Suppose that a surjection of commutative noetherian rings $A \to B$ is given.
Suppose you have a free $B$-submodule of $B^n$ for some $n$. You can easily extend this to a $A$-submodule of $A^n$ by choosing generators in $A^n$. The question is: is this a projective $A$-module, under any (possibly strong) hypotheses on $A$ and $B$? If not, is it true that it is projective at least on some open subset of $\operatorname{Spec} A$, seen as a sheaf of $A$-modules over $\operatorname{Spec} A$?
In my case, $B=\mathbb C((t))$, the ring of Laurent series.
I can prove that it is projective over an open set:
As the map $\pi:A\rightarrow B$ is surjective and $B$ is a field we have $B=A/\mathfrak{m}$ for some maximal ideal $\mathfrak{m}$ of $A$.
Now lets $b_1,\dots, b_n$ be elements in $A^n$ such that the restriction of $\pi:A^n\rightarrow B^n$ to $b_1A+\dots+b_nA$ is an isomorphism. Then $\pi(b_1),\dots, \pi(b_n)$ should be a basis of $B^n$. Then, as $B^n\cong A_\mathfrak{m} ^n\otimes_{A_\mathfrak{m}} A_\mathfrak{m}/\mathfrak{m}A_\mathfrak{m}$, by Nakayama lemma we see that (the image of) $b_1,\dots,b_n$ form a basis of the module $A_\mathfrak{m}^n$.
Now consider the exact sequence
$$\begin{align*} 0\rightarrow K\rightarrow \bigoplus_{i=1}^n e_iA&\rightarrow A^n\\ e_i&\mapsto b_i \end{align*}$$ where $K$ is the kernel of the map in the right. By tensor this sequence with $\otimes_A A_\mathfrak{m}$, as the $b_i$ become a basis after tensoring, the map in the right become an isomorphism, hence $K\otimes_A A_\mathfrak{m}=0$ and then $\mathfrak{m}\notin \mathrm{Supp}(K)$. As $K$ is finitely generated ($A$ is noetherian) the support of $K$ is closed, and as $\mathfrak{m}\notin \mathrm{Supp}(K)$ it must be a proper closed subset. Take a principal open subset set $\mathrm{Spec}(A_f)\subseteq \mathrm{Spec}(A)$ disjoint from $\mathrm{Supp}(K)$. By tensoring the exact sequence above we see that $K\otimes_A A_f=0$ and hence the map $$\bigoplus_{i=1}^n e_iA_f\rightarrow A_f^n$$ is an isomorphism. Therefore the module $(b_1A+\dots+b_nA)\otimes_A A_f=b_1A_f+\dots+b_nA_f$ is free, in particular projective.