# Singular points of an analytic variety

I am struggling with a question that was posed here before, about why a reducible analytic variety $V=V_1\cup V_2$ must be singular in $V_1\cap V_2$.

I must say I didn't really figure out the suggestions there, but I thought of a different reasoning: basically, a point in $V$ is non-singular if it has a neighborhood homeomorphic to $\mathbb{C}^n$. Now if the intersection is a point $V_1\cap V_2 = \{p\}$, and $p$ is non-singular, then a neighborhood $U$ must have a homeomorphism $\phi(U)=\mathbb{C}^n$. In this case $\phi(V_1-p)\cup\phi(V_2)$ is a disjoint union covering $\mathbb{C}^n$, and so $\mathbb{C}^n$ is not connected.

Does this proof make sense? And can it be generalized for larger $V_1\cap V_2$? (I'm wondering if the intersection must always have a point with a neighborhood that shares points with both sets).

1. $$M_1\cap M_2$$ is an analytic set of $$D$$ and therefore it is an analytic set of $$M_1$$. Since $$M_1\supset U\cap M_1=U\cap M_2\subset M_2,$$ $$U\cap M_1\subset M_1\cap M_2$$. So, the analytic set $$M_1\cap M_2$$ contains a nonempty open subset in $$M_1$$, by connectedness, $$M_1=M_1\cap M_2$$. Do the same to $$M_2$$.

2. If $$P$$ is not a singular point of $$A$$, there is a connected neighborhood $$U$$ of $$P$$ such that $$A\cap U$$ is complex submanifold of $$U$$. Choosing our coordinate system the question reduced to a local case:

The union of two proper varieties of $$\mathbb{C}^n$$ can not contain a neighborhood of origin.

Suppose that the two varieties at origin are given by zeros of $$(f_1,\ldots,f_r)$$ and $$(g_1,...,g_k)$$ respectively. Then the union at the origin is given by the zeros of $$(f_i g_j)$$. Therefore, by the identical principle, the zeros of $$f_ig_j$$ contains a neighborhood of origin if and only if $$f_i$$ is always zero or $$g_j$$ is always zero. Since the two varieties are proper subsets, this completes the proof.

I am also stuck with this question and I am not sure if this will help. Referring to a book "Holomorphic functions and integral Representations in several complex variables" by Michael Range, the two following exercise problems (unfortunately!) can help you answer your question:

[Page 31: E.2.13] Let $$M_{1}$$ and $$M_{2}$$ be closed connected complex manifolds of the region $$D\subseteq \mathbb{C}^{n}$$. If there exists $$U$$ a neighbourhood of $$P\in M_{1}\cap M_{2}$$ with $$U\cap M_{1}=U\cap M_{2}$$, then $$M_{1}=M_{2}$$.

[Page 40: E.3.8] Let $$A_{1}$$ and $$A_{2}$$ be analytic sets, $$P\in A_{1}\cap A_{2}$$. If for each $$U$$ neighbourhood of $$P$$, $$U\cap A_{1}\neq U\cap A_{2}$$, then $$P$$ is a singular point of $$A=A_1\cup A_2$$.

Remark: I think in E.3.8, it is implicitly implied that $$A_{1}\neq A_{2}$$.

Therefore, if $$V_{1}\cap V_{2}$$ is not contained in $$V_{s}$$, then we may find $$z\in V_{1}\cap V_{2}$$ which is regular. Then by E.3.8, there exists a neighbourhood $$U$$ of $$z$$ such that $$U\cap V_{1}=U\cap V_{2}$$. By $$E.2.13$$, therefore $$V_{1}=V_{2}=V$$ which is not possible.

So we are left with finishing two questions. Hope someone can help!

N.B. This is the best I can come up with. It might be wrong also, so exercise caution!