Vanishing of $H^0(X,\mathcal{L})$ and $H^0(X,\mathcal{L}^{-1})$ We have this fact: 

If $X$ is a smooth projective variety, and $\mathcal{L}$ is a line bundle, and $\mathcal{L}^{-1}$ its dual, then $H^0(X,\mathcal{L})$ and $H^0(X,\mathcal{L}^{-1})$ cannot both be nontrivial (except $\mathcal{L}=\mathcal{O}_X)$. 

This can be seen by looking at the linear system. 
My question is: is the above statement true if one takes away the projectivity assumption?
My guess is this is not true. I believe one can cook up some example on the blow up of $\mathbb{A}^2$ at a point.
One more question: I believe the reason for the first fact is that if a meromorphic function have zeros, it must also have poles, but this requires compactness, right?
 A: You only need $X$ to be compact and connected for this result to hold. I will write $\mathcal {L}^*$ instead of $\mathcal {L}^{-1}$.
There is a map 
$$H^0(X, \mathcal {L})\otimes H^0(X, \mathcal{L}^*) \to H^0(X, \mathcal{L}\otimes \mathcal{L}^*) \cong H^0(X, \mathcal{O}) \cong \Gamma(X, \mathcal{O}) \cong \mathcal{O}(X)$$
given by $f\otimes \alpha \mapsto \alpha(f) \in \mathcal{O}(X)$, where $(\alpha(f))(x) = \alpha_x(f_x)$. As $X$ is compact, $\alpha(f)$ is constant.
If $\mathcal{L}$ is trivial, then so is $\mathcal{L}^*$ and hence $H^0(X, \mathcal{L}) \cong H^0(X, \mathcal{L}^*) \cong H^0(X, \mathcal{O}) \cong \mathcal{O}(X) \cong \mathbb{C}$. 
Suppose now that $\mathcal{L}$ is not trivial, yet admits a non-zero section $f$; as $\mathcal{L}$ is not trivial, there is $x \in X$ such that $f_x \in \mathcal{L}_x$ is zero. For any section $\alpha$ of $\mathcal{L}^*$ we have $(\alpha(f))(x) = \alpha_x(f_x) = \alpha_x(0) = 0$. As $\alpha(f)$ is a constant function and has value zero at $x$, it must be the constant zero function. Therefore $\{x \in X \mid \alpha_x = 0\}\cup\{x \in X \mid f_x = 0\} = X$. As both of these sets are closed, at least one of them contains an open set. By the identity theorem, together with the fact that $X$ is connected, either $\alpha = 0$ or $f = 0$, but $f$ is a non-zero section, so $\alpha = 0$. Therefore $\mathcal{L}^*$ does not admit a nowhere-zero section.
Likewise, if $\mathcal{L}^*$ is non-trivial and admits a non-zero section, then $\mathcal{L}$ does not admit a non-zero section.
