Usbrth

Let $X$ be a complex manifold and let $f$ be a meromorphic function defined on it. Let $Div(X)$ be the group of locally finite sum of analytic irreducible hypersurfaxces of $X$.

One would like to define a divisor associated to $f$ as $$sum_i in I ord(f)|_V_iV_i$$ but I really can't see why this sum should be locally finite.

Let $h$ be analytic and non identically zero on a small disk $U = z$ and $Hol(U)$ the ring of meromorphic functions on $U$ that are analytic on some neighborhood of $0$.

Look at the set of principal ideals $S = (alpha) , alpha in Hol(U), alpha(0) = 0 $ partially ordered by inclusion. Let $v(alpha)$ be the degree of the first non-zero term in $alpha$'s power series. $v(alpha beta) = v(alpha)v(beta)$.

If $(alpha) supsetneq (beta)$ then $ beta =alphagamma$ with $gamma(0) = 0$ so $v(gamma) ge 1$ and $v(beta) ge v(alpha)+1$.

Thus the maximal elements of $S$ are well-defined : $ M = (phi) in S, not exists (varphi) in S, (varphi) supsetneq (phi)$.

From there let $h_0 = h$, while $h_j(0) = 0$, pick some $ (phi_j+1) in M$ such that $(h_j) subset (phi_j+1)$ and let $h_j+1 = frach_jphi_j+1$. Since $v(h)$ is finite the process terminates obtaining the factorization $$forall z in U, qquad h(z) = h_J(z)prod_j=1^J phi_j(z)$$ where $J le v(h)$ and $h_J(0) ne 0$.

Todo : show $v(phi_j) = 1$ so $J = v(h)$.

Find an even smaller disk $V$ where $h_J$ is non-zero and the $phi_J$ are holomorphic. The local divisor is defined as $Div(h|_V) = sum_j=1^J Z(phi_j|_V)$ where $Z(phi|_V) = z in V, phi(z)=0$.

If $f$ is meromorphic then $f = frachg$ with $g,h$ holomorphic and $Div(f|_V) = Div(h|_V)-Div(g|_V)$.

The global divisor is obtained by gluing those local divisors.

There is an open set $U$ on which your meromorphic function doesn’t have any zeros or poles, so the only nonzero coefficients come from prime divisors on the complement of $U$ in $X$. This complement is a closed set $Z$. Now for the local finiteness: Take the intersection of $Z$ with a quasicompact open set of $X$, which is Noetherian as a topological space because $X$ is locally Noetherian, as it is a complex manifold and the ring of analytic functions in finitely many variables is Noetherian. (I learned this last fact from Huybrechts’s book Complex Geometry). Every subspace of a Noetherian topological space is Noetherian, so in particular this intersection can have only finitely many irreducible components. This gives you local finiteness.