\documentstyle{amsppt}
\magnification 1200 \tolerance 5000
\def\Hom{\operatorname{Hom}} \def\End{\operatorname{End}}
\def\Alg{\operatorname{Alg}} \def\dim{\operatorname{dim}}
\def\id{\operatorname{id}} \def\ad{\operatorname{ad}}
\def\char{\operatorname{char}} \def\Sp{\operatorname{Sp}}
\def\Af{\operatorname{Af}}  \def\det{\operatorname{det}}
\def\Tr{\operatorname{Tr}}  \def\co{\operatorname{co}}
\def\ker{\operatorname{ker}} \def\Ann{\operatorname{Ann}}
\def\rad{\operatorname{rad}}

\topmatter \title Lectures on Hopf Algebras   \endtitle
\author    HANS-J\"URGEN SCHNEIDER \\  \\ {\rm Notes by Sonia Natale}
\endauthor
  \address   Mathematisches Seminar der Universit\"at
M\"unchen. Theressienstr. 39. (80333) M\"unchen. Germany
\endaddress \email hanssch\@rz.mathematik.uni-muenchen.de\endemail
\thanks Supported by a grant of DAAD-Antorchas, a Subsidio of the SECyT
 (U.N. C\'ordoba) and the CIEM. \endthanks
\leftheadtext{HANS-J\"URGEN SCHNEIDER}
\endtopmatter \document

\vskip 40pt
  \head  Preface\endhead
\vskip 20pt




These notes contain the material presented in a series of five lectures
at the University of C\'ordoba in September 1994.
The intent of this brief course was to give a quick introduction to
 Hopf algebras
and to prove as directly as possible (to me) some recent results on
finite-dimensional Hopf algebras conjectured by Kaplansky in 1975.
In particular, in the last part of the course  I gave a complete proof
 from scratch of Zhou's theorem (1994):
Any finite-dimensional Hopf algebra over the complex numbers of prime
 dimension
$p$ is isomorphic to the group algebra of the group of order $p$.

I would like to thank Nicol\'as Andruskiewitsch and the members of
 the Mathematics Department of the FaMAF (University of C\'ordoba)
for the kind invitation to visit  C\'ordoba and for their warm
hospitality. I also wish to thank the students in my class for
their attention and for most enjoyable hours after the lectures.

My special thanks are due to Sonia Natale who has written up the
 notes of a condensed course with great care and insight.


\vskip 40pt
\rightline{H.-J. Schneider}
\newpage


\head \S1 Definitions and examples \endhead

In what follows, we will consider a commutative ring $k$ (later on
$k$ will be a field); the symbols Hom and $\otimes$ will mean
$\Hom_k$  and $\otimes_k$ respectively. 

By an algebra $R$ over $k$ (or simply an algebra) we understand a
unitary, associative $k$-algebra $R$, with identity $1 = 1_{R} $.
The category of $k$-algebras will be denoted by  $\Alg_k$.  If $R$
is an algebra, then $R^{op}$ denotes the opposite algebra (i.e. the
$k$-module $R$ with multiplication $a ._{op} b = b a$).

For an algebra $R$, ${}_{R}\Cal M$ (respectively $ \Cal M_R $) will
denote the category of left (respectively right) $R$-modules. Recall
that a $k$-module $M$ is a left (respectively right) $R$-module, if
and only if, there exists an algebra map: $R @>>> \End(M)$ 
(respectively $R^{op} @>>> \End(M)$).

\bigpagebreak \remark{Remarks on representation theory}

1. Let $G$ be a group, $H : = kG$ its group algebra; $V$, $W$  in 
${}_H\Cal M$. Then $k$, $V  \otimes W$, and $V^*  = \Hom ( V, k )$
can be made into left  $H$-modules by setting: $$ \gather  g . 1 =
1, \\ g. ( v \otimes w ) = g .  v \otimes  g . w, \\  ( g. \phi  ) (
v ) = \phi ( g^{-1} . v ) , \endgather$$ for all $g \in G, v \in V,
w \in W,  \phi \in  V^* $.


We note that the algebra maps which define the module structures in
each case are given by: $$ \gather  \epsilon  : \ kG @>>>  k, \\kG
@> \Delta >> kG \otimes kG @>>> \End ( V ) \otimes \End ( W )
@>>>\End( V \otimes W ), \\ kG @>\Cal S >> kG^{op} @>\text
{transpose}>> \End( V^* ), \endgather $$ where $\epsilon (g) = 1$,
$\Delta (g) = g \otimes g$, $\Cal S(g) = g^{-1}$, $\forall g \in G$.

\bigpagebreak 2. Let us now consider a Lie algebra $\goth g$, $H =
U(\goth g)$ its universal enveloping algebra. Then the Lie algebra
maps: $$ \gather\goth g @>>> k, \quad x \mapsto 0, \\ \goth g @>>>
\goth g \times \goth g, \quad x \mapsto ( x , x ), \\ \goth g @>>>
{\goth g}^{op}, \quad x \mapsto  - x, \endgather $$              
together with the universal properties defining $H$, give rise to
algebra maps: $$\gather \epsilon :  H @>>> k, \\   \Delta :  H @>>>
H \otimes H \backsimeq U ( \goth g \times \goth g ), \\   \Cal S : 
H @>>> H^{op} \backsimeq  U ( {\goth g}^{op}  ). \endgather $$

Explicitly $$\gather \epsilon ( x ) = 0, \\ \Delta  ( x ) = 1
\otimes x + x \otimes 1, \\ \Cal S(x) = - x,\endgather$$ $x \in \goth
g$.
 
If $V$ and $W$ are left $H$-modules, with corresponding actions $H
@>>>\End (V)$ and $H @>>>\End (W)$, then the composition

$$ H @> \Delta>>  H \otimes H @>>> \End(V)  \otimes \End (W) @>>>
\End(V\otimes W) $$provides $V \otimes W$ a left $H$-module
structure. This is uniquely determined by the condition $$x . ( v
\otimes w ) =  x . v  \otimes w + v \otimes x . w, $$ $ x \in \goth
g $, $ v \in V $, $w \in  W$ .

In an analogous way, but now using the antipode, we may  let $H$ act
over $V^* $, via:

$$  H @>\Cal S>> H^{op} @>\text {transpose} >>\End ( V^* )$$

 This dual action is determined by $( x . \phi ) ( v ) = \phi ( - x
. v )$, $x \in \goth g $ , $v \in V $ ,  $ \phi \in  V^*$.


Finally, we shall consider $k$ as $H$-module via $\epsilon : H
@>>>k$, this is uniquely determined by $x . 1 = 0$, for all $x \in
\goth g$. \endremark

\bigpagebreak We want to consider algebras such that tensor products
and duals of modules are again modules, as in the examples above.

First we need the definition of coalgebra. Observe that an
associative, unitary $k$-algebra is a pair $(A,m)$, where $A$ is a
$k$-module and $m:A \otimes A @>>> A$ is a $k$-linear map, called
the \it multiplication, \rm such that:

1. The following diagram is commutative: $$\CD A \otimes A \otimes
A        @>m \otimes \id>>     A\otimes A   \\@V \id \otimes m
VV                                        @VV mV    \\A \otimes
A          @>>m >                  A        \endCD $$2. There exists
a $k$-linear map $u: k @>>> A$ such that the following diagrams commute: $$\CD
k \otimes A @>u \otimes \id >> A \otimes A @<\id \otimes u << A
\otimes k\\@VVV         @VVmV                     @VVV  \\ A      
@=                          A         @=         A,\endCD$$where the
maps $k \otimes A @>>> A$ and $A \otimes k @>>>A$ are the canonical
ones. Such a $u$ is necessarily unique. The first of these diagrams
says that the algebra A is associative and the second gives the
existence of a unit $ u(1) = 1_A $ in A.


By reversing arrows, we get the dual notion.


\definition{Definition}A \it coalgebra \rm over $k$ is a pair $ (C,
\Delta)$, where $C$ is a $k$-module and $ \Delta :C@>>>C \otimes C
$ is a $k$-linear map called the \it comultiplication, \rm such that:

1. The following diagram commutes:$$\CD C \otimes C \otimes C       
@<\Delta \otimes \id<<     C \otimes C   \\@A \id \otimes \Delta
AA                            @AA\Delta A    \\C \otimes
C            @<<\Delta <                                C.   
\endCD$$ 2. There exists a $k$-linear map $\epsilon : C @>>> k$,
such that the following diagrams commute:$$\CD k \otimes C
@<\epsilon \otimes \id << C \otimes C @>\id \otimes \epsilon >> C
\otimes k \\@AAA            @AA\Delta A                        @AAA 
\\ C              @=    C                       @=               
C.\endCD$$The map $\epsilon$ is called the \it counit \rm and is
uniquely determined by the pair $(C, \Delta )$.

The kernel of $\epsilon $ will be denoted by $C^{+}$.

If $(C, \Delta_C)$, $(D, \Delta_D)$ are coalgebras, a $k$-linear
map: $f : C @>>> D $ is said a \it coalgebra map, \rm if the
following diagrams commute:$$\CD C                      
@>f>>                           D \\@V\Delta_C
VV                                         @VV\Delta_DV  \\   
C\otimes C             @>>f \otimes f >              D \otimes
D,        \endCD$$$$\CD C        
@>f>>                                  D  \\@V\epsilon_C
VV                                          @VV\epsilon_DV\\k      
@=                                                        
k.\endCD$$\enddefinition \remark{Remark} More generally, one can
define algebras and coalgebras in monoidal categories, that is
$k$-linear categories $\Cal C$ provided with a "tensor" functor
$\otimes: \Cal C \times \Cal C @>>> \Cal C$, plus an associativity 
constraint (see below).  The opposite category ${\Cal C}^{op}$ of a
category $\Cal C$ has the same objects but the arrows are reversed:
$\Hom_{ {\Cal C}^{op} } (A,B) = \Hom_{\Cal C}(B,A)$. In this way, a
coalgebra in $\Cal C$ is the same as an algebra in ${\Cal
C}^{op}$.\endremark\example{Examples} 1. If $S$ is any set and $C =
kS$ is the free $k$-module with basis $S$, then $C$ becomes a
coalgebra if we set: $ \Delta (s) = s \otimes s $, $\epsilon(s) = 1$,
$s \in S$.

2. The universal enveloping algebra of a Lie algebra $\goth g$ is a
coalgebra with the coproduct $\Delta$ and counit $\epsilon$ just
considered.\endexample Now we dualize the definition of a module
over a $k$-algebra.

\definition{Definition}Let $C$ be a coalgebra over $k$. A \it right
comodule \rm over $C$ is a pair $(M, \Delta_M)$, where $M$ is a
$k$-module and $\Delta_M : M @>>> M \otimes C$ is a $k$-linear map
(the comodule structure), such that the following diagrams commute:
$$\CD M           @>\Delta_M >>                                     
M \otimes C \\@V \Delta_M VV            @VV \Delta_M \otimes id V
\\M \otimes C   @>>id \otimes \Delta>  M \otimes C \otimes C, 
\\\endCD$$$$\CD M                     @>\Delta_M >>      M \otimes C 
\\@VVV                                         @VV id \otimes
\epsilon V \\M \otimes k @=                             M \otimes k
.  \endCD$$\enddefinition A $k$-linear map $ \phi : M @>>> N $
between right $C$-comodules $M$, $N$, is said a \it comodule map \rm
if the following diagram commutes:$$\CD M              @> \phi
>>                             N \\@V \Delta_M
VV                                     @VV \Delta_N V \\M \otimes
C                  @>> \phi \otimes id >        N \otimes
C\endCD$$The left $C$-comodules are defined in a similar fashion. We
will denote ${\Cal M}^C$ and ${}^C \Cal M$, respectively, the
categories of right and left $C$-comodules.  Consider a $k$-module
$A$; it could happen that $A$ has both an algebra and coalgebra
structure. In case these structures "paste" well, we give $A$ a
special name:

\definition{Definition}We say that a triple $(A, m, \Delta)$ is a
\it bialgebra, \rm if $(A, m)$ is an algebra with unit $u$, $(A,
\Delta)$ is a coalgebra with counit $\epsilon$ and $\Delta: A@>>>A
\otimes A$, $\epsilon: A@>>>k$ are algebra maps. A $k$-linear map $
\phi : A @>>> B $, where  $A$ and $B$ are bialgebras is said a \it
bialgebra map \rm if it is both an algebra and a coalgebra
map.\enddefinition

\remark{Remarks} 

1. In the definition $A \otimes A$ is
considered with the  natural algebra structure. 

In general the tensor product of two algebras $A$ and $B$ has a
natural algebra structure determined by $$(a \otimes b) ( c
\otimes d) = ac \otimes bd, \quad \forall a, c \in A, b,d \in B.$$ 
Equivalently, the multiplication $m_{A\otimes B}$ is the
composition $$A\otimes B\otimes A\otimes B @> \id \otimes
\tau \otimes \id >> A\otimes A\otimes B \otimes B @>m_A
\otimes m_B >> A \otimes B.$$
Here
$\tau$ denotes the "twist" map: $ \tau : a \otimes b \mapsto b
\otimes a$.

