[proofplan] Choose holomorphic representatives of the given germs on a small neighbourhood of the origin and first dispose of the degenerate case in which all ideal generators vanish as germs. In the non-degenerate case, the local analytic Lojasiewicz inequality converts the containment of the common zero germ of the generators in the zero germ of the target function into a quantitative lower bound for the generator tuple. After shrinking to a pseudoconvex ball, a sufficiently large power of the target function satisfies Skoda's $L^2$ division integrability condition, with the weighted quotient interpreted by a measurable representative across the negligible common zero set. Skoda division then writes that power as a holomorphic linear combination of the generators, and the equality of the radical with the vanishing ideal follows by applying this implication in one direction and by evaluating powers pointwise in the other. [/proofplan]

[step:Choose representatives and encode the common zero set]Let $U \subset \mathbb C^n$ be an open neighbourhood of $0$, and choose holomorphic representatives \begin{align*} G_j: U &\to \mathbb C, \qquad 1 \leq j \leq m,\\ F: U &\to \mathbb C \end{align*} of the germs $g_j$ and $f$. Define the holomorphic map \begin{align*} G: U &\to \mathbb C^m\\ z &\mapsto (G_1(z),\dots,G_m(z)). \end{align*} After replacing $U$ by a smaller open neighbourhood of $0$, the hypothesis that $f$ vanishes on $V(\mathfrak a)$ means \begin{align*} F(z)=0 \qquad \text{for every } z \in U \text{ such that } G_1(z)=\cdots=G_m(z)=0. \end{align*} For $z \in U$, write \begin{align*} |G(z)| = \left(\sum_{j=1}^{m} |G_j(z)|^2\right)^{1/2}. \end{align*} Throughout the proof, $\mathcal L^{2n}$ denotes Lebesgue measure on $\mathbb C^n\cong \mathbb R^{2n}$.[/step]

[guided]The objects in the theorem are germs, so the first step is to replace them by honest holomorphic functions on some neighbourhood of $0$. We choose an [open set](/page/Open%20Set) $U \subset \mathbb C^n$ containing $0$ and representatives \begin{align*} G_j: U &\to \mathbb C, \qquad 1 \leq j \leq m,\\ F: U &\to \mathbb C. \end{align*} The generators are grouped into one holomorphic map \begin{align*} G: U &\to \mathbb C^m\\ z &\mapsto (G_1(z),\dots,G_m(z)). \end{align*} The zero germ $V(\mathfrak a)$ is represented, after possibly shrinking $U$, by the common zero set of the functions $G_1,\dots,G_m$. Therefore the hypothesis that $f$ vanishes on $V(\mathfrak a)$ becomes the concrete statement \begin{align*} F(z)=0 \qquad \text{whenever } z \in U \text{ and } G_1(z)=\cdots=G_m(z)=0. \end{align*} We also introduce the Euclidean norm of the generator tuple, \begin{align*} |G(z)| = \left(\sum_{j=1}^{m} |G_j(z)|^2\right)^{1/2}, \end{align*} because both the Lojasiewicz inequality and Skoda division are naturally written in terms of this quantity. Finally, $\mathcal L^{2n}$ denotes Lebesgue measure on $\mathbb C^n\cong \mathbb R^{2n}$; this is the measure used in the later $L^2$ integrability condition.[/guided]

[step:Separate the case where all generators vanish as germs]If each generator $g_j$ is the zero germ, then after shrinking $U$ we have $G_j=0$ on $U$ for every $1\leq j\leq m$. Hence the zero germ $V(\mathfrak a)$ is represented by all of $U$. Since $f$ vanishes on $V(\mathfrak a)$, after shrinking $U$ again if necessary we have $F=0$ on $U$, so $f=0$ in $\mathcal O_{\mathbb C^n,0}$. Therefore $f^1=0\in \mathfrak a$. For the rest of the proof, assume that at least one generator $g_j$ is not the zero germ. After shrinking $U$ if necessary, this means that at least one representative $G_j$ is not identically zero on $U$.[/step]

