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.

up vote 1 down vote accepted

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 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 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 at 2:12

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