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Let $A$ be a Hopf algebra over a field $k$. Let $B$ be a normal subHopf algebra of $A$. Is $A$ coflat over $A//B$? An explanation would be greatly appreciated.

(A novice to Hopf algebras, I am attempting to follow the computation of the homotopy of some Thom spectra in Kochman's book. Given $F$, an $A//B$-free coresolution of $k$, Kochman states that $F \Box_{A//B} A$ is an $A$-free coresolution of $k \Box_{A//B} A \cong B$. I don't see why $-\Box_{A//B} A$ preserves exactness.)

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What is $A//B$ ? Some GIT quotient? Also, a definition of "coflat" would be nice. You mean that cotensoring with $A$ is exact? Somehow I am not really sure it is the same "coflat" as in projecteuclid.org/DPubS/Repository/1.0/… ... –  darij grinberg Dec 22 '11 at 23:46
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I can't speak for OP, but I've seen $A // B$ used to denote $A \otimes_B k$. –  Eric Peterson Dec 23 '11 at 1:07
    
Ya the definition Vitaly uses for Coflat is also the one Im an "familiar" with –  CSA Feb 23 at 23:18
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1 Answer

up vote 6 down vote accepted

I'm going to assume that your Hopf algebras are connected in which case this follows from Theorem 4.10 of Milnor-Moore (On the structure of Hopf-algebras). That result shows that $A\cong B\otimes A//B$ as a left $B$-module and right $A//B$-comodule. I should point out that this result is remarkably useful.

This means $A$ is an extended $A//B$-comodule over a field and hence it is injective in the category of $A//B$-comodules. The fact that extended coalgebras are injective (when working over a field) is an easy exercise, but you can also find the result in the context of Hopf-algebroids as A1.2.2 'in Ravenel's Complex Cobordism and Stable Homotopy.'

Since $A$ is injective the functor $-\square _{A//B} A$ is exact.

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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? –  Vitaly Lorman Dec 24 '11 at 22:44
    
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. –  Justin Noel Dec 25 '11 at 12:09
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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. –  Justin Noel Dec 25 '11 at 12:18
    
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). ] –  Justin Noel Dec 25 '11 at 12:25
    
Thanks for the detailed answer! –  Vitaly Lorman Dec 26 '11 at 20:57
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