# Is $A$ coflat over $A//B$?

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.)

• 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/… ... Commented Dec 22, 2011 at 23:46
• I can't speak for OP, but I've seen $A // B$ used to denote $A \otimes_B k$. Commented Dec 23, 2011 at 1:07
• Ya the definition Vitaly uses for Coflat is also the one Im an "familiar" with
– ABIM
Commented Feb 23, 2014 at 23:18

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.
• 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? Commented Dec 24, 2011 at 22:44
• 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. Commented Dec 25, 2011 at 12:18