# Every affine variety in $\mathbb A^n$ consisting of finitely many points can be written as the zero locus of $n$ polynomials

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!

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 $=a_1^r$.

• I believe there is some typo in the denominators, there should be $a_{j}^{1} - a_{k}^{1}$. However, how do you ensure that it is non-zero? I guess it can happen that two distinct points have the same first coordinate.... – Jan Vysoky Sep 11 '17 at 11:36
• @JanVysoky, you are right. Some $\textbf{lazy}$ people do not accept edits to this post :( Moreover, choosing another $z_1$ does not change the sign, so the multiplication by $(-1)^j$ is wrong. The last equality is also incorrect. It must be $0 = -(z_r - a_1^r)$. However, the idea is nice. 1) When $K$ is algebraically closed, then we can "rotate" the axes, so that the first coordinates of all points are different. – user128245 Mar 18 '18 at 16:59
• 2) If $K$ is not algebraically closed, we need only one polynomial. For any point $a_i$ there is a polynomial $f_i \in K[x_1, \dots , x_n]$ with exactly one root, namely $f_i(a_i) = 0$. Then $f = f_1 f_2 \cdots f_M$ has $\{ a_1, \dots, a_M \}$ as only solution. – user128245 Mar 18 '18 at 17:00

We use induction on $$n$$, the base case $$n=1$$ being trivial (note that the result is actually false for $$n=0$$; alternatively you can require the varieties to be nonempty and take $$n=0$$ as the base case).

Now suppose the result is known for $$n$$ and let $$V\subseteq \mathbb{A}^{n+1}$$ be finite. Let $$a_1,\dots,a_m$$ be all the different first coordinates of points of $$V$$ and let $$V_i=\{b\in \mathbb{A}^n:(a_i,b)\in V\}$$. By the induction hypothesis, for each $$i$$ we can choose $$n$$ polynomials $$f_{i1},\dots,f_{in}$$ whose vanishing set is $$V_i$$. For $$1\leq k\leq n$$, we can then choose a polynomial $$g_k$$ in $$n+1$$ variables such that $$g_k(a_i,y)=f_{ik}(y)$$ for each $$i$$ (here $$y$$ is an $$n$$-tuple of variables). Explicitly, if $$e_i(x)$$ is a polynomial in one variable that is $$1$$ on $$a_i$$ and $$0$$ on $$a_j$$ for $$j\neq i$$, then you can take $$g_k(x,y)=\sum_i e_i(x)f_{ik}(y)$$. Finally, we see that $$V$$ is the vanishing set of the polynomials $$g_1(x,y),\dots,g_n(x,y)$$ together with one more $$(x-a_1)\dots(x-a_m)$$.

• There's a typo: $1 \leq k \leq m$ should be $1 \leq k \leq n$. – Stephen Aug 13 at 16:00