[guided]Before using estimates involving negative powers of $|G|$, we must exclude the case in which the tuple $G=(G_1,\dots,G_m)$ is identically zero as a germ. If every $g_j$ is the zero germ, then we may shrink $U$ so that \begin{align*} G_j(z)=0 \qquad \text{for every } z\in U \text{ and every } 1\leq j\leq m. \end{align*} The common zero set of the generators is then all of $U$, so the zero germ $V(\mathfrak a)$ is the whole germ at $0$. The hypothesis that $f$ vanishes on $V(\mathfrak a)$ therefore says that, after shrinking $U$ if necessary, \begin{align*} F(z)=0 \qquad \text{for every } z\in U. \end{align*} Thus $f=0$ as a germ. Since $0\in \mathfrak a$, we get $f^1=0\in \mathfrak a$. We may therefore continue under the non-degeneracy assumption that some generator $g_j$ is not the zero germ. Shrinking $U$ if necessary, the corresponding representative $G_j$ is not identically zero on $U$.[/guided]

[step:Apply the local analytic Lojasiewicz inequality]The preceding step gives the zero-set containment $Z(G)\subset Z(F)$ as germs at $0$, where $Z(G)$ denotes the common zero germ of $G_1,\dots,G_m$ and $Z(F)$ denotes the zero germ of $F$. By the local analytic Lojasiewicz inequality for a holomorphic tuple denominator, applied to the holomorphic functions $F,G_1,\dots,G_m$ near $0$, there exist an open neighbourhood $U_0 \subset U$ of $0$, a constant $C>0$, and an integer $r \in \mathbb N$ such that \begin{align*} |F(z)|^r \leq C |G(z)| \qquad \text{for every } z \in U_0. \end{align*}[/step]

[guided]The geometric assumption says that $F$ vanishes wherever all generators vanish. In germ language, this is the containment $Z(G)\subset Z(F)$ near $0$, where $Z(G)$ is the common zero germ of $G_1,\dots,G_m$ and $Z(F)$ is the zero germ of $F$. The local analytic Lojasiewicz inequality for a holomorphic tuple denominator is precisely the analytic tool that turns this zero-set containment into a quantitative estimate. Applied to $F$ and the tuple $G=(G_1,\dots,G_m)$, it gives an open neighbourhood $U_0 \subset U$ of $0$, a constant $C>0$, and an integer $r \in \mathbb N$ such that \begin{align*} |F(z)|^r \leq C |G(z)| \qquad \text{for every } z \in U_0. \end{align*} This is exactly the estimate needed for division: it says that the possible singularity created by negative powers of $|G|$ can be controlled by taking a sufficiently high power of $F$.[/guided]

