Simply connected surfaces and torsion in Grothendieck group Let $X$ be a projective complex surface (complex manifold of dimension 2). In a paper I met the following claim: if $X$ is simply connected $\pi_1(X) \cong \{1\}$ then the torsion of the Grothendieck group vanishes $K_0(X)_{tors} \cong 0$.
How one can prove this claim? If this is well-known what is a reference?
 A: Let $F^i$ denote the subgroup generated by sheaves $S$ on $X$ with $\operatorname{codim}_X \operatorname{Supp} S \leq i$. Then $F^i\supset F^{i+1}$ and these form a filtration of $K_0(X)$. By Riemann-Roch without denominators (Fulton's Intersection Theory, Section 15.3), assuming $X$ is a smooth projective surface, we have that $F^0/F^1 \cong \Bbb Z$, $F^1/F^2 \cong Pic(X)$, and $F^2\cong CH_0(X)$.
Proposition: for $X$ simply connected, then $K_0(X)$ torsion-free is equivalent to $CH_0(X)$ torsion-free.
Proof: If $X$ is simply connected, then $Pic(X)$ is a finitely generated free abelian group. 
Any extension $0\to F^1\to F^0\to \Bbb Z\to 0$ will split, so we have that $K_0(X)\cong \Bbb Z \oplus F^1$. So it remains to see that $F^1$ is torsion-free. Since $0\to CH_0(X) \to F^1\to Pic(X)\to 0$ is exact, this shows that $F^1$ is torsion-free iff $CH_0(X)$ is, so we're done.
So the problem is equivalent to showing that $CH_0(X)$ is torsion-free. By Rojtman's theorem, the homomorphism $CH_0(X)\to Alb(X)(\Bbb C)$ is an isomorphism on torsion subgroups. Since $\dim Alb(X) = h^{1,0} = h^{0,1}$, we have that $\dim Alb(X) = \frac{1}{2}b_1 = 0$, where this last equality comes from $H_1(X)=\pi_1(X)^{ab}=0$ by simply connectedness. Thus $Alb(X)(\Bbb C)$ is torsion-free and so is $CH_0(X)$, so $K_0(X)$ is torsion-free.