Now, if $C$ and $D$
are coalgebras, then the tensor product $C \otimes D$ can be made
into a coalgebra in a natural way, with the comultiplication$$C
\otimes D   @>\Delta_C \otimes \Delta_D >>  C \otimes C \otimes D
\otimes D @>\id \otimes \tau \otimes \id>>  C \otimes D \otimes C
\otimes D.   $$   The counit is given by $$ C \otimes D @>
\epsilon_C \otimes \epsilon_D >> k \otimes k \backsimeq k . $$
One can then check that in the definition of  bialgebra the
condition of $\Delta$ and $\epsilon$ being algebra maps may be
replaced by the (equivalent) condition of $m$ and $u$ being
coalgebra maps.\endremark

2. The kernel of the counit $\epsilon $ in a bialgebra $A$ is a two
sided ideal of codimension $1$, called the \it augmentation
ideal.\rm

Examples of bialgebras are $kG$, the group
algebra of a group $G$, with the algebra and coalgebra structures
considered at the beginning (notice that we do not make use here of
the existence of inverses for elements of $G$), and the universal
enveloping algebra of a Lie algebra $\goth g$, where $\Delta$ and
$\epsilon$ are as treated earlier. In particular, any symmetric
algebra has a bialgebra structure. The next definition will allow us
to give a characterization of a bialgebra in terms of its left
modules when considered as an algebra.

\definition{Definition}A triple $( \Cal C, \otimes, I )$, where
$\Cal C$ is a category, $\otimes : \Cal C \times \Cal C @>>> \Cal C
$ is a functor called \it formal tensor product, \rm and $I$ is an
object of  $\Cal C$ called \it unit object, \rm is said a \it
monoidal category \rm if for any objects $U,V,W$ of $\Cal C$ there
exists natural isomorphisms between functors from $\Cal C \times
\Cal C \times \Cal C$ to $\Cal C$ (respectively $\Cal C$
 to $\Cal C$) $$a_{U,V,W}:(U \otimes V) \otimes W  @>>>  U \otimes
(V \otimes W),$$ $$r_V: V \otimes I @>>> V , \quad  l_V: I \otimes V
@>>> V,$$ such that the following diagrams are commutative: $$\CD (U
\otimes V)\otimes (W\otimes X) @= (U \otimes V)\otimes (W\otimes X)
\\ @Aa_{U\otimes V,W,X}AA                     @VVa_{U,V,W\otimes
X}V\\((U\otimes V)\otimes W)\otimes X  @.      U\otimes (V\otimes
(W\otimes X)) \\@Va_{U,V,W}\otimes \id VV                  @AA\id
\otimes a_{V,W,X}A \\(U\otimes (V\otimes W))\otimes X 
@>>a_{U,V\otimes W,X}> U\otimes ((V\otimes W)\otimes X),
\endCD$$$$\CD(V\otimes I)\otimes W @>a_{V,I,W}>> V\otimes (I\otimes
W) \\@Vr_{V} \otimes \id VV                            @VV \id
\otimes l_W V \\ V \otimes W          @=     V\otimes W. \endCD$$
\enddefinition A first example of a monoidal category is the
category of left $k$-modules ${}_k\Cal M$, with the tensor product
over $k$ and unit object $I = k$. The associativity and unit
constraints are just the usual isomorphisms of $k$-modules:$$\gather
a_{U,V,W} ((u\otimes v) \otimes w) = u \otimes (v \otimes w), \\
l_V(1 \otimes v) = v, \\ r_V(v \otimes 1) = v.\endgather$$

\remark{Remark} In any monoidal category one can define
algebras and coalgebras, and their modules and comodules.
However, to define bialgebras one needs in addition a
commutativity constraint, or \it braiding \rm $$c_{U,V}:
U\otimes V @>>> V\otimes U.$$
This leads to the important notion of bialgebras in braided
categories. In these notes, we shall only consider the traditional
braided category of $k$-modules, where the braiding is the usual
twist map.\endremark

Other examples of monoidal categories are the categories of left
modules over the algebras $kG$ and  $U(\goth g)$. In both cases this
structure is inherited from that of ${}_k \Cal M$. The next
proposition gives a characterization of the $k$-algebras with this
property.

\proclaim {Proposition 1.1} 
Let $(A,m)$ be a $k$-algebra and let $\Delta : A @>>> A \otimes
A$, $\epsilon : A @>>> k$ be given algebra maps.
We consider $k \in {}_A \Cal M $ via $\epsilon$.
 Let $\otimes : {}_A \Cal M \times {}_A \Cal M @>>> {}_A \Cal M
$ be the functor which associates to each pair of $A$-modules 
$M, N$ their
tensor product over $k$, $M \otimes N$, with the $A$-action:
$$A @>\Delta >> A \otimes A @>>> \End (M) \otimes \End (N) @>>>
\End (M \otimes N).$$
Then $({}_A \Cal M , \otimes , k)$ is a monoidal category, with
canonical associativity and unit constraints, if and only if $(A, m,
\Delta )$ is a bialgebra (with counit $\epsilon$).\endproclaim
\demo {Proof}

Suppose $(A, m, \Delta)$ is  a bialgebra. The coassociativity of
$\Delta$ implies that $\forall  U, V, W \in {}_A \Cal M $ the
canonical isomorphisms of $k$-modules $$(U \otimes V) \otimes
W \backsimeq U \otimes (V \otimes W), $$ are isomorphisms of
$A$-modules. Moreover, the commutativity of the diagrams
$$\CD
A\otimes k @<\id \otimes \epsilon << A \otimes A @>\epsilon
\otimes \id >> k \otimes A \\
@VVV           @A\Delta AA  @VVV \\                                      
A @= A @= A,  \endCD $$
imply, respectively, that the left and right unit constraints $$V
\otimes k @>>v \otimes 1 \mapsto v > V, \quad \text {and} \quad k
\otimes V @>>1 \otimes v \mapsto v >V, $$ are isomorphisms of
$A$-modules. So  $({}_A \Cal M , \otimes , k)$ is a monoidal
category.

Conversely, suppose that  $({}_A \Cal M , \otimes , k)$ is a
monoidal category. For the coassociativity of $\Delta $ use the
fact that $$(A\otimes A)\otimes A @>>(x\otimes y)\otimes z
\mapsto x\otimes (y\otimes z) > A\otimes (A\otimes A), $$ is an
isomorphism of $A$-modules.

Also, the canonical maps $$k\otimes A @>>> A, \quad A\otimes k
@>>> A, $$ are $A$-isomorphisms, which implies that $\epsilon$ is
the counit.\qed \enddemo


 \bigpagebreak Now we are in a position to define 
the objects which will concern us in the sequel.

\definition{Definition}We say that a bialgebra $(H, m, \Delta)$ is a
\it Hopf algebra \rm if there exists a $k$-linear map $\Cal S: H
@>>> H$, called the \it antipode, \rm such that the following
diagrams are commutative: 

$$\CD H \otimes H  @<\Delta << H  @>\Delta >> H \otimes H  \\@V\id
\otimes \Cal SVV @Vu \epsilon VV @VV \Cal S \otimes \id V \\H
\otimes H @>>m> H        @<<m<   H \otimes H
\endCD$$\enddefinition
                            \example{Examples}1. If $G$ is a
group, the group algebra $kG$ is a Hopf algebra, with antipode given
by $\Cal S(g) = g^{-1}$, $g \in G$.

 \bigpagebreak 2. The universal
enveloping algebra $U(\goth g)$ of the Lie algebra $\goth g$ is a
Hopf algebra, with antipode $\Cal S(x) = -x$, $x \in \goth g$. In
particular any symmetric algebra over $k$ is a Hopf algebra.


\bigpagebreak 3. Let $V$ be a $k$-module, then the tensor algebra
$T(V)$ over $V$ is a Hopf algebra with the usual algebra structure
and where $\Delta (v) = 1 \otimes v + v \otimes 1$, $\epsilon (v) =
0$, $\Cal S (v) = -v$, $v \in V$.

If $k$ is a field, this example is but a particular case
of the previous one, we see this as follows:

\bigpagebreak \it Tensor algebras and free Lie algebras. \rm 

Let
$k$ be a field. A Lie algebra $\goth g$ over $k$ is said to be \it
free \rm on a set $X$ if

a) $X$ generates $\goth g$ as a Lie algebra.

b) Given a Lie algebra $\goth m$ over $k$, and a map $\phi : X @>>>
\goth m$, there exists a (unique) Lie algebra morphism $\psi : \goth
g @>>> \goth m$ that extends $\phi$.  

\bigpagebreak
It is not difficult to see that given a set $X$, if such an algebra
exists, it is unique (up to isomorphism). As to its existence,
consider the $k$-space $V$ with basis $X$. Let $T(V)$ be the tensor
algebra over $V$, and call $\goth g$ the Lie subalgebra of $T(V)$
(with the bracket $[a,b] = ab - ba$) generated by $V$. 

It is clear that $X$ generates $\goth g$. If $\phi : X @>>> \goth m$
is any map, we can extend it to a $k$-linear map ${\phi}_1:V @>>>
\goth m \subseteq U(\goth m)$ (observe that here we are making use
of the PBW theorem, which use is legitimate under the assumption that
$k$ is a field). By the universal property of $T(V)$, ${\phi}_1$ has
a unique extension to an algebra map $T(V) @>>> U(\goth m)$. Call
$\psi $ the restriction of this map to $\goth g$, then $\psi $
extends $\phi $ and is clearly a Lie algebra map.

\bigpagebreak
We assert that the universal enveloping algebra of a free Lie
algebra $\goth g$ on $X$ is isomorphic to the tensor algebra over
the $k$-vector space with basis $X$.

To see this, let $\Cal A$ be an associative algebra over $k$ and let
$\omega : \goth g @>>> \Cal A$ be a Lie algebra map. Then, by the
universal property of $T(V)$, there exists a unique algebra map
$\Omega : T(V) @>>> \Cal A$ such that $\Omega (x) = \omega (x),
\forall x \in X$, but this is equivalent to saying (as $X$ generates
$\goth g$) that $\Omega (x) = \omega (x), \forall x \in \goth g$.

So $T(V) \backsimeq U(\goth g)$.
Moreover, the Hopf algebra structures in $T(V)$ when considered
as a tensor algebra and as an enveloping algebra are the same.

\bigpagebreak 4. The following example is due to Taft (1971):

Let $k$ be a field, and $N$ a natural number. Assume that there 
exists a primitive $N$-th root of unity $\xi$ in $k$. Consider the
algebra $H$ generated over $k$ by two elements $g$ and $x$ subject
to the relations: $ g^N = 1$, $x^N = 0$, $xg = \xi gx$. We claim
that there are algebra maps $\Delta: H @>>> H\otimes H$, $\Cal S : H
@>>> H^{op}$, $\epsilon : H @>>> k$ uniquely determined by $$\gather 
\Delta (g) = g \otimes g , \quad  \Delta (x) = 1 \otimes x + x
\otimes g \\\epsilon (x) = 0 , \quad \epsilon (g) = 1 \\ \Cal S(g) =
g^{-1}, \quad \Cal S(x) = -xg^{-1} \endgather $$ \endexample We work
out the details for $\Delta $ and let $\Cal S$, $\epsilon $ to the
reader. Clearly ${\Delta (g)}^N = 1$ and $\Delta (x)$, $\Delta (g)$
$\xi$-commute, i.e., $\Delta(x) \Delta (g) = \xi \Delta (g) \Delta
(x)$. For the remaining relation we need the next lemma. 

In the polynomial algebra $\Bbb Z [ \bold q]$, we consider the
 $\bold q$-binomial coefficients 
 $${\binom ni}_{\bold q} = \frac { {(n)!}_{\bold q}} { {
(n-i)!}_{\bold q} {(i)!}_{\bold q} }, \quad \text {where }
{(n)!}_{\bold q} = {(n)}_{\bold q}  \dots {(2)}_{\bold q}
{(1)}_{\bold q}, \quad \text {and } {(n)}_{\bold q} = 1 + {\bold q}
+ \dots + {\bold q}^{n-1},$$
 for $n \in \Bbb N$, $0\leq i \leq n$.

