question were affirmative, every projective module over a finitely generated $k$-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.


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...

  • 2
    $\begingroup$ To make it explicit: take $M$ to be non-principal maximal ideal in $A/I$, for example $M=(x,y)$ $\endgroup$ – Hailong Dao Mar 28 '10 at 3:15

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 $\mathbb{Z} \stackrel{\Delta}{\rightarrow} \mathbb{Z} \oplus \mathbb{Z}$ 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 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.


There is such a sequence, of sorts. Check, for example, Rosenberg's Introduction to algebraic $K$-theory.

  • $\begingroup$ Wow, that was quick. I've checked Milnor's book and the book on Weibel's website so far. I'll give Rosenberg a shot. $\endgroup$ – Timo Schürg Mar 27 '10 at 22:52

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