On a scheme $X$, the most general definition of Cartier divisor is a global section in $\Gamma(X, \mathcal{K}^{*}/\mathcal{O}^{*})$, where $\mathcal{K}^{*}$ is the sheaf of invertible elements of the sheaf of total quotient rings.

Alternatively, some texts refer to a Cartier divisor as an equivalence class of pairs $(\mathcal{L}, s)$, where $\mathcal{L}$ is an invertible sheaf on $X$ and $s$ is a non-zero rational section of $\mathcal{L}$.

I have been able to show that these definitions are equivalent in the case that $X$ is a noetherian integral scheme. But how general does this equivalence go? Are they always the same? I expect the problem may occur when you drop reducedness.

  • $\begingroup$ how do you define a rational section of a line bundle on a general scheme? $\endgroup$
    – user690882
    Commented Jul 28, 2019 at 6:29
  • $\begingroup$ Actually good question. I guess for an irreducible scheme it is fine, but I don't know how it would be defined more generally than that. $\endgroup$
    – Luke
    Commented Jul 28, 2019 at 6:36

1 Answer 1


Cartier divisors can be defined on any scheme. And to any Cartier divisor $D$ on any scheme $X$ you can associate a line bundle (invertible $\mathcal{O}_X$-module) which is usually denoted by $\mathcal{O}_X(D)$. So there is always a map $$ \begin{align} \mathrm{Div}(X) &\to \mathrm{Pic}(X) \\ D &\to \mathcal{O}_X(D), \end{align} $$ from the group of Cartier divisors to the group of line bundles. This further induces an injective map $\mathrm{DivCl}(X) \to \mathrm{Pic}(X)$, where $\mathrm{DivCl}(X)$ denotes the group $\mathrm{Div}(X)$ modulo linear equivalence. This map is surjective if $X$ is an integral scheme, but for general schemes not all line bundles come from Cartier divisors.

Now to answer your question, for any scheme $X$ and a line bundle $\mathcal{L}$ on it, you have the following natural one to one correspondence $$ \{ \text{effective cartier divisors } D \text{ such that } \mathcal{O}_{X}(D) \cong \mathcal{L} \} \leftrightarrow \{ \text{non-zero divisors of }\Gamma(X, \mathcal{L}) \}\big/\sim, $$ where $ s \sim s'$ if and only if $s' = us$ for some $u \in \Gamma(X, \mathcal{O}_X^{\times}).$ Note that we are only considering effective Cartier divisors, because the associated line bundle of any divisor has a nonzero global section if and only if it is linearly equivalent to an effective one.

  • $\begingroup$ If you want explicit details, see pages $302-305$ of Algebraic Geometry $1$ by Görtz, Wedhorn. $\endgroup$ Commented Jul 28, 2019 at 8:00

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