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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).

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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 $0$.

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