# Zero-Dimensional Subschemes of Degree 21

I'm working on the following problem from Eisenbud and Harris' Geometry of Schemes.

Consider zero dimensional subschemes of $$\mathbb{A}^4_K$$ of degree 21 such that $$V(\mathfrak{m}^3)\subset \Gamma \subset V(\mathfrak{m}^4),$$ where $$\mathfrak{m}$$ 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 a general one is not a limit of a reduced scheme.

Here's what I have so far: If $$R$$ denotes the polynomial ring $$K[x_1,x_2,x_3,x_4]$$, then we are looking for ideals $$I$$ such that $$\Gamma = V(I)$$, $$\mathfrak{m}^4 \subset I \subset \mathfrak{m}^3,$$ and $$R/I$$ has dimension $$21$$ as a $$K$$-vector space and $$0$$ Krull dimension.

Now, $$\mathfrak{m}^3$$ is generated by all forms of degree 3 in $$R$$ and $$\mathfrak{m}^4$$ is generated by all forms of degree 4 in $$R$$. There are $${3+4-1 \choose 4-1}=20$$ forms of degree 3 and $${4+4-1 \choose 4-1}=35$$ forms of degree 4. So we are looking for ideals $$I$$ with 21 generators each either a form of degree 3 or 4 such that $$R/I$$ has Krull dimension 0. I'm not sure how to do this let alone how the conclusion would follows.

• Thank you, the question has been edited. – Rdrr Apr 3 '19 at 15:44

I think the idea should be since $$I \subset \mathfrak{m}^3$$, we know that $$R/I$$ must contain (as basis elements) the 1 degree 0, 4 degree 1, and 10 degree 2 forms for a total of 15 dimensions. Then we have an additional choice of 6 basis elements coming from the degree 3 forms that we can omit (i.e. $$I$$ must contain the other 14 dimensions of degree 3 forms). So this is the choice of a dimension 6 subspace of a dimension 20 space, so is the Grassmannian $$G(6,20)$$ which has dimension $$6*(20-6) = 84$$.
• As a $K$ vector space, $R/I$ is generated by $1, x_1,x_2,x_3,x_4, \cdots$. I mean the 1-dim'l piece generated by $1$. – user113102 Apr 3 '19 at 18:57
• Oops, I thought your we're talking about a basis for $I$, not $R/I$. How should one conclude that a general subscheme of this form is not a limit of a reduced one? – Rdrr Apr 3 '19 at 19:40
• @Rdr The set of reduced points have dimension $4\times 21$ and it is an open set. So, the limits of these form a set of strictly smaller dimension that 84. – Mohan Apr 3 '19 at 20:19
• Can you explain why they have dimension $4\times 21$ and why it's open? – Rdrr Apr 3 '19 at 21:21