# Dimension of irreducible analytic varieties

The dimension of an irreducible analytic variety is defined to be the dimension of its regular locus as a complex manifold. Suppose we have an irreducible analytic subvariety $$V$$ of $$\mathbb{C}^n$$ of dimension $$k$$. My question is whether the following is true: For any $$p\in V$$, there is a neighborhood $$U$$ of $$p$$ in $$\mathbb{C}^n$$ such that $$V\cap U$$ can be given as the zero locus of $$n-k$$ holomorphic functions on $$U$$.

The case $$k=1$$ is claimed to be true in Principles of Algebraic Geometry by Griffiths & Harris (first sentence on p.130), but I don't see why. How do I prove this? Is this true for general $$d$$? What is a good reference that explains these matters? Thanks in advance!

True for $$k=1$$, false for $$k > 1$$. There exist say codimension 2 subvarieties that require 3 functions to define. The standard example is the subvariety given by $$\text{rank} \begin{bmatrix} z_1 & z_2 & z_3 \\ z_4 & z_5 & z_6 \end{bmatrix} < 2.$$ You need all three subdeterminants to define this codimension 2 (dimension 4) subvariety of $$\mathbb{C}^6$$ (not totally trivial to prove).
For $$k=1$$ what you do is to follow the construction of the Weierstrass polynomial defining the subvariety.