most recent 30 from http://mathoverflow.net 2013-05-19T05:38:34Z http://mathoverflow.net/feeds/question/30635 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/30635/checking-locally-whether-a-homomorphism-is-a-localization Thomas Nevins 2010-07-05T16:35:42Z 2010-07-05T20:24:10Z

<p>All rings below are commutative with $1$. </p> <p>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 $A$? </p> <p>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." </p>

http://mathoverflow.net/questions/30635/checking-locally-whether-a-homomorphism-is-a-localization/30670#30670 Angelo 2010-07-05T20:24:10Z 2010-07-05T20:24:10Z

<p>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 <code>$X \smallsetminus \{p\}$</code>, 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 $A$.</p>