Non-cohomological proof that the pullback of an ample bundle by a finite morphism is ample - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T04:13:49Z http://mathoverflow.net/feeds/question/62996 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62996/non-cohomological-proof-that-the-pullback-of-an-ample-bundle-by-a-finite-morphism Non-cohomological proof that the pullback of an ample bundle by a finite morphism is ample Charles Staats 2011-04-26T02:38:07Z 2011-04-27T00:05:50Z <p>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:</p> <p>Does anyone know a reasonable proof of this fact that does not go through cohomology?</p> http://mathoverflow.net/questions/62996/non-cohomological-proof-that-the-pullback-of-an-ample-bundle-by-a-finite-morphism/63004#63004 Answer by Sándor Kovács for Non-cohomological proof that the pullback of an ample bundle by a finite morphism is ample Sándor Kovács 2011-04-26T05:47:43Z 2011-04-26T09:15:58Z <p>At least for a projective variety over a field you can use Seshadri's criterion. </p> <p>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 $y\in B$, $$\frac{D\cdot B}{\mathrm{mult}_y B}\geq \varepsilon.$$</p> <p>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.</p> http://mathoverflow.net/questions/62996/non-cohomological-proof-that-the-pullback-of-an-ample-bundle-by-a-finite-morphism/63007#63007 Answer by Sándor Kovács for Non-cohomological proof that the pullback of an ample bundle by a finite morphism is ample Sándor Kovács 2011-04-26T06:27:36Z 2011-04-27T00:05:50Z <p>Here is another proof: </p> <p>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.</p> <p>We want to prove that <code>$F\otimes (f^*L)^m$</code> is generated by global sections for $m\gg 0$.</p> <p>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 $X$.</p> <p>Consider an arbitrary $y\in V$ and the image of <code>$s\in (F\otimes L^m)(U)\to(f_*F\otimes L^m)(V)\to (f_*F)_y$</code> 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.</p> <p>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 $s$. </p> http://mathoverflow.net/questions/62996/non-cohomological-proof-that-the-pullback-of-an-ample-bundle-by-a-finite-morphism/63053#63053 Answer by K Halb for Non-cohomological proof that the pullback of an ample bundle by a finite morphism is ample K Halb 2011-04-26T15:47:42Z 2011-04-26T15:47:42Z <p>Unfortuatelly, this is too long for a comment.</p> <p>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 $m\gg 0$?</p> <p>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.</p> <p>Since pullback is right exact and commutes with the tensor product, we get an induced surjective map $O_X^{(I)}=f^*O_Y^{(I)}\twoheadrightarrow f^* f_* (F\otimes (f^* L)^m)$.</p> <p>Finally the natural map $f^* f_* (F\otimes (f^* L^m))\to F\otimes (f^* L)^m$ is surjective since $f$ is affine.</p> <p>These two maps together give the desired surjection $O_X^{(I)}\twoheadrightarrow F\otimes (f^* L)^m$.</p> http://mathoverflow.net/questions/62996/non-cohomological-proof-that-the-pullback-of-an-ample-bundle-by-a-finite-morphism/63072#63072 Answer by diverietti for Non-cohomological proof that the pullback of an ample bundle by a finite morphism is ample diverietti 2011-04-26T20:36:58Z 2011-04-26T20:42:05Z <p>Here is a differential geometric point of view, which of course works only over $\mathbb C$.</p> <p>Since $L\to X$ is ample, then it carries a smooth hermitian metric $h$ of positive definite Chern curvature $i\Theta(L)>0$.</p> <p>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 <code>$i f^*\Theta(L)$</code>. Since $f$ is finite, then its differential is (more or less) injective, so that <code>$i f^*\Theta(L)$</code> is positive definite, too.</p> <p>But then, by Kodaira's projectivity criterion, $f^*L$ is ample. </p>