Ravi Vakil's Foundation of Algebraic Geometry exercise $8.1(G)$ asks the following: Suppose, $\mathcal I_{X/Y}$ is a sheaf of ideals corresponding to a closed embedding $\pi : X \to Y$.Suppose $SpecB \subset Y$ is an open affine subscheme, and $f \in B$.Show that the natural map $(I(B))_f \to I(B_f)$ is an isomorphism.(Here $I(B)=$ kernel of the morphism $\mathcal O_Y \to \pi_{*}\mathcal O_X$, evaluated on $SpecB$)
$(1)$ I have been able to solve the problem assuming that $\mathcal I_{X/Y} = $ Kernel of the morphism $\mathcal O_Y \to \pi_{*}\mathcal O_X$.But then $\mathcal I_{X/Y}$ is becoming a very particular sheaf of ideal of $Y$ namely the ideal sheaf corresponding to $X$ whereas in the question they are talking about "a" sheaf of ideal corresponding to that closed embedding.So my question is What is exactly meant by " $\mathcal I_{X/Y}$ is a sheaf of ideals corresponding to a closed embedding $\pi : X \to Y$".
$(2)$ Since $I(B)$ is an ideal,I hope localization is also an ideal and not necessarily a ring.So my question is What is exactly meant by "isomorphism between $2$ ideals".(while solving the problem I have produced a natural map which is injective and surjective )
Any help from anyone is welcome