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I need a hint on an exercise:

Let $K$ be an algebraically closed field. Prove that any finite set $\Gamma \subset KP^n$ such that not all of its points lie on the same line can be given by polynomials of degree less than $\operatorname{card}\Gamma$.

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Why exclude the colinear case ? Then polynomials of degree 1 suffice. – user18119 Oct 28 '11 at 11:48
No they don't, you need the degree at least $\operatorname{card}\Gamma$ in that case. – Alexei Averchenko Oct 28 '11 at 12:08
@Qil: consider the case of one point :) – Mariano Suárez-Alvarez Oct 28 '11 at 14:10
Dear QiL, I added a section on non-colinearity in my answer in order to address your comment. – Georges Elencwajg Oct 28 '11 at 14:36
Dear Georges, yes I see where I got confused, thanks. – user18119 Oct 28 '11 at 15:21
up vote 4 down vote accepted

Given a set of $r+1 (r\geq 2)$ points $p_0,p_1,...,p_r \in \mathbb P^n(K) $ , I' ll show by induction on $r$ that if the points are not colinear the set is an intersection of hypersurfaces of degree $r$.

Suppose $r=2$ (Initialization of induction)
Recall our points are not aligned and suppose that $p_0$ is not on the line determined by $p_1,p_2$ .
Let $L_j \; (j=0,1,2)$ be a linear form such that its hyperplane of zeros $V(L_j)$ does not contain $p_j$ but contains the other two $p_i$'s.
Then we can, as required, describe our three-element set as the intersection of three degenerate quadrics (=union of two hyperplanes):

$$ V(L_0.L_1)\cap V(L_1.L_2) \cap V(L_2.L_0)= \lbrace p_0,p_1,p_2 \rbrace $$

Suppose $r\geq 3$.
Choose (by induction hypothesis) a family $(F_i)_{i\in I}$ of homogeneous polynomials of degree $r-1$ whose zero sets $V(F_i)$ intersect in exactly the set $\lbrace p_1,...,p_r\rbrace $.
Also, choose $n$ linear forms $L_1, ...,L_n$ whose zero sets $V(L_j)$ (hyperplanes) intersect exactly in the singleton set $\lbrace p_0 \rbrace$.
Then the homogeneous polynomials $F_i.L_j \; (i\in I,\; j=1,...,r)$ of degree $r$ have zero sets $V(F_i.G_j)$ which exactly define the required initial set: $$\bigcap_{i\in I \; j=1,...,r} V(F_i.Lj)= \lbrace p_0,p_1,...,p_r \rbrace $$

The above works for all fields; algebraic closedness is irrelevant.

Why "not colinear" ?
If our set of $r+1$ points were included in a line, any hypersurface of degree $r$ containing the points would contain the line: this is essentially the elementary fact that a polynomial of degree $r$ with $r+1$ zeros must be the zero polynomial.
But then obviously we can keep intersecting such hypersurfaces till we get blue in the face: they will never cut out our $r+1$ points, but at best the line they lie on.

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Thank you! I have used slightly more explicit construction: I took three non-aligned points on $\mathbb{P}^2$ and picked a basis on $K^{3}$ that corresponds to these points. In such coordinates any homogeneous quadratic polynomial $p(Z) = \sum_{i \leq j} a_{ij} Z_i Z_j$ would vanish on these points iff $a_{00} = a_{11} = a_{22}$, so only $a_{01}$ $a_{12}$ and $a_{20}$ are left to play with, and we need to find polynomial $p$ such that $p$, $(012)p$ and $(012)^2 p$ simultaneously vanish only on these three points. $a_{01} = 1$, $a_{12} = 0$, $a_{20} = 0$ fit the bill nicely. (To be continued). – Alexei Averchenko Oct 30 '11 at 9:31
...Then because $\mathbb{P}^2$ can be cut out of $\mathbb{P}^n$ by homogeneous linear polynomials, we can easily cut out any three non-aligned points on $\mathbb{P}^n$. Then the induction step is the same as in your proof. – Alexei Averchenko Oct 30 '11 at 9:33
Now I have a follow-up question: how do you apply Bezout for the case of aligned points? I only managed to do it by passing to an affine chart, but then one point $x$ on the line $L$ containing the points in question is left out of the chart. That's not required for the proof, but I'm just curious how does one prove that the polynomial that vanishes on $L\setminus\{x\}$ also vanishes on $x$. When $K = \mathbb{C}$ it follows from continuity, but what about the general case when we generally speaking do not have a topology on $\mathbb{P}(K^{n+1})$? – Alexei Averchenko Oct 30 '11 at 9:37
Dear @Alexei, for infinite $K$ it is the fact that if a polynomial has infinitely many zeros in $K$, it must be the zero polynomial (this does not require a topology on $K$). This is false for a finite field $K$ with $q$ elements $a_1,..., a_q$. Namely, the polynomial $(T-a_2).(T-a_3).\ldots. (T-a_q)$ on $L=K$ vanishes on $L\setminus \lbrace a_1 \rbrace$ but not on $a_1$. However my explanation in "Why not colinear ?" holds also for finite fields, because it takes into account the degree of the polynomial studied. – Georges Elencwajg Oct 30 '11 at 10:11
Yes, this is all true, but I asked about the projective line on which the points lie. Actually, I just figured it out: we just need to take two different bases corresponding to two different points we take out :) Thanks again! – Alexei Averchenko Oct 30 '11 at 10:47

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