Take the 2-minute tour ×

My question pertains to exercise II-16 in Eisenbud and Harris' "The geometry of Schemes". For an algebraically closed field $K$ the question is as follows:

Consider zero-dimensional subschemes $\Gamma \subset \mathbb{A}_K^4$ of degree 21 such that $$V(m^3)\subset\Gamma \subset V(m^4)$$ where $m$ is the maximal ideal of the origin in $\mathbb{A}_K^4$. Show that there is an 84-dimensional family of such subschemes, and conclude that in general one is not a limit of a reduced scheme.

What does it mean for a family of subschemes to have dimension 84? I can only think it means up to isomorphism there are 84 such subschemes, but this doesn't seem to work.

share|improve this question
Probably this means that the inclusions $\Gamma \to \mathbb{A}$ glue to a morphism $X \to Y \times \mathbb{A}^4$, where $X$ has dimension $84$ and the fiber of $y \times \mathbb{A}^4$ is $\Gamma_y$. So you have to find a scheme $Y$ which parameterizes $\Gamma$. But this is really just a guess! –  Martin Brandenburg Jul 12 '11 at 15:27
Martin's comment is correct. Note that the family $Y$ will also have dimension $84$, and $X$ will be a finite, flat, degree $21$ cover of $Y$. Hint: $Y$ should be a certain Grassmanian... –  Daniel Erman Jul 12 '11 at 18:05
I think one should have an $84$ dimensional scheme $X$ and a closed subscheme $Y$ in $X \times \mathbb{A}^4$, so that if $\pi: Y \to X$ denotes the morphism induced from $p_1: X \times \mathbb{A}^4 \to X$ then $\pi^{-1}(x)$ for $x \in X$ (a closed point), which is naturally a closed subscheme of$\mathbb{A}^4$, should be a subscheme of the form $\Gamma$. Moreover, all these fibres should be distinct. The technical way to say this is that the corresponding Hilbert scheme has dimension at least $84$. –  ulrich Jul 12 '11 at 18:06

1 Answer 1

up vote 7 down vote accepted

Such a subscheme is given by an ideal $I$ such that $m^4 \subset I \subset m^3$. In fact, any vector subspace in $m^3$ containing $m^4$ is an ideal (this is a simple exercise). Since $\dim O/m^3 = 15$ and $\dim O/m^4 = 35$, and we are interested in subspaces $I \subset O$ such that $\dim O/I = 21$, that is in subspaces $I/m^4$ of dimension $35 - 21 = 14$ of the space $m^3/m^4$ of dimension $35 - 15 = 20$. So, the family in question is just the Grassmannian $Gr(14,20)$. Its dimension is $14(20-14) = 14\cdot 6 = 84$.

share|improve this answer
Thank you very much - I think a lack of familiarity with Grassmannians didn't help! –  Tait Jul 13 '11 at 13:43

Your Answer


