Compact subvarieties in $\mathbb{C}^n$

127 Views Asked by At

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.

2

There are 2 best solutions below

6
On

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 $X$.)

0
On

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.