I'm trying to understand the Hilbert scheme of points of a smooth complex algebraic surface, and I think I'll have a much clearer picture of it if I understand the case of $n$ points in the affine plane very concretely.

Here's the idea. Let $X$ be the complex affine plane, or $\mathbb{C}^2$ for us dummies. A point in the Hilbert scheme of points $M_{X, n}$ is an ideal $J$ in the polynomial ring $\mathbb{C}[x,y]$ for which the dimension of $\mathbb{C}[x,y]/J$ is $n$. There's a way to make the punctual Hilbert scheme into a smooth complex variety.

The easiest way to get a point in $M_{X,n}$ is to take $n$ distinct points $p_1, \dots, p_n$ in $\mathbb{C}^2$ and let $J$ consist of polynomial functions on $\mathbb{C}^2$ that vanish at all these points. This simple recipe works generically: we get an open dense set in the punctual Hilbert scheme this way. All the fun happens in the non-generic situation where some of the points $p_1, \dots, p_n$ collide.

When $n = 2$, we get more points in the Hilbert scheme of points this way: take just one point $p$, and let $J$ consist of polynomials that vanish at $p$ and whose derivative at $p$ vanishes in one chosen direction. It's easy to check that $J$ is an ideal, and the dimension of $\mathbb{C}[x,y]/J$ is 2 because we're imposing two equations. We can think of these points in $M_{X,2}$ as arising from 'pairs of infinitesimally nearby points' in the plane.

Question 1. Does this trick give all the remaining points in $M_{X,2}$?

I feel the answer must be yes. If so we can move on to this:

Question 2. What tricks do we need to get all the points in $M_{X,3}$?

Here are some tricks I'm guessing that we need:

  • We can take $J$ to consist of all polynomials that vanish at 3 distinct points.

  • We can take $J$ to consist of all polynomials that vanish at one point and vanish along with a chosen directional derivative at a second, distinct, point.

  • More interestingly, we can take $J$ to consist of all polynomials that vanish along with both their first partial derivatives at a single point. Presumably these points in $M_{X,3}$ describe ways for 3 points in the plane to collide. I imagine that in the usual topology of $M_{X,3}$ as a complex manifold they are limits of situations where we have 3 points arranged in a triangle and the triangle shrinks to zero size while remaining the same shape, with one corner not moving. Apparently the shape of the triangle doesn't matter as long as the triangle is nondegenerate, i.e. the 3 points don't lie on a line.

  • More interestingly still, we can take $J$ to consist of all polynomials that vanish at some point along with their first and second directional derivatives in a chosen direction at a single point. Apparently these points in $M_{X,3}$ describe subtler ways for 3 points in the plane to collide: namely, limits of situations where we have 3 points in a line, all moving closer and closer to a chosen point on that line.

Perhaps these tricks give all of $M_{X,3}$, but perhaps I'm leaving something out (or making a still worse mistake). Assuming I'm on the right track, the obvious question is:

Question 3. What tricks of this sort give all the points in $M_{X,n}$ for arbitrary $n$?

We seem to be getting some sort of stratification of $M_{X,n}$. How can we keep track of these strata? There might be some combinatorial way to name them all. It reminds me a bit of the Fulton–MacPherson compactification of a configuration space, but it seems different.

Perhaps everything I need to know is lurking in here:

but it will take some work to dig it out.

