# Rational functions on reduced complex varieties that extend to global holomorphic functions

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?

• 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 $\mathbb{C}^n$"? Jun 11 '19 at 15:38
• 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. Jun 11 '19 at 15:41
• 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 Jun 11 '19 at 15:54
• @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 $\mathbb{C}^n$..? Jun 11 '19 at 17:29
• @AriyanJavanpeykar On $y^2=x^3$, the function $y/x$ extends continuously to $(0,0)$, but this extension is not holomorphic. Jun 11 '19 at 23:14

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.

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 $$h \in A$$.

• 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 Jun 12 '19 at 2:21
• 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. Jun 12 '19 at 3:45
• @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? Jun 12 '19 at 8:31