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Jason Starr
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Edit. The books by Grauert and Grauert-Remmert are wonderful sources, with "correct" arguments. Even though it is a sledgehammer, resolution of singularities does quickly establish the result. For every locally finite type $\mathbb{C}$-scheme $(X,\mathcal{O}_X)$, denote by $(i_X,i_X^\#):(X^\text{an},\mathcal{O}_X^{\text{an}})\to (X,\mathcal{O}_X)$ the associated complex analytic space. For every topological space $S$, denote by $C^0_S(\mathbb{C})$ the sheaf of continuous, $\mathbb{C}$-valued functions on $S$.

Lemma 1. For every finite type $\mathbb{C}$-scheme, the associated complex analytic space is a complex manifold if and only if $(X,\mathcal{O}_X)$ is a smooth $\mathbb{C}$-scheme. In particular, if $(X,\mathcal{O}_X)$ is a smooth $\mathbb{C}$-scheme, then we have injectivity of the natural homomorphism of sheaves of $\mathbb{C}$-algebras from $\mathcal{O}_X^{\text{an}}$ to $C^0_{X^\text{an}}(\mathbb{C})$.

Proof. This is local, so we may assume that $X$ is affine. Then the result follows from the Jacobian criterion for smoothness (on the algebraic side) and the complex analytic version of the Implicit Function Theorem (whose hypothesis is the rank condition on the Jacobian as in the Jacobian criterion). For a complex manifold, there are local biholomorphisms with polydisks, whose sheaf of holomorphic functions is manifestly a subsheaf of the sheaf of continuous functions. QED

Lemma 2. For every closed immersion of smooth $\mathbb{C}$-schemes, $(j,j^\#):(Y,\mathcal{O}_Y)\to (X,\mathcal{O}_X)$, with ideal sheaf $\mathcal{I}$, the analytic ideal sheaf $\mathcal{I}\cdot \mathcal{O}_X^{\text{an}}$ equals the subsheaf of $\mathcal{O}_X^{\text{an}}$ of germs of holomorphic functions that vanish identically on the underlying set of $Y^{\text{an}}$.

Proof. Since $\mathcal{O}_X^\text{an}$ is flat as a sheaf of $i_X^{-1}\mathcal{O}_X$-algebras, the following sequence is a short exact $$0\to \mathcal{I}\cdot \mathcal{O}_X^{\text{an}} \to \mathcal{O}_X^{\text{an}} \to j^\text{an}_*\mathcal{O}_Y^{\text{an}}\to 0.$$ By the previous lemma, $\mathcal{O}_Y^{\text{an}}$ is a subsheaf of $C^0_{Y^\text{an}}(\mathbb{C})$.QED

Lemma 3. For every closed immersion $(j,j^\#):(Y,\mathcal{O}_Y)\to (X,\mathcal{O}_X)$ of finite type $\mathbb{C}$-schemes, if $(X,\mathcal{O}_X)$ is smooth and quasi-projective, and if $(Y,\mathcal{O}_Y)$ is reduced, then the conclusion from Lemma 2 holds.

Proof. By Hironaka's Resolution of Singularities Theorem, there exists a projective, birational morphism, $$(\nu,\nu^\#):(\widetilde{X},\mathcal{O}_{\widetilde{X}})\to (X,\mathcal{O}_X),$$ and a smooth closed subscheme, $$(\widetilde{j},\widetilde{j}^\#):(\widetilde{Y},\mathcal{O}_{\widetilde{Y}})\to (\widetilde{X},\mathcal{O}_{\widetilde{X}}),$$ with ideal sheaf $\widetilde{\mathcal{I}}$ such that $\nu_*\widetilde{\mathcal{I}}\cdot \mathcal{O}_X$ equals $\mathcal{I}$.

