Algebraic Surfaces Is there any example of a smooth, projective surface $S$ over $\mathbb{C}$,  with Picard group $\mathbb{Z}$ and such that $H^1(S, L)$ is not zero for some ample line bundle $L$ ?
 A: Edit. My original exact sequences were wrong.  I straightened them out now.  
Let $X$ be a smooth, projective hyper-Kähler fourfold with $\text{Pic}(X)$ isomorphic to $\mathbb{Z}$, e.g., a sufficiently general deformation of the pair $(\text{Hilb}^2_{M/\mathbb{C}},\mathcal{O}(1))$ for $M$ a polarized K3 surface and $\mathcal{O}(1)$ the natural Plücker invertible sheaf.  Let $Y\subset X$ be a smooth ample hypersurface such that $h^r(X,\mathcal{O}_X(\underline{Y}))$ vanishes for $r=1,2$.  Let $Z\subset X$ be a very general hypersurface that is "sufficiently" ample.  In particular, assume that $h^r(X,\mathcal{O}_X(\underline{Z}))$ vanishes for $r=1,2$ and $h^2(X,\mathcal{O}_X(\underline{Z}-\underline{Y}))$ vanishes. Let $i:S\hookrightarrow X$ be $Y\cap Z$, and denote by $\mathcal{I}$ the ideal sheaf of $S$ in $X$. Let $L$ be  $\mathcal{O}_X(\underline{Y})|_S$.  
There are short exact sequences, 
$$ 0 \to \mathcal{I}(\underline{Y}) \to \mathcal{O}_X(\underline{Y}) \to i_* L \to 0,$$
$$
0 \to \mathcal{O}_X(-\underline{Z}) \to \mathcal{O}_X\oplus \mathcal{O}_X(\underline{Y}-\underline{Z}) \to \mathcal{I}(\underline{Y}) \to 0.$$
By hypothesis, $h^r(X,\mathcal{O}_X(\underline{Y})) = 0$ for $r= 1,2$.  Thus, via the long exact sequence of cohomology, the following connecting map is an isomorphism, $$\delta_\Sigma^1:H^1(S,L)\xrightarrow{\cong} H^2(X,\mathcal{I}(\underline{Y})).$$  By Serre duality, $h^{2+r}(X,\mathcal{O}_X(-\underline{Z}))$ equals  $h^{2-r}(X,\mathcal{O}_X(\underline{Z}))$.  Thus, for $r=0,1$, both of these vanish.  Finally, by Serre duality, $h^2(X,\mathcal{O}_X(\underline{Y}-\underline{Z})))$ equals $h^2(X,\mathcal{O}_X(\underline{Z}-\underline{Y}))$, so both of these vanish.  Thus, the long exact sequence of the second short exact sequence gives an isomorphism,
$$
H^2(X,\mathcal{O}_X)\to H^2(X,\mathcal{I}(\underline{Y})),
$$
so also $H^1(S,L)$ is isomorphic to $H^2(X,\mathcal{O}_X)$.
Of course for the hyper-Kähler fourfold $X$, $H^2(X,\mathcal{O}_X)$ is a $1$-dimensional vector space.  
By the Lefschetz hyperplane theorem, the restriction map $\text{Pic}(X)\to \text{Pic}(Y)$ is an isomorphism (and the same holds for $Z$).  By the generalized Noether-Lefschetz theorem (as in SGA7), for $Z$ sufficiently ample and very general, the restriction map $\text{Pic}(Y)\to \text{Pic}(S)$ is also an isomorphism.  Thus $\text{Pic}(S)$ is isomorphic to $\mathbb{Z}$.
Edit. I was misremembering what is proved in SGA 7_2.  The generalized Noether-Lefschetz theorem was not proved there.  The generalized Noether-Lefschetz was proved by Kirti Joshi.
MR1299006 (96f:14005) 
Joshi, Kirti(6-TIFR-SM) 
A Noether-Lefschetz theorem and applications. 
J. Algebraic Geom. 4 (1995), no. 1, 105–135. 
https://arxiv.org/pdf/alg-geom/9305001v2.pdf
A: For example, for an irregular surface of general type of Picard number one, $H^1(K_X)\neq0$. See Kovacs' answer here Smooth projective varieties of Picard number one. 
Edit: As pointed out by @abx, in this case the picard group could not be $\mathbb{Z}$, as the surface is irregular. 
Maybe the way of construct such example is to look at certain surface with $K_X=2L$ and try to show $H^1(L)\neq 0$. 
