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I ran across a statement, the maximum principle, which states $X\subset \mathbb{C}^n$ is compact in the Euclidean topology iff $X$ is a finite set of point.

Unfortunately, a proof didn't come along with it and I'm stuck as where to go.

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2 Answers

Use the fact that on a compact space, a continuous function is bounded.

Apply this to each of the coordinate functions on $\mathbb C^n$ restricted to $X$.

Now apply Liouville (or the maximum modulus principle, if you like), to conclude that each coordinate function on $X$ is locally constant.

(See the comment below for slightly more explanation of how to apply what is a priori a theorem about functions of a complex variable to the coordinate functions on $X$.)

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Why must Liouville hold for $X$? I don't see an obvious homeomorphism between $X$ and even disjoint unions of copies of $\mathbb C^m$, but maybe there's a more general form of the theorem I'm not aware of. –  Alex Becker Jan 24 '13 at 23:59
@Alex: Dear Alex, One possible arguments for this application of Liouville/maximum principle: restrict to the smooth locus, where we do have local coordinates, and then use the maximum principle applied to those local coordinates; we finish by using the fact that the smooth locus is dense in $X$. Another variation on this argument is to use Noether normalization: this will show that $X$ is finite over a copy of $\mathbb C^n$ for some $n$, and compactness then implies that $n = 0$ (but this is slightly further from the spirit of the instruction to use the maximal principle). Regards, –  Matt E Jan 25 '13 at 20:34
Ah, thank you. That clears things up. –  Alex Becker Jan 25 '13 at 20:59
@Alex: Great! By the way, have we ever met? (I just ask because I think we're at the same institution, although in different capacities; and my memory is not what it once was.) Cheers, –  Matt E Jan 25 '13 at 21:02
Yes, I met you over the summer. I was in the REU class you lectured for. We spoke about Euclidean domains iirc and you recommended I read a paper on the ring-theoretic characterization of them. It was very interesting. –  Alex Becker Jan 26 '13 at 0:04
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Suppose $X$ is compact. Then for any $1\le i\le n$, we have some $c\in\mathbb C$ such that $X\cap V(x_i-c)=\emptyset$, as the coordinate function $x_i$ is bounded on $X$. Thus $I(X)+(x_i-c)=\mathbb C[x_1,\ldots,x_n]$, and since $I(X)+(x_i-c)$ has height at most $1$ more than $I(X)$, $I(X)$ must have height $n$. If we let $I(X)=P_1\cap\cdots \cap P_k$ be the primary decomposition of $I(X)$, we get that each $P_1$ is $M_i$-primary where $M_i$ is a maximal ideal. Since $$\begin{align}V(P_1\cap\cdots \cap P_k) &=V(\mathrm{rad}(P_1\cap \cdots\cap P_k))\\ &=V(M_1\cap\cdots\cap M_k)\\ &=V(M_1\cdots M_k)\\ &=V(M_1)\cup \cdots\cup V(M_k)\\ \end{align}$$ by the weak Nullstellensatz we get that $X$ is a finite union of points.

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