# Zero set of holomorphic functions with full rank Jacobian is an irreducible analytic germ

Let $$f_1, \cdots, f_m \in \mathcal{O}_{\mathbb{C}^n, 0}$$ be a family of holomorphic germs on $$\mathbb{C}^n$$ at $$0$$, with $$m \leq n$$. Suppose $$f_1(0) = \cdots = f_m(0) =0,$$ and that the Jacobi matrix $$J(f_1, \cdots, f_m) := \left[ \frac{\partial f_k}{\partial z_\ell} \right]_{1 \leq k \leq m, 1 \leq \ell \leq n}$$ has rank equal to $$m$$ at the point $$z = 0$$. I would like to show that the "zero set germ" $$Z(f_1, \cdots, f_m)$$ of $$f_1, \cdots, f_m$$ is an irreducible analytic germ at $$0 \in \mathbb{C}^n$$ (here, I use the same definitions of analytic sets as in the book by Griffiths and Harris).

I am not exactly sure how to proceed with this exercise. Unpacking the definitions of analytic germs, I believe that to show that $$Z(f_1, \cdots, f_m)$$ is irreducible at $$0$$ would mean to show that, in a neighbourhood of $$0$$, we cannot write $$Z(f_1, \cdots, f_m) = Z(g_1, \cdots, g_k) \cup Z(h_1, \cdots, h_\ell)$$ for holomorphic functions $$g_1, \cdots, g_k, h_1, \cdots, h_\ell$$ defined near zero. However, I am not sure how to proceed further, nor how to use that the Jacobi matrix of $$(f_1, \cdots, f_m)$$ at $$z = 0$$ has full rank.

Alternatively, we could use a different characterization of irreducibility. For an analytic set/germ $$X \subset \mathbb{C}^n$$, we can define $$I(X) := \{f \in \mathcal{O}_{\mathbb{C}^n, 0} \ \mid \ X \subset Z(f) \}$$ which is an ideal in $$\mathcal{O}_{\mathbb{C}^n, 0}$$. With these definitions, an analytic subset $$X$$ is irreducible if and only if $$I(X)$$ is a prime ideal. Hence, for my question, I would need to show that $$I(Z(f_1, \cdots, f_m))$$ is a prime ideal. However, I am not sure how to proceed here, as I do not see a nice description of the ideal $$I(Z(f_1, \cdots, f_m))$$ (I believe we can only say that $$(f_1, \cdots, f_n) \subset I(Z(f_1, \cdots, f_m))$$, but I am not sure about the equality of these two ideals).

Suppose the Jacobian matrix is full rank around $$0 \in \mathbb{C}^n$$. Then, the submersion theorem implies that the zero set $$Z$$ of $$f_1, \dots, f_r$$ defines a complex manifold in a neighborhood of $$0$$. In particular, given a small disk $$D$$ around $$0$$ in $$\mathbb{C}^n$$, the set $$D \cap Z$$ will be biholomorphic to the unit disk $$\mathbb{D}^{n - k}$$ around $$0$$ in $$\mathbb{C}^{n - k}$$. (cf. Griffiths and Harris pg. 19, 20)
Now, we claim that $$\mathbb{D}^r - V$$ is path connected for any analytic subset $$V \subset \mathbb{D}^r$$. This can be done inductively.
The case $$r = 1$$ is clear. For $$r > 1$$, let $$x,y \in \mathbb{D}^r - V$$, and choose a hyperplane $$H$$ which contains $$x$$ and $$y$$ but not all of $$V$$. (If such an $$H$$ doesn't exist, then $$V \subset \overline{xy}$$, and we're back to the base case.) Granting the existence of such an $$H$$, we see that $$H \cap \mathbb{D}^r \cong \mathbb{D}^{r - 1}$$ and $$V \cap H \subset \mathbb{D}^{r - 1}$$ is an analytic subset, so the induction hypothesis applies.
Now, suppose $$Z$$ the union of distinct analytic subsets $$Z_1 \cup Z_2$$, where neiter $$Z_1$$ or $$Z_2$$ are contained in each other, and $$0 \in Z_2 \cap Z_2$$. Then $$Z_1 \cap Z_2$$ is an analytic subset which disconnects $$Z \cap D$$ for any disk around $$0$$.
Hence, if the Jacobain matrix of $$f_1,...f_r$$ is full rank at $$0$$, the zero locus must be irreducible near $$0$$.