Let $f:X\rightarrow S$ be a flat, surjective morphism of complex spaces with reduced fibers over $S$ reduced. Q1: Is $X$ reduced too?

Q2: Is the property " reduced fiber" preserved by base change given by the normalization?

Rk: 1) We have some result of this kind in [EGA4], Cor (3.3.5) p.44.

2) We can apply [Matsumara], Cor(ii) p.189 but with the additional asumptions: $S$ is normal and $X$ locally pure dimensional.


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    $\begingroup$ For algebraic varieties over an algebraically closed fields, Q1 follows from EGA IV, Théorème 9.7.7 (the set of $s\in S$ such that $X_s$ is geometrically reduced is locally constructible in $S$) and the Corollaire you cited (which is also the Theorem in Matsumura as pointed out by Angelo). For Q2, if you ask whether $X\times_S S'\to S'$ has reduced fibers for any morphism $S'\to S$, then the answer is clearly yes because the original fibers are geometrically reduced. $\endgroup$ – Qing Liu Jan 3 '11 at 21:12

Dear kaddar, here is a partial answer.

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 and 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 Complex Analytic Geometry (Springer, LNM 538, 1976).

Edit $\;$ On the evoked relation between flat and open, let me add the following. It is not 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 is 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.


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.

I don't know what the second question means.

[Edit] 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.

  • $\begingroup$ Dear Georges, Thank you for the partial answer (I found another reference for this (with smooth base) :satz 2.4 p.245 in "Deformation komplexer Raume "of Grauert-Kerner in Math.Ann 153). Can we deduce from the case of smooth base the general case by considering the base change given by the natural injection of the smooth part $Reg(S)$ in $S$ . We must then show that the fiber product of $X$ by $Reg(S)$ over $S$ is an open, dense and reduced subspace of $X$. I think that density is due to the flatness which preserve schematic density and reducedness is due to smooth base case.... $\endgroup$ – kaddar Jan 4 '11 at 7:58
  • $\begingroup$ Thanks to Liu and Angelo for the good reference in EGA and Matsumara commutative algebra. Dear Angelo, in the second question, i want to know if a morphism with reduced fibers stay with reduced fibers after any base change $S'\rightarrow S$ between reduced complex spaces ? Liu say that the answer is yes because the notion of "geometrically reduced " and "reduced" agree in complex analytic geometric. $\endgroup$ – kaddar Jan 4 '11 at 8:11
  • $\begingroup$ It seems to me that the result is valid for $f:X\rightarrow S$ assumed to be open (or universally open in AG). For see this, we can argue as above on $Reg(S)$ by base change... $\endgroup$ – kaddar Jan 4 '11 at 13:50
  • $\begingroup$ Is the result realy true for open morphism? $\endgroup$ – kaddar Jan 5 '11 at 10:55

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