Hartshorne Example 5.2.2 Let $X = \operatorname{Spec} A$, $Y \subseteq X$ closed subscheme defined by ideal $I \unlhd A$, and let $i : Y \rightarrow X$ denote the inclusion morphism.
Then, Hartshorne claims that $i_{*}\mathcal{O}_{Y} \cong (A/I)^{\sim}$, the sheaf associated to the  $A$-module $A/I$.
I am having trouble proving this rigorously.
Firstly, as $Y$ is a closed subscheme, it follows $i^{\sharp} : \mathcal{O}_{X} \rightarrow i_{*}\mathcal{O}_{Y}$ is surjective. Therefore, we have
$i_{*}\mathcal{O}_{Y} \cong \mathcal{O}_{X} /\mathcal{I} $
where $\mathcal{I} :=$ ker $i^{\sharp}$.
Hence, I think a good idea is to prove that
$\mathcal{O}_{X} /\mathcal{I} \cong (A/I)^{\sim}$.
But I have not been able to come up with any argument that I find convincing. For example, I looked at the surjective map
$\mathcal{O}_{X} \rightarrow (A/I)^{\sim} $
and tried to prove its kernel is $\mathcal{I}$, but I couldn't come up with a rigorous answer.
Maybe my problem is that I don't really understand what the map $(i,i^{\sharp})$ is?
Any help would be much appreciated!
 A: Following the discussion in the comment section, I will try to answer my own question. My exposition may not be fully rigorous, but I think it contains the right ideas following hints given in the comments.
We have
$(Y,\mathcal{O}_{Y}) =$ $(Spec A/I, \mathcal{O}_{Spec A/I }$).
Recall,
$\mathcal{O}_{Y}(U) = \{s : U \rightarrow \coprod_{p \in U} (A/I)_{p} : \cdots  \}$ (such that "locally looks like fractions").
Also, note that
$i_{*}\mathcal{O}_{Y}(U) = \mathcal{O}_{Y}(U \cap Y) = \{s : U' \rightarrow \coprod_{p \in U'} (A/I)_{p} : \cdots  \}$
where $U' := U \cap Y \subseteq Y$ is open in $Y$.
Then, note that for principle open $D(f) \subseteq X$,
$i_{*}\mathcal{O}_{Y}(D(f)) = \mathcal{O}_{Y}(D(f)\cap Y) = \mathcal{O}_{Spec A/I}(D([f]))$.
But,
$\mathcal{O}_{Spec A/I}(D([f])) \cong (A/I)^{\sim}(D(f))$
via
$(A/I)^{\sim}(D(f)) \rightarrow \mathcal{O}_{Spec A/I}(D([f])) $
$s \mapsto s'$
where if $\pi : A \rightarrow A/I$ denotes the projection map, then
$s'(\pi(p)) := s(p)$.
Thus, this isomorphism on the principle open sets yields an isomorphism of the sheaves.
