I am reading Gathmann's free online notes on Algebraic Geometry. One exercise asks to show that
"Every affine variety in $\mathbb A^n$ consisting of finitely many points can be written as the zero locus of $n$ polynomials".
There is a hint says "interpolation". I don't know how to start with the hint.
If $n=2$, we can use interpolation to get 1 polynomial for finitely many points. But we need to show 2 polynomials instead. I am also not sure how to apply interpolation for higher dimensions. Anyone can help? Thank you!
Best Answer
Assume that the points are $a_k=(a_k^{1},a_k^2,...,a_k^n)$, for $k=1,2,...,M$.
We can use the following system
$$\begin{cases}0&=\prod_{k=1}^{M}(z_1-a_k^1)\\ 0&=\prod_{k=1}^{M}(z_2-a_k^2)+\\&+\sum_{j=1}^{M}\left[\frac{(-1)^j\prod_{k=1,k\neq j}^{M}(z_1-a_k^1)}{\prod_{k=1,k\neq j}^{M}(a_j-a_k)}\cdot(z_2-a_j^2)\cdot\left\{\prod_{k=1,k\neq j}^{M}(z_2-a_k^2)+1\right\}\right]\\ ...\\ 0&=\prod_{k=1}^{M}(z_n-a_k^n)+\\&+\sum_{j=1}^{M}\left[\frac{(-1)^j\prod_{k=1,k\neq j}^{M}(z_1-a_k^1)}{\prod_{k=1,k\neq j}^{M}(a_j-a_k)}\cdot(z_n-a_j^n)\cdot\left\{\prod_{k=1,k\neq j}^{M}(z_n-a_k^n)+1\right\}\right]\end{cases}$$
The first polynomial forces the possible values for $z_1$ as $a_1^1,a_2^1,...,a_N^1$. The role of the other polynomials is to force the values of the other variables according to the value of $z_1$.
The equations are symmetric by permutations on the index $k$. Assume without loss of generality that $z_1$ is, say $=a_1^1$. Then the $r$-th equation, for $r=2,3,...,n$, becomes
$$\begin{align}0&=\prod_{k=1}^{M}(z_r-a_k^r)-(z_r-a_1^r)\cdot\left\{\prod_{k=1,k\neq j}^{M}(z_r-a_k^r)+1\right\}\\&=(z_r-a_1^r)\end{align}$$
from where $z_r$ is forced to be $=a_1^r$.