Timeline for Is $A$ coflat over $A//B$?
Current License: CC BY-SA 3.0
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| Dec 25, 2011 at 12:25 | comment | added | Justin Noel | You could also use the A.1.1.6(b) if you know that all of your graded Hopf algebras are finite dimensional in each fixed degree (this is often the case in topology). This gives you the useful formula [ Ext^i_\Gamma( M^*, N)\cong Cotor^i(M^{**},N)\cong Cotor^i(M,N). ] | |
| Dec 25, 2011 at 12:18 | comment | added | Justin Noel | Now let's be specific. Let $C\rightarrow D\rightarrow E$ be a short exact sequence of $A//B$-comodules. Apply $-\square_{A//B} A$ to obtain a long exact sequence: [ 0\rightarrow C\square_{A//B} A\rightarrow D\square{A//B} A\rightarrow E\square_{A//B} A \rightarrow Cotor^1_{A//B}(C,A)\rightarrow Cotor^1_{A//B}(D,A)\rightarrow \cdots ] Since $A$ is injective the higher derived functors $Cotor^i_{A//B}(-,A)$ for $i>0$ are all zero. This shows that $-\square_{A//B} A$ is exact if $A$ is an injective $A//B$ comodule. | |
| Dec 25, 2011 at 12:09 | comment | added | Justin Noel | At this point I think everything is homological algebra: Fixing either one of the terms, the box product is left exact, at least over a field (or more generally if all of the tensor products used in the definition of the box product involve flat modules). Formally, this is because it is defined as a kernel of a map of modules. Since we have enough injectives we can compute the right derived functors of the box product using an injective resolution of one of the variables. If one of those variables is injective there are no higher derived functors. | |
| Dec 24, 2011 at 22:44 | vote | accept | Vitaly Lorman | ||
| Dec 24, 2011 at 22:44 | comment | added | Vitaly Lorman | Thanks for the great answer! I see why extended $A//B$-comodules over a field are injective. This means that the functor Hom$_{A//B}(−,A)$ is exact, but why does this imply that the functor $−\Box_{A//B}A$ is exact? Is there a connection between Hom and the cotensor product as in A1.1.6(b) in Ravenel's book, or is it simpler than this? | |
| Dec 23, 2011 at 12:48 | history | answered | Justin Noel | CC BY-SA 3.0 |