[step:Choose a pseudoconvex ball and a power satisfying Skoda integrability]Choose $\rho>0$ such that the closed Euclidean ball $\overline{B}(0,\rho)$ is contained in $U_0$, and set \begin{align*} \Omega = B(0,\rho) \subset \mathbb C^n. \end{align*} The function \begin{align*} \varphi: \mathbb C^n &\to \mathbb R\\ z &\mapsto |z|^2 \end{align*} is plurisubharmonic, and $\Omega=\{z\in \mathbb C^n:\varphi(z)<\rho^2\}$, so $\Omega$ is pseudoconvex. Let \begin{align*} q = \min\{n,m-1\} \end{align*} and choose an integer $N \in \mathbb N$ such that \begin{align*} N \geq r(2q+1). \end{align*} Define \begin{align*} H: \Omega &\to \mathbb C\\ z &\mapsto F(z)^N. \end{align*} Using the Lojasiewicz estimate on $\Omega$, if $G(z)\neq 0$, then \begin{align*} |H(z)|^2 |G(z)|^{-2(2q+1)} = |F(z)|^{2N}|G(z)|^{-2(2q+1)} \leq C^{2(2q+1)} |F(z)|^{2N-2r(2q+1)}. \end{align*} The exponent $2N-2r(2q+1)$ is non-negative by the choice of $N$, and $F$ is bounded on $\Omega$ because $F$ is holomorphic on a neighbourhood of $\overline{\Omega}$. Hence there exists a constant $M>0$ such that \begin{align*} |H(z)|^2 |G(z)|^{-2(2q+1)} \leq M \end{align*} for every $z \in \Omega$ with $G(z)\neq 0$. Define the measurable function \begin{align*} W: \Omega &\to [0,\infty)\\ z &\mapsto \begin{cases} |H(z)|^2 |G(z)|^{-2(2q+1)}, & G(z)\neq 0,\\ 0, & G(z)=0. \end{cases} \end{align*} Because at least one $G_j$ is not identically zero, its zero set has empty interior and, by the standard zero-set measure property for nonzero holomorphic functions, has $\mathcal L^{2n}$-measure zero. The common zero set $G^{-1}(0)$ is contained in that zero set, so $G^{-1}(0)$ also has $\mathcal L^{2n}$-measure zero. Thus redefining the quotient on $G^{-1}(0)$ does not change its $L^1$ integrability class. Since $W\leq M$ on $\Omega\setminus G^{-1}(0)$ and $W=0$ on $G^{-1}(0)$, while $\Omega$ has finite $\mathcal L^{2n}$-measure, this gives \begin{align*} \int_{\Omega} W(z)\,d\mathcal L^{2n}(z) < \infty. \end{align*} Equivalently, \begin{align*} \int_{\Omega} |H(z)|^2 |G(z)|^{-2(2q+1)}\,d\mathcal L^{2n}(z) < \infty \end{align*} with the quotient interpreted by this measurable representative.[/step]

[guided]We now prepare the exact hypotheses needed for Skoda division. First choose $\rho>0$ such that $\overline{B}(0,\rho)\subset U_0$, and define \begin{align*} \Omega = B(0,\rho). \end{align*} Skoda division requires a pseudoconvex domain. The ball is pseudoconvex because the map \begin{align*} \varphi: \mathbb C^n &\to \mathbb R\\ z &\mapsto |z|^2 \end{align*} is plurisubharmonic and \begin{align*} \Omega=\{z\in \mathbb C^n:\varphi(z)<\rho^2\}. \end{align*} Let \begin{align*} q=\min\{n,m-1\}. \end{align*} We choose $N \in \mathbb N$ so large that \begin{align*} N \geq r(2q+1), \end{align*} and define the [holomorphic function](/page/Holomorphic%20Function) \begin{align*} H: \Omega &\to \mathbb C\\ z &\mapsto F(z)^N. \end{align*} The reason for this choice is that Skoda's theorem asks for the integrability of $|H|^2$ divided by a power of $|G|$. The Lojasiewicz inequality gives $|G(z)| \geq C^{-1}|F(z)|^r$ whenever $G(z)\neq 0$. Hence \begin{align*} |H(z)|^2 |G(z)|^{-2(2q+1)} = |F(z)|^{2N}|G(z)|^{-2(2q+1)} \leq C^{2(2q+1)} |F(z)|^{2N-2r(2q+1)}. \end{align*} Because $N \geq r(2q+1)$, the exponent on the right-hand side is non-negative. Since $F$ is holomorphic on a neighbourhood of the compact set $\overline{\Omega}$, it is bounded on $\Omega$. Therefore the right-hand side is bounded by some constant $M>0$ whenever $G(z)\neq 0$. There is one remaining point: the expression with $|G(z)|$ in the denominator is not defined on the set where $G(z)=0$. We handle this by defining a measurable representative \begin{align*} W: \Omega &\to [0,\infty)\\ z &\mapsto \begin{cases} |H(z)|^2 |G(z)|^{-2(2q+1)}, & G(z)\neq 0,\\ 0, & G(z)=0. \end{cases} \end{align*} This does not alter the integrability question because the zero set has measure zero. Indeed, by the previous non-degeneracy reduction, some $G_j$ is a nonzero [holomorphic function](/page/Holomorphic%20Function). The zero set of a nonzero [holomorphic function](/page/Holomorphic%20Function) has $\mathcal L^{2n}$-measure zero, and the common zero set $G^{-1}(0)$ is contained in that zero set. Hence $G^{-1}(0)$ has $\mathcal L^{2n}$-measure zero. The function $W$ is bounded by $M$ outside $G^{-1}(0)$ and equals $0$ on $G^{-1}(0)$. Since the ball $\Omega$ has finite $\mathcal L^{2n}$-measure, the boundedness estimate implies \begin{align*} \int_{\Omega} W(z)\,d\mathcal L^{2n}(z) < \infty. \end{align*} Equivalently, \begin{align*} \int_{\Omega} |H(z)|^2 |G(z)|^{-2(2q+1)}\,d\mathcal L^{2n}(z) < \infty \end{align*} with the quotient interpreted by the representative $W$. This is the precise integrability condition that will permit division by the generators $G_1,\dots,G_m$.[/guided]

