If $V$ is an analytic variety $V=V_1\cup V_2$, then $V_1\cap V_2\subset V_{sing}$ Let $V$ be an irreducible analytic variety, and $V_1, V_2$ analytic subvarieties such that $$V=V_1\cup V_2.$$
In Griffiths-Harris book, it is mentioned that  $V_1 \cap V_2$ is a subset of the singular locus of $V$. Can you help me understand why is this true?
Thank you!
 A: If $p$ in the intersection is a smooth point then locally its a manifold. Restricting $V_1$ and $V_2$ on that manifold, 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 $(f_1,...,f_n)$ and $(g_1,...,g_k)$ respectively. Then the union at the origin is given by the zeros of $(f_ig_j)$. 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.
A: If $V$ is a union of two distinct non-empty closed subvarieties, then $V$ is not irreducible.  Ignoring that, consider the variety $y^2=x^2$ made up of the two lines $y=x$ and $y=-x$ in the plane.  They intersect at the origin $(0,0)$, and at this point the tangent space is 2-dimensional while the lines (the irreducible components) are 1-dimensional.  This means $(0,0)$ is a singular point.
What happens with the variety $y^2 = x^4$?  In general, if two varieties meet at a point the dimension of the tangent space at that point will be larger than the dimensions of the individual varieties.
