a finite variety and its "vanishing ideal"

Question

Here is a problem from Ernst Kunz's Introduction to Commutative Algebra and Algebraic Geometry:

Let $K$ be an infinite field, $V \subset \mathbb{A}^n(K)$ a finite set of points. It's ideal $\mathfrak{J}(V)$ in $K[X_1, ..., X_n]$ is generated by $n$ polynomials. Hint: Interpolation.

note: $\mathfrak{J}(V)$ consists of all $F \in K[X_1, ..., X_n]$ with $F(x) = 0$ for all $x\in V$

A similar question can be found here. The solutions given above just answers why any finite variety can be writen as the zero locus of n polynomials. But further questions: do the n polynomials constructed in the link generate the vanishing ideal? And that's my question really concerned. Is the condition "the field K infinite" necessary? How do we use it? I've tried some simple cases: $n = 1$ and $V$ is a single point. For other cases, I don't have any accessible solutions.

Anyone can help? Prefect the solution in the link or give an alternative approach to the problem. Thank you!

Answer 1

The inductive proof I gave at the linked question works to get functions which generate $\mathfrak{I}(V)$ (and it is not necessary for $K$ to be infinite). Specifically, letting $a_1,\dots,a_m$ be all the different first coordinates of points of $V$, note that $$K[x,y_1,\dots,y_n]/((x-a_1)\dots(x-a_m))\cong \prod_{i=1}^m K[y_1,\dots,y_n]$$ by the Chinese remainder theorem (with the map being given by evaluation at the $a_i$) and so $$K[x,y_1,\dots,y_n]/((x-a_1)\dots(x-a_m), g_1(x,y),\dots,g_n(x,y)) \cong\prod_{i=1}^m K[y_1,\dots,y_n]/(g_1(a_i,y),\dots,g_n(a_i,y)).$$ So if, by the induction hypothesis, you choose $g_1,\dots,g_n$ such that the polynomials $g_k(a_i,y)=f_{ik}(y)$ for $1\leq k\leq n$ generate the ideal $\mathfrak{I}(V_i)$ for $V_i=\{b\in\mathbb{A}^n(K):(a,b)\in V\}$ for each $i$ from $1$ to $n$, then $$K[x,y_1,\dots,y_n]/((x-a_1)\dots(x-a_m), g_1(x,y),\dots,g_n(x,y))$$ will end up isomorphic to a product of copies of $K$, one for each point of $V$ (with the isomorphism given by evaluation at those points). This shows that $$((x-a_1)\dots(x-a_m), g_1(x,y),\dots,g_n(x,y))=\mathfrak{I}(V),$$ since each side is the kernel of the evaluation map $K[x,y_1,\dots,y_n]\to K^V$.

Answer 2

For $a_1,\ldots,a_M \in \overline{K}^N$ our ideal is $$J = I(\{a_1,\ldots,a_m\}) = \bigcap_m I(\{a_m\})$$

$K[X_1,\ldots,X_N]/J$ is a product of fields because it is finite dimensional $K$-algebra and it contains no nilpotent element : if $f$ doesn't vanish on all the $a_m$ then so does $f^r$.

Let $J_n = J \cap K[X_1,\ldots,X_n]$. Then $K[X_1,\ldots,X_{n}]/J_n$ is a product of fields because it is a $K$-subalgebra of $K[X_1,\ldots,X_N]/J$.

For each maximal ideal $\mathfrak{M}_{n,l}$ of $K[X_1,\ldots,X_{n}]/J_n$ then $(K[X_1,\ldots,X_{n}]/J_n / \mathfrak{M}_{n,l})[X_{n+1}]$ is a one-variable polynomial ring over a field so in that ring $J_{n+1}$ is principal generated by $f_{n+1,l}(X_1,\ldots,X_{n+1})$,

and since the $\mathfrak{M}_{n,l}$ are comaximal there is some polynomial $F_{n+1}(X_1,\ldots,X_{n+1})$ which is $\equiv f_{n+1,l}(X_1,\ldots,X_{n+1}) \bmod \mathfrak{M}_{n,l}$ for each $l$ so that $J_{n+1} = (J_n,F_{n+1}(X_1,\ldots,X_{n+1}))$ and $$J = (F_1(X_1),\ldots,F_N(X_1,\ldots,X_N))$$