Suppose that $I \subset \mathbb C[z\_1,\dots, z\_n]$ is a prime ideal. Consider the ideal $I\_{hol}$ in the ring of holomorphic functions $f: \mathbb C^n\to \mathbb C$ generated by polynomials from $I$.


Is $I\_{hol}$ prime?



Do prime ideals in polynomial ring generate prime ideals in the ring of holomorphic functions?
**Edit. I added some lemmas to address the issue raised by David Speyer and the OP.** The books by Grauert and Grauert-Remmert are wonderful sources. The proofs in those books are the "correct" arguments, using "sledgehammers" as little as possible. Even though it is a sledgehammer, Hironaka's Resolution of Singularities Theorem does *quickly* establish the result. Also, I will also use the smaller sledgehammer ("club hammer"?) of GAGA. 


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 $S$. For every complex analytic space $(S,\mathcal{O}\_S)$, denote by $u\_S:\mathcal{O}\_S\to C^0\_S(\mathbb{C})$ the natural homomorphism of sheaves of $\mathbb{C}$-algebras. 


**Question 0.** For a finite type $\mathbb{C}$-scheme $(X,\mathcal{O}\_X)$, what is the kernel of $u\_{X^{\text{an}}}:\mathcal{O}\_X^{\text{an}}\to C^0\_{X^{\text{an}}}(\mathbb{C})$?


For the nilradical $\mathcal{N}\subset \mathcal{O}\_X$, certainly $\mathcal{N}\cdot \mathcal{O}\_X^{\text{an}}$ is contained in the kernel. Most of the comments for the question and for this answer focus on the problem of proving that the kernel equals $\mathcal{N}\cdot \mathcal{O}\_X^{\text{an}},$ which it does. One approach uses "analytic reducedness" of the local rings $\mathcal{O}\_{X,p}/\mathcal{N}\_p$ at $\mathbb{C}$-points $p\in X^{\text{an}}$. Not all reduced local Noetherian rings are analytically reduced. However, **excellent local rings** that are reduced are analytically reduced, cf. the Wikipedia page for excellent local rings.


<https://en.wikipedia.org/wiki/Excellent_ring#Properties>


The approach here to the kernel of $u\_{X^{\text{an}}}$ is different, using Hironaka's Resolution of Singularities.


**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 the homomorphism $u\_{X^\text{an}}$ is injective. 


**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 $\mathcal{I}$. By Serre's GAGA, also $\nu^{\text{an}}\_\*\widetilde{\mathcal{I}}^{\text{an}}$ equals $(\nu\_\*\widetilde{\mathcal{I}})^{\text{an}}$, so that the same result holds for the associated analytic spaces.


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. In particular, $\mathcal{O}\_X\to \nu\_\*\mathcal{O}\_{\widetilde{X}}$ is an isomorphism. By GAGA, also $\mathcal{O}\_X^{\text{an}} \to \nu\_\*^{\text{an}}\mathcal{O}^{\text{an}}\_{\widetilde{X}}$ 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 this second natural homomorphism is injective. **QED**


**Lemma 4.** For every finite type $\mathbb{C}$-scheme $(Y,\mathcal{O}\_Y)$, the scheme is reduced if and only if the homomorphism $u\_{Y^{\text{an}}}$ is injective. In particular, for the nilradical $\mathcal{N}\subset \mathcal{O}\_Y$, the nilradical of $\mathcal{O}\_{Y}^\text{an}$ equals $\mathcal{N}\cdot \mathcal{O}\_Y^\text{an}$. 


**Proof.** Since $\mathcal{O}\_Y^\text{an}$ is flat over $i\_Y^{-1}\mathcal{O}\_Y$, the nilradical of $\mathcal{O}\_Y^\text{an}$ contains $\mathcal{N}\cdot \mathcal{O}\_Y^\text{an}$. Thus, if $(Y,\mathcal{O}\_Y)$ is nonreduced, then $u\_{Y^\text{an}}$ is not injective. 


