Let $X$ be a projective variety and $\mathcal{I} \subset \mathcal{O}_X$ is an ideal sheaf on $X$ not equal to $\mathcal{O}_X$. I'm supposed to show $\Gamma(X,\mathcal{I}) = 0$.
My first thought was that $\mathcal{I}(X)$ is an ideal of $\mathcal{O}_X(X)$, and since $X$ is a projective variety, $\mathcal{O}_X(X) = k$, the ground field, and of course the only ideals of a field are $k$ itself and $0$. So $\mathcal{I}(X) \in \{k,0\}$.
Since $\mathcal{I} \subsetneq \mathcal{O}_X$, my first thought was that of course $\mathcal{I}(X) \subsetneq \mathcal{O}_X(X)$, so that $\mathcal{I}(X) = 0$, but I don't think this is true anymore.
Just in terms of sheaves, a proper subsheaf can agree at the level of global sections. For instance, take the sheaf of holomorphic functions on a complex domain. A proper subsheaf would be the sheaf of constant functions, but at the level of global sections they definitely agree. I'm thinking we have to use the fact that $\mathcal{O}_X$ and $\mathcal{I}$ are coherent, but I don't really see how it applies.
This is for homework, so I'd prefer a hint over a full answer.
Since $\mathcal{I}(X) \subset \mathcal{O}_X(X)$ is an ideal, we have $\mathcal{I}(X) \in \{k,0\}$. Suppose $\mathcal{I}(X) = k$. But then for any open set $U \subset X$, we have $\mathcal{I}(U) \subset \mathcal{O}_X(U)$ is an ideal.
Since $1_k = 1 \in \mathcal{I}(X)$ and $\operatorname{res}^{\mathcal{I}}_{X,U}: \mathcal{I}(X) \rightarrow \mathcal{I}(U)$ is a ring homomorphism, we have $\operatorname{res}^{\mathcal{I}}_{X,U}(1) = 1_{\mathcal{O}_X(U)} \ni \mathcal{I}(U)$. So then $\mathcal{I}(U)$ is an ideal containing a unit, so namely $\mathcal{I}(U) = \mathcal{O}_X(U)$ for any open set $U \subset X$.
But we assumed $\mathcal{I} \subsetneq \mathcal{O}_X$, a contradiction. So $\mathcal{I}(X) = 0$, as required.