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Let $M$ be a complex manifold (or a complex analityc space) and $Z$ be an analytic subset of $M$. By Noetherianity of the rings of germs of analytic functions at a point we know that $Z$ has finitely many irreducible germs at any $p \in Z$.

What about the "global" irreducible components?

How does one prove that a decomposition into irreducible analytic sets exists? (I have looked into Gunning and Rossi but couldn't find the relevant reference off-hand.)

I also have a related question: is it possible that an analytic set is connected but has infinitely many irreducible components?

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1 Answer 1

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a) Yes, a connected analytic space may have infinitely many irreducible components:

Take a disjoint family of projective lines $(P_i)_{i\in \mathbb N}$ with distinguished points $0_i $ and $1_i$.
Then identify each $1_{i}$ with $0_{i+1}$ .

b) A variant is to take in $\mathbb C^2$ the union of the line $y=0$ and the lines $x=i\; (i\in \mathbb Z)$ . That subspace is connected and has infinitely many irreducible components.

c) The decomposition of an analytic subspace into irreducibles is proved on page 217 of Łojasiewicz's Introduction to Complex Analytic Geometry

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does this complex space properly embed as a closed analytic subspace into a complex space? –  Dima Sustretov Aug 11 '12 at 22:39
    
Yes : my space $X$ in a) is a closed analytic subspace of $X\times \mathbb C$. Anyway, I have added an example embeddable into $\mathbb C^2$. –  Georges Elencwajg Aug 11 '12 at 23:08
    
Dear Georges, thank you! –  Dima Sustretov Aug 11 '12 at 23:27
    
You are welcome, Dima. –  Georges Elencwajg Aug 11 '12 at 23:53

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