One proves that ${\binom ni}_{\bold q} \in \Bbb Z [{\bold q}]$ by
induction on $n$, using the identity $${\bold q}^k {\binom
nk}_{\bold q} + {\binom n{k-1}}_{\bold q} = {\binom {n+1}k}_{\bold
q}, \tag * $$ for $1 \leq k \leq n$.

Now, if $A$ is an associative algebra over $k$ and $q \in k$, then 
${\binom ni}_q$ denotes the specialization of ${\binom ni}_{\bold
q}$ at $q$.

\bigpagebreak\proclaim{Lemma (Quantum binomial
formula)}Let $A$ be an associative algebra over $k$, $q \in k$. If
$x, y \in A$ are two elements that $q$-commute, i.e. $ xy = q yx $,
then the following formula holds for every $n \in \Bbb N$:$$(x+y)^n
= \sum_{i = 0}^n { \binom ni }_q y^i x^{n-i}.$$ \endproclaim
\demo{Proof}By induction on $n$, again using the
identity (*). \enddemo 
Now, if $\xi$ is a
primitive $N$-th root of unity, it follows from the definitions that
${\binom Ni}_{\xi} = 0$ for $0<i<N$. Then, as $1 \otimes x$ and $x
\otimes g$ $\xi$-commute, we have: 

$$\multline{\Delta (x)}^N = (1 \otimes x + x \otimes g)^N = 
\sum_{i = 0}^N { \binom Ni }_{\xi} (x \otimes g)^i (1 \otimes
x)^{N-i}  = \\ (x \otimes g)^N + (1 \otimes x)^N = x^N \otimes g^N +
1 \otimes x^N = 0.\endmultline $$ So $\Delta$ is a well defined
algebra map $\Delta : H @>>> H \otimes H$. 

Thus $H$ is a Hopf algebra (of finite dimension $N^2$,
with basis $g^ix^j$ , $0 \leq i,j \leq N-1$).


\definition{Definition}Let $H$ be a Hopf algebra and $\tau$ denote
the twist map in $H \otimes H$. We say $H$ is \it cocommutative \rm
if $\tau \circ \Delta =\Delta$. \enddefinition 

For instance, the
algebras introduced in examples 1 and 2 are cocommutative, while the
Taft algebras are not in general. If $G$ is a finite group, then the
group algebra $kG$ is a finite dimensional cocommutative Hopf
algebra (it is commutative iff $G$ is abelian). When $k$ is an
algebraically closed field of characteristic $0$, then it can be
shown that these are the only possible ones, that is: every finite
dimensional Hopf algebra over $k$ which is cocommutative is
isomorphic to a group algebra $kG$, for some finite group G. The
next example shows this is not true when the characteristic of $k$
is positive.

 \bigpagebreak \example{The $u$-algebra of a
 restricted Lie algebra} Let $k$ be a field of characteristic
$p>0$. A Lie algebra $\Cal L$ over $k$ is called a restricted
Lie algebra if  there is a map $ \Cal L @>>> \Cal L$, denoted $ a
\mapsto a^{[p]} $, $a \in \Cal L $, such that 

$$\gather  (\alpha a )^{[p]} = {\alpha}^p a^{[p]},\\
 \ad (b^{[p]})  = (\ad b)^p, \\ (a + b)^{[p]} = a^{[p]} + b^{[p]} +
\sum_{i=1}^{p-1} s_i (a,b) \endgather$$ 
for $a,b \in \Cal L$, $\alpha \in k$, where $\ad$ denotes the adjoint
representation of $\Cal L$ on itself and $ i s_i (a,b)$ is the
coefficient of $ {\lambda}^{i-1} $ in the expansion of $\ad (\lambda
a + b )^{p-1} (a)$.

A $k$-linear map $f : \Cal L @>>> \Cal A$ between restricted Lie
algebras $\Cal L$ and $\Cal A$ is a \it morphism of restricted
Lie algebras \rm if it is a morphism of Lie algebras and $
f(a^{[p]}) ={f(a)}^{[p]}$, $ \forall a \in \Cal L$.

\bigpagebreak For
instance, if $\Cal A$ is an associative $k$-algebra, we think of
$\Cal A$ as a Lie algebra by means of the natural bracket: $ [a,b] =
ab - ba $, $a,b \in \Cal A$. Then the map $ a \mapsto a^p$, $a \in
\Cal A$, makes $\Cal A$ into a restricted Lie algebra.
\endexample

Let now  $\Cal L$ be a restricted Lie algebra, $U$ its universal
enveloping algebra, and $\Cal B$ the ideal in $U$ generated by all
the elements $ a^p - a^{[p]}$, $a \in \Cal L$. Denote by $\Cal U$ the
quotient algebra $\Cal U = U / \Cal B$. Then, the natural map $\phi
: \Cal L @>>> \Cal U$ is a morphism of restricted Lie
algebras. The pair $( \phi , \Cal U )$ is universal for $\Cal L$ in
the following sense: if $\Cal A$ is an associative algebra and $f:
\Cal L @>>> \Cal A$ is a morphism of restricted Lie algebras,
then there exists a unique algebra map $F: \Cal U @>>>\Cal A$, such
that $ f = F \circ \phi $. 

$\Cal U$ is called the $u$-algebra of $\Cal L$. By the universal
property of $\Cal U$, we have that the (restricted) morphisms
$$\gather \Cal L @>>> k, \quad a \mapsto 0, \\  \Cal L @>>> \Cal L
\times \Cal L, \quad a \mapsto (a,a) , \\  \Cal L @>>> {\Cal
L}^{op}, \quad a \mapsto -a, \endgather$$ define algebra
maps:$$\gather\Delta : \Cal U @>>> \Cal U \otimes \Cal U,
\\ \epsilon: \Cal U @>>> k,
 \\ \Cal S : \Cal U @>>> {\Cal U}^{op}, \endgather$$
uniquely determined by $$\gather \Delta(a) = 1 \otimes a + a
\otimes 1, \\ \epsilon(a) = 0, \\ \Cal S (a) = -a, \endgather$$ $a
\in \Cal L$, which make it into a cocommutative Hopf algebra.


\bigpagebreak The next theorem is analogous to the PBW theorem for
Lie algebras:

\proclaim{Theorem 1.2} Let $\Cal L$ be a restricted Lie algebra
and let $\Cal U$ its $u$-algebra.Then:

1. The map $\phi : \Cal L @>>> \Cal U $ is an injective morphism
of restricted Lie algebras.

2. If $ \{ u_i \}_{i \in I}$ is an
ordered basis for $\Cal L$, then the set of monomials: $$
u_{i_1}^{k_1} u_{i_2}^{k_2} \dots  u_{i_r}^{k_r} : \quad i_1 \leq
i_2 \leq \dots \leq i_r,  \quad 0 \leq k_j \leq p-1.$$is a basis for
$\Cal U$. \endproclaim\demo {Proof} See \cite{2, Th.12, p. 191}.\enddemo

As a consequence, if $\Cal L$ has finite dimension $n$, then $\Cal
U$ is also finite dimensional, with $\dim \Cal U = p^n$.Then, $\Cal
U$ is a finite dimensional cocommutative Hopf algebra and it is \it
not \rm isomorphic to any group algebra. To see this we first
introduce some terminology:

\bigpagebreak
\definition{Definition}Let $H$ be a Hopf algebra. $h \in H$ is said
a \it group-like \rm element if $h \neq 0$ and $\Delta (h) = h
\otimes h$, and it is said a \it primitive \rm element if $\Delta
(h) = 1 \otimes h + h \otimes 1$.\enddefinition The sets of
group-like and primitive elements of $H$ are denoted respectively 
$G(H)$ and $P(H)$.

If $h \in G(H)$, then $\epsilon (h) = 1$; similarly, if 
$h \in P(H)$, then $\epsilon (h) = 0$. 

$G(H)$ is a subgroup of the group of units of $H$, and $P(H)$ is a
Lie subalgebra of $H$ with the bracket $[a,b] = ab - ba$.

\remark{Observation} If $k$ is a field, distinct group-like elements
are linearly independent, so if $G$ is a group, then the set of
group-like elements in $kG$ is precisely $G$.\endremark

\bigpagebreak
\proclaim{Lemma 1.3} Let $k$ be a field. If $H$ is a Hopf algebra
over $k$ which is generated (as an algebra) by primitive elements,
then the set of group-like elements of $H$ is trivial.\endproclaim
\demo{Proof} Let ${\{x_i\}}_{i \in I}$ denote the family of nonzero
primitive elements of $H$ and for each $n \geq 0$, let $A_n$ be the
linear span in $H$ of elements of the form ${x_{i_1}}^{k_1} \dots
{x_{i_m}}^{k_m}$, such that $k_j$ are
nonnegative integers with $k_1 + \dots + k_m = n$.

Then the collection ${\{A_n\}}_{n \geq 0}$ satisfies the next two
properties:

1. $A_n \subseteqq A_{n+1}$, $\bigcup_{n \geq 0}A_n = H$.

2. $\Delta ({A_n}) \subseteqq \sum_{i = 0}^n {A_i \otimes A_{n-i}}$.

\bigpagebreak Now, if $g \in G(H)$, because of property 1, there
exists $m$ such that $g \in  A_n $, so we can choose $m$ to be
minimal with this property. Suppose $g \notin k = A_0$, then there
exists $f \in H^*$ such that $f(A_0) = 0$ but $f(g) = 1$.

As $g \in A_m$, we may write  $$\Delta (g) = \sum_{i=0}^m {a_i
\otimes a_{m-i}}, $$ with $a_j \in A_j$, and this implies that  $$g
= < \id \otimes f, \Delta(g) > = \sum_{i=0}^{m-1} {a_i f( a_{m-i})}
\in A_{m-1}.$$ But this contradicts the minimality of $m$. Thus $g
\in k$, and so $g = 1$ as asserted.\qed  \enddemo

Lemma (1.3), together with the previous observation, show that
the $u$-algebra of a restricted nontrivial Lie algebra cannot be
isomorphic to a group algebra $kG$.

\bigpagebreak  \subhead Sigma Notation \endsubhead 

We introduce the notation, due to Sweedler, that will be used from
now on. 

If $C$ is a coalgebra, $c \in C$, then $\Delta (c)$, as an element
of $C\otimes C$ has a representation of the form $$\Delta(c) =
\sum_i {c_i \otimes c^i},$$ where $c_i, c^i$ are elements of
$C$. We indicate such an expression in the abbreviated form 
$$\Delta (c) = \sum {c_{(1)}\otimes c_{(2)}}.$$ 

Some authors use instead 
$\Delta (c) = \sum {c_1 \otimes c_2}$. In what follows we shall
omit the sumation symbol, for the sake of brevity. So that, 
$$\Delta (c) = c_{(1)}\otimes c_{(2)}.$$

If $V$ is a right comodule for $C$ with comodule structure map
${\Delta }_V$, then we write for $v \in V$ $${\Delta }_V (v) =
\sum {v_{(0)}\otimes v_{(1)}},$$ where $v_{(0)}$ represents
elements of $V$, and $v_{(1)}$ is understood to be in $C$.
 Again, we shall omit the sumation symbol.

\bigpagebreak For instance, if $C$ is a coalgebra, the
coassociativity of $\Delta $ reads, in sigma notation $$
{(c_{(1)})}_{(1)}\otimes {(c_{(1)})}_{(2)}\otimes c_{(2)} = c_{(1)}
\otimes  {(c_{(2)})}_{(1)}\otimes {(c_{(2)})}_{(2)},$$  $\forall c
\in C$, so we may indicate ${\Delta }_2(c) : = (\Delta \otimes
\id) \circ \Delta (c) = (\id \otimes \Delta) \circ \Delta (c)$ in the
form $${\Delta }_2(c) =  c_{(1)}\otimes c_{(2)}\otimes c_{(3)}.$$
Defining inductively ${\Delta}_1 = \Delta$, ${\Delta}_{n+1} : C @>>>
C^{\otimes (n+2)}$, ${\Delta}_{n+1}= (\Delta \otimes {\id}^n )
\circ {\Delta }_{n-1}$, $n \geq 2$, we see there is no ambiguity in
writting $${\Delta}_{n}(c) = c_{(1)}\otimes \dots \otimes
c_{(n+1)}.$$ 
In this vein, the composition $f \circ {\Delta }_n$ can be
expressed as $$f \circ {\Delta }_n (c) = f( c_{(1}),..., c_{(n)} ).$$
For instance, the conditions on $\epsilon$ take
the form $$\epsilon (c_{(1)})c_{(2)} = c = c_{(1)}
\epsilon(c_{(2)}).$$


Also, if $H$ is a Hopf algebra with antipode $\Cal S$, then we must have,
$\forall h \in H$ $$ \Cal S (h_{(1)})h_{(2)} = \epsilon (h) 1 =
h_{(1)} \Cal S (h_{(2)}).$$

\bigpagebreak
\subhead Convolution Product \endsubhead

\definition {Definition} Let $(C,\Delta )$ be a coalgebra, $(A,m)$
an algebra. For $f, g \in \Hom (C, A)$ we define the \it convolution
product \rm of $f$ and $g$ to be the element of $\Hom (C,A)$,
denoted $f*g$, which results from the composition $$C @> \Delta >> C
\otimes C @> f \otimes g >> A \otimes A @>m>> A.$$ \enddefinition

In sigma notation we have, for $c \in C$, $(f*g) (c) = f(c_{(1)})
g(c_{(2)})$.

