# Bijection between effective Cartier divisors isomorphic to a line bundle $\mathscr{L}$ and regular sections of $\mathscr{L}\mod\mathscr{O}_X^\times$

$$\def\sK{\mathscr{K}} \def\sO{\mathscr{O}} \def\sL{\mathscr{L}}$$(All definitions I will be using are explained in detail in Görtz, Wedhorn, Algebraic Geometry I, Ch. 11, Divisors.) Let $$X$$ be a scheme. A Cartier divisor is a global section of $$\sK^\times_X/\sO_X^\times$$, where $$\sK_X=\mathscr{S}^{-1}\sO_X$$ is the sheaf of meromorphic functions on $$X$$ (here $$\mathscr{S}\subset\sO_X$$ denotes the multiplicatively closed subsheaf of regular sections). A Cartier divisor is said to be effective if it lies in $$\Gamma(X,(\sK_X^\times\cap\sO_X)/\sO_X^\times)$$.

I am trying to understand the last step in the proof of the following result [ibid., p. 308]:

Proposition 11.34. Let $$X$$ be a scheme, let $$\mathscr{L}$$ be a line bundle on $$X$$, and let $$R_{\mathscr{L}}$$ be the set of $$s \in \Gamma(X, \mathscr{L})$$ that are regular. Then there is a natural bijection $$\left\{\begin{array}{c} \text { effective Cartier divisors } D \\ \text { such that } \mathscr{O}_X(D) \cong \mathscr{L} \end{array}\right\} \leftrightarrow R_{\mathscr{L}} / \sim,$$ where $$s \sim s^{\prime}$$ if there exists an $$a \in \Gamma\left(X, \mathscr{O}_X^{\times}\right)$$ with $$s^{\prime}=as$$.

Proof. For an effective divisor $$D$$ and an isomorphism $$\alpha: \mathscr{O}_X(D) \stackrel{\sim}{\rightarrow} \mathscr{L}$$ the corresponding section $$s \in R_{\mathscr{L}}$$ is the image of the canonical section $$s_D$$ under $$\alpha$$. This yields a well defined map from the left hand side to the right hand side. Conversely let $$s \in R_{\mathscr{L}}$$. Choose an open covering $$\left(U_i\right)$$ of $$X$$ and isomorphisms $$\mathscr{L}_{\mid U_i} \cong \mathscr{O}_{X \mid U_i}$$. Then the images of $$s_{\mid U_i}$$ define local equations $$f_i$$ for an effective divisor $$D$$, which depends only on the equivalence class of $$s$$. The section $$s$$ then defines a monomorphism $$\mathscr{O}_X \hookrightarrow \mathscr{L}$$ which extends by definition to an isomorphism $$\mathscr{O}_X(D) \stackrel{\sim}{\rightarrow} \mathscr{L}$$. $$\square$$

It is the very last sentence of the proof that confuses me a little bit: I understand that $$\sO_X\subset\sO_X(D)$$, but I don't see how it follows “by definition” that $$s:\mathscr{O}_X \hookrightarrow \mathscr{L}$$ extends to an isomorphism $$\mathscr{O}_X(D) \stackrel{\sim}{\rightarrow} \mathscr{L}$$.

I can prove it by myself but my proof is not a one-liner. My sole question is:

From the theory developed in the book up until this point, is there a direct way to make this inference?

For completion, here's the proof I devised: Pick an open cover $$X=\bigcup U_i$$ and isomorphisms $$\varphi_i:\sL|_{U_i}\cong\sO_{U_i}$$ of sheaves of modules. Denote $$D$$ to the effective Cartier divisor represented by $$(U_i,f_i)$$, where $$f_i$$ is the image of $$s|_{U_i}$$ along $$\varphi_i$$. Then we have an actual equality $$\sO_X(D)|_{U_i}=\sO_Xf_i^{-1}$$ as subsheaves of $$\sK_X$$. Denote $$t_i$$ to the image of $$1\in\sO_X(U_i)$$ along $$\varphi_i^{-1}$$. Then $$s=\varphi_i^{-1}\circ\varphi_i(s)=\varphi_i^{-1}(f_i)=f_it_i.$$ Define an $$\sO_{U_i}$$-linear morphism $$\rho_i:\sO_Xf_i^{-1}\to\sL|_{U_i}$$ that maps $$gf_i^{-1}$$ to $$gt_i$$. We claim the maps $$\rho_i$$ glue. Indeed, $$\rho_i(gf_j^{-1})=\rho_i(gf_j^{-1}f_if_i^{-1})=gf_j^{-1}f_it_i$$. Note it doesn't make sense to say that the latter equals $$gf_j^{-1}s$$, since $$f_j$$ might not be a section of $$\sO_X^\times$$ (i.e., $$f_j^{-1}$$ is a section of $$\sK_X^\times$$ that might not be in $$\sO_X$$) and $$\sL$$ is just an $$\sO_X$$-module. We overcome the trouble in the following way: we have $$\varphi_j\rho_i(gf_j^{-1})=gf_if_j^{-1}\varphi_jt_i$$, and since the latter expression lives in $$\sK_X$$, it equals $$gf_j^{-1}\varphi_j(f_it_i)=gf_j^{-1}\varphi_js=gf_j^{-1}f_j=g$$. Putting all together, $$\rho_i(gf_j^{-1})=\varphi_j^{-1}\varphi_j\rho_i(gf_j^{-1})=\varphi_j^{-1}g=gt_j$$. This shows $$\rho_i$$ and $$\rho_j$$ agree on intersections, so they glue to an $$\sO_X$$-linear map $$\rho:\sO_X(D)\to\sL$$. Then $$\rho|_{\sO_X}=s:\sO_X\hookrightarrow\sL$$, since for $$g\in\sO_X(U_i)$$, we have $$\rho_i(g)=\rho_i(gf_if_i^{-1})=gf_it_i=gs$$. Moreover, $$\rho$$ is an isomorphism, for it is locally an isomorphism: $$\varphi_i\circ\rho_i(gf_i^{-1})=\varphi_i(gt_i)=g$$, i.e., $$\varphi_i\circ\rho_i$$ is the canonical isomorphism $$\sO_{U_i}f_i^{-1}\cong\sO_{U_i}$$ of modules.

