# Hopf Algebra: An Introduction (Pure and Applied Mathematics) by Sorin Dascalescu, Constantin Nastasescu, Serban Raianu

By Sorin Dascalescu, Constantin Nastasescu, Serban Raianu

This examine covers comodules, rational modules and bicomodules; cosemisimple, semiperfect and co-Frobenius algebras; bialgebras and Hopf algebras; activities and coactions of Hopf algebras on algebras; finite dimensional Hopf algebras, with the Nicholas-Zoeller and Taft-Wilson theorems and personality thought; and extra.

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Additional info for Hopf Algebra: An Introduction (Pure and Applied Mathematics)

Example text

A morphism of algebras. Then f*(B °) C_ A° and the induced map f° : B° ~ A° is a morphism of coalgebras. 2 that f*(B °) C_ A°. In order to show that f° is a morphism of coalgebras, we have to prove that the following two diagrams are commutative. BO B° °® B fO fO fo®fo . = b*(1) °. for any b* ~ B As for the first diagram, let b* ~ B°, Aso(b*) = ~-~i b~ ®c~’. Since ¢ : A°® A° --~ (A ® A)* is injective, in order to show that (f° ® f°)ABO= /kAof° it is enough to show that ¢(fo® f°)ABO ¢AAof°. ))(x ® = ~(f°(b~))(x)(f°(c~))(y) i = ~b;(f(x))c;(f(y)) i = b*(I(z)f(y)) 38 CHAPTER1.

It can be checked immediately, looking at each component, that (~e~C~, A,e) is a coalgebra, and that this is the coproduct of the family (C~)i~ in the category k - Cog. | Before discussing the products in the category k - Cog we need the concept of tensor product of coalgebras. Let then (C, Ac, ~c) and (D, At), two coalgebras and A : C ® D --~ C ® D ® C ® D, ~ : C ® D -~ k the maps defined by A = (I®T®I)(Ac®AD), ~ = (/)(gC®~D), where T(c®d) = d®c and ¢ : k ® k --~ k is the canonical isomorphism.

29 CONSTRUCTIONS FOR COALGEBRAS We can now give an example of a coalgebra which has no grouplike elements. 5). 11), hence G(C) = Alg(Mn(k), k). On the other hand, are no algebra maps f : Mn(k) ---* k, since for such a morphism Ker(f) would be an ideal (we will use sometimes this terminology for a two-sided ideal) of Mn(k), so it would be either 0 or M,~(k). But Ker(f) = imply f injective, which is impossible because of dimensions, and Ker(f) Mn(k) is again impossible because f(1) = 1. Therfore G(C) = ~.

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