Quasicoherent sheaves: equivalence of two definitions In the case of an affine scheme $X = \operatorname{Spec} A$, a quasicoherent sheaf is a sheaf of ideals defined on a the distiguished open sets $X_f$ to be $\mathcal{I}(X_f) = \mathfrak{a}_f$ for a fixed $\mathfrak{a} \subseteq A$. This defines a $\mathcal{B}$-sheaf (?) and the ideals on arbitrary $\mathcal{O}_X (U)$ are defined in accordance with the universal property (as in Eisenbud & Harris). I include the "?" because the separatedness is clear even from the fact that the association $\mathcal{O}_X(X_f) := A_f$ is a $\mathcal{B}$-sheaf, but the gluing property is not clear without referring to the proof that $\mathcal{O}_X(X_f) := A_f$ has the glueing property (although it is probably obvious if I did...). As an aside, is there a more concrete way of constructing $\mathcal{I}(U)$, perhaps as the ideal generated by the contractions of $\mathfrak{a}_f$ for each $X_f \subseteq U$?
Anyway, I wish to show that the general definition of quasicoherent on an arbitrary scheme $X$, that is $\mathcal{I}(U_i)$ is affine quasicoherent for some affine cover $U_i \subseteq X$, implies that $\mathcal{I}$ is affine quasicoherent when restricted to any affine subset $U \subseteq X$. For reference, this is exercise I-28 in Eisenbud & Harris.
 A: There is a standard reduction to the case $U = X = \operatorname{Spec} A$ affine, $U_i = D(f_i)$. (Key word: common refinement.)
Once you are in that case, you have to solve the following commutative algebra problem:
Given a ring $A$ and elements $f_i$ generating the unit ideal, and given $A_{f_i}$-modules $M_i$ together with isomorphisms $\phi_{ij} \colon (M_i)_{f_j} \stackrel \sim \longrightarrow (M_j)_{f_i}$ of $A_{f_if_j}$-modules satisfying $\phi_{jk}\phi_{ij} = \phi_{ik}$ on $A_{f_if_jf_k}$, construct an $A$-module $M$ with isomorphisms $\psi_i \colon M_{f_i} \stackrel \sim \longrightarrow M_i$ such that $\phi_{ij} \psi_i = \psi_j$ on $A_{f_if_j}$.
The way you define $M$ is by using the sheaf property it should satisfy: it is the kernel of the map
\begin{align*}
\prod_i M_i &\to \prod_{i,j} (M_i)_{f_j}\\
(m_i)_i &\mapsto (m_i - \phi_{ji}(m_j))_{i,j}.
\end{align*}
The isomorphisms $\psi_i$ come from localising this sequence at $f_i$ and computing the kernel in two ways. On the one hand, it is $M_{f_i}$ because we just localised. On the other hand, using the cocycle condition, you can show that it is also $M_i$.
Lots of details omitted, but I hope this will get you started.