  • 3
    $\begingroup$ From a non-expert: Q1 yes, Q2 I think yes, Q3: for a Young diagram with $n$ boxes, you can consider $\mathbb{C}[x,y]/I$ where $I$ is the ideal generated by monomials $x^i y^j$ where $(i, j)$ lies outside of the given diagram. You will get some strata indexed by tuples of partitions with a total of $n$ boxes this way, but unlike for $n=1,2,3$, quite quickly this will not be all of $M_{X, n}$. $\endgroup$ Commented Jun 7, 2017 at 12:15
  • 4
    $\begingroup$ For $n = 3$ a length $n$ subscheme $Z \subset {\mathbb A}^2$ is either curvilinear (i.e., is contained inside a smooth curve $C \subset {\mathbb A}^2$), or is given by the square of the maximal ideal of a point. If you want to use an intuition from physics, this means that either 3 points collide "along a smooth curve", or they collide "uniformly from all directions". $\endgroup$
    – Sasha
    Commented Jun 7, 2017 at 20:15
  • 3
    $\begingroup$ For every $n$, $M_{X,n}$, the Hilbert scheme of closed, zero-dimensional, length $n$ subschemes in $X=\mathbb{C}^2$, is irreducible and smooth. For a proof see, e.g., Miller--Sturmfels, "Combinatorial Commutative Algebra", Chapter 18. One of the upshots of this is that everything in $M_{X,n}$ is smoothable, i.e., expressible as a limit of unions of $n$ distinct points. That's not true in $\mathbb{C}^3$; and even in $\mathbb{C}^2$ there are a LOT of ways that $n \gg 0$ points can collide... but it's true anyway. $\endgroup$ Commented Jun 7, 2017 at 21:53
  • 3
    $\begingroup$ $J = (y-x^2,xy,y^2)$ is an ideal such that $\dim \mathbb{C}[x,y]/J = 3$. But I don't think it can be described as "polynomials that vanish at [the origin] together with their first and second directional derivatives in a chosen direction." It is curvilinear. With a parametrization $(x,y) = r(t)$ of the smooth curve $y=x^2$, the ideal is given by polynomials $p$ such that $p \circ r$ and its first two derivatives vanish at $t=0$ (where $r(0) = (0,0)$). $\endgroup$ Commented Jun 7, 2017 at 22:05
  • 2
    $\begingroup$ There are notes of Haiman I like for this topic; I think it might be this: math.berkeley.edu/~mhaiman/ftp/newt-sf-2001/newt.pdf. In particular, the case of your question is addressed in section 3.1. $\endgroup$
    – dhy
    Commented Jun 8, 2017 at 5:10

2 Answers 2


If $J_n$ is an ideal with $\mathbb C[x,y]/J_n$ of dimension $n$, then there exists an ideal $J_{n-1}$ containing $J_n$, with $J_{n-1}/J_n$ one-dimensional.

This lets you understand the possibilities for the points of $M_{X,n}$ inductively if you can calculate for the ideal $J_{n-1}$ all possible subideals with quotient of dimension $1$.

You can do this using the fact that $J_{n-1}/J_n$ is necessarily isomorphic to $\mathbb C[x,y]/(x-a,y-b)$ for some $a,b$, and $J_n$ is the kernel of some map $J_{n-1} \to \mathbb C[x,y]/(x-a,y-b)$, which are classified by linear forms on the vector space $J_{n-1} / ( (x-a) J_{n-1}, (y-b) J_{n-1})$. You can calculate this space using generators for $J_{n-1}$. In particular, the dimension is the number in a minimal set of generators over $\mathbb C[[x,y]]$.

What does that mean in practice?

$J_0$ is just $\mathbb C[x,y]$. It has one generator, so all the spaces $J_0 / ((x-a)J_0,(y-b) J_0)$ are one-dimensional. For each point, we have a unique choice of ideal, which is $(x-a,y-b)$.

So $J_1$ has the form $(x-a,y-b)$. For simplicity (and wlog), let's assume $a=0,b=0$. If we pick any other point, the quotient will again be one-dimensional and so we have a unique choice. So let's pick the same point again. Then $J_1 / (x J_1, y J_1) = (x,y)/(x^2,xy,y^2)$ which is two-dimensional. So we have a projective line of possible choices of linear maps. These linear maps can be represented as derivatives in a direction, and so the associated ideal is the space of functions whose derivatives in that direction vanish.

Again wlog, the direction is along $x$. Then $J_2$ is $(x^2,y)$. It has two generators, and $J_2/ (xJ_2,yJ_2) = (x^2,y)/ (x^3, xy,y^2)$ is again two-dimensional. You have written down the ideals corresponding to two linear maps - the derivative in $y$ and the second derivative in $x$. However, there are also ideals corresponding to some linear combination of these maps, which are functions $f$ satisfying $f(0,0)=0$, $\frac{\partial f}{\partial x}=0$, $a \frac{\partial^2 f}{\partial x^2} + b \frac{\partial f}{\partial y}=0$. These are exactly what Sasha and Zach Teitler point out - polynomials vanishing to order $3$ along a smooth curve.

