Lagrange's theorem for Hopf algebras

Under what conditions is a Hopf algebra free over any of its sub-Hopf algebras?

I am reading "Hopf algebras and their actions on rings" by Susan Montgomery, specifically chapter 3. Lagrange's theorem says that if $H$ is a subgroup of $G$ then $\mathbb kG$ is free as a $\mathbb kH$ module. When one tries to extend this to general Hopf algebras, in the finite dimensional setting everything is fine by the Nichols-Zoeller theorem which says that a finite dimensional Hopf algebra $H$ is free over any sub-Hopf algebra $K$.

Unfortunately this is not true in the infinite dimensional setting. A counterexample, which is both commutative and cocommutative, due to Oberst and Schneider, is as follows: Take $F\subset E$ a Galois field extension of degree $2$ with Galois group $G=\lbrace 1,\sigma\rbrace$, and let $\sigma$ act on $\mathbb Z$ by $z\to -z$. Then $G$ acts on $E$ and $\mathbb Z$ so one can take the Hopf algebra $H=(E\mathbb Z)^{G}$. One gets a counterexample showing that $H$ is not free over the sub-Hopf algebra $(E(n\mathbb Z))^G$ when $n$ is even.

The book then mentions that, because of this, perhaps one should look at something weaker than freeness, like being faithfully flat. Here, however, I want to focus on which Hopf algebras are free over their sub-Hopf algebras. It seems that in the literature there are several instances of this kind of result. For example this seems to hold for pointed Hopf algebras. Is there an intuitive explanation of what properties of a Hopf algebra guarantee that it satisfies Lagrange's theorem? (For example, I do not see how I could have arrived naturally at the example above. A short explanation of that would be great, too.) Is there a reference?

• The basic question here is a natural one, though it might not have a single definitive answer. – Jim Humphreys Dec 19 '11 at 15:27