Sheaf on a Stein variety such that $H^{1}(X, \mathcal{F}) \neq 0$

I would like to find a non-coherent sheaf on a Stein variety $X$ such that $H^{1}(X, \mathcal{F}) \neq 0$. Does anyone know any example?

Thank you!

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Just take $\mathcal F=\mathcal O^*_X$, the sheaf of invertible holomorphic functions.
Then $H^{1}(X, \mathcal{F}) =H^{1}(X, \mathcal O^*_X) =Pic(X)$, the Picard group consisting of isomorphism classes of holomorphic line bundles on $X$.
Now from the exponential exact sequence $0\to \mathbb Z\to \mathcal O_X\to \mathcal O_X^*\to 0$ you immediately get by taking the long exact sequence of cohomology that the first Chern class yields a group isomorphism $$H^{1}(X, \mathcal O_X) =0 \to H^{1}(X, \mathcal O^*_X) \xrightarrow {c_1} H^2(X,\mathbb Z)\to H^{2}(X, \mathcal O_X)=0$$
So a very simple example of Stein manifold with $H^{1}(X, \mathcal O^*) \neq 0$ is $\mathbb C^*\times \mathbb C^*$ since $$H^{1}(\mathbb C^*\times \mathbb C^*, \mathcal O^*)\stackrel {c_1}{\cong} H^{2}(\mathbb C^*\times \mathbb C^*, \mathbb Z)=\mathbb Z$$
 So nice @Georges, but how do you know that $H^{2}(\mathbb{C}^{\ast} \times \mathbb{C}^{\ast}, \mathbb{Z}) = \mathbb{Z}$? It seems come from some topological information of the torus, and I am not familiar with that. Could you explain or give me any reference? Thank you! – rla Oct 23 '12 at 11:40 Dear rla, you can use the Künneth formula to compute the cohomology of that product. Alternatively, you might note that $\mathbb C^*\times \mathbb C^*$is homotopy-equivalent to the product $S^1\times S^1$ of two circles and you can thus reduce to the computation of the cohomology of that compact orientable real surface. – Georges Elencwajg Oct 23 '12 at 12:09