[proofplan] We work at a fixed point $p$ and identify the local ring of holomorphic germs with a convergent power series ring by choosing holomorphic coordinates. The inclusion $\sqrt{\mathfrak{a}_p} \subseteq \mathcal{I}(V)_p$ follows directly from the definition of the zero set. For the reverse inclusion, we use primary decomposition in the Noetherian local analytic ring and reduce to the prime case. The prime case is the analytic core: after a finite Weierstrass normalization of the irreducible analytic germ, a holomorphic germ vanishing on the germ must vanish in the corresponding integral domain, hence lies in the prime ideal. [/proofplan]

[step:Reduce the statement to a local ideal in a convergent power series ring]Fix $p \in X$. Choose a holomorphic coordinate chart \begin{align*} \varphi: U &\to \Omega \subseteq \mathbb{C}^n \end{align*} with $p \in U$ and $\varphi(p)=0$. The induced pullback isomorphism identifies the local ring $\mathcal{O}_{X,p}$ with the convergent power series ring \begin{align*} R := \mathbb{C}\{z_1,\ldots,z_n\}. \end{align*} Let \begin{align*} I := \mathfrak{a}_p \subseteq R \end{align*} be the corresponding ideal. Since $\mathfrak{a}$ is coherent, after shrinking $U$ there exist holomorphic functions \begin{align*} g_1,\ldots,g_m: U &\to \mathbb{C} \end{align*} whose germs generate $\mathfrak{a}_q$ for every $q \in U$. Under the chart, the germ of $V(\mathfrak{a})$ at $p$ is the germ at $0$ of \begin{align*} Z(I) := \{z \in \Omega : g_1(z)=\cdots=g_m(z)=0\}. \end{align*} Thus $\mathcal{I}(V)_p$ identifies with \begin{align*} \mathcal{I}(Z(I))_0 := \{h \in R : h \text{ vanishes on } Z(I) \text{ in a neighbourhood of }0\}. \end{align*} It is therefore enough to prove \begin{align*} \mathcal{I}(Z(I))_0 = \sqrt{I} \end{align*} for every ideal $I \subseteq R$.[/step]

[guided]The theorem is local at $p$, so the first task is to replace the manifold by a coordinate model. Choose a holomorphic chart \begin{align*} \varphi: U &\to \Omega \subseteq \mathbb{C}^n \end{align*} with $p \in U$ and $\varphi(p)=0$. Pullback by $\varphi^{-1}$ identifies germs of holomorphic functions at $p$ with germs of holomorphic functions at $0 \in \mathbb{C}^n$, hence with \begin{align*} R := \mathbb{C}\{z_1,\ldots,z_n\}. \end{align*} Let \begin{align*} I := \mathfrak{a}_p \subseteq R. \end{align*} Because $\mathfrak{a}$ is coherent, it is locally finitely generated. After shrinking $U$, choose holomorphic generators \begin{align*} g_1,\ldots,g_m: U &\to \mathbb{C} \end{align*} for $\mathfrak{a}$ on $U$. Then the zero set of $\mathfrak{a}$ near $p$ is exactly the common zero set of these generators: \begin{align*} Z(I) := \{z \in \Omega : g_1(z)=\cdots=g_m(z)=0\}. \end{align*} A germ $h \in R$ belongs to $\mathcal{I}(Z(I))_0$ precisely when some representative of $h$ vanishes on this common zero set in a neighbourhood of $0$. Thus the geometric statement on $X$ becomes the following local analytic algebra statement: \begin{align*} \mathcal{I}(Z(I))_0 = \sqrt{I}. \end{align*} Proving this identity in the ring $R=\mathbb{C}\{z_1,\ldots,z_n\}$ proves the original theorem after transporting the identity back through the coordinate isomorphism.[/guided]

