Skip to main content
6 events
when toggle format what by license comment
Dec 7, 2013 at 12:23 comment added DonD Very good. I accepted your answer, but unfortunately i don't have enough reputation to give a "+1". But this answer was really helpful.
Dec 7, 2013 at 12:19 vote accept DonD
Dec 7, 2013 at 12:10 vote accept DonD
Dec 7, 2013 at 12:19
Dec 7, 2013 at 5:02 comment added Will Sawin Yes. You just need to choose an ideal of degree $n_i$ in $\mathbb A^1_{\mathbb F_{p_i}}$. Then choose the ideal of elements that are in the $i$th ideal mod $p_i$. By the Chinese remainder theorem, adding conditions at $p_j$ for $j\neq i$ will not change the ideal modulo $p_i$, which was constructed to be $O(-n_i)$.
Dec 6, 2013 at 18:47 comment added DonD Fantastic. In the other case we can write $O_Y=\oplus F_{p_i}$ for different primes $p_i$ and the same computation works. So we have an exact sequence $F_p\rightarrow I_Y\otimes F_p \rightarrow O(-1)$. Is it possible to choose $Y$ such that we get an exact sequence: $Tor(O_Y,F_{p_i})\rightarrow I_Y\otimes F_{p_i}\rightarrow O(-n_i)$ for finitely many primes $p_i$ with numbers $n_i$?.
Dec 6, 2013 at 16:24 history answered Will Sawin CC BY-SA 3.0