In the next example, following Wigner \cite{20}, it is shown that 
the convolution product allows to give simpler proofs  
of some old results on free 
Lie algebras (cf. \cite{2, V.4}).

\bigpagebreak
\example {Example} Let $V$ be a $k$-module and consider the Hopf
algebra structure in the tensor algebra $T(V)$ indicated earlier.
Let $\phi : T(V) @>>> T(V)$ the $k$-linear map defined by: 
$$\gather
\phi (1) = 0, \\ \phi (v) = v, \\ \phi ( v_1 v_2 \dots v_n ) =
[ v_1[ v_2  [\dots v_n] \dots ]] = \ad (v_1) ( \phi (v_2 \dots v_n) ) 
 . \endgather $$
for $v, v_1, \dots , v_n \in V$, $n \geq
2$.  Call $\goth g$ the Lie subalgebra of $T(V)$ generated by $V$.
Then:

\bigpagebreak
a) $\forall x \in T_n (V)$, $\phi * \id (x) = nx$.

\bigpagebreak b) (Theorem of Dynkin, Specht and Wever). 
Suppose $\char k = 0$. If $x \in T_n (V)$, the
following statements are equivalent: 

\bigpagebreak
(i) $x \in \goth g$.

(ii) $\Delta (x) = 1 \otimes x + x \otimes 1$.

(iii) $\phi (x) = nx$.

Then in particular, if $k$ is a field of characteristic $0$, and
$\goth g$ is a free Lie algebra over $k$, the Lie algebra of
primitive elements in $U(\goth g )$ is precisely
$\goth g$. (We remark this still remains true if $\goth
g$ is not free).\endexample 

\demo {Proof}

a) By induction on $n$. The statement is trivially true if $n =
0,1$. Let $n \geq 2$. 

Assume $x = vy$, with $y \in T_{n-1}(V)$, $v \in V$. Write $$ \Delta
(y) = 1 \otimes y + \sum_i {y_i \otimes z_i},$$ where $y_i \in
T^+(V)$. In particular $\forall v \in V$, $$ \phi (v y_i) = v
\phi (y_i) - \phi (y_i) v.$$
 We have $$\phi * \id (y) = \sum_i
{\phi (y_i) z_i},$$ and so, as $\Delta (v) = v \otimes 1 + 1 \otimes
v$, $$\multline \phi * \id (x) = \phi * \id
(vy) = \phi(v_{(1)}y_{(1)}) v_{(2)} y_{(2)} = \phi (y_{(1)}) v y_{(2)}
+ \phi (vy_{(1)})y_{(2)} =  \\
\phi (1) vy + \sum_i {\phi (y_i) v z_i} + \phi (v) y +
\sum_i {\phi (v y_i) z_i} = \\ 
vy + v (\phi * \id )(y) = vy + (n-1)vy =
nx.\endmultline $$ 

\bigpagebreak b) (i) $\Rightarrow$ (ii). Follows from the
observation that the set of primitive elements of a Hopf algebra is
a Lie subalgebra. 

(ii) $\Rightarrow$ (iii). By a) we can write $ \phi * \id (x) = nx$,
but by (ii) this equals $\phi (1) x + \phi (x) 1 = \phi (x)$.

(iii) $\Rightarrow$ (i). Because $\phi (x) = nx \in \goth g$ and
$\char k = 0$.\enddemo 

\bigpagebreak \subhead Properties of the convolution product
\endsubhead

\proclaim {Proposition 1.4} Let $C$ be a coalgebra with counit
$\epsilon $, and let $A$ be an algebra with unit $u$. Then  $(\Hom
(C,A), *)$ is an algebra with unit $u \epsilon$.\endproclaim
\demo {Proof} It is easy to see that $*$ defines an associative 
multiplication in $\Hom (C,A)$. We show that $u
\epsilon $ is the unit. 
Let $f \in \Hom (C,A)$, then $\forall c \in C$, $$(f * u\epsilon
)(c) = f( c_{(1)} ) \epsilon ( c_{(2)} ) 1_A = f( c_{(1)} \epsilon (
c_{(2)} ) ) = f( c ).$$ Similarly  $u \epsilon * f = f$. \qed
\enddemo

\bigpagebreak If $B$ is a bialgebra then by proposition (1.4),
the convolution product makes $\Hom (B,B)$ into an algebra. In the
case of Hopf algebras there is a very close relation between this
algebra structure and the antipode. 

\bigpagebreak
Recall that if $(A, m_A)$, $(B,m_B)$ are algebras,  a map $f: A @>>>
B$ is an \it antialgebra map \rm if $(f \circ m_A) =
{m_B}^{op} \circ (f \otimes f) $ and $f \circ u_A = u_B$. That is, 
 $f(xy) = f(y) f(x)$, $\forall x, y \in A$ and $f(1) = 1$.

If $(C, {\Delta }_C)$, $(D, {\Delta }_D)$ are coalgebras we say that a
map $g: C @>>> D$ is an \it anticoalgebra map \rm if ${\Delta }_D
\circ g = (g \otimes g) \circ {{\Delta }_C}^{cop}$ and ${\epsilon }_D
\circ g = {\epsilon }_C$. In sigma notation, the first condition 
reads $${g(c)}_{(1)} \otimes {g(c)}_{(2)} =
g(c_{(2)} ) \otimes  g( c_{(1)} ), \quad \forall c \in C.$$  

\bigpagebreak
\proclaim {Theorem 1.5} Let $H$ be a Hopf algebra. Then the antipode
$\Cal S$ is the inverse of the identity map $\id :H @>>> H$ with
respect to the convolution product in $\Hom(H,H)$. In particular it
is unique. We have also:

a) $\Cal S$ is an antialgebra map. 

b) $\Cal S$ is an anticoalgebra map.

c) The following statements are equivalent:

 i) ${\Cal S}^2 = \id $.

ii) $x_{(2)} \Cal S( x_{(1)} ) =\epsilon (x)1, \forall x \in H$. 

iii) $\Cal S ( x_{(1)} ) x_{(2)} = \epsilon (x)1,  \forall x\in H$.

In particular, if $H$ is commutative or cocommutative then ${\Cal
S}^2 = \id$.

d) Let $H$, $K$ are Hopf algebras (with antipodes
denoted respectively by ${\Cal S}_H$ and ${\Cal S}_K$). If $\phi : H
@>>> K$ is a bialgebra map, then $\phi$ is a Hopf algebra map, i.e.
$\phi {\Cal S}_H = {\Cal S}_K \phi$.\endproclaim 

\demo {Proof}
The first assertion follows immediately from the definition of
the antipode.

To prove a), consider the algebra structure in $\Hom (H \otimes
H, H)$, given by the convolution product. Call $m : H \otimes H
@>>> H$ the multiplication in $H$. We claim that $$\Cal S \circ m =
m^{-1} = m^{op} \circ (\Cal S \otimes \Cal S).$$ Here 
$m^{-1}$ is the inverse of $m$ with respect to the convolution
product. To see this, let $x, y \in H$, then  $$\multline ((\Cal S
\circ m) * m)(x \otimes y) = \Cal S \circ m ( x_{(1)} \otimes
y_{(1)}) m (x_{(2)} \otimes y_{(2)}) =\\ \Cal S(x_{(1)} y_{(1)})
x_{(2)} y_{(2)} = \epsilon (x) \epsilon (y) 1_H = \epsilon (x \otimes
y) 1_H.\endmultline $$ Also $$(( m^{op} \circ (\Cal S \otimes \Cal S)
) * m )(x \otimes y) = \Cal S (y_{(1)}) \Cal S (x_{(1)}) x_{(2)}
y_{(2)} = \epsilon (x) \epsilon (y) 1_H  = \epsilon (x \otimes y)
1_H. $$ In a similar way $m * (\Cal S \circ m) = m * ( m^{op} \circ
(\Cal S \otimes \Cal S) ) = u \epsilon $. 

By the uniqueness of the inverse, we obtain the desired identity.

We have also $$\Cal S(1) = \Cal S * \id (1) = 1.$$

\bigpagebreak
For b)  use the identity $\Delta \circ \Cal S = {\Delta }^{-1} =
(\Cal S \otimes \Cal S) \circ {\Delta }^{cop}$ in $\Hom (H, H\otimes
H)$.     

If $x \in H$, applying $\epsilon$ to the equality $$
\epsilon (x) 1 = \Cal S (x_{(1)} ) x_{(2)}, $$ we get
$$ \epsilon (x) = \epsilon ( \Cal S( x_{(1)} \epsilon (x_{(2)} ) ) =
\epsilon \circ \Cal S (x).$$ 

\bigpagebreak
c) i) $ \Rightarrow $ ii).
Suppose ${\Cal S}^2 = \id$. If $x \in H$, using the fact that $\Cal S$
is an antialgebra map, one gets $$ x_{(2)} \Cal S(
x_{(1)} ) = {\Cal S}^2 ( x_{(2)} ) \Cal S ( x_{(1)} ) = \Cal S ( x_{(1)}
\Cal S( x_{(2)} ) )  = \Cal S ( \epsilon(x) 1) = \epsilon (x)
1. $$

 ii) $\Rightarrow $ i).
Let $x \in H$, then $$ {\Cal S}^2 * \Cal S (x) = {\Cal S}^2 (x_{(1)})
\Cal S(x_{(2)}) = \Cal S( x_{(2)} \Cal S (x_{(1)}) ) = \Cal S (
\epsilon (x) 1 ) = \epsilon (x) 1.$$
 This shows that ${\Cal S }^2 * \Cal S = u
\epsilon$. Multiplying by id on the right, we obtain i).

Thus we saw that i) $\Leftrightarrow $ ii). Similarly i)
$\Leftrightarrow $ iii), which finishes the prove of c).  

\bigpagebreak
d) 
To see this, use the identities $$ \phi {\Cal S}_H = {\phi}^{-1} =
{\Cal S}_K \phi$$ which hold in $\Hom (H,K)$.\qed
 \enddemo

\bigpagebreak \remark {Remark} Let H be a Hopf algebra.
We saw that the antipode $\Cal S$ is an antialgebra map, and
so it is an algebra map $H @>>> H^{op}$. Let $V$ be a left
$H$-module. Then the dual $k$-module $V^*$ results a right
$H$-module and we can make it into a left $H$-module, by composing
$$H @>\Cal S >> H^{op} @>>> \End (V^*).$$ 
That is, $$h.\alpha (v) = \alpha (\Cal S (h) . v), 
\quad \forall h \in H, \alpha \in V^*, v \in V.$$
We have, moreover, that the
evaluation map $ V^* \otimes V @>>> V$ is in this way a map of left
$H$-modules (where $V^* \otimes V$ is considered as an
$H$-module via $\Delta$). \endremark

\bigpagebreak
The following is a geometric approach to Hopf algebras. For more
details see reference \cite{4}.

\bigpagebreak
\subhead Hopf algebras and affine schemes \endsubhead

In what follows ${\Cal A}_k $ will denote the category of
commutative $k$-algebras. We indicate by $\goth S $, $\goth G $,
respectively, the categories of sets and groups.

For each $R \in  {\Cal A}_k$, consider the functor $\Alg (R,\_) : 
{\Cal A}_k @>>> \goth S$, which associates to each commutative
$k$-algebra $A$ the set $\Alg (R,A)$, of all algebra maps from $R$
to $A$, and to each morphism $\phi :A @>>> B$, the map 
$\Alg(R,\phi ) : \Alg(R,A) @>>> \Alg (R,B)$, given by $\Alg (R,\phi)
(f) = \phi \circ f $.

We denote this functor by $\Sp R$ and call it the \it spectrum
\rm of $R$.

\bigpagebreak
For instance, if $R = k[T_1, \dots , T_n]$, the polynomial algebra in
$n$ variables over $k$, then $\Sp R \backsimeq {\Af}_n$, where
${\Af }_n$ is the functor \it affine $n$-space \rm :
$$ {\Af}_n (A) = A^n, \quad {\Af}_n (\phi ) = {\phi }^n : A^n @>>>
B^n,$$ for $A, B \in  {\Cal A}_k$, $\phi : A @>>> B$.

