Let $X$ be a scheme, and let $\mathcal{K}$ be the sheaf of total quotient rings of $X$.

Is the data of a Cartier divisor on $X$ equivalent to the data of an invertible subsheaf of $\mathcal{K}$ for all schemes $X$ or do we need additional assumptions?

In more detail:

In most (if not all) textbooks, a Cartier divisor is defined to be a global section of the sheaf $\mathcal{K}^*/O^*$. Explicitly, it is given by the following data: an open cover $\{U_i\}$ of the scheme $X$, and for each $i$, an $f_i \in \Gamma(U_i, \mathcal{K}^*)$, such that for each $i, j$, $f_i/f_j \in \Gamma(U_i \cap U_j, O^*)$. Two such data $\{(U_i, f_i)\}$, $\{(V_j, g_j)\}$ define the same Cartier divisor if there is a common refinement $\{W_k\}$ such that $\displaystyle\frac{(f_i)|_{W_k}}{(g_j)|_{W_k}} \in \Gamma(W_k, O^*)$. Given a Cartier divisor $D$, one may construct an invertible sheaf $L(D)$ which is a subsheaf of $\mathcal{K}$ locally generated by $\displaystyle\frac{1}{f_i}$ on $U_i$.

All of this is stated in most textbooks, but the converse is usually not. It seems to me that this process can be reversed, i.e. given an invertible subsheaf $L \subseteq \mathcal{K}_X$, we choose an open cover $\{U_i\}$ of $X$ such that $L|_{U_i}$ is trivial, which allows us to choose a trivializing section $g_i \in \Gamma(U_i, L) \subseteq \Gamma(U_i, \mathcal{K})$. I believe the fact that $g_i$ generates $L|_{U_i} \cong O_{U_i}$ implies that $g_i$ is not a “zero-divisor”, therefore $g_i \in \Gamma(U_i, \mathcal{K}^*)$. The intuition is that $\mathcal{K}^*$ models the total quotient ring in which an element is either a zero-divisor or a unit.

But I am hesitant because I did not add any additional assumption on the scheme $X$ (e.g. integral, locally noetherian, etc) and after seeing a paper on misconceptions about $\mathcal{K}_X$, I feel less confident about my reasoning.

In any case, if the above reasoning stands, we have $g_i \in \Gamma(U_i, \mathcal{K}^*)$, so we obtained a Cartier divisor $\{(U_i, g_i^{-1})\}$ (the condition $g_i/g_j \in \Gamma(U_i \cap U_j, O^*)$ is easy). The above constructions seem to be inverses of each other.

That explains the title of the question: is a Cartier divisor the same thing as an invertible subsheaf of $\mathcal{K}_X$ on any scheme $X$? I hope someone can either confirm it or deny it.

Along this line of thought, it is usually stated (e.g. Hartshorne Corollary 6.14) that on any scheme $X$, the map $\text{CaCl} X \to \text{Pic} X$ from the Cartier divisor class group to the Picard group is injective, but not surjective in general.

I am wondering if the reason why it is not surjective is that one can’t always embed an invertible sheaf $L$ in $\mathcal{K}_X$?

When $X$ is integral, as in Hartshorne Proposition 6.15, $\mathcal{K}_X$ is the constant sheaf of the function field of $X$ and $L \otimes \mathcal{K}_X \cong \mathcal{K}_X$, so $L \hookrightarrow L \otimes \mathcal{K}_X \cong \mathcal{K}_X$ can be realized as a subsheaf of $\mathcal{K}_X$.

Further remarks

Regarding when an invertible sheaf $L$ is associated to a Cartier divisor $D$, I also find the following discussion useful


It seems to me that given an invertible sheaf $L$ on a scheme $X$, the following are equivalent:

(a) there exists a Cartier divisor $D$ such that $L \cong \mathcal O(D)$;

(b) $L$ admits an invertible rational section $s$ (appropriately defined);

(c) $L \otimes_{\mathcal O_X} \mathcal K_X \cong \mathcal K_X$.

(a) $\Rightarrow$ (b): assume wlog $L = \mathcal O(D) \subseteq \mathcal K$, then $1$ is an invertible rational section.

(b) $\Rightarrow$ (a): take $D= \text{div} (s)$.

(c) $\Rightarrow$ (a): for any invertible sheaf $L$, we have an injection $L \hookrightarrow L \otimes_{\mathcal O_X} \mathcal K_X$ because $L$ is locally free of rank $1$. If $L \otimes_{\mathcal O_X} \mathcal K_X \cong \mathcal K_X$, then we get an embedding $L \hookrightarrow \mathcal K_X$, therefore $L$ is isomorphic to an invertible subsheaf of $\mathcal K_X$, which, by what we have settled, is $\mathcal O(D)$ for some Cartier divisor $D$.