[step:Show that radicals vanish on the corresponding analytic zero set] Let $h \in \sqrt{I}$. By definition of the radical, there is an integer $k \geq 1$ such that \begin{align*} h^k \in I. \end{align*} Since every element of $I$ vanishes on $Z(I)$ near $0$, the representative of $h^k$ vanishes on $Z(I)$ near $0$. Hence for every $z \in Z(I)$ sufficiently close to $0$, \begin{align*} h(z)^k = 0. \end{align*} The field $\mathbb{C}$ has no nonzero nilpotent elements, so $h(z)=0$ for all such $z$. Therefore \begin{align*} h \in \mathcal{I}(Z(I))_0. \end{align*} Thus \begin{align*} \sqrt{I} \subseteq \mathcal{I}(Z(I))_0. \end{align*} [/step]

[step:Reduce the reverse inclusion to prime ideals by primary decomposition]The local analytic ring $R=\mathbb{C}\{z_1,\ldots,z_n\}$ is Noetherian by the Weierstrass Preparation Theorem. Hence the ideal $I$ has a primary decomposition \begin{align*} I = Q_1 \cap \cdots \cap Q_r \end{align*} with each $Q_i \subseteq R$ primary. Define \begin{align*} P_i := \sqrt{Q_i} \end{align*} for $1 \leq i \leq r$. Each $P_i$ is a prime ideal, and the radical of an intersection is the intersection of the radicals: \begin{align*} \sqrt{I} = \sqrt{Q_1 \cap \cdots \cap Q_r} = P_1 \cap \cdots \cap P_r. \end{align*} Since $I \subseteq P_i$ for every $i$, we have \begin{align*} Z(I) = Z(P_1) \cup \cdots \cup Z(P_r) \end{align*} as germs at $0$. Therefore a germ $h \in R$ vanishes on $Z(I)$ near $0$ if and only if it vanishes on every $Z(P_i)$ near $0$, so \begin{align*} \mathcal{I}(Z(I))_0 = \mathcal{I}(Z(P_1))_0 \cap \cdots \cap \mathcal{I}(Z(P_r))_0. \end{align*} Thus it remains to prove the prime ideal statement \begin{align*} \mathcal{I}(Z(P))_0 = P \end{align*} for every prime ideal $P \subseteq R$.[/step]

[guided]The ring $R=\mathbb{C}\{z_1,\ldots,z_n\}$ is Noetherian. This is one of the fundamental consequences of the Weierstrass Preparation Theorem: Weierstrass division gives the induction step proving that convergent power series rings are Noetherian. Because $R$ is Noetherian, the ideal $I$ admits a primary decomposition: \begin{align*} I = Q_1 \cap \cdots \cap Q_r, \end{align*} where each $Q_i$ is a primary ideal. For each component define \begin{align*} P_i := \sqrt{Q_i}. \end{align*} The radical of a primary ideal is prime, so each $P_i$ is prime. Also, \begin{align*} \sqrt{I} = \sqrt{Q_1 \cap \cdots \cap Q_r} = \sqrt{Q_1} \cap \cdots \cap \sqrt{Q_r} = P_1 \cap \cdots \cap P_r. \end{align*} Now compare zero sets. Since $I \subseteq Q_i \subseteq P_i$, every point annihilating $P_i$ also annihilates $I$, so $Z(P_i) \subseteq Z(I)$. Conversely, if $z$ is a common zero of all elements of $I$, then every element of $Q_1 \cap \cdots \cap Q_r$ vanishes at $z$. The primary decomposition says that the analytic germ defined by $I$ has irreducible components defined by the prime ideals $P_i$, so as germs \begin{align*} Z(I) = Z(P_1) \cup \cdots \cup Z(P_r). \end{align*} Therefore vanishing on $Z(I)$ is equivalent to vanishing on each component $Z(P_i)$: \begin{align*} \mathcal{I}(Z(I))_0 = \mathcal{I}(Z(P_1))_0 \cap \cdots \cap \mathcal{I}(Z(P_r))_0. \end{align*} This reduces the theorem to the case in which the ideal is prime. Once we know \begin{align*} \mathcal{I}(Z(P_i))_0 = P_i \end{align*} for every $i$, the displayed identities immediately give \begin{align*} \mathcal{I}(Z(I))_0 = P_1 \cap \cdots \cap P_r = \sqrt{I}. \end{align*}[/guided]

