$H$-Hopf modules equal the tensor products of their coinvariants with H - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:29:44Z http://mathoverflow.net/feeds/question/102697 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102697/h-hopf-modules-equal-the-tensor-products-of-their-coinvariants-with-h $H$-Hopf modules equal the tensor products of their coinvariants with H Abtan Massini 2012-07-19T20:14:08Z 2012-07-19T23:29:24Z <p>In a comment for this old <a href="http://mathoverflow.net/questions/92081/hopf-comodule-decompositions-and-coinvariant-elements" rel="nofollow">question</a>, it was said that </p> <p>> 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.) </p> <p>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?</p> <p>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?</p> http://mathoverflow.net/questions/102697/h-hopf-modules-equal-the-tensor-products-of-their-coinvariants-with-h/102699#102699 Answer by Ralph for $H$-Hopf modules equal the tensor products of their coinvariants with H Ralph 2012-07-19T21:12:01Z 2012-07-19T21:20:11Z <p>The result you mention is a classical result on Hopf modules, first proved by Larson and Sweedler. My favorite reference is </p> <p>Pareigis: When Hopf Algebras are Frobenius Algebras. J. of Alg. 18(1971), 588-596. Lemma 2. </p> <p>There you can also find the maps you describe in your question (so the answer is yes, they are inverse to each other). </p> <p>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. </p> <p>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. </p>