(a) $\Rightarrow$ (c): assume $L$ is an invertible subsheaf of $\mathcal K_X$, then the natural map $L \otimes_{\mathcal O_X} \mathcal K_X \to \mathcal K_X$ is an isomorphism.

  • $\begingroup$ Does this help answer your question? If not, can you add some detail? stacks.math.columbia.edu/tag/02AR $\endgroup$ Commented Sep 13, 2023 at 8:57
  • $\begingroup$ @DerekAllums Done $\endgroup$
    – Bun
    Commented Sep 14, 2023 at 7:52
  • 1
    $\begingroup$ @Bun Thanks, very clear - I edited your title and question slightly to reflect what I think is the real question. Let me know if I misinterpreted. $\endgroup$ Commented Sep 14, 2023 at 8:05

1 Answer 1


We will show a line bundle $L$ is isomorphic to $\mathcal{O}(D)$ if and only if it can be embedded into the sheaf of total quotient rings, which I will denote $\mathcal{K}$. (I don't like $K_X$ for this, since this usually denotes the canonical divisor.)

First, suppose there is an injective morphism $L \to \mathcal{K}$. Fix an open cover $\{U_i\}$ trivializing $L$, and let $f_i$ be the image of $1$ under $\mathcal{O}_{U_i} \stackrel{\sim}\to L_{U_i} \to \mathcal{K}|_{U_i}$.

Note: Since this map is injective, $f_i$ is not a zero divisor and hence, $f_i \in \mathcal{K}^*(U_i)$.

This will express $L|_{U_i} \cong f_i*\mathcal{O}_{U_i}$ for same rational function over $U_i$. Next, we claim $\{(U_i, f_i)\}$ define a Cartier divisor. Indeed, on the overlap, the transition functions of $L$ are given by multiplication by a nonvanshing function $g_{ij}: \mathcal{O}_{U_{ij}} \to \mathcal{O}_{U_{ij}}$. But now, this implies that over $U_{ij}$, we have that $f_j = g_{ij}f_i$. Then, by definition, $\{(U_i, f_i)\}$ is a Cartier divisor, to which $L$ is associated.

Conversely, suppose $D = \{(U_i, f_i)\}$ is a Cartier divisor. Then, for each $U_i$ we have an injective map $\mathcal{O}_{U_i} \to \mathcal{K}$ picking out $f_i$. To see this is injective, note that $f_i \in \mathcal{K}^*(U_i)$, so it is not a zero divisor. To show these glue to give a global injective map $\mathcal{O}(D) \to \mathcal{K}$, we show that this is compatible on overlaps.

This now is a straightforward computation. We have two maps $\mathcal{O}_{U_i \cap U_j} \to \mathcal{K}_{U_i \cap U_j}$ picking out $f_i|_{U_{ij}}$ and $f_j|_{U_{ij}}$, so they differ by multiplication by $f_i|_{U_{ij}}/ f_j|_{U_{ij}}$. Since this is precisely the transition function of $\mathcal{O}(D)$ these maps glue to the required injective map $L \to \mathcal{K}$.

As you said, there is some subtlety going on in this map $\operatorname{CaCl}(X) \to \operatorname{Pic}(X)$. The failure of surjectivity is the same as the failiure of every line bundle on a scheme $X$ to be embedded into the sheaf of total quotient rings. To find a scheme where this map is not surjective, it is more practical to use other properties of Cartier divisors.

There is a famous example due to Kleiman which can be found in detail in Lazarsfeld's 'Positivity in Algebraic Geometry I' on page 11, which constructs a scheme where this map is not surjective.

However, this map is often surjective. As you stated, Hartshorne proves it in the case that $X$ is integral, but it is even true when $X$ is just assumed to be projective (say, over a noetherian ring $A$). To prove this we use the following fact which I leave as an exercise.

Fact: If $L$ admits a global section $s$ which doesn't vanish at the associated points of $X$, then $L \cong \mathcal{O}(\operatorname{Div}(s))$, where $\operatorname{Div}(s)$ is the Cartier divisor $\{U_i, \varphi_i(s)\}$, where $\varphi_i$ is a trivialzation of $L$ over $U_i$.

Proof of claim:(Due to Nakai) Let $L$ be a line bundle on $X$ and $A$ an ample line bundle. By the Serre-Cartan-Grothendieck theorem $L \otimes A^{\otimes n}$ is very ample for large $n$. Then, we can find sections $s \in H^0(X, A^{\otimes n})$ and $t \in H^0(X, L \otimes A^{\otimes n})$ which miss the associated points of $X$.

It follows that $A^{\otimes n}$ is the line bundle associated to the Cartier divisor $H = \text{Div}(s)$ and $L \otimes A^{\otimes n}$ is the line bundle associated to the Cartier divisor $D = \operatorname{Div}(t)$, so that $L \cong \mathcal{O}(D - H)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .