Are schemes that "have enough locally frees" necessarily separated - MathOverflow

Are schemes that "have enough locally frees" necessarily separated

2010-05-18T13:35:13Z

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!

Answer by ACL for Are schemes that "have enough locally frees" necessarily separated

2010-05-18T15:49:02Z

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...

Answer by Hailong Dao for Are schemes that "have enough locally frees" necessarily separated

2010-05-18T23:11:26Z

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.

Answer by Philipp Gross for Are schemes that "have enough locally frees" necessarily separated

2010-05-19T12:34:24Z

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:

$X$ has affine diagonal.
X has the resolution property.
The natural map $K_0^{naive} \to K_0$ is surjective.