Associated to the short exact sequence on $\widetilde{X}^{\text{an}}$, $$0 \to \widetilde{I}\cdot \mathcal{O}_{\widetilde{X}}^{\text{an}} \to \mathcal{O}_{\widetilde{X}}^\text{an}\to \widetilde{j}_*^\text{an}\mathcal{O}_{\widetilde{Y}^{\text{an}}} \to 0,$$ there is an exact sequence on $X^{\text{an}}$, $$0 \to \nu^\text{an}_*\widetilde{I}\cdot \mathcal{O}_{\widetilde{X}}^{\text{an}} \xrightarrow{e} \nu^\text{an}_*\mathcal{O}_{\widetilde{X}}^\text{an}\to \nu^\text{an}_*\widetilde{j}_*^\text{an}\mathcal{O}_{\widetilde{Y}^{\text{an}}}.$$ Finally, by Lemma 1, also the natural map from the third term to the pushforward of $C^0_{\widetilde{Y}^\text{an}}(\mathbb{C})$ is injective. Thus, we also have injectivity of the induced homomorphism from the cokernel of $e$ to the pushforward of $C^0_{\widetilde{Y}^{\text{an}}}(\mathbb{C})$.

Since $X$ is already smooth, the natural homomorphism $\nu^\#:\mathcal{O}_X[0]\to R\nu_*\mathcal{O}_{\widetilde{X}}$ is a quasi-isomorphism compatible with arbitrary base change, including base change to Artinian, local $\mathbb{C}$-schemes in $X$. This is enough to conclude that also $\mathcal{O}_{X^{\text{an}}}[0]\to R\nu^{\text{an}}_*\mathcal{O}_{\widetilde{X}^{\text{an}}}$ is a quasi-isomorphism. Thus, the homomorphism of sheaves of $\mathbb{C}$-algebras, $\mathcal{O}_{X^\text{an}} \to \nu_*\mathcal{O}_{\widetilde{X}}^\text{an}$, is an isomorphism. Together with the previous paragraph, it follows that the natural homomorphism from $\mathcal{O}_{X^\text{an}}/\mathcal{I}\cdot \mathcal{O}_{X^\text{an}}$ to the pushforward of $C^0_{\widetilde{Y}^\text{an}}(\mathbb{C})$ is injective. This homomorphism factors through the natural homomorphism, $$\mathcal{O}_{X^\text{an}}/\mathcal{I}\cdot \mathcal{O}_{X^\text{an}} \to j^{\text{an}}_*C^0_{Y^\text{an}}(\mathbb{C}).$$ Thus, also the natural homomorphism is injective. QED

. If a product $g\cdot h$ of holomorphic functions is contained in $I_{\text{hol}}$ then it vanishes on the zero scheme $Z$ of $I$. Since $Z$ is irreducible, the smooth locus $Z_{\text{sm}} = Z\setminus Z_{\text{sing}}$ is a connected, complex manifold. The zero loci of $g$, and $h$ on $Z_{\text{sm}}$ are complex analytic subvarieties of a connected, complex manifold. If neither of these complex analytic subvarieties equals all of $Z_{\text{sm}}$, then they are each nowhere dense. In that case, also the union is nowhere dense, contradicting the hypothesis that $g\cdot h$ vanishes on $Z$. Thus, one of the factors, say $g$, vanishes identically on $Z_{\text{sm}}$. Since $Z_{\text{sm}}$ is dense in $Z$ for the analytic topology, also $g$ vanishes on $Z$.

Now you can use vanishing of cohomology of coherent analytic sheaves on Stein analytic spaces. Fix a generating set $f_1,\dots,f_r$ of the ideal $I$. Define $\mathcal{I}\subset \mathcal{O}^{\text{an}}$ to be the image of the natural homomorphism, $$(\mathcal{O}^{\text{an}})^{\oplus r} \to \mathcal{O}^{\text{an}}, \ \ (u_1,\dots,u_r)\mapsto u_1f_1 + \dots + u_rf_r.$$ Define $\mathcal{K}$ to be the kernel of that map. Then $\mathcal{K}$ and $\mathcal{I}$ are coherent analytic sheaves. Since $H^1(\mathbb{C}^n,\mathcal{K})$ is zero, the induced long exact sequence of cohomology is a short exact sequence, $$0\to H^0(\mathbb{C}^n,\mathcal{K}) \to H^0(\mathbb{C}^n,\mathcal{O}^{\text{hol}})^{\oplus r} \to H^0(\mathbb{C}^n,\mathcal{I}) \to 0.$$ Thus the element $g\in H^0(\mathbb{C}^n,\mathcal{I})$ is in the image of $I\otimes_{\mathbb{C}[z_1,\dots,z_n]} H^0(\mathbb{C}^n,\mathcal{O}^{\text{hol}}).$

Jason Starr
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