scheme without global sections Let $R$ be an arbitrary non zero ring and $X$ a $R$-scheme. Is it possible that the ring of global sections $H^0(X, \mathcal{O}_X)$ might be a zero ring? The cruical point is that despite of the $R$-scheme structure for $X$ providing the existence of ring map $R \to H^0(X, \mathcal{O}_X)$ I see no obstruction for this map to be a zero map. 

Yes. The empty scheme has zero global sections, because for any scheme $X$,  $\mathcal{O}_X(\emptyset)$ is the zero ring. 
Conversely, any scheme with zero global sections is empty. 
Proof: Suppose $X$ has zero global sections. Let $U\subseteq X$ be an affine open set, isomorphic to $\text{Spec}(A)$. Then there is a restriction map $\rho^X_U\colon \mathcal{O}_X(X)\to \mathcal{O}_X(U)$, so $A = \mathcal{O}_X(U)$ is the zero ring (because $1_A = \rho^X_U(1) = \rho^X_U(0) = 0_A$). Since the zero ring has no prime ideals, $U$ is empty. Now $X$ admits a cover by affine open sets, all of which must be empty, so $X$ is empty. 