\bigpagebreak
\definition {Definitions}

We say a functor $X: {\Cal A}_k @>>> \goth S$ is an \it affine
scheme \rm over $k$ if it is representable, i.e., if there exists 
a natural isomorphism $X \backsimeq  \Sp R$ for some $ R \in  {\Cal
A}_k$. 

A \it group scheme \rm is a functor $G : {\Cal A}_k @>>> \goth G$
which, when composed with the forgetful functor $\goth G @>>>
\goth S$, is an affine scheme over $k$.\enddefinition

\bigpagebreak
In dealing with Hopf algebras, we have the next

\bigpagebreak
\proclaim {Proposition 1.6} If $H$ is a Hopf algebra, and $A$ a
commutative algebra, then $\Alg (H,A)$ is a subgroup of the group of
units of $\Hom (H,A)$.\endproclaim

\demo {Proof} It is clear that the unit $u \epsilon $ is an algebra
map $H @>>> A$.

Let $f, g \in \Alg (H,A)$. Then for $x,y \in H$, $$\multline (f * g)
(xy) = f( x_{(1)} y_{(1)} ) g( x_{(2)} y_{(2)} ) = \\ f( x_{(1)} ) 
g(x_{(2)} ) f( y_{(1)} ) g( y_{(2)} ) = (f*g) ( x ) (f*g) ( y
).\endmultline $$  So $f*g \in \Alg (H, A)$.

If $\Cal S$ is the antipode in $H$, we claim that for $f \in \Alg
(H,A)$, $f^{-1} = f \circ \Cal S$. Note that because $\Cal S$ is an
antialgebra map and $A$ is commutative, $f \circ \Cal S \in
\Alg (H,A)$. Now, if $x \in H$, we have $$(f * f \circ \Cal S) (x) =
f( x_{(1)} ) f( \Cal S ( x_{(2)} ) ) = f( x_{(1)}  \Cal S ( x_{(2)}
) ) = f( \epsilon (x) 1_H ) = \epsilon (x) 1_A.$$ 

Similarly, $f \circ \Cal S * f = u \epsilon$. So the assertion is
proved.\qed \enddemo

\bigpagebreak
\proclaim {Corollary 1.7} 
If $H$ is a commutative Hopf algebra, then its spectrum $\Sp H$ is an affine
group scheme. \qed \endproclaim

\bigpagebreak
We have moreover, that if $R, S$ are commutative algebras, and
$\phi: R @>>> S$ is an algebra map, then $\phi $ induces
functorially a natural transformation $$ {\phi }^* : \Sp S @>>> \Sp
R,$$ in the form ${\phi }^* (\xi ) = \xi \circ \phi$.   

This observation, together with corollary (1.7), give us
a (contravariant) functor from the category of commutative Hopf
algebras into the category of affine group schemes.

In fact this functor is an antiequivalence of categories. We go now
to this point.

\bigpagebreak
 

If $X :  {\Cal A}_k @>>> \goth S$ is any functor, the set of all 
natural
transformations $X @>>> {\Af}_1$ can be given a $k$-algebra
structure by setting $$ {(f+g)}_A (x) = f_A(x) + g_A(x),$$ for $A
\in  {\Cal A}_k$, and $x \in X(A)$, defining in analogous way
$fg$ and $\lambda f$, for $\lambda \in k$. 

We denote this algebra by $k[X]$.

\remark {Observation} The universal property of the tensor product
implies that a direct product $X \times Y$ of affine schemes
is again an affine scheme with $k[X \times Y] =  k[X] \otimes
k[Y]$.\endremark

\bigpagebreak
If $X, Y : {\Cal A}_k  @>>> \goth S$ are functors and $f:X @>>> Y$ 
is a natural transformation, then $f$ induces functorially 
a $k$-algebra map $f^*: k[X] @>>> k[Y]$, by $f^* (\tau ) = 
\tau \circ f$, $\tau \in k[Y]$.

\bigpagebreak Recall that if $\Cal C$ is a category, and
$F : \Cal C @>>> \goth S$ is a functor, then Yoneda \rq s lemma
asserts that for any object $A \in \Cal C$, the collection of all
natural transformations $\tau : F @>>> {\Hom }_{\Cal C} (A, \_)$ is in
one to one correspondence with $F(A)$. This correspondence
sends $\tau \mapsto {\tau }_A ({\id }_A )$. 

Now, as a consequence of Yoneda \rq s lemma, we have that if $R, S
\in  {\Cal A}_k $ the collection of all natural transformations $\Sp
R @>>> \Sp S$ is in one to one correspondence with $\Alg (S, R)$.
More exactly, this bijection is defined by associating  to each
$\tau : \Sp R @>>> \Sp S$, the map ${\tau }_R ({\id }_R) \in \Alg (S,
R)$.

If $R \in  {\Cal A}_k$, the algebra
isomorphism $$\Alg (k[T], R) @>>> R,  \quad f \mapsto f(T), $$
gives an isomorphism of $k$-algebras $$k[\Sp R] @>>> R.$$

\bigpagebreak
Let $G$ be an affine group scheme. Then the group structures on
$G(A)$, $A \in {\Cal A}_k$, define  natural transformations
$$\gather m : G \times G @>>> G, \\ 1: \Sp k @>>> G, \\ i:G
@>>> {G}^{op}, \endgather $$ which  give rise to algebra maps
$$\gather \Delta : k[G] @>>> k[G] \otimes k[G], \\ \epsilon : k[G]
@>>> k, \\ \Cal S : k[G] @>>> {k[G]}^{op}. \endgather $$
Translation of the group axioms on $G$, say that $k[G]$ is a Hopf
algebra where $\Delta $, $\epsilon $ and $\Cal S$
are the comultiplication, the counit and the antipode,
respectively. 

\bigpagebreak
By the above, the functor $G \mapsto k[G]$, from the category of
 affine group schemes into the category of commutative Hopf 
algebras
is a quasi inverse of $H \mapsto \Sp H$, and thus the latter is an
antiequivalence of categories.

\bigpagebreak

\example {Examples} 

\bigpagebreak 1) Let $G = GL (n, \_ )$ be the functor that
associates to each commutative $k$-algebra $A$, the group $GL
(n, A)$ of all $n \times n$ matrices with entries in $A$ and
determinant $1$. Then $G$ is an affine group scheme and the Hopf
algebra that represents it is $k[X_{ij} : 1 \leq i, j \leq n ; Y] / ( Y
\det (X_{ij}) - 1 ) $,  which is isomorphic to the localization of
$k[X_{ij} : 1 \leq i, j \leq n ]$ in the powers of $\det
(X_{ij})$.  The Hopf algebra structure is given by 
$$ \gather \Delta
( X_{ij}) = \sum_{k=0}^n { X_{ik} \otimes  X_{kj} }, \quad \Delta
(Y) = Y  \otimes Y, \\ \epsilon (X_{ij}) = {\delta }_{ij},
\quad \epsilon (Y) = 1, \\ \Cal S (X_{ij}) = {(-1)}^{i+j} Y 
\det (A_{ij}),\endgather $$
 where $A_{ij}$ denotes the submatrix of
$(X_{ij})$ obtained eliminating the $j$-th row and the $i$-th column,
i.e., $\Cal S  (X_{ij})$ is the $i,j$-entry of ${(X_{ij})}^{-1}$.

\bigpagebreak 2) Consider the affine group scheme $\Bbb U( \_ ) :
{\Cal A }_k @>>> \goth G$,  that takes $A$ to its group of units
$\Bbb U(A)$. That is, $\Bbb U (\_) = GL(1, \_)$.

In this case the representing Hopf algebra is $H =
k[T,T^{-1}]$, the algebra of Laurent polynomials in $T$ over $k$,
with $$\gather \Delta (T) = T \otimes T, \quad  \epsilon (T) = 1, \\
\Cal S (T) = T^{-1}.\endgather$$ $H$ is
isomorphic to the group algebra over the additive group of
integers, $k\Bbb Z$.

\bigpagebreak 3) \bf The circular group

\rm Let  $C : {\Cal A}_k @>>> \goth G$ be the functor  
such that $$C(A) = \{ (a,b) \in A \times A : a^2 + b^2 = 1\}.$$ for
each commutative $k$-algebra $A$. $C(A)$ has a group structure
provided by 
$$(a,b)(c,d) = (ac-bd, ad + bc).$$
In fact,  $C$ is an affine group
scheme, called the \it circular group, \rm whose representing
algebra is the so called \it trigonometric algebra: \rm $H = k[s,c] /
(s^2 + c^2 - 1)$, with Hopf algebra structure determined by $$\gather
\Delta (c) = c \otimes c - s \otimes s, \quad \Delta (s) = c \otimes
s + s \otimes c, \\ \epsilon (c) = 1, \quad \epsilon (s) = 0, \\ \Cal S
(c) = c, \quad \Cal S (s) = -s.\endgather $$ \endexample

\remark {Remark} Assume that 2 is invertible in $k$. Then $C$ is
isomorphic to $\Bbb U$ if $k$ contains a square root of $-1$, $i$. 

In fact $i$ induces an isomorphism of Hopf algebras between the
representing algebras, $k[T,T^{-1}] @>>> H$, in the form $T \mapsto
c + i s$. At the group level, the isomorphism $C(A) \to A^{*}$ is
given by 
$$ (a,b) \mapsto a+ i b;  $$
the inverse is $$x \mapsto (\frac{x+x^{-1}}2, \frac{x-x^{-1}}{2i}).$$
\endremark

\bigpagebreak
\subhead The dual algebra of a Hopf algebra \endsubhead

\bigpagebreak Let $C$ be a coalgebra. Then we know that the dual
$k$-module $C^* = \Hom (C,k)$ is an algebra with the convolution
product. We will denote this product by $fg$ (instead of $f*g$), for
$f, g \in C^*$. Then we have, $\forall c \in C$ 
$$ \gathered < fg , c > = <f,c_{(1)}> <g,c_{(2)}>, \\ 
1_{C^*} = \epsilon. \endgathered \tag 1$$
Observe that the map $C^* \otimes C^* @>>> C^*$ which define
this algebra structure is obtained  by restricting to
$C^* \otimes C^* \subseteq {(C \otimes C)}^*$ the transpose of
$\Delta : C @>>> C \otimes C$, and the unit $k @>>> C^*$ is the
transpose of the counit $\epsilon : C @>>> k$.

\bigpagebreak
Let $A$ be a finite dimensional $k$-algebra. We shall identify $A^*
\otimes A^*$ with ${(A \otimes A)}^*$ via the natural isomorphism. So
the transposes of the multiplication $m:A \otimes A @>>> A$ and the
unit $u:k @>>> A$ define maps $$\Delta : A^* @>>> A^* \otimes A^*,
\quad \text {and} \quad \epsilon: A^* @>>> k.$$ It is easy to see
that $(A^*, \Delta )$ is a coalgebra with counit $\epsilon$. 

Explicitly, for $f \in A^*$ and $x, y \in A$ $$ < \Delta (f), x
\otimes y > = <f,xy>, \quad \text {and} \quad \epsilon (f) =
<f,1>.\tag 2$$
In other words, $\Delta (f) = f_{(1)}\otimes f_{(2)}$ means
$\langle f, xy\rangle = \langle f_{(1)}, x\rangle \langle f_{(2)},y
\rangle$.

\bigpagebreak
If $(A,m, \Delta)$ is a finite dimensional bialgebra, with unit $u$
and counit $\epsilon$, then $\Delta $ and $\epsilon $ are algebra
maps. This implies that their transposes ${\Delta}^*$ and
${\epsilon }^*$ are coalgebra maps (with the coalgebra structure
in $A^*$ provided by the transposes $m^*$ and $u^*$ of $m$ and
$u$, respectively). So $(A^*, {\Delta}^*, m^*)$ is also a bialgebra.

\proclaim {Proposition 1.8} Let  $H$ be a finite dimensional Hopf
algebra with antipode $\Cal S$ and let $H^*$ be the dual bialgebra.
Then $H^*$ is a Hopf algebra with antipode ${\Cal S}^*$.
We have also that if $k$ is a field, the evaluation map $H @>>>
H^{**}$ is an isomorphism of Hopf algebras. \endproclaim
\demo{Proof} Left to the reader.
\qed \enddemo  

We will denote also by $\Delta $, $\epsilon $ and $\Cal S$ for the
comultiplication, the counit and the antipode in $H^*$.

\bigpagebreak
\remark {Remarks}

1) Let $H$ be a finite dimensional Hopf algebra. It is clear from the
definitions that $H$ is commutative iff $H^*$ is cocommutative and
viceversa.

