A complex affinie variety of dimension $\ge 1$ is never compact in the classical (Euclidean) topology

Question

A complex affine variety $X$ of dimension at least $1$ is never compact in the classical topology.

This question is exercise 2.36b, on page 20, in Gathmann's notes has already been asked before (here and here), but the methods used there are not really in line with the rest of Gathmann's text. I am looking for an answer, that is coherent with the rest of text.

My idea so far is the following: We note that:

Lemma 1: Let $f$ be a complex polynomial of degree $n \ge 2$. Then $\lvert V(f) \rvert = \infty$.

Proof: Fix arbitrary $a_2,\ldots,a_n \in \mathbb{C}$. By the fundamental theorem of algebra we know that $f$ splits into linear factors, and by this decomposition we see that $f(x_1,a_2,\ldots,a_n)$ has as a root.

On the other hand we note that

Lemma 2: Let $f$ be a complex polynomial of degree $n = 1$. Then $\lvert V(f) \rvert = 1$.

From the proof Lemma 1 we can further see that

Lemma 3: Let $f$ be a complex polynomial of degree $n \ge 2$. Then $V(f)$ is unbounded.

However, I fail to see how to use that $X$ has dimension $\ge 1$ to apply these observations. (The abstract definition via ascending chains is a bit impractical.) Could you please give me a hint?

Edit 1: I am looking for a proof that is less heavy in terms of the machinery used than the ones in the questions linked above.

Edit 2: In an old version of Gathmann's notes (p.49)the exercise is stated a bit differently:

I understand that the Zariski topology on $\mathbb{A}^1$ equals the cofinite topology. But I do not see how to go on from here.

Does this answer your question? Why are affine varieties except points not compact in the standard topology on $C^n$ ? — Moishe Kohan, Jun 01 '23 at 13:10
Please provide more detail about the type of proof you're looking for - are there specific tools you're trying to use or avoid? Without this information, your post is likely to be closed as a duplicate of one of the posts you linked. — KReiser, Jun 01 '23 at 14:44
"less heavy in terms of the machinery" is pretty vague - are there specific results you're trying to use or avoid? — KReiser, Jun 01 '23 at 15:27
@MarianoSuárez-Álvarez I think you may have a gap there (or at least you need to say something different than "intersection") - any nondegenerate space curve has nontrivial intersection with a hyperplane, but the intersection with any hyperplane is a finite set of points. — KReiser, Jun 01 '23 at 15:36
@KReiser: I want to avoid the Noether normalisation and, if possible, use some basic argumentation with arguments like the ones I wrote above. — 3nondatur, Jun 01 '23 at 20:19

score 1 · Answer 1 · answered Jun 01 '23 at 13:04

I don't know if the lemmas you give are usefull for this problem (unless $X$ is a hypersurface, in which case, your problem is a direct conequence of lemma 3). However, if you want a proof, you know by Noether normalisation that there exists a finite injective morphism $\iota : \mathbb{C}[X_1,\ldots,X_d] \hookrightarrow \Gamma(X,\mathbb{A}^1)$. Let $f : X \rightarrow \mathbb{A}^d$ be the morphism such that $\iota = f^*$.

$\iota$ is injective thus $f$ is dominant. It implies that $f(X)$ has a non-empty Zariski interior, which implies that it is dense for the Euclidean topology. In particular, $X$ can't be compact, or else $f(X)$ would be, and it wouldn't be dense.

This is also basically the solution from the questions that the OP links. — KReiser, Jun 01 '23 at 15:34

A complex affinie variety of dimension $\ge 1$ is never compact in the classical (Euclidean) topology

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