# Any finite set in $k^n$ is an algebraic set.

I'm trying to show that given a field $k$, and a finite set of points $\{a^i: i = 1\dots n\} \subset k^n$ is an algebraic set or equivalently is the set of common zeros of some set of polynomials $S \subset k[x_1, \dots, x_n]$.

For the case $n = 1$, we have $Z(x_1 - a_1^1, \dots, x_n - a_n^1)$ = $\{a^i\}$. For the case $n = 2$, we have two points $V = \{a^1, a^2 \}$. Let $f_i = (x_i - a_i^1)(x_i - a_i^2)$ , then clearly $V \subset Z(f_i : i = 1\dots n)$. Let $b \in Z(f_i)$. Then now $b$ can be a mixture of $a^1, a^2$, for example $b = (a^1_1, a^2_2, a^1_3, \dots)$. So that construction doesn't work!

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## 1 Answer

Hint:$V(I)\cup V(J)=V(I\cap J)$. Then take the intersection of those maximal ideal.

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