Let $k$ be a non-algebraically closed field and $I\subset k[x_1,\dots, x_n]$ be maximal ideal. Is $V_{\bar{k}}(I)$ necessarily finite?

Question

Let $k$ be a non-algebraically closed field and $I\subset k[x_1,\dots, x_n]$ be a maximal ideal.

$\textbf{Q:}$ Is $V_{\bar{k}}(I)=\{x\in\bar{k}^n\vert \forall f\in I, f(x)=0\}$ necessarily finite?

Answer 1

Yes. Let $R=k[x_1,\cdots,x_n]$ and $\overline{R}=\overline{k}[x_1,\cdots,x_n]$. Then $\overline{R}$ is an integral extension of $R$ and both rings are normal domains, and we may apply going up and going down to see that $V_{\overline{k}}(I)$ also has dimension zero, and the dimension zero closed subsets of affine space are precisely finite collections of points.

Alternatively, pick a generating set for $I$ and note that there are finitely many polynomials each of finite degree in this set, so we may adjoin a finite number of algebraic elements over $k$ so that all of our generating polynomials factor completely. So over some finite extension, our ideal is now a finite product of ideals of the form $(x_1-a_1,x_2-a_2,\cdots)$. Clearly the base change of this to the algebraic closure has finitely many points, but it's the same as the base change up to the algebraic closure of our original variety.