5

Let $f: X\rightarrow Y$ be a morphism of schemes over a field $k$. Let $\mathcal{L}$ be an invertible sheaf on $Y$. My question is

  1. If $\mathcal{L}$ is ample, is $f^*\mathcal{L}$ ample?

  2. If 1. is not true, is there the condition that $f^*\mathcal{L}$ is ample?

user875280
  • 63
Sang Cheol Lee
  • 3,119

2 Answers2

9

Check out Hartshorne excersise III.5.7. Among other things, it proves the following:

Suppose $X$ and $Y$ are proper. If $\mathcal{L}$ is ample on $X$ and $i: Y \to X$ is a closed immersion then $i^*\mathcal{L}$ is ample on $Y$. Furthermore, if $f:X \to Y$ is a proper finite surjective morphism and $\mathcal{L}$ is any line bundle on $Y$, then $\mathcal{L}$ is ample if and only if $f^*\mathcal{L}$ is ample. More generally, putting these two facts together, we get that if $f$ is just finite and $\mathcal{L}$ is ample, then $f^*\mathcal{L}$ is ample.

The proofs of these use the cohomological criterion for ampleness (see proposition III.5.3 in Hartshorne) which says that $\mathcal{L}$ is ample on $X$ (where $X$ is proper) if and only if for every coherent sheaf $\mathcal{F}$ there exists an $n_0$ such that

$$ H^i(X,\mathcal{F} \otimes \mathcal{L}^{\otimes n}) = 0 \enspace \enspace \enspace \forall \enspace i > 0,\enspace n > n_0 $$

Putting this together with the fact that the pushforward along affine morphisms preserves cohomology, proper morphisms preserve coherence, and finite morphisms satisfy the projection formula, we get the proof.

You definitely need strong conditions like this to make sure the pullback is ample. My guess is there are counterexamples if we weaken any of the conditions though I can't think of some immediately. I'll add some if I do.

Dori Bejleri
  • 5,090
  • @SangCheolLee A nice thing to do would be to upvote Dori's answer. –  Jan 02 '14 at 04:36
  • Thanks! I'll check it out. – Dori Bejleri Jan 02 '14 at 05:41
  • 1
    More generally, when X is quasi-compact, ampleness is equivalent to ampleness of the inverse image by any closed immersion whose sheaf of ideals is nilpotent. See (EGA II, 4.6). The finiteness assumption can also be weakened I'm pretty sure, but I don't remember the right condition at the moment. –  Jan 02 '14 at 15:56
  • In your second statement, when you say $f$ is a proper surjective morphism, do you mean, proper surjective finite morphism? – adrido Oct 24 '14 at 03:00
  • Yes. Thanks! I'll edit it. – Dori Bejleri Oct 24 '14 at 03:16
7

The conditions for the pullback of an ample line bundle (at least that I am familiar of) are that $f : X \to Y$ be a finite morphism, and that $X,Y$ be Noetherian schemes.

I will now prove this without cohomology. Suppose we have a coherent sheaf $\mathscr{F}$ on $X$. Then $f_\ast \mathscr{F}$ is coherent because $Y$ is Noetherian and $f$ is finite. As $\mathcal{L}$ is ample, there is $n_0 \in \Bbb{N}$ such that for all $n \geq n_0$, $f_\ast \mathscr{F} \otimes \mathcal{L}^{\otimes n}$ is globally generated. Now fix an $n \geq n_0$; we have a surjection

$$\mathcal{O}_Y^I \to f_\ast\mathscr{F}\otimes \mathcal{L}^{\otimes n}\to0$$ which pulls back to give a surjection

$$\mathcal{O}_X^I \to f^\ast f_\ast \mathscr{F}\otimes f^\ast\mathcal{L}^{\otimes n} \to 0.$$

As $f$ is finite (and thus affine), the natural map $\mathscr{F} \to f^\ast f_\ast \mathscr{F}$ is surjective, so we get a surjection $\mathcal{O}_X^I \to \mathscr{F} \otimes f^\ast\mathcal{L}^n \to 0$, showing that $f^\ast \mathcal{L}$ is ample.

For a converse of the result above, we need that $f : X \to Y$ be finite and surjective. The hypotheses cannot be dropped as the following example will show: Take $f : \Bbb{A}^1 \to \Bbb{P}^1$ that sends $x \mapsto [1:x]$, note $f$ is not surjective. For any $ n < 0$, $\mathcal{O}(n)$ is not ample but its pullback is $\mathcal{O}_{\Bbb{A}^1}$ which is ample (the structure sheaf of a scheme is ample iff the scheme is affine).