If $K$ is finite, then every subset of $\mathbb A^n(K)$ is algebraic

Question

I'm trying to prove that if $K$ is a finite field, then every subset of $\mathbb A^n(K)$ is algebraic. I know that if $K$ is finite, then every element of $K$ is algebraic, i.e., for every $a\in K$ there is a polynomial $f\in K$ such that $f(a)=0$, but this didn't help me to solve the question. I almost sure that we have to use this to solve this question.

I need help.

Thanks in advance.

By the way, any field (even infinite fields) is algebraic over itself :) — M Turgeon, Aug 13 '13 at 20:33

score 6 · Accepted Answer · answered Aug 13 '13 at 20:26

6

A finite union of algebraic sets is an algebraic set.

answered Aug 13 '13 at 20:26

M Turgeon

10,419

since $K$ is finite, then every subset of $\mathbb A^n$ is finite, so algebraic, am I right? – user42912 Aug 13 '13 at 20:32
1

Exactly. In general, any finite subset of affine space is algebraic. – M Turgeon Aug 13 '13 at 20:33

score 4 · Answer 2 · answered Aug 13 '13 at 20:34

Here is a another approach, I'll give you a hint. Show that every map $K^n\to K$ is a polynomial. Begin by showing that you can write the indicator function at any point as a polynomial (remember you can find a polynomial that has, as roots, any subset of $K$). Then, just add up appropriate multiples of these indicator functions.

If $K$ is finite, then every subset of $\mathbb A^n(K)$ is algebraic

2 Answers2