We can go further, but the number of possibilities multiplies. If $J_3 = (x^2,xy,y^2)$, then we have a three-dimensional space of possibilities for the next linear form. We can view these linear forms as differential operators of order $2$, evaluated at $(0,0)$. These could either be differential operators along a line, or Laplacian-like ones.

If $J_3$ is something like $(x^3,y)$, then again you have a two-dimensional quotient, and the linear forms give a projective line. All but one of the points will correspond to the third derivative along a cubic curve, and that one remaining point $\frac{\partial f}{\partial y}$ will force every element of your ideal to vanish to order $2$ at $(0,0)$, making this overlap with the previous case.

  • $\begingroup$ Thanks very much! You write "$J_0$ is just $R$". Is $R = \mathbb{C}$, here? $\endgroup$
    – John Baez
    Commented Jun 9, 2017 at 11:30
  • $\begingroup$ @JohnBaez Oh, I meant $\mathbb C[x,y]$. $\endgroup$
    – Will Sawin
    Commented Jun 9, 2017 at 11:55

I'm not sure if this is a complete answer to your question, but hopefully it's helpful. First, I'm going to constantly use the fact that I can go back and forth between thinking about ideals in $\mathbb{C}[x,y]$ with codimension $n$, $n$-dimensional quotient algebras of $\mathbb{C}[x,y]$, and certain zero-dimensional closed subschemes of $\mathbb{A}^2$. Each of these quotient algebras is going to be a finite product of local algebras. Each factor gives a maximal ideal in $\mathbb{C}[x,y]$, so we can think of the algebra as being supported at finitely many points of the plane. Each of these local algebras has finite dimension as a complex vector space, and we will call this dimension the multiplicity of the subscheme at the corresponding point. The sum of the multiplicities will be $n$.

The crux of the matter is then to understand these zero-dimensional local algebras. There is another space, which I will call $\mathcal{H}^m$, which is confusingly also called the punctual Hilbert scheme, which parametrizes $m$-dimensional quotient algebras of $\mathbb{C}[x,y]$ supported at a given point. Since this is the same as $m$-dimensional quotient algebras of the power series ring $\mathbb{C}[[x,y]]$, it doesn't matter which point we're looking at, or even which (smooth) surface.

For low $m$, it's not too hard to describe $\mathcal{H}^m$. For $m=1$, the punctual Hilbert scheme consists of a single point corresponding to the ideal $(x,y)$. For $m=2$, the ideals are all of the form $(ax+by)$, which consist of polynomials with vanishing directional derivative in a specified direction. For $m=3$, there are two types of points: there's the so-called "fat point" $(x,y)^2=(x^2,xy,y^2)$, which consists of polynomials with both partials vanishing, and the curvilinear points $(ax+by,cx^2+dxy+ey^2)$ consisting of polynomials with vanishing first and second derivatives in a specified direction.

Now let's talk about the relationship between the $\mathcal{H}^n$ and your punctual Hilbert schemes. We're going to want to start with something somewhat similar to the Hilbert scheme of $n$ points, called the Chow variety of $n$ points. The Chow variety can be defined as $(\mathbb{A}^2)^n/S_n$, where $S_n$ is the symmetric group. We can think of it as parametrizing points in $\mathbb{A}^2$ with multiplicities. There's a Hilbert-Chow morphism from the Hilbert scheme of $n$ points to the Chow variety of $n$ points which sends an ideal to its support with multiplicity. The fibers of the Hilbert-Chow morphism are going to be products of punctual Hilbert schemes.

Let's see how this works for Hilbert schemes of small numbers of points. If we're dealing with two points, we have two types of points in the Chow variety. If we write points of the Chow variety additively, we have things of the form $p+q$ with $p$ and $q$ distinct, and we have points of the form $2p$. The fiber of the Hilbert-Chow morphism over $p+q$ will be a single point. The fiber over $2p$ will be $\mathcal{H}^2$.

For three points, we have $p+q+r$, $p+2q$, and $3p$, and we can completely understand the fibers of the Hilbert-Chow morphism over each of these three types of points. For four points, the only sort of newish behavior is with things like $2p+2q$, where the fiber of the Hilbert-Chow morphism is $\mathcal{H}^2 \times \mathcal{H}^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.