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I am reading Definition 1.17 of Liu on what a Cartier Divisor is:

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I have several questions concerning this definition.

Question 1: It makes sense to say to me that an element $D \in H^0(X, \mathcal{K}_X^\ast/\mathcal{O}_X^\ast)$ can be represented by $\{(U_i,f_i)\}$ where $f_i \in H^0(U_i, \mathcal{K}_X^\ast)$ and $f_i|_{U_i \cap U_j} \in f_j|_{U_i \cap U_j} \mathcal{O}(U_i \cap U_j)^\ast$. However in Liu's definition he says we may take the $f_i$ to be a quotient of two non-zero divisors of $\mathcal{O}_X(U_i)$. How does this make sense?

Question 2: He gives a definition for what it means for two Cartier divisors $\{(U_i,f_i)\}$ and $\{(V_j,g_j)\}$ to be equivalent. Taking his definition for a Cartier divisor, what does it mean to say $f_i$ and $g_j$ differ by a multiplicative factor in $\mathcal{O}_X(U_i\cap U_j)^\ast$? For example, it is not even clear to me why $f_i|_{U_i \cap V_j} $ should be an element of $\mathcal{O}_X(U_i\cap U_j)^\ast$. Or does it need to be?

Question 3: What is the canonical map $H^0(X,\mathcal{O}_X \cap \mathcal{K}_X^\ast) \to H^0(X,\mathcal{K}_X^\ast/\mathcal{O}_X^\ast)$ in the definition of an effective divisor? Is it "derived" from the map $\mathcal{K}_X^\ast \to \mathcal{K}_X^\ast/ \mathcal{O}_X^\ast$?

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  • $\begingroup$ Dear user, I agree that the condition that $f_i$ be the quotient of two non zero divisors in $\mathcal O_X(U_i)$ is a bit strange: a general section over $U_i$ of $\mathcal{K}_X^\ast$ certainly is not of this form: just think of the case where $U_i=\mathbb P^1$, or any complete variety ! Despite my great admiration for Liu, I don't think his definition should be adopted. $\endgroup$ – Georges Elencwajg Oct 13 '13 at 15:19
  • $\begingroup$ Dear @GeorgesElencwajg: In the definition of a Cartier divisor, it is not said that $f_i$ is a quotient for any covering, but some covering. So this definition is correct, right ? $\endgroup$ – Cantlog Oct 13 '13 at 17:14
  • $\begingroup$ Dear @Cantlog, I agree with your point but the definition would be quite awkward sinnce you would have to restrict yourself to quite special coverings. I find it healthier and more flexible to give oneself a divisor by just following the procedure which permits one to describe the global sections of a quotient sheaf like $\mathcal{K}_X^\ast/\mathcal{O}_X^\ast$ by choosing an arbitrary open covering of $X$ and sections of the numerator sheaf subjected to some compatibility conditions. $\endgroup$ – Georges Elencwajg Oct 13 '13 at 17:41
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    $\begingroup$ By the way, these concepts are so subtle that Grothendieck, Hartshorne and Altman-Kleiman all gave false definitions of $\mathcal K_X$ ! See here $\endgroup$ – Georges Elencwajg Oct 13 '13 at 17:51
  • $\begingroup$ @GeorgesElencwajg link appears dead to me, so here's a refreshed link and for future seekers the paper is titled Misconceptions about $\mathcal{K}_X$ $\endgroup$ – Callus Aug 4 '18 at 12:42
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Question 1: I'm having a hard time interpreting the question "How does this make sense?"—it doesn't give a very good sense of what you're struggling with. Anyway, you say that you are comfortable saying $f_i \in H^0 (U_i,\mathcal{K}_X^\ast)$. But $\mathcal{K}_X^\ast$ is exactly the sheaf of quotients of two non-zero divisors in $\mathcal{O}_X$.

What I think you're having trouble with is how the sheaf $\mathcal{K}_X^\ast / \mathcal{O}_X^\ast$ is constructed. It's a sheafification, which means that every section has a local trivialization. But what is it a sheafification of, exactly? Of the presheaf that maps every affine open $\operatorname{Spec} A$ to the quotient group $K(A)^*/ A^*$, where $K(A)$ is the fraction field—so $K(A)^\ast$ is just the set of quotients of two non-zero-divisors.

