Minimum degree of polyonial (several variables) such that it vanishes in a prescribed number of points?

Question

I saw recently the following claim. Given any 19 points in $\mathbb{R}^3$ it is always possible to find a polynomial $p(x,y,z)$ with $\deg p \leq 3$ such that it vanishes in the previous points.

My question is why is this true (for the previous particular case) and where can I find a general version of the statement (given any $m$ points in $\mathbb{R}^n$ which is the least positive integer $d$ such that I can always find a polynomial $p$ that vanishes at the points with degree $\leq d$?).

Thank you in advance!

Answer 1

I am not sure, but this is just a guess.

Given $n$ points in $\Bbb R$, you can always find a polynomial (in one variable) of degree $n$ that vanishes at the given points. This follows from the Fundamental Theorem of Algebra.

With a polynomial of degree $n$, there are $n+1$ coefficients, and the "solution polynomial" is only defined up to a constant multiple.

The general form of a three variable polynomial is

$$a_1x^3+a_2y^3+a^3z^3+a_4x^2y+a_5x^2z+a_6y^2x+a_7y^2z+a_8z^2x+a_9z^2y+a_{10}xyz+a_{11}x^2+a_{12}y^2+a_{13}z^2+a_{14}xy+a_{15}yz+a_{16}zx+a_{17}x+a_{18}y+a_{19}z+a_{20}$$

So I would guess that, since there are $20$ coefficients, you can always make it vanish at any $19$ points.

Answer 2

I will just close the problem. Thank you for the previous answer, which gave me the idea! This is just a rewriting in a formal way.

With our previous discussion we have already seen that the vector space $V$ of polynomials in three variables with degree lower or equal than $3$ has dimension $20$.

Clearly if we look for the polynomials $p$ such that $p(P_1)=0, \dots, p(P_{19})=0$ we will have $19$ linear equations involving the coefficients $a_1,\dots,a_{20}$. So, we know that at least we have a $1$-dimensional subspace in $V$ (or maybe bigger) where each polynomial of the subspace vanishes at the prescribed points $P_1,\dots,P_{19}$.

Thank you all guys!