TITLE: Checking locally whether a homomorphism is a localization
QUESTION [6 upvotes]: All rings below are commutative with $1$.  
Suppose $A\subset B$ is a subring and that $A\rightarrow A'$ is a faithfully flat ring homomorphism.  [You may assume the rings are actually ${\mathbb C}$-algebras if it helps.]  Suppose that $B' = A'\otimes_A B$ is a localization of $A'$, i.e. there is a multiplicatively closed subset $S$ of $A'$ such that $B' = S^{-1}A'$.  Must $B$ be a localization of $A$?  
I find it hard to believe that the answer is "yes."  But I'm having a mental block coming up with an example to show that it's "no."

REPLY [5 votes]: Let $A$ be the coordinate ring of a smooth affine curve $X$ over $\mathbb C$, and let $p$ be a point of infinite order in the class group of $A$. Let $B$ be the coordinate ring of $X \smallsetminus \{p\}$, and let $C$ be the coordinate ring of an open subscheme $U$ of $X$ containing $p$ such that $p$ is principal in $U$. Set $A' = B \times C$. Then it it is easy to see that $A' \otimes_A B$ is a localization of $A'$, while $B$ is not a localization of $A$.