Is a 'generic' variety nonsingular? Or singular?

Question

I'd like to know whether there's some coherent meaning of 'generic' for which one can say that a 'generic' variety over an algebraically closed field $K$, say, is nonsingular or singular. We could replace variety here with scheme or whatever but I'd like things to be well-behaved enough that nonsingular has a sensible definition; I don't know exactly how well-behaved that is.

For example, given a singular variety $X$, we might ask whether $X$ falls into some natural family of objects admitting a moduli space $\mathcal{M}$ such that an open dense subset of the $K$-points of $\mathcal{M}$ correspond to nonsingular objects. Of course, when the question is formulated like this, both answers may be correct: there might also be a nonsingular variety $Y$, even quite closely related to $X$, such that $Y$ is part of a different natural family almost all of whose objects are singular. This is the kind of behaviour I'd like to know about. But from here a natural question is certainly: pick a Hilbert scheme $\operatorname{Hilb}_{\mathbb{P}^n}^P$. Are the subschemes of $\mathbb{P}^n$ it parametrises generically singular/nonsingular? Or are those subschemes generally so vicious that that question doesn't even make sense?

Finally, I don't really know anything about deformation theory, but it seems plausible that my question admits a rigorous statement and solution in that language. If anyone knows anything along these lines, I'd also be grateful to hear about that. For instance, is there a singular variety which one cannot perturb into a nonsingular one?

Answer 1

Executive summary: If you look at the whole Hilbert scheme associated to a given polynomial, the locus of points corresponding to nonsingular (which I take to mean smooth) subschemes can sometimes be very small in terms of dimension and number of irreducible components. So in this sense, most subschemes are singular.

Details: The Hilbert scheme $\operatorname{Hilb}^P_{\mathbf{P}^n}$ associated to a given Hilbert polynomial $P$ is connected (a theorem of Hartshorne), but in general it has many irreducible components, each with its own generic point. Thus there are several different "generic" closed subschemes with the same Hilbert polynomial, each a member of a different family.

The locus of points in the Hilbert scheme corresponding to smooth (=nonsingular) subschemes of $\mathbf{P}^n$ is a Zariski open subset, which implies that it is Zariski dense in the union of the components that it meets, but there are often other components of the Hilbert scheme all of whose points correspond to singular subschemes.

Because the Hilbert scheme need not have a single generic point, one might ask: How many of these generic points parametrize singular subschemes, and what are the dimensions of the corresponding components of the Hilbert scheme?

As a case study, consider the Hilbert scheme $H_{d,n}$ of $d$ points in $\mathbf{P}^n$, i.e., the case where $P$ is the constant polynomial $d$. Points of $H_{d,n}$ over a field $k$ correspond to $0$-dimensional subschemes $X \subseteq \mathbf{P}^n$ of length $d$, or in other words, such that $\dim_k \Gamma(X,\mathcal{O}_X) = d$. Each smooth $X$ with this Hilbert polynomial is a disjoint union of $d$ distinct points. These smooth $X$'s correspond to points of an irreducible subscheme of $H_{d,n}$, and the closure of this irreducible subscheme is a $dn$-dimensional irreducible component $R_{d,n}$ of $H_{d,n}$. Sometimes $H_{d,n}=R_{d,n}$, which means that every $X$ is smoothable. But for each fixed $n \ge 3$, Iarrobino observed that $\dim H_{d,n}$ grows much faster than $\dim R_{d,n}$ as $d \to \infty$. (He proved this by writing down large families of $0$-dimensional subschemes, like $\operatorname{Spec} (k[x_1,\ldots,x_n]/\mathfrak{m}^r)/V$, where $\mathfrak{m}=(x_1,\ldots,x_n)$ and $V$ ranges over subspaces of a fixed dimension in $\mathfrak{m}^{r-1}/\mathfrak{m}^r$.) This shows that $H_{d,n}$ is not irreducible for such $d$ and $n$, and that the ``bad'' components all of whose points parametrize singular subschemes can have much larger dimension than the one component in which a dense open subset of points parametrize smooth subschemes. With a little more work, one can show that the number of irreducible components of $H_{d,n}$ can be arbitrarily large (and as already remarked, the components themselves can have larger dimension than $R_{n,d}$). So in this sense, one could say that for $n \ge 3$, most $0$-dimensional subschemes in $\mathbf{P}^n$ are singular.

