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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."

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Map $f:{\rm{Spec}}(B) \rightarrow {\rm{Spec}}(A)$ is homeo. onto image since true after fpqc base change (EGA IV$_4$, 2.6.2(iii)), so if localize at prime $P$ of $A$ in image then $B$ replaced with local ring at corresponding point $Q$ (since loc. at prime is injlim of rings of fns on base of opens around pt in Spec). Replacing $A'$ by local ring at prime $P'$ above one at which we localized on $A$ gives $B'_ {P'} = A'_ {P'} \otimes_ {A_P} B_ Q$ and this is local. Since $B'$ is loc. of $A'$ at mult. set, their local rings match. Hence, $A_P = B_Q$. Then...??? – BCnrd Jul 5 '10 at 18:24
Tom, since you mention $\mathbf{C}$-algebras, it sounds like you may be happy to impose some finiteness hypotheses on the situation (otherwise can't imagine what being a $\mathbf{C}$-algebra is meant to do). So are you happy to assume some rings here are noetherian, or anything else? More specifically, is this an idle question or does it come up somewhere? – BCnrd Jul 5 '10 at 18:36
Hi Brian, I'm willing to impose noetherian if necessary for understanding. No, it's not idle, I'm curious whether "essentially of finite type" is a flat-local condition. – Thomas Nevins Jul 5 '10 at 18:39
up vote 5 down vote accepted

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 $A$.

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Great---thanks, Angelo! – Thomas Nevins Jul 6 '10 at 14:00

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