Does $k[X]$ always have a countable basis? ($X$ is an affine variety)

Question

Let $X$ be an affine variety over $k$ an algebraically closed field. I was wondering does $k[X]$ always have a countable basis? I think it does... but I was wondering how can one show this? Thank you.

Answer 1

Yes. If $X$ is a closed subvariety of $\mathbb{A}^n$, then $k[X]$ is a quotient of $k[x_1,\dots,x_n]$. The ring $k[x_1,\dots,x_n]$ has the monomials in $x_1,\dots,x_n$ as a basis, and there are only countably many of them. So $k[X]$ is a quotient of a countable-dimensional vector space and hence also countable-dimensional.