7
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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 Jun 11 at 15:38
  • 1
    $\begingroup$ 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 Jun 11 at 15:41
  • 3
    $\begingroup$ 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 Jun 11 at 15:54
  • 1
    $\begingroup$ @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 Jun 11 at 17:29
  • 2
    $\begingroup$ @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 Jun 11 at 23:14
2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $h \in A$.

$\endgroup$
  • $\begingroup$ 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 Jun 12 at 2:21
  • 2
    $\begingroup$ 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 Jun 12 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 Jun 12 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.