For a commutative ring $A$ and an $A$-module $M$, $\mathbb{V}(M)$ is defined to be $\text{Spec}(S^*M)$, where $S^*M = \bigoplus_k S^kM$ is the symmetric algebra. Thus $\mathbb{V}(M)$ is an affine scheme over $\text{Spec}(A)$.
I tried to solve the following exercise:
Let $(X, \mathcal{O}_X)$ be a noetherian scheme, and $\mathcal{F}$ a coherent $\mathcal{O}_X$-module. Cover $X$ by open affines, i.e. $X = \bigcup_{i=1}^n U_i$ with $U_i = \text{Spec}A_i$. Then $\mathcal{F}|_{U_i} \cong \tilde{M_i}$ for some finite $A_i$-module $M_i$.
Show that there exists a (unique?) scheme $f: \mathbb{V}(\mathcal{F}) \rightarrow X$, such that $f^{-1}(U_i) \cong \mathbb{V}(M_i)$.
Obviously this look like a simple glueing of the schemes $\mathbb{V}(M_i)$. But we failed to show that they are isomorphic on $U_i \cap U_j$.
Define $\chi_i: \mathbb{V}(M_i) \rightarrow U_i$ to be the map from the definition above, which comes from the inclusion $\iota: A \hookrightarrow SM$. Thus for any prime ideal $\mathfrak{p} \subset SM$ we have $\chi_i(\mathfrak{p}) = \mathfrak{p} \cap A$.
If $a_i \in A_i$, then we have the standard open $D(a_i) \subset \text{Spec}(A_i) \subset X$, and
$$\chi_i^{-1}(D(a_i)) = D(\iota(a_i)) = \text{Spec}((SM_i)_{\iota(a_i)}) = \text{Spec}(S(M_i)_{\iota(a_i)}).$$
But if we consider two open affines $U_i = \text{Spec}(A_i)$ and $U_j = \text{Spec}(A_j)$, I'm not certain that it is possible to cover the intersection by sets of the form $V = D(a_i) = D(a_j)$ for $a_i \in A_i, a_j \in A_j$, and thus I don't know how to prove that $\mathbb{V}(M_i)$ and $\mathbb{V}(M_j)$ are isomorphic on the intersection $U_i \cap U_j$.
Another thought I had was that this looks a bit like Exercise II 5.17 in Hartshorne's Algebraic Geometry. Maybe I will try to think about this first.