It is a standard fact that for any finite morphism of proper Noetherian $A$-schemes ($A$ being Noetherian), the pullback of an ample line bundle is ample. The usual proof of this fact is via Serre's cohomological criterion for ampleness. However, since the statement seems, on its face, to have nothing to do with cohomology, I thought the following question worth asking:


Does anyone know a reasonable proof of this fact that does not go through cohomology?



Non-cohomological proof that the pullback of an ample bundle by a finite morphism is ample
Unfortuatelly, this is too long for a comment.


Can't we directly show that for every coherent sheaf $F$ on $X$ we have that $F\otimes (f^\*L)^m$ is generated by global section for $m\gg 0$?


Since $f$ is finite and $L$ is ample, we have that $f\_\*F\otimes L^m$ is generated by global sections for $m\gg 0$. So there is a surjection $O\_Y^{(I)} \twoheadrightarrow f\_\* F\otimes L^m$. Note that $f\_\* F\otimes L^m=f\_\* (F\otimes f^\* (L^m))$ by the projection formula.


Since pullback is right exact and commutes with the tensor product, we get an induced surjective map $O\_X^{(I)}=f^\*O\_Y^{(I)}\twoheadrightarrow f^\* f\_\* (F\otimes (f^\* L)^m)$.


Finally the natural map $f^\* f\_\* (F\otimes (f^\* L^m))\to F\otimes (f^\* L)^m$ is surjective since $f$ is affine.


These two maps together give the desired surjection $O\_X^{(I)}\twoheadrightarrow F\otimes (f^\* L)^m$.


