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 $\mathcal{O}(1)?$


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 $0$.


1 Answer 1


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).


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