[step:Use Skoda division to express $f^N$ in the ideal]Apply Skoda's $L^2$ division theorem with parameter $\alpha=2$ to the pseudoconvex domain $\Omega$, the holomorphic tuple $G_1,\dots,G_m$, and the [holomorphic function](/page/Holomorphic%20Function) $H$. The hypotheses are satisfied by the pseudoconvexity of $\Omega$, the holomorphicity of $G_1,\dots,G_m,H$ on $\Omega$, and the integrability estimate \begin{align*} \int_{\Omega} |H(z)|^2 |G(z)|^{-2(2q+1)}\,d\mathcal L^{2n}(z) < \infty. \end{align*} Therefore there exist holomorphic functions \begin{align*} H_j: \Omega &\to \mathbb C, \qquad 1 \leq j \leq m, \end{align*} such that \begin{align*} H(z)=\sum_{j=1}^{m}G_j(z)H_j(z) \qquad \text{for every } z \in \Omega. \end{align*} Since $H=F^N$, passing to germs at $0$ gives \begin{align*} f^N=\sum_{j=1}^{m} g_j h_j \end{align*} in $\mathcal O_{\mathbb C^n,0}$, where $h_j$ is the germ of $H_j$ at $0$. Hence $f^N \in \mathfrak a$.[/step]

[guided]We now invoke Skoda's $L^2$ division theorem. The theorem applies on a pseudoconvex domain to holomorphic functions $G_1,\dots,G_m$ and a [holomorphic function](/page/Holomorphic%20Function) $H$, provided the weighted integral \begin{align*} \int_{\Omega} |H(z)|^2 |G(z)|^{-2(\alpha q+1)}\,d\mathcal L^{2n}(z) \end{align*} is finite for some $\alpha>1$, where $q=\min\{n,m-1\}$. We take $\alpha=2$. The required integral is exactly \begin{align*} \int_{\Omega} |H(z)|^2 |G(z)|^{-2(2q+1)}\,d\mathcal L^{2n}(z), \end{align*} and the previous step proved that it is finite. The domain $\Omega$ is pseudoconvex, and all functions involved are holomorphic on $\Omega$, so the hypotheses of Skoda division are satisfied. Skoda division gives holomorphic functions \begin{align*} H_j: \Omega &\to \mathbb C, \qquad 1 \leq j \leq m, \end{align*} with \begin{align*} H(z)=\sum_{j=1}^{m}G_j(z)H_j(z) \qquad \text{for every } z \in \Omega. \end{align*} Since $H(z)=F(z)^N$, this identity says \begin{align*} F(z)^N=\sum_{j=1}^{m}G_j(z)H_j(z) \qquad \text{for every } z \in \Omega. \end{align*} Taking germs at $0$ converts this analytic identity into the ring identity \begin{align*} f^N=\sum_{j=1}^{m}g_j h_j \end{align*} in $\mathcal O_{\mathbb C^n,0}$, where $h_j$ is the germ represented by $H_j$. Therefore $f^N$ belongs to the ideal $(g_1,\dots,g_m)=\mathfrak a$.[/guided]

