# Do I correctly understand the constructions involved in definition of Cartier divisor?

Let $(X,\mathcal O)$ be a ringed space, where $\mathcal O$ is a sheaf of unital commutative integrity domains. Let $\widetilde{\mathcal M}_U$ be the field of fractions of ring $\mathcal O_U$ for any open $U$ and $\widetilde{\mathcal M}$ be the presheaf given by $\widetilde M_U$ over an open $U$. Sheafification $\mathcal M$ of $\widetilde{\mathcal M}$ is called the sheaf of meromorphic functions on $X$. Its sections over an open $U$ will be denoted by $\mathcal M_U$. Then $\mathcal O_U$ is naturally included in $\mathcal M_U$. Now let $\mathcal O^\times$ and $\mathcal M^\times$ be the sheaves of multiplicative groups of invertible elements of $\mathcal O$ and $\mathcal M$. Again, $\mathcal O_U^\times \subset \mathcal M_U^\times$. We can consider the sheaf $\mathcal M^\times / \mathcal O^\times$. Its global sections are called Cartier divisors.

If I'm not mistaken, any section of $\mathcal M$ over an open $U$ is given by an open cover $\{U_i\}$ of $U$ together with a family of elements $\{f_i / g_i \mid f_i, g_i \in \mathcal O_{U_i}$, $g_i \neq 0 \}$ and such that $$f_i|_{U_i \cap U_j} g_j|_{U_i \cap U_j} - g_i|_{U_i \cap U_j} f_j|_{U_i \cap U_j} = 0$$ if $U_i \cap U_j \neq \varnothing$.Thus the inclusion $\mathcal O_U \to \mathcal M_U$ for any $f \in \mathcal O_U$ is given by the data of the single set $U_1 = U$ and of a single fraction with $f_1 = f$, $g_1 = 1_{\mathcal O_U}$.

Then any section of $\mathcal M^\times / \mathcal O^\times$ over an open $U$ is given by an open cover $\{U_i\}$ of $U$ together with a family of elements $\{f_i / g_i \cdot \mathcal O_U^\times|_{U_i} \mid f_i, g_i \in \mathcal O_{U_i}$, $f_i, g_i \neq 0 \}$ each of which represents a set and such that $$f_i|_{U_i \cap U_j}/ g_i|_{U_i \cap U_j} \cdot \mathcal O^\times_U |_{U_i \cap U_j} = f_j|_{U_i \cap U_j} / g_j|_{U_i \cap U_j} \cdot O^\times_U |_{U_i \cap U_j} \tag{1}$$ if $U_i \cap U_j \neq \varnothing$. Are these considerations correct?

• I think you need some hypothesis on the sheaf $\mathcal O$ to make $\widetilde{M}$ a presheaf. Namely, if $V$ is an open subset of $U$, it is not clear for me that $\mathcal O_X(U)\to \mathcal O_X(V)$ induces a map on the fields of fractions. Otherwise, everything is correct. – Cantlog Mar 24 '14 at 20:54
• @Cantlog is right: mathoverflow.net/a/28566/450 – Georges Elencwajg Mar 24 '14 at 21:00
• @Cantlog according to discussion in comments with Michael Joyce, not everything is correct. In partucular, one needs to replace $\mathcal O^\times_U|_{U_i \cap U_j}$ in (1) by $\mathcal O^\times_{U_i \cap U_j}$ but I don't understand why. – Appliqué Mar 24 '14 at 21:23
• @Nimza: I was confused with your notation. Usually $\mathcal O_U^\star$ is the sheaf on $U$, in which case $\mathcal O_U^\star|_{U_i\cap U_j}$ is just $\mathcal O_{U_i\cap U_j}^\star$. With your notation, yes, you need the invertible elements on $U_i\cap U_j$ because you want to compare two sections of $\mathcal M^\star$ on $U_i\cap U_j$. – Cantlog Mar 24 '14 at 22:00

You must also be careful to keep in mind that two different collections $(\mathcal{U_i}, f_i / g_i)$ and $(\mathcal{V_j}, f'_j / g'_j)$ will define the same Cartier divisor if $f_i / g_i$ and $f'_j / g'_j$ restrict to the same rational function on $U_i \cap V_j$ (provided it is non-empty) up to multiplication by a nowhere zero, every regular function on $U_i \cap V_j$.
• Thank you, just one question about your last phrase: this "nowhere zero regular" function on $U_i \cap V_j$ must always be a restriction of some nowhere zero regular function on the whole $U$, right? – Appliqué Mar 24 '14 at 20:59
• No, it does not have to be a restriction. For example the point $\{ 0 \} \subset \mathbb{P}^1$ is a Cartier divisor given by the open cover $(U_0 = \{ x_0 / x_1 \neq 0 \}, x_0 / x_1 )$ and $(U_1 = \{ x_1 / x_0 \neq 0 \}, 1)$. Then $x_0 / x_1 = 1 \cdot(x_0 / x_1)$, but the function $x_0 / x_1$ is only nowhere zero and regular on $U_0 \cap U_1$, not on either $U_0$ or $U_1$ individually. Similar things happen when you compare two different representatives for the same Cartier divisor. – Michael Joyce Mar 24 '14 at 21:05
• Ok, then there is some problem in my definition in the question, because when we factor $\mathcal M_U^\times$ by $\mathcal O_U^\times$ we consider the data defining elements of $\mathcal M_U^\times$ modulo multiplication by elements of $\mathcal O_U^\times$, but the latter elements are regular nowhere zero functions on the whole $U$. – Appliqué Mar 24 '14 at 21:13
• See cantlog's comment. In some cases, you might be better off writing $\mathcal{O}_X(U)$ where you are writing $\mathcal{O}_U$. Also, be careful to remember that $(\mathcal{M} / \mathcal{O})(U)$ is not simply the quotient group $\mathcal{M}(U) / \mathcal{O}(U)$. This might be causing your confusion. – Michael Joyce Mar 24 '14 at 22:38
• A really important example to study with regard to quotient sheaves is the en.wikipedia.org/wiki/Exponential_sheaf_sequence. Note that $\mathcal{O}_M^* / \text{exp}(\mathcal{O}_M) = 0$, even though $\mathcal{O}_M^*(M) / \text{exp}(\mathcal{O}_M(M))$ may not be zero, for example if $M = \mathbb{C}^*$, since there is no everywhere defined analytic complex logarithm. – Michael Joyce Mar 25 '14 at 15:28