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Suppose $A$ is an integral domain and a finite type $\mathbb{C}$-algebra. Let $X := \text{Spec}(A)$ and $K := \text{Frac}(A)$ be the fraction field. Suppose $h \in K$ is a rational function that extends to a global complex analytic function on $X(\mathbb{C}).$ Can we conclude that $h \in A$?

If $A$ is integrally closed then, (it seems to me that) just the fact that $h$ extends continuously to $X(\mathbb{C})$ suffices to conclude that $h \in A$ and if $A$ is not necessarily normal, I realize that a continuous global extension is not sufficient to draw the required conclusion. Hence I'm interested in what happens in the absence of normality/regularity hypotheses and when we additionally require a holomorphic global extension?

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    $\begingroup$ I am guessing that we embed $X$ into $\mathbb{C}^n$ by choosing generators for $A$ and then the definition of "holomorphic"on $X$ is "restriction of a holomorphic function from $\mathbb{C}^n$"? $\endgroup$
    – David E Speyer
    Commented Jun 11, 2019 at 15:38
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    $\begingroup$ Maybe the following works? Let $\phi$ be a holomorphic function on $X^{an}$. The assumption is that for some dense open affine subset $U\subset X$, we have that $\phi|_U = g^{an}/h^{an}$, where $g$ and $h$ are regular functions on $X=Spec A$ (with no common factors say). Note that this implies that $h \phi - g$ is a holomorphic function which is the zero function on $U$. Now you use that $A$ is an integral domain (or $X$ is an integral scheme) to say that $h\phi - g =0$ on $X$. We conclude that $\phi = g/h$ on $X$. This then forces $\phi$ to be regular. $\endgroup$
    – Ariyan Javanpeykar
    Commented Jun 11, 2019 at 15:41
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    $\begingroup$ Side comment: You might be interested in knowing that $A$ is integrally closed in the ring of (global) holomorphic functions $\mathcal{H}(X^{an})$ on $X^{an}$; see for instance Prop. 2.2 in arxiv.org/pdf/1806.09338.pdf $\endgroup$
    – Ariyan Javanpeykar
    Commented Jun 11, 2019 at 15:54
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    $\begingroup$ @DavidESpeyer The definition of holomorphic I had in mind was just that $h$ is a global section of the structure sheaf of the complex analytification $X^\text{an}$.. I think in terms of coordinates this would translate to being a holomorphic function in some open neighbourhood of the analytic subset $X(\mathbb{C})$ in $\mathbb{C}^n,$ but not necessarily a global restriction from $\mathbb{C}^n$..? $\endgroup$
    – Abhishek
    Commented Jun 11, 2019 at 17:29
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    $\begingroup$ @AriyanJavanpeykar On $y^2=x^3$, the function $y/x$ extends continuously to $(0,0)$, but this extension is not holomorphic. $\endgroup$
    – David E Speyer
    Commented Jun 11, 2019 at 23:14

2 Answers 2

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Let $A$ be a noetherian integral domain, $K$ its field of fractions, and $f \in K$. Assume that for each maximal ideal $\frak m$ of $A$ the element $f \in K \subseteq K\otimes_{A}\hat{A}_{\frak m}$ is in $\hat{A}_{\frak m} \subseteq K\otimes_{A}\hat{A}_{\frak m}$ (here $\hat{A}_{\frak m}$ denotes the completion of $A$ at $\frak m$). Then $f \in A$.

Here is the proof. It is enough to show that $f \in A_{\frak m}$ for all maximal ideals $\frak m$; hence we can assume that $A$ is local. Set $\hat K = K \otimes_A \hat A$; then $\hat K$ contains both $K$ and $\hat A$, and the statement is that $A = K \cap\hat A \in \hat K$.

So, $f \in K \cap\hat A$. By the easy part of descent theory, it is enough to show that $f \otimes 1 = 1 \otimes f \in \hat A \otimes_A \hat A$. But $f \otimes 1 = 1 \otimes f \in \hat K \otimes_K \hat K$, because $f \in K$, and $\hat A \otimes_A \hat A$ injects into $K \otimes_A (\hat A \otimes_A \hat A) = \hat K \otimes_K \hat K$.

By the way, I really don't like to interact with anonymous users, so I would appreciate it if you sent me a private email telling me who you are. You can find my email address on my homepage.

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The answer is yes. Ariyan Javanpeykar has contributed the hard part; here are the easy parts.

Let $\tilde{A}$ be the integral closure of $A$ in $\mathrm{Frac}(A)$ and let $\tilde{X} = \mathrm{Spec}(\tilde{A})$. Since the map $\tilde{X} \to X$ is continuous, the pull back of $h$ to $\tilde{X}$ is a continuous function. As the OP notes, this means that $h \in \tilde{A}$. So $h$ is integral over $A$.

On the other hand, Javanpeykar and Kucharczyk show that $A$ is integrally closed in the ring of holomorphic functions on $X$. We assumed that $h$ is a holomorphic function, and we have just shown that $h$ is integral over $A$. So $h \in A$.

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  • $\begingroup$ Perfect, thank you very much! I was about to post this myself with a proof in the integrally closed case.The missing link was the integral closedness result of Javanpeykar and Kucharczyk. I would also be interested in knowing any proof that bypasses the reduction to the normal case. But for now, this is great $\endgroup$
    – Abhishek
    Commented Jun 12, 2019 at 2:21
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    $\begingroup$ One can also prove this with descent theory, without reducing to the normal case. And there is a formal version, that works for any noetherian domain. $\endgroup$
    – Angelo
    Commented Jun 12, 2019 at 3:45
  • $\begingroup$ @Angelo That's a nice suggestion! Is the idea that by descent theory it suffices to prove this for some fppf (fpqc/etale) cover of $X$? Any suggestions for which covers of $X$ I should try to look at, where it would be easier to prove the result? Also, what would the statement in the formal scheme setup be? $\endgroup$
    – Abhishek
    Commented Jun 12, 2019 at 8:31

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