Timeline for What is the support of a coherent sheaf on $X\times Y$ if it is invariant by tensoring a very ample line bundle on $X$?
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Dec 5, 2016 at 6:05 | comment | added | t3suji | I don't have a reference, but it is pretty straightforward: enough to consider the case $S=Spec(R)$, and then a coherent sheaf $F$ is given by a finitely generated graded $R[x_0,\dots,x_n]$-module $M=\bigoplus M_d$; for large enough $D$, $M_D$ generates $M_{\ge D}:=\bigoplus_{d\ge D} M_d$, so the annihilator ideal of $M_D$ in $R$ is contained in the annihilator ideal of $M_{\ge D}$. | |
Dec 5, 2016 at 2:41 | comment | added | Zhaoting Wei | @t3suji Is it a standard result in algebraic geometry? | |
Dec 4, 2016 at 22:16 | comment | added | t3suji | The claim is that for any (quasi-compact noetherian) scheme $S$ and any coherent sheaf $F$ on $S\times{\mathbb P}^n$, we have $p(supp(F))=supp(p_*(F(n)))$ for $n\gg 0$. Here $p$ is the projection onto $S$. | |
Dec 4, 2016 at 16:25 | comment | added | Zhaoting Wei | @t3suji I'm sorry why this gives what I want? Could you give some more details? | |
Dec 4, 2016 at 7:25 | comment | added | t3suji | Note that if $F$ has this property, then so does $F\otimes p_Y^* E$ for any sheaf $E$ on $Y$. In particular, you know that $dim(supp (p_*(F\otimes p_Y^*E)))=0$ for any $E$. Taking $E$ to be a high power of an ample line bundle gives you what you want. | |
Dec 4, 2016 at 3:53 | history | asked | Zhaoting Wei | CC BY-SA 3.0 |