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
 A: 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.
A: This can be also viewed as a consequence of a general fact due to Grothendieck (which can be proved by noetherian induction).
Let $X$ be a noetherian topological space of dimension $n$. Then for all $i>n$ and all sheaves $\mathcal{F}$ of abelian groups on $X$ we have $\mathrm{H}^i(X,\mathcal{F}) = 0$.
For this fact c.f. Hartshorne's Algebraic Geometry, chapter III, Theorem 2.7. The proof uses the fact (as a basis of induction) that on zero dimensional neotherian spaces (on finite discrete spaces) all higher cohomologies of sheaves vanish. So the best approach to your question is showing that sheaves on zero dimensional noetherian spaces are flabby. 
