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?

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).
