# Why do $V_+(x_0)$ and $\mathcal{O}(1)$ represent linearly equivalent cartier divisors on $\mathbb{P}^n_k$?

This is the gap in Harthsorne's Corollary II.6.17. I'm following the advice of this answer from before on the same question. I'm going to call $$X = \mathbb{P}_k^n = \operatorname{Proj} k[x_0, \dots, x_n]$$ for a field $$k$$. Following the proof of II.6.11, to find the resulting cartier divisor it suffices to show that the divisor $$V(x_0)$$ is a locally principal. I'm not sure how to find the cartier divisor of $$\mathcal{O}(1)$$ so showing it is equivalent to the one formed by $$V(x_0)$$ is confusing.

What is the Cartier divisor associated to $$\mathcal{O}(1)$$ and why is it linearly equivalent to the carrier divisor associated to the hyperplane $$V(x_0)$$.

My Attempt:

Each $$D_+(x_i) \cong \operatorname{Spec} k[x_0/x_i, \dots, x_n/x_i]$$ is the spectrum of a UFD so it $$V(x_0) \cap D_+(x_i)$$ principal with its corresponding ideal in $$k[x_0/x_i, \dots, x_n/x_i]$$ being $$(x_0/x_i)$$ for $$i \neq 0$$ and $$(1)$$ for $$i = 0$$. (I'm not too sure about this step but I'm pretty sure it works.)

Hence, its cartier divisor is $$D = \{(D_+(x_i), x_0/x_i)\}$$ with the corresponding open cover $$D_+(x_i)$$. Next, we want to find the Cartier divisor $$D'$$ associated to $$\mathcal{O}(1)$$. On each $$D_+(x_i)$$, its global sections are the degree 1 part of the localization $$k[x_0, \dots, x_n]_{x_i}$$. I'm not sure what the generator here would be and I'm not sure how to use the fact that $$\mathcal{O}(1)$$ is generated by global sections. I think it would be $$x_i$$ since the canonical isomorphism $$S_{(f)} \to S(n)_{(f)}$$ is given by $$s \mapsto f^n s$$. [see II.5.12(a)]

If $$D' = \{(D_+(x_i), 1/x_i)\}$$ is indeed the associated carrier divisor, then $$D - D' = \{(D_+(x_i), x_0)\}$$, and I'm not so sure why this is principal. Namely, $$x_0$$ is not in the function field which is made of fractions of the same degree.

How would you resolve this and find the correct cartier divisor associated to $$\mathcal{O}(1)?$$

Thanks!

EDIT: This answer alleges to answer my question as well but I don't understand how the last paragraph leads to the conclusion.

I will also add that my expression of $$D'$$ does not really define a Cartier divisor... after all, $$1/x_i$$ is of degree $$-1$$ and hence not an element of the function field of $$\mathbb{P}^n_k$$, since the function field is made up of rational functions of degree $$0$$.

As in the proof of II.6.13(a), the process of finding the cartier divisor relies on an embedding $$\mathscr{L} \hookrightarrow \mathscr{K}$$ of the line bundle into the total quotient sheaf.
Define an embedding $$\mathcal{O}(1) \hookrightarrow \mathscr{K}$$ by $$D_+(f)$$ by $$g/f^k \mapsto (1/x_0) * g/f^k$$. Since you're multiplying by the same value on each open set these morphism glue together to a well defined embedding.
Now we follow the process in the proof. On $$U = D_+(x_i)$$, the map $$\mathcal{O}_X(U) \to \mathcal{O}(1)(U)$$ is explicitly $$f/x_i^k \mapsto x_i * f/x_i^k$$ using the quoted isomorphism $$S_{(x)} \to S(n)_{(x)}$$ mentioned in the question. Hence, under this morphism $$1 \mapsto x_i$$. Then we compose with the embedding into $$\mathscr{K}(U) = K$$ the function field to obtain the representative $$x_0/x_i$$ on each open set $$D_+(x_i)$$ where $$i \neq 0$$. On $$D_+(x_0)$$, the same process yields the representative $$1$$.
As such the correct Cartier divisor of $$\mathcal{O}(1)$$ is $$D' = \{(D_+(x_i), x_0/x_i)\}$$. Then, $$D = D'$$ without the need to use linear equivalence.
The the choice of $$D'$$ is not unique since there are many other embeddings $$\mathcal{O}(1) \hookrightarrow \mathscr{K}$$ which would yield a linearly equivalent but different Cartier divisor (eg, take $$x_1$$ rather than $$x_0$$ in the defined map above).