# Is an irreducible component of analytic set just a connected component of its regular points?

Say $$Z$$ is an anlatyic set in a complex manifold $$M$$. Is it true that irreducible components of $$Z$$ are just the closures in $$Z$$ of connected components of $$Z\backslash sing(Z)$$?

I need this for a paper I am reading. In general, can someone reccomend a good book about complex analytic geometry? I don't want to waste a lot of time figuring these things out without at least a good source for all these notions.

Thank you! Benny

[1] The answer to your question is yes.There is an important theorem on decomposition analytic subset into irreducible components:(Cf E.M.Chirka's Complex Analytic Sets chapter 5.4.Theorem1)

Let $$A$$ be an analytic subset of a complex manifold $$\Omega$$.Then:

(1)every irreducible component $$A$$ has the form $$\overline S$$,where $$S$$ is a connected component of $$regA=A\backslash singA$$;

(2)if $$reg A=\bigcup_{j\in J}S_j$$ is the decomposition into connected connected components ($$J$$ is finite or countable, $$S_j\bigcap S_k=\varnothing$$ for $$j\not= k$$),then $$A=\bigcup_{j\in J}\overline {S_j}$$,and this is the decomposition of $$A$$ into ieeducible components;

(3) the decomposition of $$A$$ into irreducible components is locally finite,i.e.for every compact $$K\subset\Omega$$ there is only a finite (or empty) set of indices $$j\in J$$ such that $$K\bigcap\overline {S_j}$$ is nonempty.

[2] Demailly's Complex Analytic and Differential Geometry is wonderful.I also recommend E.M.Chirka's Complex Analytic Sets,it's worth reading and full of a lot of classical theorems.

• Thank you for your answer! In the case $Z$ is an algebraic set in $M=\mathbb{C}^{n}$, is it true that the irreducible components in the sense of zariski topology coincide with the irreducible components in the analytic sense? Commented Dec 11, 2019 at 19:34