5
$\begingroup$

The question is essentially the title. In other words, is there some universal property that the Amitsur complex for a morphism of rings $\phi:A\to B$ satisfies as a cosimplicial ring, or cosimplicial $A$-algebra? Suppose that the limit of the Amitsur complex for $\phi:A\to B$ is isomorphic to $A$, then given another cosimplicial $A$-algebra $C^\bullet$ whose limit is isomorphic to $A$, can we say that $C^\bullet$ is isomorphic to the Amitsur complex? Or that it at least admits a map to the Amitsur complex?

In particular I'm interested in the case that all of this is happening in some model category (or quasicategory) so the limits above are homotopical and the isomorphisms are just equivalences, but I just want to know if this is true in any setting.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ If $A$ and $B$ are commutative, the Amitsur complex is the initial cosimplicial $A$-algebras whose 0th term is a $B$-algebra. $\endgroup$
    – Marc Hoyois
    Commented Feb 24, 2016 at 14:45
  • $\begingroup$ Thanks @MarcHoyois do you know to what degree this holds for, say, ring spectra, commutative or not? $\endgroup$
    – Jonathan Beardsley
    Commented Feb 24, 2016 at 15:17
  • 3
    $\begingroup$ I was thinking of ring spectra. I'm just saying that the Cech nerve is the 0-coskeleton. But the Cech nerve interpretation only works for commutative things (where the tensor product is the coproduct). $\endgroup$
    – Marc Hoyois
    Commented Feb 24, 2016 at 15:51
  • $\begingroup$ @MarcHoyois ah okay. Yeah, I think I do know how to do this in that case, since you can iteratively build the Amitsur complex. I recall Clark Barwick saying at some point that there's a unique map from the free monoidal category with an algebra to associative ring spectra that picks out the Amitsur complex, and that it's unique. But I don't know where this is written down. $\endgroup$
    – Jonathan Beardsley
    Commented Feb 24, 2016 at 16:11

0

Reset to default

You must log in to answer this question.