• How about the following. Tensor $s:{\cal O}_X\to {L}$ by ${L}^\vee$. Then you get $0\to {L}^\vee\to {\cal O}_X\to {\cal O}_D\to 0.$ In particular, ${L}^\vee$ is isomorphic to the kernel of the map ${\cal O}_X\to {\cal O}_D$, which is the ideal sheaf of $D$. But this ideal sheaf is ${\cal O}(D)^\vee={\cal O}(-D)$ by definition, so you get what you want. Commented Oct 26, 2023 at 10:35
• @DamianRössler How would one show that $\operatorname{Im}(L^\vee\to\mathcal{O}_X)=\mathcal{I}_X(D)$? Commented Oct 26, 2023 at 11:04
• I give a more elementary approach below. Commented Oct 26, 2023 at 14:33
• My approach is essentially the same as yours. My main point is that if you have an inclusion ${\cal O}_X(D)\subseteq K_X$ globally (and this is done in the book), then you obtain what you want once you have constructed the isomorphism locally and shown that it is unique. Commented Oct 26, 2023 at 14:37

Let $$R$$ be a ring and let $$M$$ be a $$R$$-module which is free of rank one. Let $$a:R\to M$$ be an injective map of $$R$$-modules. Let $$\phi:M\to R$$ be an isomorphism. Let $$K$$ be the localisation of $$R$$ at all the non-zero divisors of $$R$$. Let $$r_\phi:=\phi(a(1)).$$ Let $$R_M$$ be the $$R$$-submodule of $$K$$ generated by $$r^{-1}_\phi.$$ Note that this makes sense, because $$r_\phi$$ is a non-zero divisor since $$a$$ is injective. Also, $$R_M$$ does not depend on $$\phi$$, because a different choice of $$\phi$$ would multiply $$r_\phi$$ by a unit of $$R$$. Note also that we have a natural inclusion $$R\subseteq R_M.$$
I claim that $$a$$ extends uniquely to an isomorphism of $$R$$-modules $$A:R_M\simeq M.$$
Indeed we may define $$A(\lambda/r_\phi)=\lambda\phi^{-1}(1).$$ We then have $$A(1)=A(r_\phi/r_\phi)=\phi(a(1))\phi^{-1}(1)=\phi^{-1}(\phi(a(1)))=a(1)$$ so this extends $$a$$. The map $$A$$ is an isomorphism since $$A(1/r_\phi)$$ is a basis of $$M$$ and $$1/r_\phi$$ is a basis of $$R_M$$. It is unique because a different extension would differ from $$A$$ by a multiplication by a non trivial unit of $$R$$, and this is impossible because $$A$$ coincides with $$a$$ on $$R$$ and $$a$$ is injective. In particular, $$A$$ does not depend on $$\phi.$$
Now consider an injective map of $${\cal O}_X$$-modules $${\cal O}_X\to L$$, where $$L$$ is a line bundle. Let $$D$$ be the associated effective Cartier divisor. The above construction gives maps $${\cal O}_X(D)|_U\simeq L|_U$$ for any open affine subscheme where $$L$$ is trivial. Since these isomorphisms are unique, they glue to a global isomorphism.
• Alternative way to see uniqueness of $A$: if $A'$ is another extension of $a$, then $r_\phi A(\lambda/r_\phi)=\lambda a(1)=r_\phi A'(\lambda/r_\phi)$. Since $r_\phi$ is a non-zerodivisor and $M$ is free of rank 1, $r_\phi A(\lambda/r_\phi)=r_\phi A'(\lambda/r_\phi)$ implies $A(\lambda/r_\phi)=A'(\lambda/r_\phi)$. Commented Oct 27, 2023 at 7:39