Timeline for Exponentials in the opposite category of finite separable algebras
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 20, 2013 at 6:24 | vote | accept | Fujita Tomomi | ||
| Nov 19, 2013 at 23:20 | comment | added | Marguax | @Fujita: In geometric terms, the special case of the "subobject classifier" $2^Y$ for a given finite etale $X$-scheme $Y$ is a representing object for the functor ${\mathbf{Of}}(Y)$ assigning to any $X$-scheme $T$ the set of open and closed subschemes of $Y \times_X T$ (since monic maps in the category of finite etale $X$-schemes are open and closed immersions). See Lemma 18.5.3 in EGA IV$_4$ for a nice discussion of representing this latter functor more generally for $Y \rightarrow X$ any finite locally free morphism (though it is not a "subobject classifier" when $Y$ isn't etale over $X$). | |
| Nov 19, 2013 at 20:24 | comment | added | Marguax | Dear Mariano: To see explicitly where your suggestion goes awry, note that it would provide an algebra map from $C$ into $C^B$, hence from $C \otimes_K B$ into $C$, and so from $B$ into $C$ (chase $1 \otimes b$). But there is usually no such map. In effect, what is "missing" is to adequately combine the Galois actions on the sets of primitive idempotents of $B \otimes K_s$ and $C \otimes K_s$ (which is what underlies the construction in my answer). | |
| Nov 19, 2013 at 16:29 | comment | added | Marguax | This guess is unfortunately wrong. The answer has to make more effective use of the fact that $B$ is an etale $K$-algebra. See my answer below. | |
| Nov 19, 2013 at 16:22 | answer | added | Marguax | timeline score: 12 | |
| Nov 19, 2013 at 15:19 | answer | added | Paul Taylor | timeline score: 2 | |
| S Nov 19, 2013 at 13:02 | history | suggested | Paul Taylor |
added linear logic tag to get the attention of the experts
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| Nov 19, 2013 at 12:51 | review | Suggested edits | |||
| S Nov 19, 2013 at 13:02 | |||||
| Nov 19, 2013 at 12:49 | answer | added | Paul Taylor | timeline score: 4 | |
| Nov 19, 2013 at 7:02 | comment | added | Mariano Suárez-Álvarez | A natural guess is $C^B=B^*\otimes C$ with $B^*$ the dual space of $B$ with some algebra structure; since $B$ is separable, there are special isomorphisms $B\to B^*$, and one can transport the algebra structure to $B^*$ (and probably $\mathrm{op}$ it?) | |
| Nov 19, 2013 at 6:50 | history | asked | Fujita Tomomi | CC BY-SA 3.0 |