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 separatedness. (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 Illusie. 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!



Are schemes that "have enough locally frees" necessarily separated
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.


