7
$\begingroup$

I'm trying to solve Exercise III 12.4 from Hartshorne's Algebraic geometry. There is a flat projective morphism $f: X \to Y$ of schemes of finite type over an algebraically closed field $k$. Also $Y$ is assumed to be integral, and all fibers are integral schemes.

Now suppose $\mathcal{F}$ is an invertible sheaf on $X$, that is trivial on each fiber $X_y$. I was able to show that $f_*\mathcal{F}$ is an invertible sheaf on $Y$ (this is essentially Cor 12.9), and now I would like to show that the natural map $f^*f_* \mathcal{F} \to \mathcal{F}$ is an isomorphism, for which it is enough to show that it is surjective, because both sheaves are locally free.

$\endgroup$
2
  • $\begingroup$ @KReiser Is it sufficient to choose $U \subset Y$ such that $f_* \mathcal{F}|_U \cong \mathcal{O}_Y|_U$? If so, why? $\endgroup$ Commented May 5, 2020 at 16:19
  • $\begingroup$ My previous comment was not as helpful as it should have been, and I am sorry. I have posted a solution, do let me know if it resolves your issues. $\endgroup$
    – KReiser
    Commented May 6, 2020 at 6:03

1 Answer 1

6
$\begingroup$

Since whether $f^*f_*\mathcal{F}\to\mathcal{F}$ is surjective is a local condition, we can check it on a cover $X$ by open subsets of the form $f^{-1}(U)$ where $U\subset Y$ is open and $f_*\mathcal{F}|_U$ is free. So it suffices to treat the case where $Y=\operatorname{Spec} R$ is affine and $f_*\mathcal{F}=\mathcal{O}_Y$.

Now let's recall some facts about the natural map $f^*f_*\mathcal{F}\to\mathcal{F}$. First, the natural map is the image of $id_{f_*\mathcal{F}}\in \operatorname{Hom}_{\mathcal{O}_Y}(f_*\mathcal{F},f_*\mathcal{F})$ under the isomorphism of $\mathcal{O}_Y(Y)=R$-modules given by the adjunction $$\operatorname{Hom}_{\mathcal{O}_Y}(f_*\mathcal{F},f_*\mathcal{F})\cong \operatorname{Hom}_{\mathcal{O}_X}(f^*f_*\mathcal{F},\mathcal{F}).$$

But since $f_*\mathcal{F}=\mathcal{O}_Y$, we have $\operatorname{Hom}_{\mathcal{O}_Y}(f_*\mathcal{F},f_*\mathcal{F})=\operatorname{Hom}_{\mathcal{O}_Y}(\mathcal{O}_Y,\mathcal{O}_Y)=\mathcal{O}_Y(Y)=R$, and $id_{f_*\mathcal{F}}=id_{\mathcal{O}_Y}=1\in R$. On the other hand, the pullback of the structure sheaf is the structure sheaf, so $\operatorname{Hom}_{\mathcal{O}_X}(f^*f_*\mathcal{F},\mathcal{F})=\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X,\mathcal{F})=\mathcal{F}(X)$, so $\mathcal{F}(X)=R$. As the $R$-linear endomorphisms of $R$ are exactly given by multiplication by an element of $R$, the endomorphisms that are isomorphisms are exactly multiplication by a unit. So we see that $id_{f_*\mathcal{F}}$ must be sent to the map $\mathcal{O}_X\to\mathcal{F}$ which picks out an invertible element of $\mathcal{F}(X)$, that is, a non-vanishing global section. So $\mathcal{F}$ is trivial and the natural map $f^*f_*\mathcal{F}\to\mathcal{F}$ is an isomorphism.

$\endgroup$
5
  • $\begingroup$ I don't see why the global section $s \in \mathcal{F}(X)$ should be non-vanishing. Your reasoning appears a bit circular, we don't know yet that $\mathcal{O}_X \to \mathcal{F}$ is an isomorphism. $\endgroup$ Commented May 6, 2020 at 7:12
  • $\begingroup$ Also, this would prove the general implication $f_* \mathcal{F} \cong \mathcal{O}_Y \implies \mathcal{F} \cong \mathcal{O}_X$, without any conditions on $Y$, $X$ or $f$. That makes me a bit suspicious, although I would like to be convinced if it's true. $\endgroup$ Commented May 6, 2020 at 7:15
  • 1
    $\begingroup$ @red_trumpet $s\in \mathcal{F}(X)=R$ is the image of the global section $1\in \mathcal{O}_X(X)=R$ under an $R$-linear isomorphism of $R$. Since $\operatorname{Hom}_R(R,R)=R$ for commutative rings, with composition of maps corresponding to multiplication of the corresponding elements of $R$, we see that all the isomorphisms are given by multiplication by units. So $s$ is a unit, which means it's not in any maximal ideal, which is equivalent to saying it doesn't vanish anywhere. $\endgroup$
    – KReiser
    Commented May 6, 2020 at 8:04
  • 1
    $\begingroup$ We did also use the fact that $\mathcal{F}$ was a line bundle in here - the conclusion that $\mathcal{F}$ is trivial if one can produce a nowhere vanishing global section is essential and is only true for line bundles. $\endgroup$
    – KReiser
    Commented May 6, 2020 at 8:07
  • 1
    $\begingroup$ No, $\mathcal{O}_X(X)=R$. By Stein factorization combined with the fact that $f:X\to Y$ is projective with geometrically irreducible fibers, we have $f_*\mathcal{O}_X=\mathcal{O}_Y$. $\endgroup$
    – KReiser
    Commented May 6, 2020 at 8:12

You must log in to answer this question.

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