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I'm a topologist who is learning a bit of algebraic geometry for fun (to pass the time during my Christmas break), and I'm stuck on Exercise II-13 of Eisenbud-Harris's "The Geometry of Schemes".

Let $\mathcal{H}$ be the set of finite subschemes of degree $3$ of $\mathbb{A}_k^2$ which are supported at the origin. In other words, $\mathcal{H}$ is the space of ideals in $k[x,y]/(x,y)^3$ whose underlying vector space is $3$-dimensional. This vector space is a subspace of the $5$-dimensional vector space $V$ whose basis is $\{x,y,x^2,xy,y^2\}$, so $\mathcal{H}$ is identified with a subspace of the Grassmannian of $3$-dimensional subspaces of $V$.

The exercise in question asserts that $\mathcal{H}$ is isomorphic to a $2$-dimensional cubic cone in $\mathbb{P}^3_k$ whose vertex is the ideal spanned by $\{x^2,xy,y^2\}$. Can anyone help me prove this?

EDIT: I've figured out more, but not enough. Here's what I've got. As above, let $V$ be the vector space with basis $\{x,y,x^2,xy,y^2\}$, so our ideals are $3$-dimensional subspaces of $V$. Let $I_0$ be the subspace of $V$ spanned by $\{x^2,xy,y^2\}$, so $I_0 \in \mathcal{H}$. Every $I \in \mathcal{H}$ such that $I \neq I_0$ can be written as $I_L$ via the following construction:

  1. Let $L$ be a line in $V$ that does not lie in $I_0$.
  2. Pick some nonzero $\vec{v} \in L$.
  3. Define $I_{L}$ to be the subspace spanned by $\{\vec{v},x \vec{v}, y \vec{v}\}$. Here multiplication by $x$ and $y$ works in the obvious way where cubic terms are set to $0$.

Now let $W \subset V$ be the span of $\{x,y\}$. We get an injective map $f\colon \mathbb{P}(W) \rightarrow \mathcal{H}$ taking $L \in \mathbb{P}(W)$ to $I_L$. What is more, for every $L \in \mathbb{P}(W)$ there is a subvariety $C_L \subset \mathcal{H}$ with the following properties:

  1. $C_L$ consists of $I_0$ together with all $I_{L'}$ such that $L'$ is a line that projects to $L$ under the natural projection $V \rightarrow W$.

  2. $C_L \cong \mathbb{P}^1$.

  3. Every two distinct $C_L$ intersect only at $I_0$.

  4. Every point of $\mathcal{H}$ is in some $C_L$.

This all makes it look a lot line $\mathcal{H}$ is a cone on something resembling $\mathbb{P}^1$, but I can't quite get the description in the problem. Can anyone help me?

Adam Smith
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1 Answers1

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OK, I found an answer. I'll sketch it here in case it is useful for anyone else.

The short answer: just work out the Plucker embedding!

To simplify our notation, I will write $a=x^2$ and $b=xy$ and $c=y^2$. The vector space $V$ thus has for a basis $\{x,y,a,b,c\}$. Consider the Plucker embedding

$$\phi\colon \text{Gr}(3,V) \rightarrow \mathbb{P}(\wedge^3 V).$$

Define $W$ to be the subspace of $V$ spanned by $\{x,y\}$. For $L \in \mathbb{P}(W)$, define $I_L \in \text{Gr}(3,V)$ to be the following subspace of $V$. Write $L = [\lambda,\nu]$ in homogeneous coordinates with respect to the basis $\{x,y\}$. Then $I_L$ is the span of the vectors

$$\{\lambda x+\nu y, \lambda a + \nu b, \lambda b + \nu c\}.$$ The second two vectors here are the result of multiplying $\lambda x+\nu y$ by $x$ and $y$. We thus have $I_L \in \mathcal{H}$. The image $\phi(I_L)$ is then the line in $\wedge^3 V$ spanned by the following element:

\begin{align*} &(\lambda x+\nu y) \wedge (\lambda a + \nu b) \wedge(\lambda b + \nu c) \\ &= \lambda^3(x \wedge a \wedge b) + \lambda^2 \nu(x \wedge a \wedge c+y \wedge a \wedge b) + \lambda \nu^2(x \wedge b \wedge c+y \wedge a \wedge c) + \nu^3(y \wedge b \wedge c). \end{align*}

From this, we see that the images of the various $I_L$ under $\phi$ are precisely the rational cubic in the projectivation of the $4$-dimensional subspace of $\wedge^3 V$ spanned by the vectors

$$\{x \wedge a \wedge b, x \wedge a \wedge c+y \wedge a \wedge b, x \wedge b \wedge c+y \wedge a \wedge c, y \wedge b \wedge c\}.$$

Letting $I_0 \in \mathcal{H}$ be the span of $\{a,b,c\}$, we have that

$$\phi(I_0) = a \wedge b \wedge c.$$

As I discussed in the question, every point of $\mathcal{H}$ is contained in a unique $\mathbb{P}^1$ containing $I_0$ and $I_L$ for some $L \in \mathbb{P}(W)$. To complete the proof, it is enough to show that $\phi$ takes these $\mathbb{P}^1$'s to the lines in $\mathbb{P}(\wedge^3 V)$ connecting $\phi(I_0)$ and $\phi(I_L)$. Making a linear change of coordinates, it is enough to deal with the case where $L = [1,0]$ with respect to the homogeneous coordinates associated to $\{x,y\}$. We thus have that

$$I_L = \langle x,a,b \rangle \quad \text{and} \quad I_0 = \langle a,b,c \rangle.$$

The $\mathbb{P}^1$ in $\mathcal{H}$ connecting these consists of

$$I_{[\nu,\mu]} = \langle \nu x + \mu c, a, b\rangle$$

for $[\nu,\mu] \in \mathbb{P}^1$. The image $\phi(I_{[\nu,\mu]})$ in $\mathbb{P}(\wedge^3 V)$ consists of the line spanned by

$$(\nu x + \mu c) \wedge a \wedge b = \nu (x \wedge a \wedge b) + \mu(a \wedge b \wedge c).$$

The desired result follows.

Adam Smith
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