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?


2 Answers 2


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:

Lemma 2.2.12 (De Jong-Pfister, "Local Analytic Geometry"). Suppose the Nullstellensatz holds for all prime ideals $\mathfrak{p} \subset K[x_1, \ldots, x_n]$. Then the Nullstellensatz holds for all ideals $I \subset K[x_1, \ldots, x_n]$.

(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.$$\mathbb{C} = \mathbb{C}[x_1, \ldots, x_n]/I \implies V(I) \neq \emptyset.$$

Theorem 2.2.9 (Noether Normalization, De Jong-Pfister, "Local Analytic Geometry"). Let $K$ be a field with infinitely many elements, $A = K[x_1, \ldots, x_n]/I$ a finitely generated $K$-algebra. Then after a general linear coordinate change, there exists a number $r \le n$, and an inclusion$$K[x_1, \ldots, x_r] \subset A,$$such that if $A$ is a finitely generated $K[x_1, \ldots, x_r]$-module. If moreover $I \neq (0)$, then $r < n$.

If $V = V(I)$, we also say that the projection on the first $r$ coordinates$$\pi: V \to K^r$$is a Noether normalization of $V$.

Theorem 2.2.8 (Projection Theorem, De Jong-Pfister, "Local Analytic Geometry"). Consider a nonzero ideal $I \subset K[x_1, \ldots, x_n]$. Suppose that $I$ contains an element $f_0$ which is regular in $x_n$, that is$$f_0 = x_n^d + a_1x_n^{d - 1} + \ldots + a_0$$for certain elements $a_i \in K[x_1, \ldots, x_{n - 1}]$. Let $p: K^n \to K^{n - 1}$ be the map which sends $(x_1, \ldots, x_n)$ to $(x_1, \ldots, x_{n - 1})$. Define moreover $I' := I \cap K[x_1, \ldots, x_{n - 1}]$, and put $X' = V(I')$. Then$$p(X) = X'.$$

$\text{}$2. Hilbert's Nullstellensatz for analytic varieties.

(a) and (b) are the same:

Lemma 3.4.5 (De Jong-Pfister, "Local Analytic Geometry"). Suppose the Nullstellensatz holds for prime ideals. Then the Nullstellensatz holds for all ideals.

(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.

Theorem 3.3.20 (De Jong-Pfister, "Local Analytic Geometry"). Suppose that $\mathfrak{p} \subset \mathbb{C}\{x_1, \ldots, x_n\}$ is a prime ideal. Then we can get a Noether normalization$$\mathbb{C}\{x_1, \ldots, x_k\} \subset \mathbb{C}\{x_1, \ldots, x_n\}/\mathfrak{p},$$with the property that the class of $x_{k + 1}$ in $Q(\mathbb{C}\{x_1, \ldots, x_n\}/\mathfrak{p})$ is a primitive element of the field extension$$Q(\mathbb{C}\{x_1, \ldots, x_k\}) \subset Q(\mathbb{C}\{x_1, \ldots, x_n\}/\mathfrak{p}).$$We call a Noether normalization with the above property a primitive Noether normalization.

Theorem 3.4.14 (Local Parametrization Theorem, De Jong-Pfister, "Local Analytic Geometry"). Let $(X, 0) = (V(\mathfrak{p}), 0)$ for $\mathfrak{p} \subset \mathbb{C}\{x_1, \ldots, x_n\}$ a prime ideal and $\mathbb{C}\{x_1, \ldots, x_k\} \subset \mathbb{C}\{x_1, \ldots, x_n\}/\mathfrak{p}$ be a primitive Noether normalization (cf. 3.3.20). Let $\pi: \mathbb{C}^n \to \mathbb{C}^{k + 1}$ be the projection on the first $k + 1$ coordinates, $p: \mathbb{C}^{k + 1} \to \mathbb{C}^k$ be the projection on the first $k$ coordinates. Then we have the following statements.

(1) $\mathfrak{p} \cap \mathbb{C}\{x_1, \ldots, x_{k + 1}\} = (P)$, for $P \in \mathbb{C}\{x_1, \ldots, x_k\}[x_{k + 1}]$ a Weierstrass polynomial.

(2) Let $\Delta$ be the discriminant of $P$ with respect to $x_{k + 1}$. Then

(2a) there exist $q_{k + 2}, \ldots, q_n \in \mathbb{C}\{x_1, \ldots, x_k\}[x_{k + 1}]$ such that $Q_j := \Delta x_j - q_j \in \mathfrak{p}$ for $j = k + 2, \ldots, n$;

(2b) there exist $f_1, \ldots, f_s \in \mathbb{C}\{x_1, \ldots, x_k\}[x_{k + 1}, \ldots, x_n]$ such that:

(2bi) $(f_1, \ldots, f_s) = \mathfrak{p}$ in the ring $\mathbb{C}\{x_1, \ldots, x_n\}$,

(2bii) $f_j \in \mathbb{C}\{x_1, \ldots, x_k\}[x_{k + j}]$ are Weierstrass polynomials for $j = 1, \ldots, n - k$ and $P, Q_{k + 2}, \ldots, Q_n \in \{f_1, \ldots, f_s\}$,

(2biii) $(P, Q_{k + 2}, \ldots, Q_n)\mathbb{C}\{x_1, \ldots, x_k\}_\Delta[x_{k + 1}, \ldots, x_n] = (f_1, \ldots, f_s)\mathbb{C}\{x_1, \ldots, x_k\}_\Delta[x_{k + 1}, \ldots, x_n]$.

(3) There is an open neighborhood $U$ of $0$ in $\mathbb{C}^k$, such that $P, \Delta, Q_{k + 2}, \ldots, Q_n, f_1, \ldots, f_s$ converge on $U \times \mathbb{C}^{n - k}$ and for$$\begin{align} X & = \{x \in U \times \mathbb{C}^{n - k} : f_1(x) = \ldots = f_s(x) = 0\} \\ Y & = \{y \in U \times \mathbb{C}^{n - k} : P(y) = Q_{k + 2}(y) = \ldots = Q_n(y) = 0\} \\ X' & = \{x' \in U \times \mathbb{C} : P(x') = 0\} \\ D & = \{x \in U: \Delta(x) = 0\} \end{align} $$the following statements hold:

(3a) $Y \setminus (p \circ \pi)^{-1}(D) = X \setminus (p \circ \pi)^{-1}(D)$;

(3b) $\pi: X \setminus (p \circ \pi)^{-1}(D) \to X' \setminus p^{-1}(D)$ is biholomorphic;

(3c) $X \setminus (p \circ \pi)^{-1}(D)$ is a complex submanifold of $U \times \mathbb{C}^{n - k}$;

(3d) the map $p \circ \pi: X \to U$ is surjective and finite and $(p \circ \pi)^{-1}(0) = \{0\}$.

(4) The map $p \circ \pi: (X, 0) \to (\mathbb{C}^k, 0)$ is finite and surjective.

(5) The map $\pi: (X, 0) \to (V(P), 0)$ is finite and surjective.

  • $\begingroup$ I will read it in some time. Thank you for the detailed answer. Will definitely reply by the end of the week. $\endgroup$ Aug 3, 2017 at 7:29
  • $\begingroup$ I am unable to follow the argument fully because of my lack of understanding of commutative algebra, especially Noether normalization. I am accepting this answer though. For if this weren't correct, nobody would have the imagination to invent it. $\endgroup$ Aug 6, 2017 at 15:18

Another proof of Analytical Hilbert's Nullstellensatz is available freely in J.-P. Demailly - Complex Analytic and Differential Geometry (click), chapter II, section 4.


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