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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.

http://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

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# M - affine and H^1(M,Z)=H^2(M,Z) = 0 imply (?) Pic^algebraic(M) = 0. Note: in algebraic category NO exponential sequence

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.