\bigpagebreak
2) The grouplike elements in
$H^*$ are the algebra maps $H @>>> k$.  The primitive elements
are the derivations of $H$, i.e., the linear maps $D:H @>>> k$ such
that $D(ab) = \epsilon (a) D(b) + D(a) \epsilon (b)$, $a, b \in H$.

\bigpagebreak
3) In case $A$ is not a finite dimensional algebra, then the
inclusion $A^* \otimes A^* \subseteq {(A \otimes A)}^*$ is proper,
so we can no more consider the coalgebra structure in $A^*$ as above.
In this case we take the \it finite dual \rm of $A$, which is defined
as $$A^o = \{f \in A^*: \exists  I \subseteq A, \text {ideal of finite
codimension, with }f(I) = 0 \}.$$
It can be shown that if $H$ is a Hopf algebra then $H^o$ is also a
Hopf algebra \it dual \rm to $H$, in the sense that it
satisfies equations (1) and (2).\endremark


\bigpagebreak 
\example {Examples}

\bigpagebreak 1) Let $G$ be a group, and consider the Hopf
algebra $kG$. Then the dual Hopf algebra ${kG}^*$ may be
identified with the algebra of functions over $G$, $k^{G}$, where
the algebra structure is pointwise multiplication. If $k$ is a
field and $G$ is a \it finite \rm group, then there is an isomorphism
of Hopf algebras $kG \backsimeq {k^{G}}^*$, given by the evaluation
map $g \mapsto E_g$, where $E_g(h) = h(g)$, $g \in G, h \in k^{G}$.

\bigpagebreak 2) Let $A = M_n(k)$ be the algebra of all $n
\times n$ matrices with entries in $k$. Identifying $A$ with $A^*$
via the trace (i.e., $< X,Y > = \Tr (XY) $, $X, Y \in A$), we get a
coalgebra structure in $A$. The comultiplication, $\Delta$, is
given by $$\Delta (E_{ij}) = \sum_{k=1}^n { E_{kj}\otimes E_{ik} },
$$ where $E_{ij}$  denotes the matrix having all entries equal to
zero, except for a $1$ in the entry $ij$. The counit is $\epsilon =
\Tr : A @>>> k$.  

Observe that $A$ is not a bialgebra with this comultiplication.

\bigpagebreak 3) Let $H$ be the Taft algebra over the field $k$. 
Then $H$ is isomorphic to $H^*$. This can be seen as follows.

Let $G \in H^*$ the algebra map defined by $$ G(g) = {\xi }^{-1},
\quad G(x) = 0,$$
 and let $X$ be the $k$-linear map $ X : H @>>> k$, such that
$$\gather X (g^i x) = 1, \quad  0 \leq i \leq N-1, \\ X(g^i x^j ) = 0,
\quad 0 \leq i,j\leq N-1, j\neq 1.\endgather
$$ Then the map $g \mapsto G$, $x \mapsto X$, gives an
isomorphism of Hopf algebras $H @>>> H^*$.

\endexample


\newpage \head \S2 Hopf modules and integrals \endhead
This section covers the main results of Larson and Sweedler's
fundamental paper \cite{8}. 

 \definition {Definition} Let $H$ be a Hopf algebra
over $k$. A $k$-module $V$ is called a (right)\it Hopf module \rm for
$H$, if it satisfies the three conditions:

1) $V$ is a right $H$-module.

2) $V$ is a right $H$-comodule.

3) \it Compatibility condition: \rm ${\Delta }_V (v.x) = v_{(0)}
. x_{(1)} \otimes v_{(1)} x_{(2)}$, $\forall $ $x \in H, v \in V$. 

If $V$, $W$ are Hopf modules, a $k$-linear map $f : V @>>> W$ is
said a \it Hopf module map, \rm if it is both a module and a
comodule map.\enddefinition

\bigpagebreak \remark {Remark}
If ${\Delta }_V : V @>>> V
\otimes H$ is the comodule structure map on $V$, then condition
3) says that ${\Delta }_V$ is a morphism of $H$-right modules,
with the right $H$-action on $V \otimes H$ given by $\Delta$: $ (v
\otimes g).h = v. h_{(1)} \otimes g h_{(2)}$. \endremark

\bigpagebreak We denote by ${\Cal M}_H^H$ the category of right
$H$-Hopf modules. 

If $V$ is a right Hopf module, the \it invariant \rm and \it covariant
\rm submodules of $V$ are defined, respectively, to be $$V^H =
\{v \in V : v . h = \epsilon (h) v, \forall h \in H\}, $$ and $$V^{\co H }
= \{ v \in V : {\Delta}_V (v) = v \otimes 1\}.$$ 

\bigpagebreak  
Given a $k$-module $W$, the tensor product $W \otimes H$ can be
made into a  $H$-Hopf module by setting $$\gather 
(w\otimes h) . g = w \otimes hg, \\  {\Delta }_{W \otimes H} (w
\otimes h) = w \otimes h_{(1)} \otimes h_{(2)}. \endgather $$
$\forall {} w \in W, {}h, g \in H$. Such Hopf modules will be
called {\it trivial\/}.

The following theorem asserts that all Hopf modules are trivial.

\bigpagebreak 
\proclaim {Theorem 2.1. Fundamental Theorem of Hopf modules
(Larson - Sweedler, 1969)}  Let $V$ be a right $H$-Hopf module. Then
the multiplication map $$\rho : V^{\co H} \otimes H @>> v \otimes h
\mapsto v . h > V,$$ is an isomorphism of Hopf modules (where
$V^{\co H} \otimes H$ has the trivial Hopf module structure).
\endproclaim \demo {Proof}

We first show that $\phi : V @>>> {V}^{\co H}$, given  by
$\phi (v) = v_{(0)}. \Cal S( v_{(1)} )$ is a well defined linear map.
 
For this, let $v \in V$ and calculate $$\multline {\Delta }_V ( \phi
(v) ) = {\Delta }_V ( v_{(0)}. \Cal S( v_{(1)} ) ) = v_{(0)}. \Cal S
(v_{(3)} ) \otimes v_{(1)}.\Cal S ( v_{(2)} ) = \\ v_{(0)}. \Cal S
(v_{(2)} ) \otimes \epsilon (v_{(1)}) 1 =  v_{(0)} .\Cal S (v_{(1)}
) \otimes 1 = \phi (v) \otimes 1.\endmultline $$ So $\phi (v) \in
V^{\co H} $ as we wanted. 

Now we claim that $(\phi \otimes \id) {\Delta }_V : V @>>> V^{\co H }
\otimes H$ is the inverse of $\rho$.

Let $v \in V$, then $$\rho \circ (\phi \otimes \id ){\Delta }_V (v) 
= \phi
(v_{(0)}).v_{(1)} = v_{(0)} .\Cal S (v_{(1)} )v_{(2)} = v_{(0)}
\epsilon (v_{(1)} ) = v.$$
If $v \in V^{\co H}$, $h \in H$, we have $$(\phi \otimes \id )
{\Delta }_V \circ
\rho (v \otimes h) = \phi ( (v.h)_{(0)} )\otimes (v.h)_{(1)} = \phi
(v .h_{(1)} )\otimes  h_{(2)} = v.h_{(1)} \Cal S( h_{(2)} )\otimes
h_{(3)} = v \otimes h.$$
This finishes the proof, since both maps are morphisms of right
$H$-Hopf modules. \qed \enddemo

\bigpagebreak
\definition {Definition}
Let $H$ be a Hopf algebra. The $k$-linear spaces of left and right
integrals in $H$ are defined, respectively, as follows:  
$$ {\Cal I}_l (H) = \{ h \in H : xh = \epsilon (x) h , \forall x \in H \}, $$
and $${\Cal I }_r (H) = 
\{ h \in H : hx = \epsilon (x) h , \forall x \in H \}. $$

\bigpagebreak $H$ is called \it unimodular \rm if ${\Cal I}_l (H) =
 {\Cal I }_r (H)$.\enddefinition

\bigpagebreak

\bigpagebreak
\remark {Observation} With this definition, for instance  the
space of (left) integrals in $H$ is the (left) ideal of invariants
with respect to the action of $H$ on itself by left multiplication. 
It is clear that it is also a right ideal. \endremark

\bigpagebreak
\example {Examples}

1) Let $H$ be finite dimensional. Then $\phi \in H^*$ is a left
integral, if and only if, $$ h_{(1)} < \phi , h_{(2)} > = < \phi , h > 1_H,
$$ $\forall h \in H$.

\bigpagebreak
2) Let $G$ be a finite group, and let $kG$ its group algebra. Then
$kG$ is unimodular, with ${\Cal I }_l = {\Cal I }_r = k(\sum_{g \in
G} g )$.

(The definition of left integral in $kG = {k^{G}}^*$, which is the
space of all distributions on $G$, coincides with that of left
invariant measure, i.e., those linear functionals $\alpha $ on $k^{G}$
such that $< \alpha , \phi > = < \alpha , L_g \phi > $, $\forall \phi \in
k^{G}$, $g \in G$. Here $L_g$ denotes the left translation by $g$ in
$k^{G}$: $< L_g \phi, h > = \phi (gh)$).

\bigpagebreak
3) Let $H$ be the Taft algebra of dimension $N^2$ over the field
$k$. Then the spaces of left and right integrals are respectively
$${\Cal I}_l (H) = k( \sum_{j=0}^{N-1} {g^j x^{N-1} } ),$$ and 
$${\Cal I}_r (H) = k( \sum_{j=0}^{N-1} {{\xi}^j g^j x^{N-1} } ).$$
\endexample

\bigpagebreak
Let $H$ be a Hopf algebra, and let $V$ be a right
$H$-comodule with comodule structure map ${\Delta }_V$. Then $V$
is a left $H^*$-module via $$ H^* \otimes V @>\id \otimes
{\Delta }_V>> H^* \otimes V \otimes H @>\id \otimes \tau >> H^*
\otimes H \otimes V @>{< , >} \otimes \id >> k \otimes V @>>> V, $$
where $< , > : H^* \otimes H @>>> k$ is the evaluation map: $ h^*
\otimes h \mapsto <h^*, h>$.

Explicitly, $$ h^* . v = <h^*,v_{(1)}> v_{(0)}, \tag *
$$ $v \in V$, $h^* \in H^*$.

\bigpagebreak 
A left $H^*$-module $V$ with the property that there exists a
right $H$-comodule structure on $V$ such that (*) holds is called
\it rational. \rm

For finite dimensional $H$, it is possible to show that all left
$H^*$-modules are rational.

\bigpagebreak
Let $\dim H$ be finite. The above allows us to consider $H^*$ as a
right $H$-comodule as we indicate now.

$H^*$ is a left $H^*$-module via left multiplication. So it is a
rational $H^*$-module. We can then consider the  right coaction
$\rho : H^* @>>> H^* \otimes H$,  such that $$\rho (f) = f_{(0)}
\otimes f_{(1)} \quad \Leftrightarrow \quad <p , h_{(1)}> <f ,
h_{(2)}> = <p, f_{(1)}> <f_{(0)} , h >,\tag 1 $$ $\forall p \in H^*$,
$h \in H$. 

\bigpagebreak
On the other hand, we have that $H^*$ is a left (respectively
right) $H$-module via the transpose of right (respectively left)
multiplication in $H$.  This action is denoted by $h
\rightharpoonup h^*$ (respectively $h^* \leftharpoonup h$), for
$h \in H$, $h^* \in H^*$.

By means of the antipode, we get left and right actions of $H$ on
$H^*$: $$h \rightharpoondown h^* = h^* \leftharpoonup \Cal S(h),
\quad \text {and} \quad  h^* \leftharpoondown h = \Cal S(h) 
\rightharpoonup h^*. $$

These actions are determined by
$$< h^* \leftharpoondown h , g > = < h^* , g \Cal S(h)  >, \quad <h
\rightharpoondown h^*, g > = <h^* , \Cal S(h) g >, $$
 for $h^* \in H^*$, $g, h \in H$.  

