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Here is a proposition from Griffith and Harris, Principles of Algebraic Geometry.

Let $V$ be an analytic variety on a complex manifold $M$, let $V^*$ be its points of smooth points. If $V^*$ is connected, then $V$ is irreducible.

The proof in the book is very short. It goes like this:

Suppose $V=V_1\cup V_2$ is a decomposition of $V$. Then $V_1\cap V_2$ is contained in the set of singularities of $V$. Hence $V^*$ can not be connected.

I cannot figure out why $V$ must be singular at the intersection of $V_1$ and $V_2$. This is wrong when the $V_i$'s are allowed to be reducible. For example, $V$ consists of three lines on the plane, $V_1$ is the first two lines and $V_2$ is the last two. But in this case the choice of $V_i$'s has some kind of redundancy.

To make it simple, let's assume that it has precisely two irreducible components and try to derive a contradiction. I am already convinced that, in your notation, $V\cap U$ is irreducible when $V\cap U$ is smooth. What I am afraid of is a picture like this: for example, $V_1$ is the line $y=0$ on the plane and $V_2$ looks like the graph of a bump function. I know this would never happen, because analytic varieties are "rigid". But I don't know how to exclude this possibility.

Can anybody give me a hand? Thanks!

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If $V=V_1 \cup V_2$ is a decomposition, then the local ring $\mathcal O_{p,V}$ at a point $p \in V_1 \cap V_2$ is not an integral domain, hence cannot be a regular local ring, so $p$ is not a smooth point of $V$. –  Parsa Apr 7 '12 at 21:02
    
@Parsa: Why cannot $\mathcal O_{p,V}$ be an integral domain? Why can not $V_1$ and $V_2$ overlap at some smooth point? –  Andrew Apr 8 '12 at 15:41
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Try looking at an example. Let $V_1 = \{ y = 0 \}$ and $V_2 = \{ y = x^2 \}$ and consider the point $p = (0,0)$ where they meet. You might expect the tangent space of $V_1 \cup V_2$ at $p$ to be one-dimensional (the line $y = 0$), but use the definition of tangent space and you'll see that the tangent space becomes two dimensional. Once you understand this example well, you'll see how to generalize it to prove the assertion in Griffiths-Harris. –  Michael Joyce Apr 9 '12 at 18:44

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