Algebraic subset of $\mathbb C^n$ is finite if it is countable

Question

Let $X\subset \mathbb C^n$ be Zariski closed and countable. Why is $X$ finite?

So far I noticed that it is true for $n=1$, because if $X$ is algebraic, then it is the zero set of a polynomial and thus finite.
Also, the claim is not true for the euclidean topology, because lattices are closed.

Maybe a compactness argument works?

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This is a follow up to my question, where I asked about a more general setting, but actually had this in mind.
There, it was pointed out, that the assertion is true and a strategy was proposed in the comments. However, I do not understand it.
I think its best to ask this as a separate question.

Answer 1

I'll expand on my comments. Let $ V=V(f_1, \dots, f_n)$ be your Zariski closed subset, where $f_i \in \mathbb{C}[x_1,\dots,x_n]$. Suppose V has dimension $m > 0$. By a geometric version of Noether's Normalization lemma (you'll find it in Ideals, Varieties, and Algorithms) there exists a surjective map $\pi: V \to \mathbb{C}^m$. So a positive dimensional $V$ cannot be countable. Now if $V$ is zero dimensional, this means the dimension of the complex vector space $\mathbb{C}[x_1,\dots,x_n]/(f_1,\dots, f_n)$ is finite. So any $1,x_i, x_i^2,\dots $ must have linear relationship between them, so $(f_1, \dots, f_n) \cap \mathbb{C}[x_i] \neq {0}$ for all $i$. This implies $V$ must be finite.

Answer 2

This follows from Noether normalization, which says (among other things) that if $V$ is an irreducible affine variety of dimension $d$, then there is a surjective morphism $V\to\mathbb{A}^d$. So if $X$ is a variety of positive dimension over $\mathbb{C}$, this immediately implies it is uncountable. If $X$ has dimension $0$, then each irreducible component of $X$ is just a point and so $X$ must be finite.