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  1. What is the classification up to isomorphism of subschemes of $\mathbb{A}_K^2$ of dimenion $0$ and degrees $4$ and $5$ with support at the origin?
  2. Which are isomorphic as schemes over $\text{Spec}\,K$?

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As in here, it suffices to classify all local rings $A$ of vector space dimension $4$ and $5$ over an algebraically closed field, subject to the condition that $A$ has Zariski tangent space at most two. As in the linked, consider the filtration induced by the powers of the maximal ideal. Let $h_i$ be the dimenion of $m^i/m^{i + 1}$. Note that$$h_0 = 1; \quad h_i \le h_{i + 1},\text{ }i \ge 1; \quad \sum_i h_i = l;$$where $l$ is the length. The middle inequality follows from Nakayama. If $l = 4$, the only possibilities are$$(1, 1, 1, 1), \quad (1, 2, 1),$$and if $l = 5$, the only possibilities are$$(1, 1, 1, 1, 1), \quad (1, 2, 1, 1), \quad (1, 2, 2).$$In the first case, pick an element $x \in m$ which generates $m/m^2$. Then $x^i$ generates $m^i/m^{i + 1}$, and so $A = \text{Spec}\,k[x]/\langle x^l\rangle$. Otherwise, pick two elements $x$ and $y$ whose images generate $m/m^2$. If $l = 4$, there are then two possibilities, up to isomorphism. Either $x^2$ or $xy$ generates $m^2$. We have$$A = k[x, y]/\langle x^3, xy, y^2\rangle, \quad A = k[x, y]/\langle x^2, y^2\rangle,$$in the former and latter cases, respectively. Now suppose $l = 5$, and we have the middle option. This case is similar to the length four case$$A = k[x, y]/\langle x^4, xy, y^2\rangle.$$Finally, consider the last option. In this case, the elements of $m^2/m^3$ can be thought of as a pencil of conics. It is an instructive exercise left to the reader to see that there are two such pencils$$\lambda x^2 + \mu y^2, \quad \lambda x^2 + \mu xy.$$The corresponding algebras are$$A = k[x, y]/\langle xy, x^3, y^3\rangle, \quad A = k[x, y]/\langle y^2, x^3\rangle.$$

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