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In The homogeneous ideal of $2n$ points in general position in $\mathbb{P}^n$, we let $\Gamma$ be a set of $d=2n$ points in general position in $\mathbb{P}^n$, and we want to show that the associated homogeneous ideal $I(X)$ is generated by homogeneous polynomials of degree $2$. If $\Gamma$ is a set of $d\leq kn$ points of $\mathbb{P}^n$ in general position, where $k\geq 2$, how could I show that $\Gamma$ vanishes for a family of homogeneous polynomials of degree $\leq k$?

The main problem I have is that I want to prove the result using Hilbert polynomials and resolutions only, my professor told me there is a very easy way to prove it using such tools but I don't see how.

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  • $\begingroup$ For the second part, you can use exactly the same argument, with a partition $\Gamma = X_1 \cup \dot X_k$ with $n$ points each, use $k$ hyperplanes $H_1, \dots, H_k$ and so one. $\endgroup$ – user171326 May 4 '17 at 14:00
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    $\begingroup$ Thanks @N.H. , I figured I could follow more or less the same procedure, but my goal is to show the result using Hilbert polynomials & resolutions. $\endgroup$ – GSF May 4 '17 at 14:03
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In my humble opinion: the following proof is not "very easy" (cit.); but it is the one that I know (via Hilbert polynomials).


Let $I$ be the homogeneuos ideal of $\mathbb{K}[x_0,\dots,x_n]$ associated to $X=\{P_0,P_1,\dots,P_m\}$; where $0\leq m\leq kn$, $k\geq2$ and $P_0,P_1,\dots,P_m$ are points in general position; I set $S(X)=\mathbb{K}[x_0,\dots,x_n]_{\displaystyle/I}$.

Without loss of generality, one can assume that $X\cap V_{+}(x_0)=\emptyset$; where $V_{+}(x_0)$ is the hyperplane of equation $x_0=0$. Let $$ I=(f_1,\dots,f_r),\,J=(f_i(1,x_1,\dots,x_n)\mid i\in\{1,\dots,n\}), $$ then $S(X)\cong\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/J}$.

Remark. Since $S(X)$ is a Noetherian ring of Krull dimension $0$, it is an Artinian ring; in particular, $S(X)$ is a finite-dimensional $\mathbb{K}$-vector space.

Let $h_X$ the Hilbert function of $X$ and let $\chi_I$ the Hilbert polynomial of $I$; because $\dim_{Krull}X=0$ then $\chi_I$ is a constant, by theory: $$ \exists d_0\in\mathbb{N}_{\geq0}\mid\forall d\in\mathbb{N}_{\geq d_0},\,h_X(d)=\dim_{\mathbb{K}}S(X)_d=\chi_I; $$ defined $$ \mathbb{K}[x_1,\dots,x_n]_{\leq d}=\{f\in \mathbb{K}[x_1,\dots,x_n]\mid\deg f\leq d\},\,J_{\leq d}=\{f\in J\mid\deg f\leq d\}, $$ let $B_d$ a base of $J_{\leq d}$ as $\mathbb{K}$-vector space: by remark this is a finite set, then $$ ^hB_d=\left\{x_0^df\left(\frac{x_1}{x_0},\dots,\frac{x_n}{x_0}\right)\in I_d\mid f\in B\right\} $$ is a base of $I_d$ as $\mathbb{K}$-vector space; in particular: $$ \dim_{\mathbb{K}}J_{\leq d}=\dim_{\mathbb{K}}I_d; $$ from all this: $$ \forall d\geq d_0,\,\chi_I=h_X(d)=\dim_{\mathbb{K}}\mathbb{K}[x_1,\dots,x_n]_{\leq d\displaystyle/J_{\leq d}}. $$ Considering the canonical epimorphism $$ \pi:\mathbb{K}[x_1,\dots,x_n]\twoheadrightarrow\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/J}, $$ it induces the inclusion $$ i:\mathbb{K}[x_1,\dots,x_n]_{\leq d\displaystyle/J_{\leq d}}\hookrightarrow\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/J}; $$ let $E$ be a base of $\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/J}$, then $E_{\leq d}=\{f\in E\mid\deg f\leq d\}$ is a system of generators for $\mathbb{K}[x_1,\dots,x_n]_{\leq d\displaystyle/J_{\leq d}}$, that is one can define an epimorphism $$ \epsilon:\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/J}\twoheadrightarrow\mathbb{K}[x_1,\dots,x_n]_{\leq d\displaystyle/J_{\leq d}} $$ where $\ker\epsilon=E_{>d}=\{f\in E\mid\deg f>d\}$.

Because these are finite-dimensional $\mathbb{K}$-vector spaces, they are isomorphic; and moreover: $$ \forall d\geq d_0,\,\chi_I=h_X(d)=\dim_{\mathbb{K}}S(X), $$ in other words, $I$ can be generated by homogeneous polynomials of degree at most $d_0$.

Important remark. Until this point: I have not used the fact that $X$ is a (finite) set of points in general position!

Let $X=Y_1\cup\dots\cup Y_k$ a partion in sets with at most $n$ points, then any $Y_i$ is contained in an hyperplane $H_i$; let $f_i$ be the linear polynomial vanishing on $H_i$, since $F=f_1\cdot\dots\cdot f_k\in I$ and $\deg F=k$ one has $k\geq d_0$; in other words, $I$ can be generated by polynomials of degree at most $k$.

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