Vague question. Is there anything special about degenerations of smooth projective varieties (separating them from arbitrary projective schemes)?

Precise setup. Let $f:X\to Y$ be a projective flat morphism of algebraic schemes (say, over $\mathbb{C}$), where $Y$ is an irreducible variety. Suppose that the generic fiber of $f$ is smooth and connected. Consider special fibers $X_y=f^{-1}(y)$, $y\in Y$. Of course, $X_y$ may be singular, reducible, and/or non-reduced. Zariski's main theorem implies $X_y$ is connected. Can we say anything else?

Specific Questions. Can $X_y$ have embedded components? Can $X_y$ have non-constant regular functions? (Since $X_y$ is connected, such a regular function would have to be nilpotent.)

Reformulation. Consider a Hilbert scheme of closed subschemes in ${\mathbb P}^N$ (with fixed Hilbert polynomial). There is an open subset in it corresponding to smooth connected projective varieties. The question concerns the closure of this set.

Remark. I ran across this in a concrete situation, where the goal is to understand $Rf_*O_X$. But it seems that the question is quite natural, so perhaps the corresponding statements or counterexamples are well known... Any comments would be helpful!

  • $\begingroup$ It seems the coherent cohom is better behaved than etale cohom for degenerations, but it wouldn't hurt to consider $Rf_*Q_l$ (or $Rf_*Z$) too. When $f$ is a log smooth morphism between log smooth varieties (e.g. semi-stable degen), the log de Rham complex is well-behaved (e.g. one still has Poincare lemma and de Rham isom; see relevant works of F. Kato and Nakayama etc.), and the results are better stated on the space $X^{log}$ assoc. to $X.$ Hopefully one gets some info about $Rf_*O_X$ when running the Hodge-de Rham spectral seq. $\endgroup$
    – shenghao
    Feb 20, 2011 at 15:06
  • $\begingroup$ Note that Zariski's main theorem applies only when the generic fiber is geometrically connected (otherwise consider an étale cover). $\endgroup$
    – Qing Liu
    Feb 20, 2011 at 16:15

3 Answers 3


Q1: Sure, you can have embedded components. Project a curve in $P^3$ into a plane. Where the image crosses itself, you get embedded points.

You can improve connected to "set-theoretically equidimensional, and connected in codimension 1".

I think the real moral should be that you shouldn't keep track just of the scheme structure when you're done degenerating a smooth variety, but of e.g. the induced log structure, and consider a moduli space of varieties + that extra structure. I should probably ask a question about that!

  • $\begingroup$ Thanks a lot! Of course, having embedded points also means there will be non-trivial regular functions. (I wonder if I should decouple the two questions: suppose there are no embedded components: can there still be non-trivial functions?) $\endgroup$
    – t3suji
    Feb 20, 2011 at 0:57

A partial answer: if $X$ is normal and $Y$ is smooth of dimension $1$, then $X_y$ only have constant regular functions ($X\to Y$ is cohomologically flat in relative dimension $0$). This is proved in Raynaud: Spécialisation du foncteur de Picard, Prop. 6.4.2 (use the characteristic 0 hypothesis here, otherwise it is false even when $X$ is also smooth.) It is also proved in the begining of ''Surfaces fibrées en courbes de genre deux'', Lecture Notes in Math. 1137 (1985) by Gang Xiao.

Add: and of course in this situation $X_y$ has no embedded point as $X$ is (S$_2$).

  • $\begingroup$ Thanks a lot, this is very interesting. From a quick glance at Raynaud's text, it seems that Chapter 6 might have some other relevant statements --- I'll see how much I can get form it... $\endgroup$
    – t3suji
    Feb 20, 2011 at 16:34

It seems that you are asking about smoothability of singularities. Some singularities are smoothable some are not.

I don't think there is a general criterion that tells you how to decide whether a specific singularity is smoothable or not and it may not be that a singularity that seems bad is necessarily not smoothable while one that looks OK is.

Hypersurface singularities are smoothable. I bet you can see why.

The simplest example I know of an innocent looking non-smoothable singularity is a cone over an abelian variety of dimension at least $2$. The fact that this is not smoothable follows from that it is a Du Bois singularity and hence by a result of Schwede if it were smoothable, the total space would have rational singularities, which is CM, and then all fibers are CM, but this singularity is not. The same argument shows that any projective variety (or variety with isolated singularities) with DB but not CM singularities give examples of what cannot be the limit of smooth projective varieties.

Another set of examples is provided by quotient singularities. They are rigid and hence non-smoothable in dimension at least $3$, but there are two-dimensional quotient singularities that are smoothable, for instance $\big(\mathbb A^2/(x,y)\sim (-x,-y)\big)\simeq (x^2=yz)\subset \mathbb A^3$ is a hypersurface singularity.

If you put more conditions on $f$ that obviously limits further the possible singularities you can get.

EDIT incorporated Karl's comment into the example given.

  • 1
    $\begingroup$ Is it even known if dimension 1 singularities are smoothable? $\endgroup$ Feb 20, 2011 at 3:07
  • $\begingroup$ Thanks for the answer. I am somewhat confused about the last paragraph. Isn't the constant term of the Hilbert polynomial the Euler characteristic of the structure sheaf rather than dimension of the space of global sections? After all, the example from Allen Knutson's answer is a curve degenerating into a curve with an embedded point; it does have non-trivial regular functions (supported at the embedded point). What I am wondering about now is whether it is possible to have non-trivial regular functions without embedded components. $\endgroup$
    – t3suji
    Feb 20, 2011 at 4:44
  • $\begingroup$ You are totally right. That was silly. Sorry. First I actually gave an example and then I convinced myself that it was wrong. Yes, of course, Allen's example gives you that. On the other hand I am not sure that you can have one that's supported on the entire fiber. $\endgroup$ Feb 20, 2011 at 6:30
  • $\begingroup$ @Allen: I think you are right. It seems that in some sense this question is harder for smaller dimension or at least that in higher dimension one can ask for the fibers to be normal or having higher depth. $\endgroup$ Feb 20, 2011 at 9:09
  • 5
    $\begingroup$ To Allen: There are examples of 1-dimensional singularities which are not smoothable. This was shown by David Mumford in "Pathologies IV". $\endgroup$
    – Angelo
    Feb 20, 2011 at 9:42

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