On pg. 13 of Milnor's Singular Points of Complex Hypersurfaces, the author seems to be using the following statement without proof:
Let $V\subseteq \mathbf C^m$ be a complex variety (that is, an irreducible algebraic set), and $f_1, \ldots, f_k\in \mathbf C[z_1,\ldots, z_m]$ be polynomials which generate the prime ideal $I(V)$. Let $h:\mathbf C^m\to \mathbf C$ be a complex analytic function which vanishes on a neighborhood of a point $p\in V$. (Here our neighborhood is NOT taken in the Zariski topology, but in the topology induced on $V$ from the Euclidean topology on $\mathbf C^m$). Then some power $h^s$ of $h$ can be written as $a_1 f_1+ \cdots + a_k f_k$, where $a_1, \ldots, a_k$ are germs of analytic functions (at the point $p$).
Milnor refers to this as the Locally Analytic Nullstellensatz and mentioned a book by Gunning and Rossi (pg. 90) for a proof. Does anybody know a proof of this?
Ask yourself: What does it mean that the polynomials $f_1, \ldots, f_k$ in $\mathbb{C}[z_1, \ldots , z_m]$ generate the prime ideal $I(V)$?
Understand this, and you will understand what you are asking for.
$\text{}$1. Hilbert's Nullstellensatz for affine varieties.
Let $I \subseteq \mathbb{C}[x_1, \ldots, x_n]$ be an ideal then $\sqrt{I} = \mathcal{I}(V(I))$.
Idea of proof:
(a) Reduction to the weak Nullstellensatz:$$I \subsetneqq \mathbb{C}[x_1, \ldots, x_n] \implies V(I) \neq \emptyset.$$We have always $\sqrt{I} \subseteq \mathcal{I}(V(I))$. Let$$f \in \mathcal{I}(V(I)), \quad L = \langle I, yf - 1\rangle_{\mathbb{C}[x_1, \ldots, x_n, y]}.$$Then $V(L) = \emptyset$ since $p = (\overline{p}, \overline{y}) \in V(L)$ implies $g(\overline{p}) = 0$ for all $g \in I$ and $\overline{y}f(\overline{p}) = 1$. This implies $\overline{p} \in V(I)$ and $f(\overline{p}) = 0$, a contradiction. Using the weak Nullstellensatz this implies $L = \langle 1\rangle$, i.e.$$1 = \sum \xi_i g_i + \zeta(yf - 1),\quad I = \langle g_1, \ldots, g_m\rangle_{\mathbb{C}[x_1, \ldots, x_n]}, \quad \xi_i, \zeta \in \mathbb{C}[x_1, \ldots, x_n, y].$$If we put $y = 1/f$ we obtain$$1 = \sum \xi_i\left(x, {1\over f}\right)g_i.$$Clearing denominators gives $f^N \xi_i(x, 1/f)\in \mathbb{C}[x_1, \ldots, x_n]$ for some $N \implies f^N \in I$, i.e. $f \in \sqrt{I}$.
(b) Reduction to the case that $I$ is prime:
(c) We use theorem about Noether normalization and the projection theorem to see that after a generic coordinate change$$\mathbb{C}[x_1, \ldots, x_r] \subset \mathbb{C}[x_1, \ldots, x_n]/I$$is finite and the corresponding map $V(I) \to \mathbb{C}^r$ is surjective. This implies $V(I) \neq \emptyset$ if $r > 0$, i.e. the weak Nullstellensatz. If $r = 0$ then $\mathbb{C}[x_1, \ldots, x_n]/I$ is a finite extension of $\mathbb{C}$, i.e.$$\mathbb{C} = \mathbb{C}[x_1, \ldots, x_n]/I \implies V(I) \neq \emptyset.$$
$\text{}$2. Hilbert's Nullstellensatz for analytic varieties.
(a) and (b) are the same:
(c) We use the theorem about Noether normalization and instead of the projection theorem the local parametrization theorem which gives that the projection $X \to U$ is finite and surjective, $X$, $U$ suitable representatives of $(V(I), 0)$ respectively $(\mathbb{C}^r, 0)$ if (after a linear coordinate change) $$\mathbb{C}\{x_1, \ldots, x_r\} \subset \mathbb{C}\{x_1, \ldots, x_n\}/I$$is a primitive Noether normalization. As above we obtain $(V(I), 0)$ is not empty.