# For what schemes $X$ are Cartier divisors the same thing as invertible subsheaves of $\mathcal{K}_X$?

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

https://mathoverflow.net/questions/53567/why-is-line-bundle-appropriate-rational-section-not-a-standard-kind-of-diviso

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.

• Does this help answer your question? If not, can you add some detail? stacks.math.columbia.edu/tag/02AR Commented Sep 13, 2023 at 8:57
• @DerekAllums Done
– Bun
Commented Sep 14, 2023 at 7:52
• @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. Commented Sep 14, 2023 at 8:05

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 $$L \cong \mathcal{O}(D - H)$$.