# There exist invertible sheaf which is not isomorphic to $\mathcal{O}_{X}(D)$ for any $D$ Cartier divisor.

This is an example (example 1.3) of the Hartshorne's book Ample Subvarieties of algebraic varieties.

The proposition (by Kleiman) is the following:

Proposition There is a complete scheme $$X'$$ and an invertible sheaf $$L'$$, such that $$L'$$ is not isomorphic to $$\mathcal{O}(D')$$ for any Cartier divisor $$D'$$ on $$X'$$.

I have questions about the proof, which I quoted in bold letters as (a), (b), (c), (d) and (e). Down after I explain my reasoning about each question.

Here (Hartshorne's book), scheme proper over $$k$$ and complete scheme are equivalents. $$k$$ is algebraically closed. A non-singular scheme is assumed to be irreducible.

Proof

Let $$X$$ be a complete non-singular, non-projective 3-fold containing two disjoint irreducible curves $$C_1$$ and $$C_2$$ such that $$C_1+C_2$$ is algebraically equivalent to zero. (Hartshorne give a reference and an example for this afirmation, such an X exists).

If $$D$$ is any prime divisor which intersects $$C_1$$ properly (and there are such, since $$X$$ is non-singular), then $$D$$ must contain $$C_2$$, because $$(D\cdot C_2) = -(D\cdot C_1) < 0$$. (a)

Let $$P$$ be a point of $$C_2$$. Define a new scheme $$X'$$, which has the same topological space as $$X$$, and the same structure sheaf everywhere except P. At $$P$$, define $$\mathcal{O}_{X',P} =\mathcal{O}_{X,p}\oplus k$$, with multiplication defined by $$(a,\lambda)(b,\mu) = (ab, a\lambda+b\mu)$$.

Then the only non-zero divisors of $$\mathcal{O}_{X',P}$$ are units, so no Cartier divisor on $$X'$$ can contain $$P$$. (b)

In fact, it is easly noted that there is a one-to-one correspondence $$Div (X') = \{D \in Div (X)\ \ /\ \ P\notin D\} .$$ On the other hand, we have an exact sequence $$0\rightarrow k \rightarrow \mathcal{O}_{X'}^* \rightarrow \mathcal{O}_X^* \rightarrow 0\ \ \ \ \mathbf{(c)}$$ so $$Pic (X') \cong Pic (X)$$. (d)

Let $$D$$ be a divisor on $$X$$ which intersects $$C_1$$ properly. Let $$L = \mathcal{O}_X(D)$$, and let $$L'$$ be the corresponding invertible sheaf on $$X'$$.Then L' is not isomorphic to $$\mathcal{O}_{X'}(D')$$ for any Cartier divisor $$D'$$ on $$X'$$.(e) $$\blacksquare$$

So I think the next for the items

(a) $$D\cdot C_1>0$$, since $$D$$ intersects $$C_1$$ properly. As $$C_1+C_2$$ is algebraically equivalent to zero then its is linearly equivalent to zero, $$D\cdot(C_1+C_2)=0 \Rightarrow D\cdot C_2=-D\cdot C_1 < 0$$. If $$C_2 \not\subset D$$ so $$D\cdot C_2 \geq 0$$ and this is a contradiction, hence $$C_2\subset D$$.

(b) Then the local equations for $$D$$ is an $$f\in \mathcal{O}_{X',P}^*$$, an unity, and this can not let be because $$\{x\ \ /\ \ f(x)=0\}=\emptyset$$.

(c) I think that in the sequence I should be change $$k$$ by $$k^*$$. The sequence $$0\rightarrow k^* \rightarrow \mathcal{O}_{X',x}^* \rightarrow \mathcal{O}_{X,x}^* \rightarrow 0$$ is exact for any $$x\neq P$$ by the natural inclusions. The sequence $$0\rightarrow k^* \rightarrow \mathcal{O}_{X',P}^* \rightarrow \mathcal{O}_{X,P}^* \rightarrow 0$$ is exact in $$x=P$$ by the maps $$\lambda\mapsto (0,\lambda)$$ and $$(a,\lambda)\mapsto a$$.

