I have a question about a theorem and example in Hartshorne's "Deformation Theory" p. 65. The theorem reads:
Theorem 8.9 Let $Y_0$ be an ACM closed subscheme of codimension 2 of $X_0=\mathbb{P}^n_k$, and assume $\dim Y_0\ge1$. Using notation (6.1) [note: this means we have an exact sequence $$0\to J\to C'\to C\to0$$ where $C$ and $C'$ are local Artin rings with residue field $k$ and $J$ is an ideal with $\mathfrak{m}_{C'}J=0$], suppose we are given a closed subscheme $Y$ of $X=\mathbb{P}^n_C$, flat over $C$ and with $Y\times_Ck=Y_0$. Then:
(a) There is an $r\times(r+1)$ matrix $\varphi$ of homogeneous elements of $R=C[x_0,\ldots,x_n]$ whose $r\times r$ minors $f_i$ generate the ideal $I$ of $Y$, and give a resolution $$0\to\bigoplus R(-b_i)\xrightarrow{\varphi}\bigoplus R(-a_i)\xrightarrow{f}R\to R/I\to0.$$
(b) For any lifting $\varphi'$ of $\varphi$ to $R'=C'[x_0,\ldots,x_n]$, taking $f'$ to be the $r\times r$ minors gives an exact sequence $$0\to\bigoplus R'(-b_i)\xrightarrow{\varphi'}\bigoplus R'(-a_i)\xrightarrow{f'}R'\to R'/I'\to0$$ and defines a closed subscheme $Y'$ of $X'=\mathbb{P}^n_{C'}$, flat over $C'$, with $Y'\times_{C'}C=Y$.
(c) Any lifting $Y'$ of $Y$ to $X'$, flat over $C'$ with $Y'\times_{C'}C=Y$, arises by lifting $\varphi$ as in (b).
This theorem is followed by:
Example 8.9.1 The conclusions of (8.9) are false in the case of a scheme $Y_0$ of dimension 0 in $\mathbb{P}^n$. For example, let $Y_0$ be a set of three collinear points in $\mathbb{P}^2$. Then there is a linear form in the ideal of $Y_0$. But as you deform the points in the direction of a set of three noncollinear points of $\mathbb{P}^2$, the linear form does not lift. So the deformations of $Y_0$ cannot all be obtained by lifting the elements of the corresponding matrix $\varphi$.
I think I understand why this is a counterexample in dimension 0. Now suppose we replace this $Y_0$ with the cone over $Y_0$ in $\mathbb{P}^3$. Now it has dimension 1, and is still ACM of codimension 2, so (8.9) applies. Yet it seems like the argument in (8.9.1) still works. Instead of three collinear points, we have three coplanar coincident lines, so there is a linear form in the ideal of the lines. Yet if we deform in the direction of three noncoplanar lines, still coincident at the cone point (just take the cone over the deformation from (8.9.1)), the linear form does not lift. But this would be a counterexample to the theorem. What is wrong with this argument?
I figured out the mistake. The point is that the you can no longer deform the three coplanar lines in the direction of three noncoplanar lines because this is not a flat family. In fact, part of Exercise III.9.5 in Hartshorne's "Algebraic Geometry" is to show that the cone over a flat family is not necessarily flat.
In this case, we can find a flat family $X$ of subschemes of $\mathbb{P}^2_k$ over the base $\mathbb{A}^1_k$ such that the fiber over $0$ consists of three collinear points and every other fiber consists of three noncollinear points. If we let $X'$ be the restriction of $X$ to $\mathbb{A}^1\setminus\{0\}$, then the cone $C(X')$ will still be flat, because all the fibers have the same Hilbert polynomial. But the flat limit of $C(X')$ at $0$ has an embedded point at the cone point, which is not the same as the cone $C(X_0)$. Thus $C(X)$ is not a flat family.
For a concrete example, consider the family of ideals $I_t=(z(x-y),z(tx-z),y(tx-z))$. For $t\neq0$, this is the saturated ideal of the (noncollinear) points $(1:0:0)$, $(0:1:0)$, and $(1:1:t)$. The limit of this family is $I_0=(z(x,y,z),xy(x-y))$, whose vanishing locus consists of the three collinear points $(1:0:0)$, $(0:1:0)$, and $(1:1:t)$. But $I_0$ is not saturated (we can see this even without calculating $I_0$, as the saturated ideal of three collinear points should contain a linear form, while the ideals $I_t$ are generated by quadrics). This is not a problem in $\mathbb{P}^2_k$ where the ideal $(x,y,z)$ is irrelevant, but when we pass to the cone, the associated prime $(x,y,z)$ becomes an embedded point.