In particular, $H^*$ is a right $H$-module via 
$h^* \leftharpoondown h$. If $H$ is
finite dimensional, recalling the right coaction $\rho$ of $H$ on
$H^*$ given by (1), we find that $H^*$ is both a right module and
comodule for $H$. Moreover, we have:

\bigpagebreak
\proclaim {Lemma 2.2 (Larson - Sweedler, 1969)}
Let $H$ be a finite dimensional Hopf algebra. Then $H^*$ is a right
Hopf module, with action $ \leftharpoondown $ and coaction $\rho$.
\endproclaim
\demo {Proof}

We must show that $\forall f \in H^*$, and $\forall h \in H$, $\rho (f
\leftharpoondown h )= f_{(0)} \leftharpoondown  h_{(1)} \otimes
f_{(1)} h_{(2)}$.  That is, we must see that $\forall p \in H^*$, and
$\forall x \in H$, $$<p, x_{(1)}><f \leftharpoondown h, x_{(2)}> =
<f_{(0)} \leftharpoondown h_{(1)}, x><p, f_{(1)} h_{(2)}>.$$
Now, $$\multline <f_{(0)} \leftharpoondown h_{(1)}, x><p, f_{(1)}
h_{(2)}> = <f_{(0)}, x \Cal S(h_{(1)})><h_{(2)} \rightharpoonup p,
f_{(1)}> = \\ <h_{(3)} \rightharpoonup p, x_{(1)} \Cal
S(h_{(2)})><f, x_{(2)} \Cal S (h_{(1)}) > = <p, x_{(1)} <\epsilon ,
h_{(2)}>> < f, x_{(2)} \Cal S (h_{(1)})> = \\ <p, x_{(1)}><f, x_{(2)}
\Cal S(h)> = <p, x_{(1)}><f \leftharpoondown h,
x_{(2)}>. \qed\endmultline$$  \enddemo

\bigpagebreak
In what follows $k$ will be a field. The next theorem, due to
Larson and Sweedler (1969), is a consequence of 
theorem (2.1) and lemma (2.2).

\bigpagebreak
\proclaim {Theorem 2.3 (Larson - Sweedler, 1969)} Let $H$ be a finite
dimensional Hopf algebra over $k$. Then 

1) $\dim {\Cal I}_l (H) = \dim {\Cal I}_r (H) = 1$.

2) The antipode $\Cal S$ is bijective, and $\Cal S ({\Cal I}_l) =
{\Cal I}_r$.

3) For $0 \neq \lambda \in {\Cal I}_l (H^*)$, the map $H @>>> H^*$,
given by $h \mapsto h \rightharpoonup \lambda$, is a left linear
isomorphism. \endproclaim

\demo{Proof} 
1) Consider the Hopf module structure on $H^*$ given by lemma (2.2).
By theorem (2.1), ${(H^*)}^{\co H} \otimes
H \backsimeq H^*$. And, as $H$ is finite dimensional, we get $\dim
{(H^*)}^{\co H} = 1$.

It remains to observe that ${(H^*)}^{\co H} = {\Cal I }_l (H^*)$,
which is clear from the definitions.  Then replacing $H^*$ by
$H$ (again using the finite dimensionality of $H$), we get $\dim
{\Cal I }_l (H) = 1$. The remaining equality will follow from 2).  

\bigpagebreak
2) By theorem (2.1), the map$${\Cal I }_l (H^*) 
\otimes H @>> \lambda \otimes h \mapsto \lambda
\leftharpoondown h > H^*,$$ is an isomorphism.

Suppose now that $0 \neq \lambda \in {\Cal I }_l (H^*)$, and let $h
\in H$ such that $\Cal S(h) = 0$. Then $$0 = \Cal S(h)
\rightharpoonup \lambda = \lambda \leftharpoondown h.$$
Thus $\lambda \otimes h = 0$, and so $h = 0$.

This shows that $\Cal S$ is injective. Now, as $H$ is finite
dimensional, it is bijective.  

Finally, using this it is an easy calculation to show that $\Cal S
({\Cal I}_l) = {\Cal I}_r$.

\bigpagebreak
3) Again using theorem (2.1), plus the fact that 
$\dim {\Cal I}_l (H^*) = 1$, we get that for any $0 \neq \lambda \in 
 {\Cal I}_l (H^*)$, $$H^* = \lambda \leftharpoondown H = \Cal S(H)
\rightharpoonup \lambda.$$ Now, as $\Cal S$ is bijective, it
follows that $H^* = H \rightharpoonup \lambda$, which proves 3).
\qed \enddemo

\bigpagebreak From now on we will write $h f := h \rightharpoonup
f $ and $f h : = f \leftharpoonup h$, for $h \in H$, $f\in H^*$. 

\bigpagebreak
In what follows we consider a class of algebras that contains the
finite dimensional Hopf algebras. 

\bigpagebreak

\subhead Frobenius algebras \endsubhead

Let $k$ denote a field. If $A$ is a $k$-algebra,  the \it left \rm
(respectively \it right\rm ) \it regular representation \rm of $A$
is the module structure in $A$ given by left (respectively right)
multiplication. It is denoted by ${}_A A$ (respectively $A_A$).

Recall that if $M$ is a right $A$-module, then the dual space $M^*$
is a left $A$-module by $$< a . \phi , m > = < \phi , m . a >, $$ $a
\in A$, $\phi \in M^*$, $m \in M$.

In particular, this holds if we take $M$ to be the right regular
module for $A$, ${}_AA$.

\bigpagebreak
Let $A$ be a finite dimensional algebra (say $\dim A = n$), $\phi \in
A^*$, and $r_i, l_i \in A$, $1 \leq i \leq n$. Then $\phi $ is called a
\it Frobenius homomorphism, \rm with \it dual bases \rm $(r_i,
l_i)$ if one of the following (equivalent) conditions holds:

a) $\forall x \in A$, $x = \sum_i {r_i <\phi ,l_i x>}$.

b) $\forall x \in A$, $x = \sum_i {<\phi ,x r_i> l_i}$. 

\bigpagebreak
\proclaim {Proposition 2.4} Let $A$ be a finite dimensional algebra,
and let $f \in A^*$. Then the following statements are equivalent:

1) The map ${}_A A @>>> {A_A}^*$, given by $x \mapsto xf$ is
an isomorphism of left $A$-modules.

2) There exist $r_i, l_i \in A$, $1 \leq i \leq n$ ($n = \dim A$), such
that $f$ is a Frobenius homomorphism with dual bases $(r_i,l_i)$.

3) The map $A_A @>>> { {}_A A }^*$, given by $x \mapsto fx$ is an
isomorphism of right $A$-modules.\endproclaim

\demo {Proof}

1) $\Rightarrow $ 2). Let $(l_i)$ be a $k$-basis for $A$, and let
$(f_i)$ its dual basis. 

By 1), there exist $r_i \in A$, such that $f_i = r_i f$. Then, $\forall
x \in A$, $$x = \sum_i {<f_i,x >l_i} =  \sum_i {<r_i f,x >l_i} = \sum_i
{<f,x r_i> l_i}.$$

2) $\Rightarrow $ 1). Suppose $fx = 0$ for some $x \in A$. Then
$<fx,y> = <f, xy> = 0$, $\forall y \in A$. Hence $$x =  
\sum_i {<f,x r_i> l_i}
 = 0.$$

So the map $x \mapsto fx$ is injective, and as $A$ is finite
dimensional it is bijective.

Similarly one shows that 2) $\Leftrightarrow $ 3), but here using
the condition a) in the definition of dual bases for Frobenius
homomorphisms. \qed \enddemo

\bigpagebreak
\definition {Definition} Let $A$ be a finite dimensional
$k$-algebra. If $A$ satisfies any of the equivalent conditions 1)
-- 3) of the proposition above, then $A$ is called a \it Frobenius
algebra. \rm \enddefinition 

\bigpagebreak
\remark {Remark} By theorem (2.3), we know
that if $H$ is a finite dimensional Hopf algebra, then $H$ is a
Frobenius algebra.

We have moreover, that if $0 \neq \lambda$ is any left integral in
$H^*$, then the maps ${}_H H @>>> {H_H}^*$, $h \mapsto h \lambda$,
and $H_H @>>> { {}_H H }^*$, $h \mapsto \lambda h$ are
isomorphisms of left (respectively right) $H$-modules.

We may then choose $\Lambda \in H$ such that $\lambda \Lambda
= \epsilon$. Such $\Lambda$ is necessarily a right integral in $H$
as the following argument shows.

\bigpagebreak
If $I$ is a right integral in $H$, then $\lambda I = <\lambda ,I>
\epsilon$. And by the injectivity of $h \mapsto \lambda h$, 
$<\lambda ,I> \neq 0$ if $I \neq 0$. 

So we can choose $I$ to be such that $<\lambda ,I> = 1$, and then 
$\lambda I = \epsilon$.

Again using the injectivity of $h \mapsto \lambda h$, we find that
$\Lambda = I $ is a right integral in $H$. 

We saw also that the condition $\lambda \Lambda = \epsilon$ is
equivalent to $<\lambda ,\Lambda > = 1$.
\endremark

\newpage
\head \S3 Finite Hopf algebras \endhead

In what follows, $k$ will denote a field. Let $H$ be a Hopf
algebra. 

\bigpagebreak
The following are conjectures made by Kaplansky (1975) (see
appendix):

1. If $R$ is a Hopf subalgebra of $H$, then $H$ is a free
$R$-module.   

2. If $\dim H $ is finite and $H$ is semisimple, then the square of
the antipode is the identity.

3. If $\dim H = p$, $p$ a prime, then $H$ is commutative and
cocommutative.

\bigpagebreak
Of these, (1) is known to be false in general \cite{12}, though it is true if
$H$ is finite dimensional. As to (2) and (3), they are known over
fields of characteristic $0$.

The main results we treat in this section and the next are partial
answers to these questions.

\bigpagebreak
 A finite
dimensional Hopf algebra $H$ over $k$ will be called \it finite.
\rm By the \it order \rm of $H$ we will mean its dimension.

\bigpagebreak
The results in $\S2 $ allowed us to assert that $H$ is a Frobenius
algebra. Now we treat the connection between this structure and
the Hopf algebra structure in $H$ following \cite{13}.

\bigpagebreak
\proclaim {Theorem 3.1} Let $0 \neq \lambda \in H^*$ be a left
integral, and let $\Lambda \in H$ be such that $\lambda \Lambda =
\epsilon $. Then $\lambda $ is a Frobenius homomorphism with dual
bases $(\Cal S ( {\Lambda }_{(1)}), {\Lambda }_{(2)} )$.\endproclaim

\demo {Proof}Let $x\in H$. Then 
$$\Cal S ( {\Lambda }_{(1)}) \langle \lambda, {\Lambda }_{(2)} x\rangle = 
\Cal S ( {\Lambda }_{(1)}) {\Lambda }_{(2)} x_{(1)} \langle \lambda, {\Lambda }_{(3)}
x_{(2)}\rangle = 
x_{(1)} \langle \lambda, \Lambda x_{(2)}\rangle = x \langle \lambda, \Lambda\rangle =
x. \qed 
$$\enddemo

\bigpagebreak In an enterily similar fashion, one can see that if 
$\gamma \in H^*$
is a nonzero right integral, then there exists a left integral
$\Gamma \in H$ such that $\Gamma \gamma = \epsilon$. We also 
have that $\gamma $ is a Frobenius homomorphism in $H$ with
dual bases $({\Gamma }_{(1)} , \Cal S ({\Gamma }_{(2)}) )$.

From now on we fix a nonzero left (respectively right) integral
$\lambda \in H^*$ (respectively $\gamma $), and call $\Lambda $
(respectively $\Gamma $) the right (respectively left) integral
in $H$ such that $<\lambda , \Lambda > = 1$ (respectively
$<\gamma , \Gamma > = 1$), without further comment. 


 
\bigpagebreak
\remark {Remark}
If $A$ is a Frobenius algebra, and $f \in A^*$ is a Frobenius
homomorphism with dual bases $(r_i, l_i)$, then in $A \otimes A$
we have the identity $$ \sum_i {x r_i \otimes  l_i } = \sum_i {r_i
\otimes l_i x }.$$ $\forall x \in A$.

Indeed, these have the same image in $\End (A)$ under the map
$$A \otimes A @>> a\otimes b \mapsto a\otimes fb > A\otimes
A^* \backsimeq \End (A).$$ Here, as usual, the linear isomorphism
$A\otimes A^* @>>> \End (A)$ is given by $(a\otimes \phi)(x) =
<\phi , x> a$. \endremark

This remark motivates the following

\bigpagebreak
\definition {Definition} 

A $k$ algebra $A$ is called \it separable \rm if  there exist $r_i,
l_i \in A$, $1 \leq i \leq n = \dim A$, such that: 

1)$ \sum_i {r_i l_i } = 1$.

2) $\forall x \in A$, $\sum_i {x r_i \otimes l_i } = \sum_i {r_i
\otimes l_i x}$ in $A\otimes A$.\enddefinition

\bigpagebreak
We say a Hopf algebra $H$ is semisimple if it is semisimple as an
algebra, i.e., if every finite dimensional left  $H$-module is
completely reducible.

\bigpagebreak
\remark {Remark} For a finite dimensional $k$-algebra $A$, the
condition of being semisimple is equivalent to $\rad (A) = 0$.
Where $\rad (A)$ denotes the \it radical \rm of $A$, which by
definition is the sum of all nilpotent left ideals in $A$. 
(See \cite{1, $\S25 $, p. 163}).

