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Let $X$ be an arbitrary scheme of finite type over $\mathbb{C}$ and $f:X \rightarrow \mathbb{A}_{\mathbb{C}}^1$ be a flat morphism. I suppose that the special fiber have some nice property, I would like to know if in that case the generic fiber has the same property. More precisely, I have in mind one of the following things:

1) the special fiber is reduced;

2) the special fiber is irreducible, but not necessarly reduced;

3) the special fiber is normal/smooth.

Thank you by adance for your answers or your references (there is things like that in the book of Matsumura but not these precise things).

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Are you asking if one special fiber having this property implies that the general fiber does, or if all special fibers having this property implies that the general fiber does? –  Will Sawin Apr 26 '12 at 20:07
    
To simplify I suppose that there is only one special fiber. I want to know if good properpties for this fiber are given to the other fibers (the generic fibers). –  sabrebooth Apr 26 '12 at 20:33
    
Just a minor nitpick on terminology. There is only one generic fibre in this case; it's the fibre over the generic point of $\mathbf{A}^1_{\mathbf{C}}$. All other fibres are called "special" fibres. –  Harry Apr 26 '12 at 22:52
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2 Answers 2

Hi, here are some references taken from Algebraic geometry I. Schemes. by Gortz and Wedhorn, Appendix E. These are generally scattered in EGAIV as far as I remember. Anyway, Kazuma Shimomoto recently pointed out to me this appendix.

EDIT: I originally misread what Gortz and Wedhorn were saying, and got the references wrong. Here are corrected references:

EDIT: As Mohan and Laurent Moret-Bailly point out, you really need the properness. Otherwise, imagine that the bad locus (for example, the singular/non-normal locus) is like a hyperbola that goes to infinity near the special fiber.

1). This is Appendix E.1(11). They reference EGAIV, 12.2.1

2). I don't see this in their list... You could try checking out EGA.

3). Normality is Appendix E.1(20). They reference EGAIV 12.2.4. Smoothness is Appendix E.1(18), again see EGAIV 12.2.4.

Of course, they are talking about geometric normality etc, but in your case, the residue field at the special point is $\mathbb{C}$, so if the special fiber is normal it is geometrically normal.

In particular, they state the openness of the statements for a proper flat map.

EDIT: A different approach: Let me also state a different approach to some of these questions.

I say that an open property $P$ deforms if for a local ring $(R, \mathfrak{m})$ and if there exists a regular element $0 \neq f \in \mathfrak{m}$ such that $R/f$ satisfies property $P$ then $R$ also satisfies property $P$.

Regularity, Cohen-Macaulayness, Gorensteinness all deform for more or less obvious reasons. Normality is pretty easy and I know it's in EGA somewhere... (it's also in a paper of R. Heitmann where he actually shows the statement for semi-normality, it follows from regularity and S_n computations). In fact, if $R/f$ is reduced or integral, then so is $R$.

Just for completeness, let me mention that rational singularities also deform (a result of Elkik). Log canonical and log terminal singularities do not deform in this way unless additional assumptions on $R$ are made (for example, if $R$ is Gorenstein). In general, see Inversion of adjunction. A recent theorem of Kovacs and myself proves it for Du Bois singularities (see the arXiv). As far as I know, this is an open question for weak-normality.

Anyway, why is this relevant? Well, let's say we have an open condition (like being reduced, being smooth, being normal, being Cohen-Macaulay, having rational singularities). Then suppose that $\pi : Y \to C$ is a flat map over a smooth pointed curve $0 \in C$. If the special fiber $Y_0$ saties a property $P$, and $P$ deforms, then $Y$ satisfies property $P$ in a neighborhood of $Y_0$. Indeed, work locally, modding out by the equation defining the special fiber is just modding out by some regular element.

$\bullet$ If $\pi$ is now proper, then this implies that the non-$P$-locus has closed image in $C$ under $\pi$. In particular, it only messes up a few fibers and the generic fiber is $P$ (in fact the total space $Y$ is also $P$ is one is willing to remove a couple points from $C$). At this point, you can use Bertini's theorem to obtain that the "nearby" closed fibers of $\pi$ also satisfy $P$.

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In general, these type of questions can go wrong if the map is not proper. For example, both 2 and 3 can be made to go wrong unless the map is proper. –  Mohan Apr 27 '12 at 6:26
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And 1) also fails in the non-proper case (e.g. the special fiber may be empty). –  Laurent Moret-Bailly Apr 27 '12 at 6:38
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I think that the answer for 2) is negative. Let $C$ be the union of the axises in the plane and $p:C \to \mathbb A ^1$ be given by $p(x,y)=x+y$. the fiber of $0$ is "irreducible" but non-reduced (i.e. its reduction is irreducible), but the generic fiber is reducible.

Another counterexample is $X=[(x,y,z) \in \mathbb A ^3|,x(x-yz)=0 ]$ and $\phi:X\to \mathbb A^1$ is defined by $\phi(x,y,z)=z$. Here the generic fiber is not only not irreducible but also not locally irreducible.

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