Let 7 points in $\mathbb {P}^3$ in general linear position, i.e. no 3 are on a line, no 4 are one a plane, etc.
If I denote with $R $ the ring associated to this variety show that
$H_R (0)=1$
$H_R (1)=4$
$H_R (d)=7$, when $d\geq 2$
$H_R (d) $ is the Hilbert function
Looks very intuitive.
I found this question in "Geometry of syzygies" by Eisenbud.
Say $Z$ is a set of $7$ points in linearly general position in $\mathbb{P}^3$ and $I=I_Z$ is the ideal of $Z$. Write $S = k[x_0,\dotsc,x_3]$ and $R=S/I$.
Since $Z$ is nonempty, there are no nonzero constants in $I$, that is $\dim I_0 = 0$, so $H_R(0) = \dim S_0 = 1$.
Since $Z$ does not lie on any hyperplane, $\dim I_1 = 0$, so $H_R(1) = \dim S_1 = 4$.
For all $d \geq 0$ we have $H_R(d) \leq 7$ (this is only interesting for $d \geq 2$, but we might as well state it for all $d \geq 0$). Indeed, each point of $Z$ imposes at most one linear condition on degree $d$ forms, so there are at most $7$ conditions; then $\dim I_d \geq \dim S_d - 7$, so $\dim R_d = \dim S_d - \dim I_d \leq 7$. In a little more detail, choose any basis for the space $S_d$ of degree $d$ forms (for example, you can choose the basis of degree $d$ monomials, if you like). Every degree $d$ form can be written as a linear combination of basis elements; the coefficients give coordinate functions on $S_d$. For each point of $Z$, substituting the coordinates of the point into the chosen basis polynomials gives a linear condition on the coefficients of a degree $d$ form, for that form to vanish at that point.
For example, say one of the points has homogeneous coordinates $[1:2:3:4]$, and choose the basis for $S_2$ consisting of monomials $x_0^2,x_0 x_1, x_0 x_2, x_0 x_3, x_1^2, \dotsc, x_3^2$. Write a quadratic form as $$ Q = a_1 x_0^2 + a_2 x_0 x_1 + a_3 x_0 x_2 + a_4 x_0 x_3 + a_5 x_1^2 + \dotsb + a_{10} x_3^2 . $$ Then $Q$ vanishes at $[1:2:3:4]$ if and only if $$ a_1 + 2a_2 + 3 a_3 + 4 a_4 + 4 a_5 + \dotsb + 16 a_{10} = 0. $$
We get a system of $7$ linear equations. A priori, the equations might fail to be independent, which is why we can (at this point) only say that the $7$ points of $Z$ impose at most $7$ conditions on degree $d$ forms.
If the points are general, then the equations are independent. All we need is for at least of the $7 \times 7$ minors of the matrix of this system of linear equations, to be nonzero. (For $d=2$, there are $7$ equations in $10$ unknowns, the coefficients of a quadratic form, so it corresponds to a $7 \times 10$ matrix.) But the nonvanishing of that minor is a nonlinear condition. Can we deduce it from the linearly general position hypothesis?
Yes. For each point $P$ in $Z$ we claim the following:
Claim: For $d \geq 2$ there is a degree $d$ form vanishing at all points of $Z-P$, and not vanishing at $P$.
Proof of claim: Divide the $6$ points of $Z-P$ into two groups of $3$ (in any way). Each set of $3$ points determines a plane; we get a pair of planes. Let $F$ be the degree $d$ form defining the union of those $2$ planes, plus any other hypersurface of degree $d-2$ not passing through $P$. $\square$
This means that there is an element of $S_d$ whose coefficients satisfy $6$ of the equations in our system of linear equations, but don't satisfy the $7$th equation. Therefore the system of $7$ linear equations is independent.
This proves that $\dim I_d$ really is equal to $\dim S_d - 7$, so $\dim R_d = 7$, for $d \geq 2$.