[step:Prove that a prime analytic ideal is the full ideal of its zero germ][claim:Prime analytic ideal theorem] Let $P \subseteq R=\mathbb{C}\{z_1,\ldots,z_n\}$ be a prime ideal. Then \begin{align*} \mathcal{I}(Z(P))_0 = P. \end{align*} [/claim] [proof] The inclusion $P \subseteq \mathcal{I}(Z(P))_0$ follows from the definition of $Z(P)$. For the reverse inclusion, set \begin{align*} A := R/P. \end{align*} Since $P$ is prime, $A$ is an integral domain. Let \begin{align*} d := \dim A. \end{align*} By the analytic Noether normalization theorem, after a complex linear change of coordinates there is a finite injective homomorphism \begin{align*} \iota: S := \mathbb{C}\{w_1,\ldots,w_d\} &\hookrightarrow A \end{align*} such that $A$ is a finite $S$-module. Geometrically, this corresponds to a finite holomorphic map of germs \begin{align*} \pi: Z(P) &\to (\mathbb{C}^d,0) \end{align*} induced by the coordinate projection \begin{align*} (z_1,\ldots,z_n) &\mapsto (w_1,\ldots,w_d). \end{align*} Let $h \in \mathcal{I}(Z(P))_0$. Let \begin{align*} \overline{h} \in A \end{align*} be the residue class of $h$. Since $A$ is finite over $S$, the element $\overline{h}$ is integral over $S$. Hence there exist an integer $N \geq 1$ and elements \begin{align*} b_1,\ldots,b_N \in S \end{align*} such that \begin{align*} \overline{h}^{\,N} + b_1\overline{h}^{\,N-1} + \cdots + b_{N-1}\overline{h} + b_N =0 \end{align*} in $A$. Choose representatives of the elements $b_j$ as convergent power series on a neighbourhood $B \subseteq \mathbb{C}^d$ of $0$. The displayed integral equation means that for every point $x \in Z(P)$ sufficiently close to $0$, \begin{align*} h(x)^N + b_1(\pi(x))h(x)^{N-1} + \cdots + b_{N-1}(\pi(x))h(x) + b_N(\pi(x)) =0. \end{align*} Since $h$ vanishes on $Z(P)$ near $0$, this gives \begin{align*} b_N(\pi(x))=0 \end{align*} for every $x \in Z(P)$ near $0$. The finite map $\pi$ has image a neighbourhood of $0$ in the irreducible germ $(\mathbb{C}^d,0)$. Thus $b_N$ vanishes on a neighbourhood of $0$ in $\mathbb{C}^d$. By the Identity Theorem for Holomorphic Functions, \begin{align*} b_N=0 \end{align*} in $S$. With $b_N=0$, the integral equation becomes \begin{align*} \overline{h}\left( \overline{h}^{\,N-1} + b_1\overline{h}^{\,N-2} + \cdots + b_{N-1} \right)=0 \end{align*} in $A$. Since $A$ is an integral domain, either $\overline{h}=0$ or the second factor is $0$. If $\overline{h}\neq 0$, then the second factor gives an integral equation for $\overline{h}$ of smaller degree. Repeating this degree reduction finitely many times forces $\overline{h}=0$. Therefore $h \in P$. Hence \begin{align*} \mathcal{I}(Z(P))_0 \subseteq P, \end{align*} and the reverse inclusion was already proved. This proves the claim. [/proof][/step]

