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A set of $d$ points $X\subset \mathbb P^r$ is said to be in general position if there are no more than $2$ points of $X$ on any line, no more than $3$ points on any $2$-plane,..., no more than $r$ points in an $(r-1)$-plane.

Let $X$ be the union of $d$ general points in $\mathbb P^r$. Is it true that $$H_X(s)=\min\left\{d,\binom{r+s}r\right\}\tag{$\star$}\label{star}?$$

For example, if $X$ consists of $3$ non-collinear points in $\mathbb{P}^2$, then $H_X(s)=3$ for all $s\in\mathbb N$ [1]. If $X$ consists of $4$ points that are not collinear, then $$H_X(s)=\begin{cases}3 & s=1\\ 4& s \geq 2\end{cases}.$$ One can also see that $(\ref{star})$ holds for a set of $7$ (or less than $7$) general points in $\mathbb P^3$.

Now let $X\subset \mathbb P^3$ be the union of $8\,(\gt 2\cdot3+1)$ general points. According to $(\ref{star})$, we should have $H_X(2)=8$. But I don't see it and I suspect it is not true. Probably it has to do with the following fact:

Let $X$ be a set of $d\le 2r+1$ points in $\mathbb{P}^r$ in linearly general position. Then the space of quadratic forms vanishing on $X$ is $\binom{r+2}{r}-d$ dimensional (i.e. $H_X(2)=d$) [2].

By the way, I saw $(\ref{star})$ in the following paper: Generators for the Homogeneous Ideal of $s$ General Points in $\mathbb P^3$.

Can someone tell me what's going on? Is $(\ref{star})$ true?


Edit: I think I'm misinterpreting generic $d$-position as general position. This paper discusses the two notions on page number 534. So it looks like there are counterexamples to the above equality.

cqfd
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Consider the set $X=\{[1:1:1],[-1:1:1],[2:4:1],[-2:4:1],[3:9:1],[-3:9:1]\}\subset\mathbb P^2$ of $6$ points. It is easy to see that $X$ consists of points in general position (draw a diagram). Moreover, $X$ is a complete intersection of a quadric and a cubic, and its ideal $I_X$ is $\langle x^2-yz,(y-z)(y-4z)(y-9z)\rangle$. Since $ \dim_k (I_X)_2=1$, we get $H_X(2)=6-1=5$. But $5\neq \min\{6,\binom{2+2}2\}=6.$

cqfd
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