We know that 5 points in general position determine a conic section uniquely. I wonder if there is a generalization. Consider the equation $F(x_1,\dots, x_n) =0$ where $F$ is a polynomial of degree $d$ (over $R$). Is there a (sharp) bound on the number of points $(x_1,...,x_n)$ (in general position) which uniquely determines $F$ ?

13$\begingroup$ Well, basically the reason why 5 general points determines a conic is that the space of quadratic polynomials in three variables is 6 dimensional (it is spanned by $x^2,y^2,z^2,xy,xz,yz$), and scaling the quadratic does not change the zeroset. More generally the space of degree $d$ polynomials in $\mathbb{P}^n$ has dimension ${n+d \choose d}$ and so ${n+d \choose d}1$ general points determine the hypersurface uniquely. $\endgroup$– J.C. OttemMay 31, 2011 at 23:15
1 Answer
The answer has essentially been given by J. C. Ottem in a comment. I just put it here (with a couple details) so that the question is "answered".
The space of degree $d$ polynomials in n+1 variables has dimension $\binom{n+d}{d}$ (coefficient count), so hypersurfaces of degree d in $\mathbb{P}^n$ are parameterized by a projective space of dimension $N:=\binom{n+d}{d}1$. Asking that the hypersurface goes through a given point is a linear equation on the coefficients of polynomials.
If the base field (or domain) is infinite, then the conditions imposed by general points are independent. You prove this by induction on the number of points: assume that hypersurfaces of degree d in $\mathbb{P}^n$ through $k1 < N1$ general points are parameterized by $\mathbb{P}^{Nk+1}$. EDIT: Choose one of these hypersurfaces $X$; it does not contain all of the points in $\mathbb{P}^n$, as the base field is infinite. So if the $k$th point is out of $X$ (which we can assume as the points are general) the parameter space of hypersurfaces containing all $k$ points is strictly conained in $\mathbb{P}^{Nk+1}$, which means the last condition is linearly independent and defines a hyperplane $\mathbb{P}^{Nk}$.
So $N$ is the number of general points that uniquely determine a hypersurface of degree $d$.

$\begingroup$ @quim Thanks. I still have problem understanding your argument. When you wrote "Since there are infinitely many points in $P^n$, not all of them belong to one of the given family of hypersurfaces", what family do you refer to ? Our setting is that we have $N$ general points given and want to show that the conditions imposed are independent. These points are given, so I am a bit confused with the next sentence "Since there are infinitely many points,.... Taking one....." as well (as we don't have the freedom to take an arbitrary point). $\endgroup$– vanvuJun 1, 2011 at 13:40

$\begingroup$ I edited the answer, I hope it is clearer now. The points are assumed to be in general position, otherwise the result would be false. $\endgroup$– quimJun 1, 2011 at 14:30

$\begingroup$ @quim So all we need is that there is one $X$ in the family that does not contain the $k$th point ? or we need it to hold for all $X$'s ? (the later cannot be true) So would the precise claim be: There is one $X$ going through the first $k1$ points and does not contain the $k$th point ? You seems to suggest any $X$ may work ? $\endgroup$– vanvuJun 1, 2011 at 15:30

$\begingroup$ Exactly, it is enough that one X does not contain the kth point to ensure that the kth condition is relevant. Fix a set of k1 points such that the set of hypersurfaces X of degree n through them is a $\mathbb{P}^{Nk+1}$. Let $Z\subset \mathbb{P}^n$ be the base locus (ie, the intersection of all those hypersurfaces). Then every point q in the complement of $Z$ works as $k$th point, in the sense that the condition "going through q" is independent of the previous conditions (because at least one X misses q). Since Z is a Zariski closed proper subset, you are done. $\endgroup$– quimJun 1, 2011 at 16:35

$\begingroup$ So, once the first k1 points are fixed (and assuming they impose independent conditions) the "general position" hypothesis on the kth can be made precise by requiring that q does not belong to Z. If you want, it is possible from this to obtain an explicit (somewhat ugly) description of the Zariskiclosed proper subset V of $(\mathbb{P}^n)^k$ of kuples of points that impose k nonindependent linear conditions. Then, not being in V is exactly what "in general position" means, for this particular result. $\endgroup$– quimJun 1, 2011 at 16:40