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Let $\{[1:0:0],[0:1:0],[0:0:1],[1:1:1] \} = \{p_1,p_2,p_3,p_4\}$ be four points in the projective space $\mathbb{P}^2$. For every $p_i$, show there is a homogenous polynomial $f_i$ such that $$\left\{\begin{matrix} f_i(p_j) \neq 0 & i= j\\ f_i(p_j) = 0 & i \neq j \end{matrix}\right..$$

Now if $I$ is the ideal associated with these four points, then the Hilbert function $H(R/I,t) = 4$ for all $t > 1$.

I was thinking $f_i = x_1x_2x_3 + x_i^3$ might work, but the last point $[1:1:1]$ is killing me here.

user26857
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ferric duffe
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1 Answers1

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Just take $f_1=(x-y)(x-z),\quad f_2=(y-z)(y-x),\quad f_3=(z-x)(z-y), \quad f_4=xz .$

Edit
For completeness, let me mention that:
a) The homogeneous ideal of the set $S=\{p_1,p_2,p_3,p_4\}$ is $$I(S)=\langle z(x-y),x(y-z)\rangle \subset k[x,y,z] $$

b) The Hilbert polynomial of the set $S$ is the constant polynomial $$H(t)=4\in \mathbb Q[t]$$ c) The Hilbert function of the set $S$ is the function $h: \mathbb N\to \mathbb N$ characterized by $$h(0)=1, h(1)=3 \operatorname {and} h(t)=4 \operatorname {for} t\geq 2.$$

Georges Elencwajg
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