ampleness of invertible sheaves 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

*

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


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