(Note: I think that what I wrote there is technically correct, though it's not as simple as I'm suggesting to actually define this presheaf. One needs to work with $\mathcal{K}_X$ as the sheafification of $U\mapsto \mathcal{O}_X(U)[S(U)^{-1}]$, where $S(U)$ is the collection of elements of $\mathcal{O}_X(U)$ which are non-zero-divisors at every stalk in $U$. This is important for ensuring that the representatives $f_i$ actually restrict to quotients of regular elements.)

$\mathcal{K}_X^\ast / \mathcal{O}_X^\ast$ doesn't necessarily look like that on a random affine open. But for any section $s$, $X$ will have a covering by affine opens $\{U_i\}$ such that $s|_{U_i}$ is of that form.

Question 2: It's not necessarily true that $f_i|_{U_i \cap V_j} \in \mathcal{O}_X (U_i \cap V_j)^*$, or even that $f_i|_{U_i \cap U_j} \in \mathcal{O}_X (U_i \cap U_j)^*$. Remember, the latter is in the fraction field, but, for $f_i$ and $f_j$ to be compatible on $U_i \cap U_j$, they must lie in the same coset of the quotient, which is the same as saying: $f_i|_{U_i \cap U_j} \in f_j|_{U_i \cap U_j} \mathcal{O}(U_i \cap U_j)^\ast$ (plus the same thing with $i$ and $j$ reversed).

If you're having trouble seeing how this interacts with the sheafification, you can think about it entirely on stalks—because the stalk of $\mathcal{K}_X^\ast / \mathcal{O}_X^\ast$ really is a quotient group. And you can show that $f_i|_{U_i \cap U_j} \in f_j|_{U_i \cap U_j} \mathcal{O}(U_i \cap U_j)^\ast$ holds if and only if it holds at each stalk of $U_i \cap U_j$.

The comparison between $f_i$ and $g_j$ is identical; he's just indicating that the condition is the same as for $f_i$ and $f_j$. The reasoning is identical as well.

Question 3: Yes. The quotient map $\mathcal{K}_X^\ast \to \mathcal{K}_X^\ast /\mathcal{O}_X^\ast$ you understand, and $\mathcal{O}_X \cap \mathcal{K}_X^\ast \to \mathcal{K}_X^\ast$ is the inclusion of a subsheaf (the one consisting only of those quotients of non-zero-divisors which already lie in the sheaf $\mathcal{O}_X$, which can be understood in a number of different intuitive ways). And a morphism of sheaves induces a morphism of global sections.

This is all another way of saying: a Cartier divisor is effective if we can choose its representatives $f_i$ to all be regular functions.

For example, on $\mathbb{P}^1_k$, with projective coordinates $x$ and $y$, you could construct an effective Cartier divisor as follows: On the open $x\neq 0$, take the regular function $y/x$. On $y\neq 0$, take the regular function $x/y$. On the intersection $U$, $\mathcal{K}_X(U) = k(x/y)$ and $\mathcal{O}_X(U) = k[x/y,y/x]$. But $x/y$ and $y/x$ are both regular functions, so they both lie in the coset $k[x/y,y/x]^\ast$, hence are compatible.

(If you think of the group of Cartier divisors of $\mathbb{P}^1$ up to equivalence as $\mathbb{Z}$, then the one I just described is $2$.)

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  • $\begingroup$ There was a typo in question 1. I have corrected it. $\endgroup$ – user38268 Oct 13 '13 at 12:08
  • $\begingroup$ I think for (1) the thing I don't understand is that an element of $\mathcal{K}_X(U_i)$ is not literally a quotient of two regular elements. It is the sheaf associated to the presheaf $U_i \mapsto \text{Tot}\mathcal{O}_X(U_i)$. $\endgroup$ – user38268 Oct 13 '13 at 13:32
  • $\begingroup$ @user 38268: you are perfectly right. See my comment under your question. $\endgroup$ – Georges Elencwajg Oct 13 '13 at 15:19
  • $\begingroup$ @user38268: sections of $K_X^*(U)$ are not necessarily of quotient of regular elements (even when $U$ is affine), but for any such a section, this becomes true after refining the covering. This comes essentially from the definition of the sheafification. $\endgroup$ – Cantlog Oct 13 '13 at 19:48
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    $\begingroup$ @Cantlog There's the extra issue that a non-zero-divisor may become a zero divisor on a smaller affine open. So it's important to note that these are specifically quotients of elements which are regular at each stalk. $\endgroup$ – Slade Oct 13 '13 at 20:13

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