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Jan 16, 2013 at 17:05 history edited Timo Schürg CC BY-SA 3.0
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Jan 14, 2013 at 14:02 comment added Timo Schürg I think this true, at least with finiteness ass. By the way, I'd be surprised if your question has a complete answer in terms of only ordinary rings. As evidence, one can define something like L-finite maps having perfect rel.cot.complex. Then this notion is completely unrelated to being of finite presentation. On the contrary, if you look at maps that are homotopically of finite presentation, then you can again detect these on the cotangent complex (if you add $\pi_0(B)$ finitely presented over $\pi_0(A)$). In short, I think it's hard to say something about the cot.complex using ord. rings.
Jan 14, 2013 at 11:19 comment added anon Ah, thanks. This observation strongly supports the (well-advertised) point of view that a simplicial commutative ring is just a ring with additional "nilpotent" data. It also raises the following question: if $A \to B$ is an $L$-trivial map of ordinary rings that is additionally an isomorphism modulo nilpotents, then is $A \simeq B$?
Jan 14, 2013 at 10:51 history answered Timo Schürg CC BY-SA 3.0