Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
3  
There is also a non-cohomological proof in EGA II, (5.1.12), which shows that for an arbitrary quasi-compact scheme $Y$, an ample line bundle $L$ on $Y$ and a quasi-affine morphism $f\colon X \to Y$ the pullback $f^*L$ is again ample. –  Torsten Wedhorn Apr 30 '11 at 5:46
add comment

4 Answers

up vote 13 down vote accepted

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 $O_X^{(I)}\twoheadrightarrow F\otimes (f^* L)^m$.

share|improve this answer
    
This is the sort of thing I was hoping for. –  Charles Staats Apr 26 '11 at 21:44
    
Very nice! Follow up discussion <a href="math216.wordpress.com/2011/03/31/…;. –  Ravi Vakil Apr 30 '11 at 22:29
    
Thank you very much :-) –  K Halb May 2 '11 at 14:39
add comment

Here is a differential geometric point of view, which of course works only over $\mathbb C$.

Since $L\to X$ is ample, then it carries a smooth hermitian metric $h$ of positive definite Chern curvature $i\Theta(L)>0$.

Now let $f\colon Y\to X$ be any finite morphism. Then $f^*L$ inherits a smooth hermitian metric which is just the pull-back oh $h$ and its Chern curvature is given by the pull-back $i f^*\Theta(L)$. Since $f$ is finite, then its differential is (more or less) injective, so that $i f^*\Theta(L)$ is positive definite, too.

But then, by Kodaira's projectivity criterion, $f^*L$ is ample.

share|improve this answer
1  
Diverietti, sorry, but are you claiming that the curvature of the pullback is strictly positive everywhere, even at the ramification points (I don't see why), or are you saying this not an essential problem? I believe it must the second possibility, but perhaps some extra work is needed? –  Donu Arapura Apr 26 '11 at 21:25
    
I intended (but indeed I didn't say) that it is not an essential problem! You definitely need some extra work, but it is quite straightforward: after all, at those points, it is positive semi-definite, so a small local perturbation will do the job. I didn't want to write a heavy answer, since there were already plenty of complete proofs! –  diverietti Apr 26 '11 at 21:30
    
Anyway, what I wrote suffices to conclude (from a differential geometric point of view) that $f^*L$ is semi-ample and big. It seems to me that at this point it is sufficient to observe that it does not contract anything to conclude that it is ample, right? –  diverietti Apr 26 '11 at 21:35
    
Sure, that's fine. I just wanted to make sure I understood. –  Donu Arapura Apr 26 '11 at 21:42
add comment

Here is another proof:

Let $f:X\to Y$ be finite, $L$ an ample line bundle on $Y$, and let $F$ be a coherent sheaf on $X$. We may assume that $X$ and $Y$ are irreducible. Assume that $X$ is generically reduced.

We want to prove that $F\otimes (f^*L)^m$ is generated by global sections for $m\gg 0$.

Let $x\in X$ and $s\in F_x$. By noetherian induction we may assume that $x\in X$ is general and hence $X$ is locally irreducible and we may assume that $s$ is not torsion. Choose a (small enough irreducible) neighborhood $x\in U\subseteq X$ such that we may assume that $s\in (F\otimes L^m)(U)$. Let $Z=X\setminus U$. Since $f$ is proper, $f(Z)\subsetneq Y$ is closed. Let $V=Y\setminus f(Z)$, a non-empty open set. Then $f^{-1}V\subseteq U$ is a non-empty and hence dense open set in $X$.

Consider an arbitrary $y\in V$ and the image of $s\in (F\otimes L^m)(U)\to(f_*F\otimes L^m)(V)\to (f_*F)_y$ via restriction. By assumption $L$ is ample, so $f_*F\otimes L^m$ is generated by global sections for $m\gg 0$, so there exists a $\sigma\in H^0(Y, f_*F\otimes L^m)$ such that its image in $(f_*F)_y$ is the above element.

Observe that $\sigma\in H^0(X, F\otimes (f^*L)^m)$ and the above means that its restriction to $U$ agrees with $s$ on a non-empty open subset. But then it has to agree everywhere on $U$, in particular its germ at $x$ has to be the same as the original $s$.

share|improve this answer
4  
Dear Sandor, I'm a bit confused by the claim of the last paragraph. What if $F$ has a torsion piece, of which $s$ was a section, and the open set on which $\sigma$ and $s$ coincide was contained in the complement of the support of that torsion piece (so in fact $\sigma$ is just zero). In short, to go from agreement on a dense open subset to agreement on all of you, don't you need to know that this non-empty open subset meets every component of the support of $s$? Best wishes, Matt –  Emerton Apr 26 '11 at 11:58
1  
Matt, (as always) you are completely right. I knew there was an issue with torsion, but did not have time to flash it out. I think it is relatively easy to handle that case via noetherian induction. Given that in the mean time other solutions were produced, I don't feel motivated to work out the details, but I think this works. Thanks for pointing that out. Best, –  Sándor Kovács Apr 27 '11 at 0:08
add comment

At least for a projective variety over a field you can use Seshadri's criterion.

Let $f:X\to Y$ be finite and $D\subset Y$ an ample (Cartier) divisor. By Seshadri's criterion there exists an $\varepsilon>0$, such that for any proper curve $B\subseteq Y$, and any $y\in B$, $$ \frac{D\cdot B}{\mathrm{mult}_y B}\geq \varepsilon.$$

Now consider any proper curve $C\subseteq X$ and any $x\in X$. Let $B=f_*C$ and $y=f(x)$. Since $f$ is finite, $B$ is an actual curve. (I.e., if it weren't finite, $B$ could be zero as a $1$-cycle.) Then since $f^*D\cdot C=D\cdot f_*C$ and ${\mathrm{mult}_xC}\leq {\mathrm{mult}_yB}$, we have $$ \frac{f^*D\cdot C}{\mathrm{mult}_xC}\geq \frac{D\cdot B}{\mathrm{mult}_y B}\geq \varepsilon,$$ which implies that $f^*D$ is ample by Seshadri's criterion again.

share|improve this answer
3  
I think one can also make a proof using the Nakai-Moishezon criterion along these lines. Let $f:X\to Y$ be the finite morphism. We need only check that $f^*D$ has positive self-intersection, and that ${f^*D}^{dim Z}.Z>0$ for any integral subscheme $Z\subset X$. But these items both follow from the projection formula. –  J.C. Ottem Apr 26 '11 at 9:25
    
Perhaps I'm missing something, but to use Seshadri's criterion, don't you also have to assume $X$ and $Y$ are smooth? –  Charles Staats Apr 26 '11 at 13:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.