Flat cover by a locally Noetherian scheme - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:33:12Z http://mathoverflow.net/feeds/question/83529 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83529/flat-cover-by-a-locally-noetherian-scheme Flat cover by a locally Noetherian scheme hadimath 2011-12-15T15:52:01Z 2011-12-17T16:44:41Z <p>Les S be a scheme. Does there exist a faithfully flat morphism T to S with T a locally Noetherian scheme?</p> http://mathoverflow.net/questions/83529/flat-cover-by-a-locally-noetherian-scheme/83602#83602 Answer by Laurent Moret-Bailly for Flat cover by a locally Noetherian scheme Laurent Moret-Bailly 2011-12-16T08:06:22Z 2011-12-17T16:44:41Z <p>Note that if $T\to S$ is also quasicompact, then $S$ must be locally noetherian: this boils down to the well-known fact that if $A\to B$ is a faithfully flat ring homomorphism and $B$ is noetherian, then so is $A$.</p> <p>This proves that Jason's example above is indeed a counterexample. More generally, any non-noetherian <em>local</em> scheme $S$ is a counterexample: if $T\to S$ is faithfully flat, there is an open affine $U\subset T$ which covers $S$.</p> <p>EDIT: In fact, here is a complete answer ($S$ is any given scheme):</p> <p>(1) The following are equivalent:<br> (1a) There exists a locally noetherian scheme $T$ and a faithfully flat and quasicompact morphism $T\to S$.<br> (1b) $S$ is locally noetherian.</p> <p>(2) The following are equivalent:<br> (2a) There exists a locally noetherian scheme $T$ and a faithfully flat morphism $T\to S$.<br> (2b) For each $s\in S$, the ring <code>$\mathcal{O}_{S,s}$</code> is noetherian.</p> <p>Proof: exercise. To show that (2b) implies (2a), take $T=\coprod_{s\in S}\mathrm{Spec}(\mathcal{O}_{S,s})$.</p>