1
$\begingroup$

[I am not a native English speaker, so my sentences may sound strange. ]

I'm studying about complex analytic spaces. For meromorphic functions, I don't know how to define their principal divisors because I don't know the definition of the discrete valuation on analytic space.

Let $X$ be a normal complex analytic space. It seems that a prime divisor $D$ is defined as an irreducible codim 1 closed subspace of $X$. If it were an algebraic variety, we could define the valuation of meromorphic function $f$ on $D$ because its structure ring $\mathcal{O}_D =\mathcal{O}_X/\mathcal{I}$ is 1-dimensional regular Noetherian local ring.

On complex analytic space, $\mathcal{O}_D $ :Noetherian local ring is true, but maybe not even UFD. So HOW define these?

Improve this question
$\endgroup$
9
  • $\begingroup$ What's your definition of normal? Doesn't that imply that $X$ is regular in codimension 1? $\endgroup$
    – Sándor Kovács
    Commented Aug 17, 2022 at 17:58
  • $\begingroup$ $X$ is normal when every $\mathcal{O}_{X,x}$ is normal ring(integral and integrally closed in its fraction field). $X$ is regular in codim 1 is true. But I thought it doesn't mean every $D$ is always regular. $\endgroup$
    – map
    Commented Aug 18, 2022 at 1:26
  • $\begingroup$ I'm also unclear that such $\mathcal{O}_{D,x}$ is 1-dim(in the sense of Krull dimension). $\endgroup$
    – map
    Commented Aug 18, 2022 at 1:33
  • $\begingroup$ Welcome new contributor. I do not fully understand what you are asking. Do you want an alternative definition of principal divisor that manifestly works for complex analytic spaces (e.g., the pushforward via the projection to $X$ of the pullback from the projective line of the "tautological" principal divisor with respect to the closure of the graph of the meromorphic function in the product of the projective line and $X$)? Or are you asking how to extend the algebraic approach to also work for complex analytic spaces? $\endgroup$
    – Jason Starr
    Commented Aug 18, 2022 at 14:48
  • $\begingroup$ @map: You are right, $D$ does not need to be regular, but you don't need that. What you need is that $X$ is regular at the generic point of $D$. Also, it is not $\mathscr O_{D,x}$ that should be 1-dimensional, but $\mathscr O_{X,\delta}$ where $\delta$ is the generic point of $D$. The fact that $D$ has codimension 1 should mean that that ring is 1-dimensional. $\endgroup$
    – Sándor Kovács
    Commented Aug 18, 2022 at 23:03

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .