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.
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 $6*(20-6) = 84$.