What exactly does the Hilbert scheme of points parametrize?

Question

The Hilbert scheme of points is defined as $$ \text{Hilb}^n(X) = \{ I \subset \mathbb{C}[x,y] \text{ such that } \text{dim}_{\mathbb{C}}/I = n \} $$ or, in words, the Hilbert scheme of points consists of all ideals $I$ of polynomials on $x,y$ with complex coefficients $ \mathbb{C}[x,y]$ such that when we mod out the latter with the first, $ \mathbb{C}[x,y]/I$ the resulting space is $n$ dimensional.

I struggle to understand this definition. Why does this definition require only polynomials in two variables and not, say, three? Is it possible to give me an example? I.e., can you give me some polynomial $\in \mathbb{C}[x,y]$ and ideal of it and show me how the quotient space is $n$ dimensional?

Also, what does the Hilbert-Chow morphism $$ \text{Hilb}^n(X) \to Sym^n(X) $$ actually mean? I struggle to undestand what for example would mean $$ \text{Hilb}^n{\mathbb{C}}^2 \to Sym^2\mathbb{C}^2 = \mathbb{C}^2/S_2 $$

Any help would be highly appreciated.

Answer 1

The Hilbert scheme $\text{Hilb}^n(\mathbb C^2)$ just parametrizes the set of $n$ points in the plane.
The definition is crystal clear when the points are distinct but what does "$n$ points" mean when the said points are not distinct?
Grothendieck's scheme theory gives a nice answer to this question in terms of plane subschemes $S\subset \mathbb A^2$ whose algebra $\mathcal O_S(S)$ of global sections is non-reduced but still has dimension $n$ over $\mathbb C$.
There is a canonical Chow morphism $$\pi: \text{Hilb}^n(\mathbb C^2)\to \operatorname {Sym}^n (\mathbb C^2)$$ mapping for example any $S\in \text{Hilb}^n(\mathbb C^2)$ with support the single point $P$ (beware that there are incredibly many such!) to the class $[P,\dots,P]\in \operatorname {Sym}^n (\mathbb C^2)$ of the $n$-tuple $(P,\dots,P)\in (\mathbb C^2)^n$ .
One of the great virtues of this morphism is that it provides a very natural desingularization of the singular variety $\operatorname {Sym}^n (\mathbb C^2)$, since the Hilbert scheme $\text{Hilb}^n(\mathbb C^2)$ is smooth (and connected by Hartshorne's thesis).

Answer 2

This is the easiest non trivial example I know and I consider it the most important to try to visualise and understand both Hilbert scheme of points and Hilbert-Chow morphism.

Suppose you want to understand $\rm{Hilb}^2(\mathbb{A}^2)$. First of all you may want to understand what $\rm{Sym}^2(\mathbb{A}^2)$ is. By definition $\rm{Sym}^2(\mathbb{A}^2)=\rm{Spec}\,\mathbb{C}[x,y,z,w]^{\mathfrak{G}_2}$, where $\mathfrak{G}_2=\{e,\sigma\}$ is the symmetric group with two elements and $\rm{Spec}\,\mathbb{C}[x,y,z,w]^{\mathfrak{G}_2}$ is the ring of invariant elements. Here I am assuming that you are a bit familiar with Geometric Invariant Theory, otherwise I suggest you to look at some standard references about that (Mumford, Newstead, etc..). Working and messing with this data you should finally get to the conclusion that $$ \rm{Sym}^2(\mathbb{A}^2)=\rm{Spec}\,\mathbb{C}[a,b,c,d,e]/I$$ where $I=(e^2-abe-4cd+b^2c+a^2c)$. How visualise this? If you fix the $a$ and $b$ variables, then $\rm{Sym}^2(\mathbb{A}^2)=\rm{Spec}\,\mathbb{C}[a,b][c,d,e]/I$ where now we can recognise $I$ as the ideal of a quadric $Q\subset\mathbb{A}^3=\rm{Spec}\,\mathbb{C}[c,d,e]$ (recall that we are now supposing that $a$ and $b$ are fixed). In fact, by direct calculations, we can further say that $Q$ is actually a cone. The moral of this picture is that locally $\rm{Sym}^2(\mathbb{A}^2)$ looks like a product $\mathbb{A}^2\times Q$, where the verteces of these cones lie on the diagonal $\Delta$ inside $(\mathbb{A}^2)^2$.

Now, in $\rm{Hilb}^2(\mathbb{A}^2)$ we have only two kind of ideals. The first kind are not so interesting, corresponding to ideals of the form $(x-a,y-b)\cap(x-a',y-b')$ where at least one between $a,a'$ and $b,b'$ is distinct, i.e. distinct points, which are sent into the point $((a,b),(a',b'))\in\rm{Sym}^2(\mathbb{A}^2)$ through the Hilbert-Chow morphism. The second kind of ideals are the ideals of length 2, which can be described in the following way. Fix a curve $C\subset\mathbb{A}^2$ and a point $p\in C$. We can consider the limit (whatever it means, if you are interested you can look at Harris-Eisenbud 'The geometry of schemes', somewhere in Chapter 2 if I am right) of another point $q\in C$ moving on $C$ and approaching $p$. This limit is the non reduced scheme whose support consists of the single point $p$, but its regular sheaf recall the direction of the crash, i.e. recall the tangent direction of the curve $C$ in the point $p$. This element will be sent in the point $(p,p)\in\Delta\subset\rm{Sym}^2(\mathbb{A}^2)$.

So we have now a good description of what $\rm{Hilb}^2(\mathbb{A}^2)$ is and how the Hilbert-Chow morphism works. To conclude we need to note that the set of tangent directions through a point is canonically identified with the projective line $\mathbb{P}^1$. The Hilbert-Chow morphism sends any ideal corresponding to one of these directions to the same point on the diagonal $\Delta$. Hence we are saying that the Hilbert-Chow morphism attach a copy of $\mathbb{P}^1$ upon each point on the diagonal (which is the singular locus of the symmetric product). Other way said, the Hilbert-Chow morphism is the blow-up of $\rm{Sym}^2(\mathbb{A}^2)$ along the diagonal.