# If $D$ is a Weil divisor, how do I get the invertible sheaf $\mathcal{O}_S(D)$?

In Beauville's Complex Algebraic Surfaces, chapter $$1$$, the author takes a smooth variety $$S$$ (over $$\Bbb{C}$$) and mentions the correspondence between Weil divisors modulo linear equivalence and invertible sheaves modulo isomorphisms: $$D\mapsto \mathcal{O}_S(D)$$

I know that a Cartier divisor $$D=\{(U_i,f_i)\}\in\text{CaDiv}(S)$$ provides an invertible sheaf defined simply by: $$\mathcal{O}_S(D)\big|_{U_i}:=\frac{1}{f_i}\mathcal{O}_S\big|_{U_i}$$

Which is how we define a map $$\text{CaCl}(S)\to\text{Pic}(S)$$.

I've already found in different authors an explicit isomorphism $$\text{WCl}(S)\to\text{CaCl}(S)$$, so technically I'm able do describe $$D\mapsto \mathcal{O}_S(D)$$ by the composition $$\text{WCl}(S)\to\text{CaCl}(S)\to\text{Pic}(S)$$.

However, the construction of $$\text{WCl}(S)\to\text{CaCl}(S)$$ uses a somewhat subtle argument involving local rings in order to find the functions $$f_i$$, so we need at least one paragraph to explain the whole composition honestly.

The way Beauville mentions the map $$D\mapsto\mathcal{O}_S(D)$$ makes it seem natural, so I wonder if there's a simpler description that I don't know about.

• I believe definition 1.2 here answers your question. You can describe this line bundle explicitly as "rational functions on $S$ with specified zeros/poles according to the divisor." Commented Aug 3, 2020 at 4:38
• @Stahl, thank you! If you would like to write it as an answer, I'll be happy to accept it. Commented Aug 3, 2020 at 17:58

Let $$X$$ be a Noetherian, integral separated scheme, regular in codimension $$1,$$ and let $$K(X)$$ be its function field. If you want to weaken these hypotheses, you can -- but you will need to carefully replace $$K(X)$$ by the sheaf of rational functions $$\mathcal{K}_X$$ on $$X.$$
Suppose that $$D = \sum_{Y} n_Y[Y]$$ is a Weil divisor on $$X.$$ We may define the line bundle $$\mathcal{O}_X(D)$$ explicitly to be the set of invertible rational functions $$f$$ on $$X$$ (i.e., elements of $$K(X)^\times$$) such that $$f$$ has zeros and poles prescribed by the divisor $$D$$ in the following sense: $$\mathcal{O}_X(D) := \{f\in K(X)^\times\mid (f) + D\geq 0\}.$$ Here, $$(f)$$ is the divisor associated to $$f$$: $$(f) := \sum_{Y}\nu_Y(f)[Y],$$ where $$\nu_Y : \mathcal{O}_{X,\eta}\to\Bbb{Z}$$ is the valuation on the DVR $$\mathcal{O}_{X,\eta},$$ the stalk of $$\mathcal{O}_X$$ at the generic point of $$Y.$$ Recall that we say that $$f$$ has a zero along $$Y$$ of order $$\nu_Y(f)$$ if $$\nu_Y(f)>0,$$ and that $$f$$ has a pole along $$Y$$ of order $$-\nu_Y(f)$$ if $$\nu_Y(f) < 0.$$
So, $$\mathcal{O}_X(D)$$ consists of rational functions on $$X$$ such that $$\nu_Y(f)\geq -n_Y$$ for all $$Y.$$ That is to say, if $$n_Y < 0,$$ then $$f$$ must have a zero along $$Y$$ of order at least $$-n_Y,$$ and if $$n_Y > 0,$$ $$f$$ is allowed to have a pole along $$Y$$ of order at most $$n_Y.$$
• Is there a reason to write "$[Y]$" instead of just "$Y$"? Commented Aug 3, 2020 at 19:52
• @rmdmc89 Just to remember that we're working in the group of formal sums on the set of codimension 1 closed integral subschemes $Y$ and not get confused about taking some "sum of subschemes." It's just notation, though. Commented Aug 3, 2020 at 20:24