Timeline for How do ideal sheaves behave on the special fibers of the projective line over the integers?
Current License: CC BY-SA 3.0
6 events
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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 |