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Let $X \to S$ be a morphism of schemes. What properties of morphisms satisfy: if all fibers $X_s \to k(s)$ over closed points $s \in S$ satisfy the property, then the morphism $X \to S$ satisfies the same property?

For example, is this true (maybe under additional conditions; connected base et cetera) for proper morphisms? Separated morphisms? Smooth morphisms?

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    $\begingroup$ Certainly not for proper or separated. For proper, take any finite morphism of curves and remove a point from the source. For separated, take the map form the affine line with the doubled origin to the affine line that "squashes" the two origins together. $\endgroup$ – M P Sep 5 '11 at 15:38
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    $\begingroup$ Also, the normalization of a node on a curve seems to be a counterexample for smoothness. $\endgroup$ – M P Sep 5 '11 at 15:42
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It might be helpful to split your question into two parts:

  1. When is it true that if the fibers $X_s \to {\rm Spec} \kappa (s)$ satisfy a certain property for all $s \in S$, $X \to S$ satisfies this property?

  2. When is it true that if the fibers $X_s \to {\rm Spec} \kappa (s)$ satisfy a certain property for all closed $s \in S$, all fibers satisfy this property.

The answer to the first question is almost never positive if you do not make additional assumptions. The two main cases, where I know of a positive answer, are:

Ia. The first one is Francesco's answer: If $X \to S$ is flat (EDIT: and locally of finite presentation) and all fibers are smooth (or etale), then $X \to S$ has the same property. There are variants for other types of singularities.

Ib. If $X \to S$ is proper and all fibers are finite, then $X \to S$ is finite (Grothendieck's version of Zariski's main theorem).

There are variants, as if $f\colon X \to S$ is proper, flat and of finite presentation and $X_s \to {\rm Spec} \kappa(s)$ is an isomorphism for one $s \in S$, then there exists an open neighborhood $U$ of $s$ such that $f^{-1}(U) \to U$ is an isomorphism.

The answer to the second question is much more often positive. There is the following general principle: Assume that $X$ is of finite presentation over $S$. Let $C$ be the set of points $s \in S$ where $X_s \to {\rm Spec} \kappa(s)$ has a certain "decent" property $P$. Usually "decent" should at least imply that you have a property of schemes over a field that is stable under extension of the base field; thus "irreducible" is not a decent property, but "geometrically irreducible" is.

Then $C$ is constructible. $C$ is open if $X$ is in addition proper and flat over $S$.

Of course this is not true for all properties (e.g., for the property of being an affine morphism) but it does hold for many properties. Many statements of this kind are proved in EGAIV, mainly §9 and §12. There is also a rather long list in the appendix of my book with Görtz.

Therefore to answer the second question it suffices to answer the following question.

  1. Let $C$ be a constructible (resp. open) subset of a scheme $S$ containing the subset $S_0$ of all closed points of $S$. Is $C = S$?

In general the answer is no because there are schemes that do not contain any closed point. If $C$ is constructible, then the answer is positive if $S$ is of finite type over a field or, more generally, if $S$ is Jacobson (e.g. if $S$ is of finite type over a Dedekind ring with infinitely many prime ideals - as the ring of integers). In this case $C \mapsto C \cap S_0$ yields a bijection between constructible subsets of $S$ and constructible subsets of $S_0$.

If $C$ is open, then to get a positive answer it suffices to have a scheme where every point has a closed point in its closure. This is true for quasi-compact schemes (easy to see using that this is certainly true for affine schemes) or locally noetherian schemes (not that easy to see).

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    $\begingroup$ In Ia, you should also require the morphism to be locally of finite presentation. I stumbled upon this when I thought of unramified morphisms where you do not need flatness, but the condition on the fibers is still not sufficient on its own. $\endgroup$ – user2035 Sep 7 '11 at 7:58
  • $\begingroup$ Thanks, a-fortiori. You are of course right. I made the correction. Otherwise there are certainly counterexamples: Let R be a discrete valuation ring with field of fraction K and let A be the ring of polynomials f in K[T] with f(0)∈R. Then A is R-torsion free hence flat, but A is not noetherian and thus not of finite type over R. The fiber of SpecA over the special (resp. general point) of R is an affine space of dimension 0 (resp. dimension 1) and thus they are certainly smooth. It is actually kind of fun to "sketch" ${\rm Spec} A$. $\endgroup$ – Torsten Wedhorn Sep 7 '11 at 9:05
  • $\begingroup$ @Torsten Wedhorn: (if you ever happen to read this) do you have references for these facts (or should I be able to prove them myself?)? perhaps your wonderful book (part two of which we are all waiting for... :)? $\endgroup$ – Yosemite Sam Apr 16 '12 at 23:21
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This is true for smooth morphism (essentially by definition). But notice that in the definition of smooth morphism one also requires flatness. In other words, a flat morphism is smooth if and only if the fibers over closed points are so. See [Liu, Algebraic Geometry and Arithmetic curves, Definition 3.35 page 142]. Dropping flatness assumption the statement is no longer true, see MP comment about the normalization of a nodal curve.

This is also false for proper morphisms: just consider the case where $X \to S$ is an open immersion. But it is true under additional assumptions: see again Liu's book, Remark 3.28 page 107 and the references given there.

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  • $\begingroup$ Liu uses this definition only if $S$ is locally noetherian - for a good reason. There are non-locally noetherian schemes that do not have any closed point. $\endgroup$ – Torsten Wedhorn Sep 5 '11 at 17:43
  • $\begingroup$ On second thoughts, I think there is a problem with the statement as above even in the noetherian case: Let $R$ be a discrete valuation ring and let $A = R[T,U,V]/(T(\pi U - 1), T(\pi V - 1))$, $\pi$ a generator of the maximal ideal of $R$. Then ${\rm Spec} A$ is faithfully flat over $R$ and the fiber over the unique closed point of ${\rm Spec} R$ is isomorphic to an affine space of dimension $2$ over $R/\pi$. But the generic fiber is connected with two irreducible components. In particular it is not smooth. $\endgroup$ – Torsten Wedhorn Sep 5 '11 at 17:56
  • $\begingroup$ @Francesco: in the reference you cited, a flat morphism is smooth if all fibers are smooth at their closed points (and not "smooth over closed points"). One can restrict to closed fibers if the morphism is proper. @Torsten: one can also juste take a singular variety over the field of fractions of $R$. $\endgroup$ – Qing Liu Sep 5 '11 at 23:31
  • $\begingroup$ Ah, of course you are right. I did not think carefully enough. I just used my favorite example that a faithfully flat surjective morphism of finite type between equi-dimensional connected schemes has not necessarily equi-dimensional fibers - which somehow contradicts the philosophical principle that "flat" should mean "continuously varying fibers". But of course the problem is, that the morphism is not proper and thus there might be "holes" in the special fiber. $\endgroup$ – Torsten Wedhorn Sep 6 '11 at 5:13
  • $\begingroup$ I agree that it is not enough to have $X$ and $S$ equidimensional to get equidimensional fibers (with faithfully flat morphism of finite type). But this is also moral because usually the properties of $X, S$ don't transfer to the fibers (e.g. regularity). If we suppose moreover that $S$ is irreducible and universally catenary, and the generic fiber of $X\to S$ is equidimensional, then it is true that the fibers are equidimensional. $\endgroup$ – Qing Liu Sep 6 '11 at 7:23

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