# Mapping Cartier Divisors to Picard Group

I have some questions about the steps in proof of Prop. 7.1.18 in Liu's "Algebraic Geometry" (page 257):

Here we have the Cartier-Divisor $D \in H^0(X,\mathcal{K}^*_X /\mathcal{O}^* _X) = \mathcal{K}^*_X /\mathcal{O}^* _X (X)$ represented by $\{(U_i, f_i)_i\}$ (sheafification property) and the map $\rho: D \to \mathcal{O}_X (D)$ where $\mathcal{O}_X (D) \subset \mathcal{K}_X$ is defined by $\mathcal{O}_X(D)|U_i = f^{-1}_iO_X |U_i$ (more detailed description of used symbols: see images below).

My questions (refer to red tagged lines):

1.: If $D$ is the $Ker$ of $\rho$, why it's image under $\rho$ has the shape $f\mathcal{O}_X$ for $f \in \mathcal{O}_X(D)$? (note: $H^0(X, \mathcal{F}) = \mathcal{F}(X)$)

2.: Why if we have an invertible subsheaf $\mathcal{L} \subset \mathcal{K}_X$ with covering $\{U_i\}_i$ such that $\mathcal{L}|U_i$ is free this section is generated by a $f_i \in \mathcal{K}'_X(U_i)$ and futhermore $f_i \in \mathcal{K}'_X(U_i)^*$? (therefore locally invertible)

Here the used definitions:

Cartier Divisors:

The sheaf $\mathcal{K_X}$:

The $Pic(X)$ group:

• This is a good question, but a little too advanced for me. Ask it on MO, and link that question here so I can follow along.
– Alex
Commented Oct 8, 2017 at 2:31
• Have not enough points to post images in MO :( Commented Oct 8, 2017 at 2:36
• Would you do it? Commented Oct 8, 2017 at 2:50
• @Alex : Mathoverflow is for research problems, which is clearly not the case here. Commented Oct 8, 2017 at 8:36

1. Ask yourself what it means for a divisor $D$ to be in the kernel of $\rho$. It means that $\rho(D)\cong \mathcal{O}_X$. Since $\mathcal{O}_X$ is generated by a single global section, $1$, it must be the case that $\rho(D)$ is also generated by a single global section, ie the image of $1$ under the isomorphism $\rho(D)\cong \mathcal{O}_X$. The image of $1$ is exactly $f$.
2. Picking $U_i$ such that $\mathcal{L}|_{U_i}$ is free is equivalent to choosing an isomorphism $\mathcal{L}|_{U_i}\cong \mathcal{O}_{U_i}$. If $\mathcal{L}\subset \mathcal{K}$, then this inclusion will descend to the restriction of these sheaves to $U_i$. Since $\mathcal{O}_{U_i}$ is generated by a single global section, $\mathcal{L}|_{U_i}$ must be as well, and since $\mathcal{L}|_{U_i}\subset \mathcal{K}|_{U_i}$, the generating global section must be a global section of $\mathcal{K}|_{U_i}$. This implies that there exists some $f_i\in \mathcal{K}|_{U_i}(U_i)$ such that $\mathcal{L}|_{U_i}\cong f_i\mathcal{O}_{U_i}$.
To see that $f_i$ is actually invertible, consider the following. Let $X$ be a scheme, $f,g\in \mathcal{K}(X)$, $\mathcal{F}=f\mathcal{O}_X$, and $\mathcal{G}=g\mathcal{O}_X$. Then $\mathcal{F}\otimes_{\mathcal{O}_X}\mathcal{G}=fg\mathcal{O}_X$. Since $\mathcal{L}|_{U_i}=f_i\mathcal{O}_{U_i}$ is invertible with inverse $\mathcal{L}|_{U_i}^{-1}=f_i'\mathcal{O}_{U_i}$, and $$\mathcal{L}|_{U_i}\otimes_{\mathcal{O}_{U_i}} \mathcal{L}|_{U_i}^{-1}=f_if_i'\mathcal{O}_{U_i}=\mathcal{O}_{U_i}$$ ie $f_if_i'=1$, so each of $f_i,f_i'$ are units.