For more details about $H_{d,n}$, including explicit examples of nonsmoothable $0$-dimensional schemes, see the following articles and the references cited therein:

The moduli space of commutative algebras of finite rank

Hilbert schemes of 8 points

(Warning: my notation $H_{d,n}$ is different from the notation of those articles.)

Answer 2

Here are some elementary remarks which would lead me to bet that a variety, of which I am told nothing, is nonsingular. I want to emphasize that I am not answering your question very satisfactorily (but see damiano's comment to the question) : for that we will have to wait for real experts.

0) At the most primitive level, if $k$ is a field (not necessarily algebraically closed) and $f(X)\in k[X]$ is a polynomial, the subscheme $V(f)\subset \mathbb A_k^1 $ it defines is smooth (actually étale) over $Spec(k)$ unless the discriminant of $f$ is zero. So you have generic smoothness.

1) Bertini's theorem. Suppose $K$ is algebraically closed and let $X\subset \mathbb P_K^n$ be a smooth subvariety. Then for an open dense set of hyperplanes in the dual projective space of hyperplanes $ \check{\mathbb P}^n_K $, the intersection $X\cap H$ is smooth.

2) Bertini's theorem: a variant. Suppose that $K$ is an algebraically closed field of characteristic zero.Then, given a dominant morphism $f:X\to Y$ with $X$ smooth, there exists a dense open subset $U\subset Y$ such that all fibers $f^{-1} (y)\subset X \;\; (y\in U)$ are smooth.

And here is a remarkable article, on both Bertini's biography and his theorems, by a masterful expositor: Steve Kleiman.

http://arxiv.org/PS_cache/alg-geom/pdf/9704/9704018v1.pdf

Answer 3

Regarding your last question, the answer is yes. There are singularities, called nonsmoothable, which you cannot deform into smooth varieties.

There is a discussion of the history of this concept and some example in this article of Greuel and Steenbrink in the book Singularities of PSPM, edited by Orlik.

Answer 4

It is also possible to construct singular varieties that cannot be globally smoothed, but such that every singularity is locally smoothable. For instance, in Catanese's paper "Everywhere non-reduced moduli spaces" you can find examples of surfaces of general type whose only singularities are ordinary double points (i.e. singular points locally given by $x^2+y^2+z^2=0$) and such that every small deformation preserves the singularities. The reason is that these examples are hypersurfaces $S$ in a weighted projective space $\mathbb{P}$, and all their singularities come from those of $\mathbb{P}$. Since one proves that every small deformation of $S$ is again a hypersurface in the same weighted projective space, it follows that one cannot get rid of the singularities. As a consequence of this strange behaviour it follows that, called $\widetilde{S} \to S$ the minimal desingularization of $S$, the moduli space $\mathcal{M}_{\widetilde{S}}$ is non-reduced (but $\mathcal{M}_S$ is generically smooth!)

Answer 5

(This is an addendum to Andrea's answer and link.) Greuel and Steenbrink quote Severi's postulate from 1909 that all singular varieties should be smoothable; in fact, this had already been disproved, $20$ years earlier (implicitly, it is true), by del Pezzo, with his theorem that if $S$ is a smooth linearly normal (meaning, embedded by a complete linear system) surface in $\mathbb P^d$ of degree $d$, then $d\le 9$. (Much more is true, of course.) So the cone $T$ over a linearly normal elliptic curve of degree $d\ge 10$ is not smoothable in $\mathbb P^d$ (from where it's easy to see that neither can $T$ be smoothed in the abstract, since it is embedded by its complete anti-canonical linear system).