Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let me motivate my question a bit.

Thm. Let $X$ be a locally noetherian finite-dimensional regular scheme. If $X$ has enough locally frees, then the natural homomorphism $K^0(X)\longrightarrow K_0(X)$ is an isomorphism.

A locally noetherian scheme has enough locally frees if every coherent sheaf is the quotient of a locally free coherent sheaf, $K^0(X)$ denotes the Grothendieck group of vector bundles on $X$ and $K_0(X)$ denotes the Grothendieck group of coherent sheaves on $X$.

The above theorem is shown as follows.

By the regularity (and finite-dimensionality!) of $X$, we can construct a finite resolution by a standard procedure. (Surject onto the kernel at each stage with a vector bundle.) Then the "Euler characteristic" associated to this resolution is inverse to the natural morphism.

Now, I was looking through the literature (Weibel's book basically) and I saw that this theorem appears with the additional condition of separability. (Edit: This is not necessary. The point is that noetherian schemes that have enough locally frees are semi-separated.)

Example. Take the projective plane $X$ with a double origin. Then $K^0(X) \cong \mathbf{Z}^3$ whereas $K_0(X) \cong \mathbf{Z}^4$.

Example. Take the affine plane $X$ with a double origin. Then $K^0(X) \cong \mathbf{Z}$, whereas $K_0(X) \cong \mathbf{Z}\oplus \mathbf{Z}$.

So I figured I must be missing something...Thus, I ask:

Q. Are locally noetherian schemes that have enough locally frees separated?

EDIT.

The answer to the above question is "No" as the example by Antoine Chambert-Loir shows.

From Philipp Gross's answer, we conclude that a noetherian scheme which has enough locally frees is semi-separated. This means that, for every pair of affine open subsets $U,V\subset X$, it holds that $U\cap V$ is affine. Note that separated schemes are semi-separated and that Antoine's example is also semi-separated.

Taking a look at Totaro's article cited by Philipp Gross, we see that a regular noetherian scheme which is semi-separated has enough locally frees. (Do regular semi-separated and locally noetherian schemes have enough locally frees?)

This was (in a way) also remarked by Hailong Dao. He mentions the result of Kleiman and independently Illuzie. Recently, Brenner and Schroer observed that their proof works also with $X$ noetherian semi-separated locally $\mathbf{Q}$-factorial. See page 4 of Totaro's paper. In short, separated is not really needed but semi-separated is.

Thus, we can conclude the following.

Suppose that $X$ is a regular and finite-dimensional scheme.

If $X$ has enough locally frees, then $K^0(X) \longrightarrow K_0(X)$ is an isomorphism. For example, $X$ is noetherian and semi-separated.

Anyway, thanks to everybody for their answers. They helped me alot!

share|improve this question

3 Answers 3

up vote 9 down vote accepted

The property that every coherent sheaf admits a surjection from a coherent locally free sheaf is also known as the resolution property.

The theorem can be refined as follows:

Every noetherian, locally $\mathbb Q$-factorial scheme with affine diagonal (equiv. semi-semiseparated) has the resolution property (where the resolving vector bundles are made up from line bundles).

This is Proposition 1.3 of the following paper:

Brenner, Holger; Schröer, Stefan Ample families, multihomogeneous spectra, and algebraization of formal schemes. Pacific J. Math. 208 (2003), no. 2, 209--230.

You will find a detailed discussion of the resolution property in

Totaro, Burt. The resolution property for schemes and stacks. J. Reine Angew. Math. 577 (2004), 1--22

Totaro proves in Proposition 3.1. that every scheme (or algebraic stack with affine stabilizers) has affine diagonal if it satisfies the resolution property.

The converse is also true for smooth schemes:

Proposition 8.1 : Let $X$ be a smooth scheme of finite type over a field. Then the following are equivalent:

  1. $X$ has affine diagonal.
  2. X has the resolution property.
  3. The natural map $K_0^{naive} \to K_0$ is surjective.
share|improve this answer
    
As Philipp says, the correct condition is having affine diagonal. As far as I know, there are no known examples of notherian schemes (or even algebraic stacks) with affine diagonal that do not have the resolution property. –  Angelo May 19 '10 at 19:49
    
Are you saying that probably all noetherian semi-separated schemes have enough locally frees? That is, the locally Q-factorial condition is probably not needed? –  Ari May 19 '10 at 20:31
1  
I am saying that I don't know. I don't even know enough to guess. –  Angelo May 19 '10 at 21:05

Actually the implication should be reversed: a separated regular Noatherian scheme has enough locally frees (this is Exercise 6.8, Chapter III Hartshorne). So the hypothesis is certainly needed for the proof.

EDIT: The statement in Hartshorne assumes X is integral, but this is not needed: see SGA 6, 2.2.3, 2.2.4, 2.2.5 and 2.2.7.1 (page 168-172 here ). In particular, you need separatedness and locally factorial (which follows from regular) to show that any coherent sheaf is a quotient of a direct sum of line bundles (for precise statement and example see 2.2.6 and 2.2.6.1).

In summary, the statement of the theorem is : For $X$ a regular, Noetherian, separated scheme one has $K_0(X) \cong K^0(X)$.

The answer to your question is no, as pointed out by Antoine.

share|improve this answer
    
The exercise of Hartshorne actually also assumes the scheme to be integral (i.e., connected in our case). I don't see why this implies that the hypothesis of separability is certainly needed for the proof. In fact, the example by Chambert-Loir of the affine line clearly shows that it is NOT needed. The question really isn't anymore about schemes being separable if they have enough locally frees, for that's not true. But it's actually about: Where does one use the separability of a scheme in the proof of the above theorem? (Looking at the sketch of the proof I gave, it really isn't clear.) –  Ari May 19 '10 at 10:29
    
I think I finally got it. I edited the question if you're interested. What you're saying can be improved a little if I'm not wrong. –  Ari May 19 '10 at 18:06

This doesn't answer your question but it is may be worth noticing that any vector bundle $E$ on the affine plane $X$ with doubled origine is trivial. Indeed, the inverse image of $E$ via the two natural maps $u_i\colon A^2\rightarrow X$ are vector bundles on $A^2$, so are trivial. The glueing condition on $X\setminus\{o_1,o_2\}$ ($o_1,o_2$ are the two origins) is an automorphism of the trivial line bundle on $A^2\setminus\{o\}$, hence extends to an automorphism on $A^2$ by Hartogs. This implies that the initial vector bundle is trivial.

By the way, the affine line with doubled origin certainly has enough locally frees...

share|improve this answer
    
This almost answers my question. I overlooked the simple example of the affine line with a doubled origin which clearly has enough locally frees. Thank you for that. Thus, schemes that have enough locally frees are not necessarily separated. It is pretty weird though, because I can't seem to understand where the separability of X is used in the proof of the above theorem... –  Ari May 18 '10 at 22:00
    
Separated implies a weaker condition called "semi-separated" that the intersection of two affines is affine. The line with the doubled origin is semi-separated. This allows it to avoid the Hartog issues of the plane with the doubled origin. Probably regular schemes have enough locally frees iff they are semi-separated. But you should instead redefine $K^0$ to be about perfect complexes, which is a local condition on the complex. –  Ben Wieland May 18 '10 at 23:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.