In a comment for this old question, it was said that

> There is a theorem that Hopf modules are (up to isomorphism) the tensor products of their coinvariants with H. (This theorem is usually worded in terms of a category equivalence. I don't know a good reference.)

I'm guessing that this means that given a (right) $H$-comodule $V$, for $H$ a Hopf algebra, equipped with a right $H$ action for which $\Delta_R(vh) = v_{(0)} g_{(0)} \otimes v_{(1)}g_{(1)}$, we have an isomorphism $$ V \simeq V_{\text{inv}} \otimes H $$ Now it's easy to see that we have a surjective map $$ V_{\text{inv}} \otimes H \to V, ~~~ v \otimes h \mapsto vh. $$ How does one show that this is an isomorphism?

EDIT: Wait, I think this is actually obvious: The map $$ V \to V_{\text{inv}} \otimes H, ~~~~~ v \mapsto v_{(0)} S(v_{(1)}) \otimes v_{(2)}, $$ seems to have the multiplication map as its inverse. So this gives us the isomorphism. Yes?

  • 1
    $\begingroup$ A good reference is Sweedler's book on Hopf algebras. $\endgroup$ Jul 19 '12 at 20:55
  • $\begingroup$ Havce you tried computing the two compositions of the maps you wrote in the question and seeing if they are identities? $\endgroup$ Jul 19 '12 at 20:57

The result you mention is a classical result on Hopf modules, first proved by Larson and Sweedler. My favorite reference is

Pareigis: When Hopf Algebras are Frobenius Algebras. J. of Alg. 18(1971), 588-596. Lemma 2.

There you can also find the maps you describe in your question (so the answer is yes, they are inverse to each other).

Another reference is Sweedler's book (mentioned already by Mariano), Theorem 4.1.1. However, I think in the book the base ring is always a field, while Pareigis works over a comm. ring.

Somewhere in the book "Brzezinski, Wisbauer: Corings and Comodules" I read that the result in question also follows from a more general theorem on comodules over corings. But I don't know details.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.