Cohomology of zero dimensional sheaf Let,$E$ be a zero dimensional sheaf on an algebraic surface $S$ over the field of complex numbers,i.e $dim(Supp(E)) =0$
Then my question is the following : Is it true that $H^{1}(E) =0$ or in general $H^{i}(E) =0$ for all $i \geq 1$?
Any help from anyone is welcome

I develop my comment here. I assume your sheaf is coherent as you are interested in algebraic surfaces.
Thus let $E$ be a coherent sheaf on an algebraic variety. Assume $Supp(E)=\{x_1, \dots, x_n\}$ is finite. Write $E(x_j)$ the stalk of $E$ at $x_j$. The restriction map $E\to E^{'}:=\oplus_j E(x_j)$ is an isomorphism of sheaves as it is an isomorphism on each stalk.
Let $U\in X$ be an open and let $x_{i_1},\dots x_{i_k}$ be the subset of $x_i$'s lying in $U$. Then the sheaf $E^{'}$ verifies
$$E^{'}(X) = E(x_1)\oplus \cdots \oplus E(x_n)$$
and
$$E^{'}(U)= E(x_{i_1})\oplus \cdots \oplus  E(x_{i_k})$$
and the transition map $E^{'}(X) \to E^{'}(U)$ is given by the projection
$$E'(X) \to E'(U), (a_1,\dots, a_n) \mapsto (a_{i_1},\dots, a_{i_k})$$
which is clearly surjective.
Finally you can check that if $E'(X)\to E'(U)$ is surjective for all open subset $U$ then the sheaf is flabby.
