Exercise II-11 from Eisenbud-Harris, subscheme of dimension $0$, degree $3$, supported at origin isomorphic to what?

Question

Suppose that $K$ is algebraically closed, and let $Z = \text{Spec}\,K[x_1, \ldots, x_n]/I \subset \mathbb{A}_K^n$ be any subscheme of dimension $0$ and degree $3$, supported at the origin. How do I get started showing that $Z$ is isomorphic either to $X = \text{Spec}\,K[x]/(x^3)$ or to $Y = \text{Spec}\,K[x, y]/(x^2, xy, y^2),$ and that $X$, $Y$ are not isomorphic to each other?

Answer 1

These two schemes are certainly not isomorphic to each other, as the first has one dimensional tangent space and the second two dimensional. Clearly any such scheme must be affine, so it is just a case of classifying local rings $A$ of vector space dimension three over an algebraically closed field $k$. Let $m$ be the maximal ideal. Then $m$ indues a filtration of $A$,$$A \supset m \supset m^2 \supset m^3 \ldots.$$Since $A$ is an algebra over $k$, $A/m \cong k$ has dimension one. There are only two possibilities. $m/m^2$ has dimension one, in which case $m^3 = 0$ and $m^2$ has dimension one, or $m/m^2$ has dimension two and $m^2 = 0$. In the former case, pick any element which maps to a nonzero element of $m/m^2$. Then $x^2$ generates $m^2$, and $x^3 \in m^3$ must be zero. This is the first algebra. Otherwise, pick two elements $x$ and $y$ of $A$ such that their images generate $m/m^2$. As any element of $m^2$ is zero,$$x^2 = xy = y^2 = 0,$$and this gives the second algebra.