(d) So if I write the long exact sequence of cohomology $$\cdots \rightarrow H^1(X,k^*)\rightarrow H^1(X,\mathcal{O}_{X'}^*)\rightarrow H^1(X,\mathcal{O}_{X'}^*)\rightarrow \cdots$$ hence $$H^1(X,\mathcal{O}_{X'}^*)\cong H^1(X,\mathcal{O}_{X}^*)$$ because $$H^i(X,k^*)=0 \ \forall \ i>0$$ since $$k^*$$ is a constant sheave on $$X$$ and this is flasque.

(e) If $$L'\cong \mathcal{O}_{X'}(D')$$ then $$D'\sim D$$ and $$\mathcal{O}_{X}(D)\in Div(X')$$ which is a contradiction because $$p\in D$$ since $$D$$ intersects $$C_1$$ properly ( item (a)).

If my reasoning is correct, because is important the algebraic equivalence to zero of $$C_1+C_2$$? and I don't know why it is necessary to $$X$$ to be non projective, although in the Hartshorne's book the note before this example says that for every projective scheme the map $$Div(X)\rightarrow Pic(X)$$ is sobrejective (by Nakai).

I would really appreciate any explanation for my questions and a revision for my reasonings for the proof of the proposition.

Thanks.

Nice write-up of an interesting proposition. There are parts I also don't understand, so this is more working through it with you than really answering your question.

for a., it's not true that algebraic equivalence implies linear equivalence -- but it is true that algebraic equivalence implies equality of intersection numbers, which is what Hartshorne is using.

b. yes, exactly.

c. I agree, it should be $$k^*$$ and not $$k$$. We can check that a sequence of sheaves is exact on stalks, and it's clear that this one is.

d. Right on, yes.

e. There's something I don't get here. We know that $$D$$ intersects $$C$$ properly, but maybe $$D'$$ doesn't. It seems that we are using something explicit about the correspondence between invertible sheaves on $$X$$ and $$X'$$.

Two more things I don't get (I'll use Roman numerals and maybe you can answer me):

i. Hartshorne just defines the new scheme $$X'$$ like it isn't a problem. But it's not true in general that if $$R$$ is a ring and $$P$$ is a prime, there's another ring $$S \to R$$ with a prime $$Q$$ pulling back to $$P$$ such that $$S_Q$$ is whatever ring you like and $$\text{Spec} (R \setminus P) = \text{Spec} (S \setminus Q)$$ .

ii. Where are we using that the ambient space $$X$$ is non-projective? It's easy to construct projective three-folds with disjoint irreducible algebraically equivalent curves.

• I see the answer to my question ii. now; we want $C_1$ to be $-C_2$, which perhaps can't happen on a projective three-fold. – hunter Oct 11 '18 at 13:06
• Thanks for your answer, its very helpful. Can you explain me more about your answer for your question ii), please? Why $C_1$ can not be $-C_2$ over a projective 3-fold? I don't understand your question i) and what means $Spec(R\P)$? Spec(R\P)=Spec(R)\P ? Hartshorne defines $X'=(X,\mathcal{O}_{X'})$ where $\mathcal{O}_{X',x}=\mathcal{O}_{X,x}$ for $x\neq P$ and $\mathcal{O}_{X',P}=\mathcal{O}_{X,P}\oplus k$ for $x= P$, and I assumed it as true (that X' is well defined), but I still trying to convince myself of that. Is your question i) about this? – Protágoras Oct 12 '18 at 1:40
• @Protagoras. Yes. My question in (i) is: a scheme needs to have an open cover by spectra of rings. I'm not convinced that one can just change the local ring at a point without some argument. For (ii), I don't know that this is true, just speculating. – hunter Oct 12 '18 at 2:12