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?



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


