In this exercise 5.6. d) in chapter 2 from "Algebraic Geometry" by Hartshorne one has to show that for any ideal $\mathfrak a \subset A$ of a noetherian ring $A$ and $A$-module $M$ the sheaf $\Gamma_{\mathfrak a}(M)^{\sim}$, we get from the submodule $\Gamma_{\mathfrak a}(M) = \{ m\in M|\mathfrak a^nm = 0 $ for some $n\} $, is isomorphic to $\mathscr H_Z^0(\mathscr F)$, where $\mathscr F = M^{\sim}$ and $Z = V(\mathfrak a)$ and $\mathscr H_Z^0(\mathscr F)$, defined in exercise 1.20. as the sheaf we get by only taking sections of $\mathscr F$ with support in $Z$, from this exercise we know that this is not only a presheave but a sheave.
Now to my question. I think I have done the proof right: As one can gather from the hint, one shows that $\mathscr H_Z^0(\mathscr F)$ is quasi-coherent, as it is a closed subsheave of $\mathscr F$, so $\mathscr H_Z^0(\mathscr F)(U)$ is a sub-module of $\mathscr F(U)$, and then because the sections of both $\mathscr H_Z^0(\mathscr F)$ and $\Gamma_{\mathfrak a}(M)^{\sim}$ are zero on $U = X - Z$($X=Spec(A)$) and we can show that $\Gamma_{\mathfrak a}(M)\cong \Gamma_Z(\mathscr F)$, the two sheaves have to be isomorphic.
But it somehow eludes me why it is necessary to look at powers of $\mathfrak a$ in the definition of $\Gamma_{\mathfrak a}(M)$. Why is that necessary?
Intuitively it makes a bit of sense that else we maybe wouldn't get enough elements, but I can't really explain why or even if that is actually the case. I would also appreciate an example why we need those powers and an explanation where it is needed in the proof and where i went wrong as i didn't use it.
Using quasicoherence, we just need to show that the global sections of $\Gamma_{\mathfrak{a}}(M)$ and $\mathscr{H}^0_Z(\mathscr{F})$ are isomorphic, where $\mathscr{F} = \tilde{M}$. We're also assuming $X = \operatorname{Spec} A$ where $A$ is a noetherian ring.
We just need to compute the global sections of $\mathscr{H}^0_Z(\mathscr{F})$. By definition, these are $m \in \Gamma(X, \mathscr{F}) = M$ such that $\operatorname{Supp} m \subset Z = V(\mathfrak{a})$. Now, since $\operatorname{Supp} m = V(\operatorname{Ann} m)$ by part (a), we know that $m \in \Gamma(X, \mathscr{H}^0_Z(\mathscr{F}))$ if and only if $V(\operatorname{Ann} m) \subset V(\mathfrak{a})$. This is then equivalent to $\mathfrak{a} \subset \sqrt{\operatorname{Ann} m}$, which is equivalent to $\mathfrak{a}^n m = 0$ for some $n > 0$, by the noetherian hypothesis.
This last step using the radicals is where the powers of $\mathfrak{a}$ come into play.