Assume M affine algebraic manifold over C and we know that H^1(M,Z)=H^2(M,Z) = 0.


**Question** Does it imply Pic^algebraic(M) = 0 ? 


Pic^algebraic means group of algebraic line bundles = H^1(M, O^\*) in Zariskky topology.


This is follow up to:


[Are there other ways to show Pic(G)is trivial when G is a simple-connected semisimple algebraic groups over C?](https://mathoverflow.net/questions/96987/are-there-other-ways-to-show-picgis-trivial-when-g-is-a-simple-connected-semisi)


**Question 2** If it is true what should be analogue of this statement for arbitrary algebraic closed field ? Can we substitute topological cohomology by etale ? 


Let me point out the following subtlety: 


If I ask about analytic line bundles, then the answer is YES, by exponential sequence
argument and vanishing of H^i(M, O) for affine manifolds.
If manifold would be compact then by GAGA they coincide, but for non-compact algebraic and analytic are essentially different. If you take elliptic curve and drop out 1 point - you get affine curve - so by exponential sequence Pic^analytic=0, while Pic^algebraic = curve itself+point (as far as I remember). This was quite strange and surprising for me when I learn it. 


In the comments Daniel Loughran suggested that Kummer sequence might work,
I think it is reasonable, but I am not so experience with etale-cohomology, 
so I am asking



M - affine and H^1(M,Z)=H^2(M,Z) = 0   imply (?) Pic^algebraic(M) = 0.  Note: in algebraic category NO exponential sequence  
Let us assume that we work over an algebraically closed base field $k$, and fix a prime $\ell$ different from the characteristic of $k$.
First of all, the Kummer-sequence
$$0\rightarrow \mu\_{\ell^n}\rightarrow \mathbb{G}\_m\xrightarrow{\ell^n}\mathbb{G}\_m\rightarrow 0$$
for $\ell$ prime to the characteristic of $k$, shows that $Pic(X)$ contains no $\ell$-power torsion. In particular, if $char(k)=0$, then $Pic(X)$ is torsion free.


Next, recall that for étale cohomology we have $H^1(X,\mathbb{Z}\_\ell(1))=\hom^{cont}(\pi^{et}\_1(X),\mathbb{Z}\_{\ell})$. Hence, if $H^1(X,\mathbb{Z}\_\ell(1))=0$, then the maximal abelian pro-$\ell$-quotient of $\pi\_1^{et}$ is $0$. 


Now assume that we have a smooth proper scheme $X'$ containing $X$ as a dense open subset. In your characteristic $0$ situation this can always be achieved. Then the abelian maximal pro-$\ell$-quotient of $\pi\_1^{et}(X)$ surjects onto the maximal abelian pro-$\ell$ quotient of $\pi\_1(X')$, so this group is also trivial. Again, using the Kummer sequence, this implies that $Pic(X')$ contains no $\ell$-power torsion. Since $X'$ is proper, there is a Picard scheme $Pic\_{X'/k}$, such that $Pic(X')=Pic\_{X'/k}(k)$. In particular, if $Pic^0\_{X'/k}$ denotes the connected component of the origin, then $Pic^0\_{X'/k}(k)$ has no $\ell$-power torsion. But $Pic^0\_{X'/k}$ (or rather its reduced closed subscheme, if $k$ has positive characteristic) is an abelian variety. It follows that $Pic^0\_{X'/k}$ is a $0$-dimensional abelian variety (as otherwise there would be nontrivial $\ell$-torsion). This implies that $Pic(X')=NS(X')=Pic\_{X'/k}(k)/Pic^0\_{X'/k}(k)$, which is a finitely generated group without prime-to-$p$-torsion, i.e. if $char (k)=0$, then $Pic(X')$ is free of finite rank.


Since $X'$ was smooth, we have a surjection $Pic(X')\twoheadrightarrow Pic(X)$, so $Pic(X)$ is free of finite rank (because we already knew it is torsion free).


Next, the Kummer sequence gives an exact sequence
 $$0\rightarrow Pic(X)\xrightarrow{(-)^{\ell^n}}Pic(X)\xrightarrow{c\_{1,\ell^n}}H^2(X,\mathbb{Z}/\ell^n\mathbb{Z}(1))$$
for every $n$, so $Pic(X)/\ell^n\subset H^2(X,\mathbb{Z}/\ell^n\mathbb{Z}(1))$, and passing to the limit gives $Pic(X)\otimes \mathbb{Z}\_\ell\subset H^2(X,\mathbb{Z}\_\ell(1))$, because we know that $Pic(X)$ is finitely generated. 


Thus, if we assume that $H^2(X,\mathbb{Z}\_\ell(1))=0$, then $Pic(X)=0$.