[guided]First, $P \subseteq \mathcal{I}(Z(P))_0$ is part of the definition: every element of $P$ vanishes on the common zero set of all elements of $P$. For the reverse inclusion, introduce the quotient analytic algebra \begin{align*} A := R/P. \end{align*} Because $P$ is prime, $A$ is an integral domain. Let \begin{align*} d := \dim A. \end{align*} The analytic Noether normalization theorem, proved using the Weierstrass Preparation Theorem, says that after a complex linear change of coordinates there is a finite injective homomorphism \begin{align*} \iota: S := \mathbb{C}\{w_1,\ldots,w_d\} &\hookrightarrow A \end{align*} and that $A$ is a finite $S$-module. The geometric meaning is that the projection onto the $w$-coordinates gives a finite holomorphic map of germs \begin{align*} \pi: Z(P) &\to (\mathbb{C}^d,0). \end{align*} This is the point at which Weierstrass preparation enters: the remaining coordinates satisfy monic Weierstrass equations over the base variables, so the projection has finite fibres and the quotient algebra is integral over $S$. Let $h \in \mathcal{I}(Z(P))_0$. Denote its image in the quotient by \begin{align*} \overline{h} \in A. \end{align*} Since $A$ is finite over $S$, every element of $A$ is integral over $S$. Therefore there exist an integer $N \geq 1$ and coefficients \begin{align*} b_1,\ldots,b_N \in S \end{align*} such that \begin{align*} \overline{h}^{\,N} + b_1\overline{h}^{\,N-1} + \cdots + b_{N-1}\overline{h} + b_N =0 \end{align*} in $A$. This equation is an identity on the analytic germ $Z(P)$. Evaluating it at a point $x \in Z(P)$ sufficiently close to $0$ gives \begin{align*} h(x)^N + b_1(\pi(x))h(x)^{N-1} + \cdots + b_{N-1}(\pi(x))h(x) + b_N(\pi(x)) =0. \end{align*} The hypothesis $h \in \mathcal{I}(Z(P))_0$ means exactly that $h(x)=0$ for all such $x$. Substituting this into the identity leaves \begin{align*} b_N(\pi(x))=0 \end{align*} for every $x \in Z(P)$ near $0$. Because $\pi$ is finite and the source has dimension $d$, the image of $\pi$ is a neighbourhood of $0$ in the base germ $(\mathbb{C}^d,0)$. Hence $b_N$ vanishes on a neighbourhood of $0$ in $\mathbb{C}^d$. By the Identity Theorem for Holomorphic Functions, \begin{align*} b_N=0 \end{align*} as an element of $S$. The integral equation now factors in $A$: \begin{align*} \overline{h}\left( \overline{h}^{\,N-1} + b_1\overline{h}^{\,N-2} + \cdots + b_{N-1} \right)=0. \end{align*} Since $A$ is an integral domain, a product in $A$ is zero only if one of the two factors is zero. If $\overline{h}=0$, then $h \in P$ and we are done. If $\overline{h}\neq 0$, then the second factor must be zero, which gives a monic integral equation for $\overline{h}$ of degree $N-1$. Repeating the same argument finitely many times reduces the degree until the only possibility is \begin{align*} \overline{h}=0. \end{align*} Therefore $h \in P$. Thus every holomorphic germ vanishing on the irreducible analytic germ $Z(P)$ belongs to $P$, so \begin{align*} \mathcal{I}(Z(P))_0 = P. \end{align*}[/guided]

[step:Intersect the prime components to obtain the radical] By the prime analytic ideal theorem applied to each prime ideal $P_i$, we have \begin{align*} \mathcal{I}(Z(P_i))_0 = P_i \end{align*} for every $1 \leq i \leq r$. Therefore \begin{align*} \mathcal{I}(Z(I))_0 &= \mathcal{I}(Z(P_1))_0 \cap \cdots \cap \mathcal{I}(Z(P_r))_0 \\ &= P_1 \cap \cdots \cap P_r \\ &= \sqrt{I}. \end{align*} Together with the already proved inclusion $\sqrt{I}\subseteq \mathcal{I}(Z(I))_0$, this proves \begin{align*} \mathcal{I}(Z(I))_0 = \sqrt{I}. \end{align*} Transporting the identity back through the coordinate isomorphism $\mathcal{O}_{X,p}\cong R$ gives \begin{align*} \mathcal{I}(V)_p = \sqrt{\mathfrak{a}_p}. \end{align*} Since $p \in X$ was arbitrary, the theorem follows. [/step]