Conversely, assume that $(Y,\mathcal{O}\_Y)$ is reduced. To prove that $u\_{Y^{\text{an}}}$ is injective and that $\mathcal{N}\cdot \mathcal{O}\_Y^{\text{an}}$ equals the entire nilradical, it suffices to work locally. Locally there are closed immersions of $(Y,\mathcal{O}\_Y)$ into affine space. Thus, the result follows from the previous lemma. **QED**


**Lemma 5.** For every finite type, affine $\mathbb{C}$-scheme $(X,\mathcal{O}\_X)$, for every surjection of coherent $\mathcal{O}\_X$-sheaves, $\phi:\mathcal{F}\to \mathcal{G}$, the induced map $\phi^{\text{an}}(X^{\text{an}}):\mathcal{F}^{\text{an}}(X^{\text{an}})\to \mathcal{G}^{\text{an}}(X^{\text{an}})$ is surjective.


**Proof.** Probably there is a more direct proof, but this also follows from vanishing of the higher sheaf cohomology of coherent analytic sheaves on a Stein complex analytic space. Since $\mathcal{O}\_{X^\text{an}}$ is flat over $i\_X^{-1}\mathcal{O}\_X$, the kernel of $\phi^{\text{an}}$ equals the coherent analytic sheaf associated to the kernel of $\phi$. Since $X$ is affine, the complex analytic space $X^{\text{an}}$ is Stein. By vanishing of the higher cohomology of coherent analytic sheaves on a Stein complex analytic space, the first sheaf cohomology of $\text{Ker}(\phi^{\text{an}})$ vanishes. Surjectivity of $\phi^{\text{an}}(X^{\text{an}})$ follows by the long exact sequence of sheaf cohomology. **QED**


**Proposition.** For a finite type $\mathbb{C}$-scheme $(X,\mathcal{O}\_X)$ that is smooth and affine, for every closed subscheme $(Z,\mathcal{O}\_Z)$ that is integral and with ideal sheaf $\mathcal{I}$, also the ideal $I^\text{hol}:=\mathcal{I}(X)\cdot \mathcal{O}\_X^\text{an}(X^{\text{an}})$ is a prime ideal in $\mathcal{O}\_X^\text{an}(X^\text{an})$.


**Proof.** If a product $g\cdot h$ of holomorphic functions is contained in $I^\text{hol}$ then it vanishes on $Z^\text{an}$, since each element of $I(X)$ vanishes on $Z^\text{an}$. Since $Z$ is irreducible, the smooth locus $Z\_{\text{sm}} = Z\setminus Z\_{\text{sing}}$ is a dense open subscheme that is an integral, smooth $\mathbb{C}$-scheme. The zero loci of $g$, and $h$ on $Z^\text{an}\_{\text{sm}}$ are complex analytic subvarieties of a connected, complex manifold. If neither of these complex analytic subvarieties equals all of $Z^\text{an}\_{\text{sm}}$, then they are each nowhere dense. In that case, also the union is nowhere dense, contradicting the hypothesis that $g\cdot h$ vanishes on $Z^\text{an}$. Thus, one of the factors, say $g$, vanishes identically on $Z^\text{an}\_{\text{sm}}$. Since $Z^\text{an}\_{\text{sm}}$ is dense in $Z^\text{an}$ for the analytic topology, also $g$ vanishes on $Z^\text{an}$. By Lemma 3, $g$ is a section of $(\mathcal{I}\cdot \mathcal{O}\_X^\text{an})(X^\text{an})$.


Fix a finite generating set $f\_1,\dots,f\_r$ of the ideal $I = \mathcal{I}(X)$. By Lemma 5, the induced map, $$(\mathcal{O}\_X^{\text{an}})(X^\text{an})^{\oplus r} \to (\mathcal{I}\cdot\mathcal{O}\_X^{\text{an}})(X^\text{an}), \ \ (u\_1,\dots,u\_r)\mapsto u\_1f\_1 + \dots + u\_rf\_r,$$ is surjective. By construction, the image is in $I^{\text{hol}}$. Therefore $I^{\text{hol}}$ equals all of $(\mathcal{I}\cdot\mathcal{O}\_X^{\text{an}})(X^\text{an})$. Thus, $g$ is an element of $I^{\text{hol}}$. **QED** 


