I have some problems to understand a solution I found of following exercise from Hartshorne's Algebraic Geometry, Ex 9.5 in Chapter III, part (a):
Very Flat Families. For any closed subscheme $X \subset \mathbb{P}^n$, we denote by $C(X) \subset \mathbb{P}^{n+1}$ the projective cone over $X$ (I, Ex. 2.10).
(a) Give an example to show that if $\{X_t\}$ is a flat family of closed subschemes of $\mathbb{P}^n$, then $\{ C(X_t)\}$ need not be a flat family in $\mathbb{P}^{n+1}$.
A proposed solution of this exercise I found works as follows:
Consider the flat (?) family $X_t$ defined by $(1:0:0), (0:1:0)$, and $(1:1:t)$ in $\mathbb{P}^2$.
These points are only on a line together at $t = 0$.
For $t \neq 0$, then $I_{X_t} = \langle xz-txy, yz -txy, z^2-t^2xy \rangle$ . Let $\{X'_t\}$ be the restriction of $\{X_t\}$ to $\mathbb{A}^1\setminus\{0\}$, then the cone $\{ C(X_t')\}$ will still be flat since all fibers have same Hilbert polynomial.
And some calculations show that the flat limit of $\{ C(X_t')\}$ at $0$ has embedded point at cone, which is not the same as $C(X_0)$.
I have a basic question about first line: I not understand why the observation that the family $\{X_t\}$ of $X_t \subset \mathbb{P}^2$ defined as $(1:0:0), (0:1:0)$, and $(1:1:t)$ is flat not contradicts to Theorem 9.9 from Hartshorne:
Theorem 9.9: Let $T$ be an integral noetherian scheme. Let $X \subset \mathbb{P}^n_T$ be a closed subscheme. For each point $t \in T$, we consider the Hilbert polynomial $P_t \in \mathbb{Q}[z]$ of the fibre $X_t$ considered as a closed subscheme of $\mathbb{P}^n_{k(t)}$· Then $X$ is flat over $T$ if and only if the Hilbert polynomial $P_t$ is independent of $t$.
Doesn't it not contractict to the flatness of family $X_t$ from above?
Indeed this family sits as closed subscheme $X_t \subset \mathbb{P}^2 \times\mathbb{A}^1$,
therefore if it is flat by the Theorem all fiberes $X_{t_0}=(1:0:0), (0:1:0)$, and $(1:1:t_0)$ should have
same Hilbert polynomial. On the other hand there is a fact that
The Hilbert function of three points in $\Bbb P^2$ is $$\begin{cases} 0 & n<1 \\ 2 & n=1 \\ 3 & n>1\end{cases}$$ if they're collinear, or $$\begin{cases} 0 & n<1 \\ 3 & n\geq 1\end{cases}$$ if they're not collinear.
(see for example Here (in german)
By construction of $X_t$ only $X_0$ consists of three collinear point,
every $X_t$ with $t \neq 0$ not. Therefore fibers $X_0$ and $X_t$ should have
have different Hilbert functions. On the other hand this contradicts
to Theorem 9.9.
Could anybody help me to resolve my confusion?