Flat map with reduced fibers. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T08:45:36Z http://mathoverflow.net/feeds/question/51004 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51004/flat-map-with-reduced-fibers Flat map with reduced fibers. kaddar 2011-01-03T10:46:59Z 2011-01-03T21:34:12Z <p>Hi.</p> <p>Let $f:X\rightarrow S$ be a flat, surjective morphism of complex spaces with reduced fibers over $S$ reduced. Q1: Is $X$ reduced too?</p> <p>Q2: Is the property " reduced fiber" preserved by base change given by the normalization?</p> <p>Rk: 1) We have some result of this kind in [EGA4], Cor (3.3.5) p.44. </p> <p>2) We can apply [Matsumara], Cor(ii) p.189 but with the additional asumptions: $S$ is normal and $X$ locally pure dimensional.</p> <p>Thanks.</p> http://mathoverflow.net/questions/51004/flat-map-with-reduced-fibers/51023#51023 Answer by Georges Elencwajg for Flat map with reduced fibers. Georges Elencwajg 2011-01-03T14:56:40Z 2011-01-03T21:34:12Z <p>Dear kaddar, here is a partial answer.</p> <p>According to a theorem of Douady, a flat map $f:X\to S$ between complex analytic spaces is always open . So if you assume that the fibers of $f$ are reduced <em>and</em> that your reduced space $S$ is actually smooth (i.e. is a manifold), then $X$ is indeed reduced : this follows from the Proposition on page 158 of Gerd Fischer's <em>Complex Analytic Geometry</em> (Springer, LNM 538, 1976).</p> <p><strong>Edit</strong> $\;$ On the evoked relation between flat and open, let me add the following. It is <em>not</em> true that an open morphism $f:X\to S$ of complex spaces is flat: the simplest counter-example is the immersion of a simple point into a double point i.e. the morphism of schemes $Spec \;\mathbb C \to Spec \; \mathbb C[\epsilon ] \quad(\epsilon^2=0)$ seen analytically. However if $X$ and $S$ are complex manifolds then it <em>is</em> true that $f$ open implies $f$ flat (Fischer, same page 158) and so for morphisms between manifolds you have the easy to remember equivalence flat=open, which helps understand the notoriously unintuitive notion of flatness.</p> http://mathoverflow.net/questions/51004/flat-map-with-reduced-fibers/51050#51050 Answer by Angelo for Flat map with reduced fibers. Angelo 2011-01-03T19:44:35Z 2011-01-03T20:12:57Z <p>This is a consequence of the following result: if $A \to B$ is a flat local homomorphism of local rings, $A$ and $B/\mathfrak{m}_AB$ are reduced, then $B$ is reduced. Keeping in mind that reduced is equivalent to $R_0$ and $S_1$, this follows from Theorem 23.9 in Matsumura's Commutative Ring Theory.</p> <p>I don't know what the second question means.</p> <p> I just noticed that the hypotheses in the cited theorem are actually stronger, so that it would imply the result for schemes but not for analytic spaces, at least not immediately. I need to think about it some more.</p>