[step:Identify the radical with the vanishing ideal]Let the vanishing ideal of the [zero germ](/page/Zero%20Germ) $V(\mathfrak a)$ be \begin{align*} I(V(\mathfrak a)) = \{f \in \mathcal O_{\mathbb C^n,0}: f \text{ vanishes on } V(\mathfrak a)\}. \end{align*} The previous steps prove that $I(V(\mathfrak a))\subset \sqrt{\mathfrak a}$, where $\sqrt{\mathfrak a}$ denotes the [radical](/page/Radical%20of%20an%20Ideal) of $\mathfrak a$. Conversely, let $\phi \in \sqrt{\mathfrak a}$. Then there exists $k\in \mathbb N$ such that $\phi^k\in \mathfrak a$. Choose representatives \begin{align*} \Phi: U &\to \mathbb C,\\ A_j: U &\to \mathbb C, \qquad 1 \leq j \leq m, \end{align*} on a neighbourhood $U$ of $0$ such that \begin{align*} \Phi(z)^k=\sum_{j=1}^{m}G_j(z)A_j(z) \end{align*} near $0$. If $z\in V(\mathfrak a)$, then $G_j(z)=0$ for every $j$, hence \begin{align*} \Phi(z)^k=0. \end{align*} Since $\mathbb C$ has no non-zero nilpotent elements, $\Phi(z)=0$. Thus $\phi$ vanishes on $V(\mathfrak a)$, so $\sqrt{\mathfrak a}\subset I(V(\mathfrak a))$. Combining the two inclusions gives \begin{align*} \sqrt{\mathfrak a} = \{f \in \mathcal O_{\mathbb C^n,0}: f \text{ vanishes on } V(\mathfrak a)\}. \end{align*}[/step]

[guided]Define the vanishing ideal of the [zero germ](/page/Zero%20Germ) by \begin{align*} I(V(\mathfrak a)) = \{f \in \mathcal O_{\mathbb C^n,0}: f \text{ vanishes on } V(\mathfrak a)\}. \end{align*} The argument above proves the first inclusion. Namely, if $f\in I(V(\mathfrak a))$, then $f$ satisfies the hypothesis of the theorem, so there exists $N\in\mathbb N$ such that $f^N\in\mathfrak a$. This is exactly the definition of $f\in\sqrt{\mathfrak a}$, where $\sqrt{\mathfrak a}$ is the [radical](/page/Radical%20of%20an%20Ideal) of $\mathfrak a$, and therefore \begin{align*} I(V(\mathfrak a))\subset \sqrt{\mathfrak a}. \end{align*} For the reverse inclusion, take an arbitrary element $\phi\in\sqrt{\mathfrak a}$. By definition of the radical, there exists $k\in\mathbb N$ such that \begin{align*} \phi^k\in\mathfrak a. \end{align*} This membership means that, after choosing representatives on a sufficiently small neighbourhood $U$ of $0$, there are holomorphic functions \begin{align*} \Phi: U &\to \mathbb C,\\ A_j: U &\to \mathbb C, \qquad 1 \leq j \leq m, \end{align*} where $\Phi$ represents $\phi$, such that \begin{align*} \Phi(z)^k=\sum_{j=1}^{m}G_j(z)A_j(z) \end{align*} for all $z$ in a possibly smaller neighbourhood of $0$. Now let $z$ be a point of the representative of $V(\mathfrak a)$ in this neighbourhood. By definition of the zero germ, each generator vanishes at $z$: \begin{align*} G_j(z)=0 \qquad \text{for every } 1\leq j\leq m. \end{align*} Substituting this into the displayed identity gives \begin{align*} \Phi(z)^k=\sum_{j=1}^{m}0\cdot A_j(z)=0. \end{align*} The field $\mathbb C$ has no non-zero nilpotent elements, so $\Phi(z)=0$. Hence $\phi$ vanishes on $V(\mathfrak a)$, and therefore \begin{align*} \sqrt{\mathfrak a}\subset I(V(\mathfrak a)). \end{align*} Combining the two inclusions yields \begin{align*} \sqrt{\mathfrak a} = \{f \in \mathcal O_{\mathbb C^n,0}: f \text{ vanishes on } V(\mathfrak a)\}. \end{align*}[/guided]