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Ted E
Ted E
9:59 PM
If $f:R\to S$ is a ring homomorphism, where $R,S$ are not necessarily commutative. What conditions can be placed on $f$, such that the restriction of $S$-modules to $R$-modules via $f$ preserves projectivity? These clearly exist, since we have the identity map, and yet inclusion and quotients can fail to satisfy this $\Bbb Z\to \Bbb Q$ and $\Bbb Z\to \Bbb Z/p\Bbb Z$ for example.

Originally I was thinking I could make some categorical argument, using the adjunction $\otimes_R S\dashv \text{res}_f$, but that can only help me restate that projective implies flat.
 
S. carmeli
S. carmeli
@TedE isn't S being projective as an R-module necessary and sufficient? I mean, S is projective as an S-module so it is clearly necessary but on the other hand a projective is a retract of a free and if S is projective as an R module so does every free S-module. Or maybe this works only for commutative or something?
 
Ted E
Ted E
11:05 PM
@S.carmeli Great, thanks!
 
Adrian Clough
Adrian Clough
I’m reading Higher Galois Theory by Hoyois. There, a torsor in a topos $\mathcal{X}$ is defined to be a pair $(A,\chi)$, where $A \in \mathcal{S}$ together with a map $1 \to \pi^*A$. This is claimed to be equivalent to a functor $P: A \to \mathcal{X}$ whose colimit is $1 \in \mathcal{X}$ via descent. I’m for the life of me not able to see how to exhibit this equivalence.
The relevant notion of descent here would seem to be $\mathcal{X} \simeq \mathcal{X}_{/1} \simeq \mathcal{X}_{\mathrm{colim}\,P} \simeq \lim_{\alpha \in A} \mathcal{X}_{/P_\alpha}$, but this doesn’t seem to help. Could someone point me in the right direction?
 

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