Prove that over an infinite field, a finite set of points in $\mathbb{ A}^n$ can be obtained as vanishing set of n polynomials

Asked Apr 17 '16 at 08:39

Active Apr 17 '16 at 09:18

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While reading Ernst Kunz's commutative algebra book, I came across this problem:

Let $K$ be an infinite field, and $V \in\mathbb{ A}^n (K)$ (the affine n-space) be a finite set of points. Show that the ideal of $V$, i.e. {$f(x_1,\dots,x_n) : f(p) = 0 \forall p\in V$} is generated by exactly n polynomials.

For $|V| = 1$ this is obvious, since the ideal $(x_1 - a_1, ..., x_n - a_n)$ works. The book says to use interpolation as a hint, but I have no clue how to use it (to be honest, being a pure math student I haven't handled interpolation in my life). Thanks in advance.

edited Apr 17 '16 at 09:18

asked Apr 17 '16 at 08:39

Hajime_Saito

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This doesn't give the sharp bound, though. – Captain Lama Apr 17 '16 at 08:47
@DietrichBurde: What I meant to say was that the number of polynomials is the same as the dimension of the affine n-space. I have edited my question a little to make it clearer. – Hajime_Saito Apr 17 '16 at 09:19

Prove that over an infinite field, a finite set of points in $\mathbb{ A}^n$ can be obtained as vanishing set of n polynomials

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