7
$\begingroup$

Let $X$ be a normal connected complex analytic space, $x\in X$ a point, $f$ a nonzero holomorphic function vanishing at $x$. Denote by $U\subseteq X$ the locus where $f$ is nonzero. Suppose that $\pi:U'\to U$ is a finite degree covering space. This gives $U'$ the structure of an analytic space. Let $A$ be the ring of pairs $(V, h)$ where $V$ is a neighborhood of $x$ in $X$ and $f$ is a bounded holomorphic function on $\pi^{-1}(V\cap U)$, where for $V'\subseteq V$ we identify $(V, h)$ with $(V', h|_{V'})$. This is a subring of the stalk of $j_* \pi_* \mathcal{O}_{U'}$ at $x$, where $j:U\to X$ is the inclusion. It contains $\mathcal{O}_{X, x}$ (the ring of germs of holomorphic functions at $x$).

Question. Is $A$ a finitely generated $\mathcal{O}_{X, x}$-module?

In other words, can we "normalize $X$ inside $U'$", that is, extend $U'\to U$ to a finite map $X'\to X$?

$\endgroup$
5
+200
$\begingroup$

See Theorem 3.4 in Dethloff and Grauert's "Seminormal Complex Spaces" (observe that in your setting $A$ is empty, so the theorem becomes rather easier). This seems to be exactly what you're looking for, if I understand your question correctly. I believe that this is a modern version of very classical theorems of Grauert, Remmert, and Stein.

For convenience I include the statment of the theorem, which is actually slightly stronger than what is required. Dethloff and Grauert prove:

Theorem 3.4. Let $N$ be a normal complex space, and let $B\subset N$ be a nowhere dense analytic subset. Let $\pi: Y\to (N\setminus B)$ be an analytically branched covering with a critical locus $A \subset (N\setminus B)$. Assume that $A\cup B\subset N$ is analytic. Then $\pi: Y\to (N\setminus B)$ can be extended uniquely to an analytically branched covering $\pi: X\to N$, which is uniquely determined up to equivalence of analytically branched coverings.

Note that Theorem 1.3 and Theorem 3.3 and the remarks following it in the same notes shows that in this case $X$ must be normal, so this does indeed answer the question in its entirety.

$\endgroup$

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.