Timeline for Are closed subfunctors complementary to open subfunctors?
Current License: CC BY-SA 3.0
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| Jul 25, 2011 at 13:14 | comment | added | amaanush | ...to an $A$-valued point of $\mathcal{F}$), the unique reduced ideal in A would be the nil-radical of A, and not the extension of $\{0\}$, which is $\{0\}$! In other words, the association above, as you describe do not associate $I_A \subseteq A$ functorially, thus $\mathcal{F}(Spec (A/I_A)) \subseteq \mathcal{F}$ functorially, ie. does not give a subfunctor. While it may still be possible to augment your method to correctly get an ideal in "$\mathcal{O}_\mathcal{F}$". Please let me know if I have understood you correctly. thanks. | |
| Jul 25, 2011 at 12:54 | comment | added | amaanush | I am inclined to think that by "compatible", you mean a functorial association (of course, only then do they stitch together to give a closed subfunctor)-- but that's exactly where the problem is! Given an extension $A\to B$, the pullback of $Spec (A/I)$ to $Spec (B)$ is given by the extension $J = I^e \subseteq B$, but this only has $J \subseteq I_B$, or in fact $\sqrt $J = $I_B$, and need not coincide with $I_B$. For example, even if $\mathcal{F}$ is representable, by, say $Spec (R)$, and $Z$ actually corresponds to $|spec R/{0}| = Spec (R)$, for a non-reduced R-algebra A (corresponding .. | |
| Jul 22, 2011 at 8:05 | history | answered | Martin Brandenburg | CC BY-SA 3.0 |