If $N\otimes_A B$ and $N\otimes_A B'$ are projective modules, what about $N$? Let $A$ be a domain with fraction field $K$ and let $B$ and $B'$ be two sub-$A$-algebras of $K$ such that $B\cap B'=A$.
Is this true that if $N$ is a finitely generated $A$-module such that $N\otimes_A B$ and $N\otimes_A B'$ are projective modules over $B$ and $B'$ respectively, then $N$ is projective?
Many thanks!
 A: Let $k$ be a field, $A=k[X, Y],$ $B=k[X^{\pm 1}, Y]$ and $B'=k[X, Y^{\pm 1}]$, and take $N=k=A/(X, Y)$.
Then $N \otimes_A B =B/(x, y)B=0$, since $X$ is a unit of $B$, and similarly $N \otimes_A B'=0$. Thus, $N \otimes_A B, N \otimes_A B'$ ae projective, yet $N$ is clearly not.
If you do not like the fact that the projectives in the example are $0$, you can replace $N$ by $N \oplus A$, the argument will work the same since tensor product commutes with direct sums.

A bit more dissection:
There is something to be said about the apparent strangeness of this: Given a finitely presented module $N$ over $A$, what is true is that if $N$ is projective locally on the spectrum, i.e. $N \otimes_A B_i$ is a projective $B_i$-module where $B_i$ are principal localizations of $A$ such that $\bigcup_i \mathrm{Spec}\,B_i=\mathrm{Spec}\,A,$ then $N$ is indeed projective. The trouble is that even in the situation above, the assumption $B \cap B'=A$ does not guarantee that $\mathrm{Spec}\,B, \mathrm{Spec}\,B'$ cover $\mathrm{Spec}\,A$ in this way: the point $(X, Y)$ is not in $\mathrm{Spec}\,B \cup \mathrm{Spec}\,B'$, so $B$'s does not see that some torsion is happening over that point.

Regarding the extra assumption "$N$ is torsion-free" in comments:
Assuming that $N$ is torsion-free will not help. In the counterexample above, you can take instead $N=(X, Y),$ which is torsion-free but not projective (you can e.g. show that projective ideals over $A=k[X, Y]$ are exactly the principal ones, as in this answer of Georges Elencwajg). Then, using flatness of $B$ over $A$, one has that $(X, Y)\otimes_A B=(X, Y)B=B$, and similarly $N \otimes_A B'=B'$. So over $B, B'$, $N$ becomes free, hence projective, but $N$ itself is not.
