Explicit description of the inverse image sheaf of an ideal sheaf. $\DeclareMathOperator{\Spec}{Spec}$
Let $f: \Spec A \to \Spec B$ be a morphism of affine schemes and $f^\#: B \to A$ be the corresponding ring homomorphism. Let $\mathcal{I} \subseteq \mathcal{O}_{\Spec B}$ be an ideal sheaf on $\Spec B$, which corresponds to an ideal $I \subseteq B$. Is there an explicit description of the sheaf $f^{-1}\mathcal{I}$ over $\Spec A$? 
It's known that $f^\ast \mathcal{I}$ corresponds to the $A$-module $I \otimes_B A$. I also know that $f^{-1} \mathcal{I} \cdot \mathcal{O}_{\Spec A}$ corresponds to the ideal $f^\#(I)A \subseteq A$, which can be also described as the image $\mathrm{Im}(I \otimes_B A \to B \otimes_B A \cong A)$ in $A$.
However, it seems hard to describe $f^{-1}\mathcal{I}$, or even $f^{-1}\mathcal{O}_{\Spec B}$, for arbitrary morphism $f$.
Any comment would be appreciated.
 A: $F=f^{-1} \mathcal{O}_{\mathrm{Spec}(B)}$ is a sheaf of rings on $\mathrm{Spec}(A)$ with stalks $F_{\mathfrak{p}}=B_{f(\mathfrak{p})}$. Thus, for open subsets $U \subseteq \mathrm{Spec}(A)$ there is a canonical embedding $F(U) \hookrightarrow \prod_{\mathfrak{p} \in U} B_{f(\mathfrak{p})}$. The image may be described as follows: $(s_\mathfrak{p})_{\mathfrak{p} \in U}$ lies in the image if and only if for all $\mathfrak{p} \in U$ there is an open neighborhood $\mathfrak{p} \in V \subseteq U$, some basic-open subset $D(b) \subseteq \mathrm{Spec}(B)$ with $f(V) \subseteq D(b)$  and some element $\frac{u}{b^k}$ in $B[b^{-1}]$ such that for all $\mathfrak{q} \in V$ the element $s_{\mathfrak{q}} \in B_{f(\mathfrak{q})}$ is the image of $\frac{u}{b^k}$ under the localization map $B[b^{-1}] \to B_{f(\mathfrak{q})}$. It should be clear how to modify this to get a description of $f^{-1} \mathcal{I}$, which is an ideal sheaf of $F$. These element descriptions are quite complicated. When you want to work with inverse image sheaves, you will just have to use adjunction $f^{-1} \dashv f_*$, where $f_*$ is the direct image functor.
