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The question was already answered here, but I do not have much knowledge of Algebra, so I am trying to check my more analytical proof.

Arguing for irreducible affine varieties: since $X$ is affine over $\mathbb{C}$, we have an inclusion $i\colon X \rightarrow \mathbb{A}_{\mathbb{C}}^n$ and projections $\pi_k \colon \mathbb{A}_{\mathbb{C}}^n \rightarrow \mathbb{A}_{\mathbb{C}}^1$ for any $k=1,\dots,n$. The compositions are holomorphic maps. Then by the open mapping theorem in several variables we have that $(\pi_k \circ i)(X)$ is either a point or open. However, in the second case we would have that $(\pi_k \circ i)(X)$ is closed and open in $\mathbb{C}$, so it is $\mathbb{C}$, which is not compact. This is a contradiction, since holomorphic maps are continuous and thus send compact spaces in compact spaces, and $X$ is compact. Therefore $\pi_k \circ i$ is constant for every $k$. So $X$ is a point.

I followed a hint given in the notes of Algebraic Geometry by Andreas Gathmann, but I am not sure if the Open mapping theorem in several variables is the most elegant way to solve this - please, comment if the proof does not sound correct. On the other hand, Noether normalization theorem looks quite abstract to me. Does anyone know a simpler solution to this problem?

Thanks.

Gibbs
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    If your variety is smooth, your proof works fine. But in the presence of a singular set $\Sigma\subset X$ there is no notion of holomorphicity on all of $X$. You may still apply your idea to show non-compactness of $X\backslash \Sigma$, but that is clear anyways. – Jan Bohr Dec 11 '17 at 17:15
  • Yes, it is smooth. Thanks for the observation. – Gibbs Dec 11 '17 at 19:39

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Your proof is correct. Here is another idea : take an hyperplane $H$ such that $H \cap X$ is not empty. By induction, it is enough to shows that curves are never compact. But it's easy to see that an analytic curve $C \subset \Bbb C^2$ is never bounded, so we are done.

Nicolas Hemelsoet
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