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How does one prove that an Analytic set $V$ in $C^n$ is irreducible if the set of regular points $V^*$ is connected?

Proceeding by contradiction, if we assume that $V$ is in fact reducible and if $V ={V_1} \cup{V_2}$ is the decomposition, then it suffices to show that $V_1\cap V_2 \subset V_s$ where $V_s$ is the set of the singular points in $V$. I am unable to prove this. Any suggestions would be welcome!

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    $\begingroup$ Product rule?... $\endgroup$
    – Thierry Zell
    Commented Jan 1, 2011 at 15:51
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    $\begingroup$ Can you elaborate? $\endgroup$
    – Poincare-Lelong
    Commented Jan 1, 2011 at 19:07
  • $\begingroup$ Dear unknown, your statement (and Griffiths-Harris's) should be made more precise. Indeed, if $V$ is reducible, it can be be decomposed into irreducibles but there might be more than two irreducible components. Actually there might be infinitely many such components.For example, think of a comb i.e. in $\mathbb C^2$ the union of the horizontal $x$-axis and the vertical lines with integral first coordinate (to be continued) $\endgroup$
    – Georges Elencwajg
    Commented Jan 2, 2011 at 1:36
  • $\begingroup$ (continuation) And if you just write $V=V_1\cup V_2$ without bothering whether the $V_i$'s are irreducible, the statement is false: just add a smooth point $s$ of $V$ to each of $V_1$ and $V_2$ and look at $V= W_1 \cup W_2 $ with $W_i=V_i \cup \{s\}$. The point $s$ is in $W_1 \cap W_2 $ and yet is a smooth point of $V$. I have modified your question in my answer below in order to take these remarks into account. $\endgroup$
    – Georges Elencwajg
    Commented Jan 2, 2011 at 2:02

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Dear unknown, here is a sketch of proof of your question ( which I have modified to make it more accurate, as explained in my comments to your original post .)

Statement If $V=V_1 \cup V_2$ with $V_1, V_2$ irreducible and distinct from $V$, then the intersection $V_1 \cap V_2$ consists of singular points of $V$.

Sketch of proof Suppose there is a point $v\in V_1\cap V_2$ which is holomorphically non singular on $V$, i.e. holomorphically smooth. Then the germ of analytic space $V_v$ would have a decomposition $V_v=(V_1)_v \cup (V_2)_v$ . But this is absurd because the germ of an analytic space at a smooth point is irreducible. This boils down to the fact that the local ring of a smooth point of an analytic space is an integral domain, which is clear since it is a a ring of convergent power series $\mathbb C \{z_1,\ldots, z_n\}$.

By the way, judging from your notation, I suppose you extracted this question from Griffiths-Harris. I find their treatment a little cavalier , since indeed they give no explanation at all for their assertion, which is actually not quite correct, as explained in my comments to your question.

If you want full and details, I recommend the brothers Kaup's book Holomorphic Functions of Several Variables (de Gruyter Studies in Mathematics 3), where they prove that a reduced complex space is irreducible iff its smooth points form a connected open subset (49.7 Corollary, page 194).

And, last but not least, happy New Year to you and all our friends of MathOverflow !

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    $\begingroup$ A corollary of this nice answer is the same result in the algebraic case, since by the argument in Shafarevich II.2.2, the local ring of a variety at a simple point embeds as a subring of a power series, hence is also a domain. $\endgroup$
    – roy smith
    Commented Jan 1, 2011 at 21:05

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