Eisenbud-Harris Exercise II-14, limit scheme isomorphic to triple point and remembers both tangent line, osculating $2$-plane to subscheme Let $C$ be the subscheme of $\mathbb{A}_K^n$ given by the ideal$$J = (x_2 - x_1^2, x_3 - x_1^3, \ldots).$$A closed point in $C$ is of the form $f(t) = (t, t^2, t^3, \ldots, t^n)$, for $t \in K$; that is, it has ideal $(x_1 - t, x_2 - t^2, \ldots)$. Consider for $t \neq 0$ the three-point subscheme$$X_t = \{f(0), f(t), f(2t)\} \subset C.$$


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*How do I see that the limit scheme as $t \to 0$ is$$X_0 = \text{Spec}\,K[x_1, \ldots, x_n]/(x_2 - x_1^2, x_1x_2, x_3, x_4, \ldots, x_n)$$and is isomorphic to the triple point $\text{Spec}\,K[x]/(x^3)$ above?

*How do I see that $X_0$ is not contained in the tangent line to $C$ at the origin, and that rather, the smallest subspace of $\mathbb{A}_K^n$ in which $X_0$ lies is the osculating $2$-plane$$x_3 = x_4 = \ldots = x_n = 0$$to $C$, while the tangent line to $C$ is the smallest subspace of $\mathbb{A}_K^n$ containing the subscheme defined by the square of the maximal ideal in the coordinate ring of $X_0$? Thus, in this sense, $X_0$ "remembers" both the tangent line and the osculating $2$-plane to $C$.

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*The ideal of the union of the three points has generators$$c_{ijk} = x_i(x_j - t^j)(x_k - (2t)^k).$$In particular, the ideal $I$ of the limit contains all triple products of monomials, so that $m^3 \subset I$. Now, consider the difference $c_{ijk} - c_{ikj}$, where $k < j$. Dividing by $t^k$ and then setting $t$ equal to zero in the above expression, we see that $x_ix_j \in I$. Thus, $I$ contains $x_ix_j$ if $ij \ge 2$. Now, consider triple differences$$ac_{ijk} + bc_{jki} + cc_{ikj},$$where we choose $a$, $b$, and $c$ so as to kill the first two terms, expanding in powers of $t$. In this case, if $i < j < k$, the next term will be $t^\alpha x_k$, for an appropriate power of $t$. But then $x_k \in I$, for $k > 2$. Finally, the difference $c_{211} - c_{121}$ shows that $x_2 - x_1^2 \in I$. As the quotient$${{k[x_1, x_2, \ldots, x_n]}\over{\langle x_2 - x_1^2, x_1x_2, x_3, x_4, \ldots, x_n\rangle}}$$has length three, the result follows.

*The tangent line is given by $\langle x_2, x_3, \ldots, x_n\rangle$, which is not contained in $I$, since $x_2 \notin I$. On the other hand, the truncation of $X_0$ does contain the tangent line. Thus, the smallest linear space containing $X_0$ has dimension at least two, and must contain the tangent line. Since $I$ contains the ideal of the osculating two plane given by $\langle x_3, x_4, \ldots, x_n\rangle$, we are done.