We also have the identity $\rad (A) = \Cal J$, where $\Cal J$
denotes the \it Jacobson radical \rm of $A$, i.e., $\Cal J :=
$intersection of all maximal left ideals of $A$. In particular,
as $\Cal J$ is a nilpotent ideal of $A$ ($A$ being finite
dimensional), every element of $\Cal J$ is nilpotent. We will use
these facts later on. \endremark

\bigpagebreak
\example {Example}

Consider a finite group $G$,  $kG$ its group algebra. Recall the
well known Maschke theorem, which asserts that $kG$ is
semisimple iff the characteristic of $k$ does not divide the order
$|G|$ of $G$.

Now, $|G| = \sum_{g \in G} {<\epsilon , g>} = <\epsilon , \sum_{g \in
G} g >$. 

As $\sum_{g \in G} g $ spans the (one dimensional) space of left
integrals in  $kG$, Maschke theorem is equivalent to the
statement " $kG$ is semisimple iff $< \epsilon , {\Cal I}_l(kG) >
\neq 0$". \endexample

\bigpagebreak The next theorem generalizes this result.



\bigpagebreak
\proclaim {Theorem 3.2. Maschke Theorem for Hopf algebras}

Let $H$ be a finite dimensional Hopf algebra. Then the following
statements are equivalent.

1) $H$ is semisimple.

2) $H$ is separable.

3) $<\epsilon , {\Cal I}_l(H) > \neq 0$. \endproclaim

\demo{Proof}

Observe that as $\Cal S ({\Cal I}_l(H) ) = {\Cal I}_r(H)$ and
$\epsilon \Cal S = \epsilon$, then $<\epsilon , {\Cal I}_l(H) > = 0
\Leftrightarrow <\epsilon , {\Cal I}_r(H) > = 0$.

3) $\Rightarrow $ 2).

Take $(r_i, p_i) = (\Cal S ( {\Lambda }_{(1)}), {\Lambda }_{(2)} )$.

We have that $$\sum_i {r_i
p_i} = \Cal S ( {\Lambda }_{(1)} ) {\Lambda }_{(2)} = <\epsilon ,
\Lambda > 1, $$by 3), $t = <\epsilon , \Lambda >
\neq 0$. 

Then changing $p_i$ by $l_i = t^{-1}p_i$,  and using theorem
(3.1), we find that $(r_i , l_i)$ makes $H$ into a separable
algebra.  

\bigpagebreak
2) $\Rightarrow$ 1).

Let $Y$ be a finite dimensional $H$-module, and let $X$ be a
submodule of $Y$.

Let $\pi : Y @>>> X$ be any $k$-linear projection, and 
define $\widehat
{\pi } : Y @>>> Y$ in the form $\widehat {\pi } (y) = \sum_i {r_i .
\pi (l_i . y) }$.

We claim that $\widehat {\pi } $ is an $H$-linear projection.

As $X$ is an $H$-submodule, $\widehat {\pi }(Y) \subseteq X$.

By 2), $\sum_i {h r_i \otimes l_i } = \sum_i {r_i \otimes l_i h}$ in
$H\otimes H$, $\forall h \in H$. This implies that $$\sum_i {h r_i .
\pi (l_i . y)} = \sum_i {r_i \pi (l_i h . y)}, \quad \forall h \in H,  y \in
Y.$$ Hence, $$\widehat {\pi } (h . y) = \sum_i {r_i . \pi (l_i h. y) } = 
\sum_i { h r_i . \pi (l_i . y) } = h . \widehat {\pi } (y),$$ 
$\forall h \in H$, $y \in Y$. So $\widehat {\pi }$ is $H$-linear.  

Finally, $$\widehat {\pi }(x) = \sum_i {r_i \pi (l_i . x)} = 
 \sum_i {r_i  l_i . x} = (\sum_i {r_i  l_i}) . x  = 1 . x = x,$$ $\forall
x \in X$. So the claim is proved.

\bigpagebreak
1) $\Rightarrow $ 3).

As $H$ is semisimple, we have that any short exact sequence of
left $H$-modules $$0 @>>> U @>\mu >> V @>\pi >> W @>>> 0,$$ splits.
That is, there exists an $H$-linear map $\nu : W @>>> V$, such that
$\pi \nu = {\id}_W$.

Applying this to the short exact sequence $$0 @>>> \ker \epsilon
@>>> H @>\epsilon >> k @>>> 0,$$ we get in particular that $<\epsilon
,\nu (1) > = 1$.

Now,  if $h \in H$, as $\nu $ is $H$-linear $$h \nu (1) = \nu
(<\epsilon , h>) = <\epsilon , h> \nu (1) .$$ This shows that $\nu
(1)$ is a left integral in $H$. So $<\epsilon , {\Cal I }_l (H)
>\neq 0$. Then $ <\epsilon , {\Cal I }_r (H) >\neq 0$, and as
$\Lambda $ is a nonzero right integral,  being $\dim {\Cal I }_r (H)
= 1$, we must have $< \epsilon , \Lambda > \neq 0$. \qed
\enddemo

\bigpagebreak

\remark {Observation} Replacing $H$ by $H^*$ in theorem (3.2) we
get that $H^*$ is semisimple iff $< \lambda , 1 > \neq 0$, for some
left integral $\lambda \in H^*$.\endremark

\bigpagebreak
Recall that for a finite dimensional vector space $V$ over $k$, we
have a canonical isomorphism $V^* \otimes V @>>>\End (V) $,
given by $$(\phi \otimes v) (w) = <\phi , w > v, \quad v , w \in V ,
\phi \in V^*.$$ Via this identification,  if $\Tr : \End (V) @>>> k$
denotes the \it trace \rm map, i.e., $\Tr (F)$ := trace of $F$, we
find that $\Tr ( \phi \otimes v ) = <\phi , v>$, for $v \in V$, $\phi
\in V^*$.

\bigpagebreak
\proclaim {Lemma 3.3} Let $H$ be a finite Hopf algebra,
and let $F$ be an endomorphism of $H$. Then the trace of $F$ is
given by $$\Tr (F) = < \lambda , F( {\Lambda }_{(2)} ) \Cal S (
{\Lambda }_{(1)} ) >.$$ \endproclaim

\demo {Proof}

Let $F \in \End (H)$. We know that for all $x \in H$, $$F(x) =
<\lambda , F(x) \Cal S ({\Lambda }_{(1)}) > {\Lambda }_{(2)}.$$ So
via the identification $H^* \otimes H \backsimeq \End (H)$, $F$
corresponds to $$< \lambda , F(\_) \Cal S ({\Lambda }_{(1)} ) >
\otimes {\Lambda }_{(2)},$$ then 
$$\Tr (F) = < \lambda , F( {\Lambda }_{(2)} ) \Cal S ({\Lambda }_{(1)}
) >.$$ \qed 
\enddemo

\bigpagebreak
In a finite Hopf algebra the trace of the square of the antipode
plays a very important role, as we will  see soon. 

\bigpagebreak
For an element $h \in H$, call $L_h \in \End (H)$ the endomorphism
of $H$ given by $L_h (x) = hx$,  $x \in H$.

This defines a $k$-linear map ${\Tr }_H : H @>>> k$, in the form: 
${\Tr }_H (h) = \Tr (L_h)$, $h \in H$.
 
This map will be of central importance later on.  As a consequence
of lemma (3.3), we have the next proposition.

\bigpagebreak
\proclaim {Proposition 3.4} Let $H$ be a finite Hopf
algebra, then:

1) $\Tr ({\Cal S}^2 )= <\epsilon , \Lambda > < \lambda , 1>$.

2) If ${\Cal S }^2 = \id$, then ${\Tr }_H = <\epsilon , \Lambda >
\lambda$.\endproclaim

\demo {Proof}

1) Taking $F = {\Cal S }^2$ in the lemma above, we get $$\Tr
({\Cal S}^2) = <\lambda ,{\Cal S}^2 ({\Lambda }_{(2)}) \Cal
S ( {\Lambda }_{(1)} ) > = <\lambda , \Cal S ( {\Lambda }_{(1)} \Cal S
( {\Lambda }_{(2)} )) > = <\epsilon , \Lambda > < \lambda , 1>.$$

2) As ${\Cal S }^2 = \id$, we have that 
$ h_{(2)} \Cal S ( h_{(1)} ) = < \epsilon ,
h > 1$, $\forall h \in H$. 

Now take $F = L_h$, $h \in H$, then $${\Tr }_H (h) = \Tr (L_h) = <
\lambda , h {\Lambda }_{(2)} \Cal S ( {\Lambda }_{(1)} ) > =
<\epsilon , \Lambda > <\lambda , h>.$$ \qed \enddemo

\bigpagebreak
\proclaim {Corollary 3.5} 

a) $H$ and $H^*$ are semisimple, if and only if $\Tr ({\Cal S }^2 )
\neq 0$.

b) If ${\Cal S}^2 = \id$ and $\char k$ does not divide the
order of $H$, then $H$ and $H^*$ are semisimple. \endproclaim

\demo {Proof} It is immediate.\qed \enddemo 

\bigpagebreak
Indeed, if the characteristic of $k$ is zero, then the converse of
2) holds. 

\remark{Remark}\footnote{This application was added by S. Natale.}
Combining Theorem 1.5 c) and Corollary (3.5), we conclude that a
cocommutative finite Hopf algebra $H$ is semisimple and cosemisimple,
in characteristic 0. Therefore, if $k$ is algebraically closed, $H$
is a group algebra. We recover in this way part of the Fundamental
theorem for cocommutative  Hopf algebras in characteristic 0.
\endremark

\bigpagebreak
In what follows, we develop a formula  for
${\Cal S}^4$, that is valid over an arbitrary field $k$, and
that will imply this last assertion. 

It will also prove  the finiteness of the order of the antipode for
finite Hopf algebras.

\bigpagebreak
\subhead The order of the antipode. Radford\rq s 
formula for ${\Cal S}^4$ 
\endsubhead

\bigpagebreak
If  $0 \neq t \in H$ is a right integral and $h \in H$, then it is
clear that $ht$ is again a right integral. Thus, as the space of
right integrals is one dimensional, we may find an $\alpha = \alpha
(h) \in k$ such that $ht = \alpha (h) t$.

This defines an element $\alpha \in H^*$. We have,
moreover, that $\alpha \in \Alg (H,k) = G(H^*)$. For, if $x, y \in
H$, then $$< \alpha , xy > t = xy t = < \alpha , y><\alpha ,x>t,$$and
as $t \neq 0$, it must be $<\alpha ,xy > = <\alpha , x><\alpha , y>$.

In analogous way, given a right integral $0 \neq I \in H^*$, we may
find an element $a \in H$ such that $h_{(1)} <I, h_{(2)} > = <I,h>
a$, $\forall h \in H$.

Changing the roles of $H$ and $H^*$ in the preceding discussion,
it results that $a \in G(H)$ is a grouplike element in $H$. 

It is clear that $\alpha $ and $a$ are independent of the choice of
the nonzero right integrals $t$ and $I$.  So they are determined
exclusively by $H$. 

\bigpagebreak
\definition {Definition} Let $a \in G(H)$ and $\alpha \in \Alg(H,k)$
as above. Then $a$, $\alpha $ are called the \it modular elements
\rm of $H$.\enddefinition

\bigpagebreak
Observe that as $a$ and $\alpha $ are grouplike elements, then they
are units (of $H$ and $H^*$ respectively), and  $\Cal S(a) =
a^{-1}$, $\Cal S (\alpha) = {\alpha }^{-1}$.

\bigpagebreak
\remark {Remarks} 

1) It follows from the definitions that $H^*$ is
unimodular iff $a = 1$, and $H$ is unimodular iff $ \alpha =
\epsilon $. 

\bigpagebreak
2) If $H$ is semisimple, then $H$ is unimodular. 

To see this, let $0 \neq t$ be a right integral in $H$. So that
$<\epsilon , t> \neq 0$, by the semisimplicity of $H$. 

If $h \in H$, then apply $\epsilon$ to the defining equation 
$$ht = <\alpha , h> t.$$ Then $\alpha = \epsilon $,
and thus $H$ is unimodular. \endremark

\bigpagebreak
Let $\gamma \in {\Cal I }_r (H^*)$ be the Frobenius
homomorphism of $H$ considered earlier. Then $\gamma $
determines an isomorphism $H @>>> H^*$,  $h \mapsto h\gamma $.

Now, for fixed $x \in H$, we can consider the element of $H^*$
defined by $y \mapsto <\gamma ,xy>$, $y \in H$.

There exists then $\rho  = \rho (x) \in H$ such that
$\forall y \in H$ $$<\gamma , xy > = <\rho (x) \gamma , y > =
<\gamma , y \rho (x)>.$$ In other words, $\gamma x = \rho (x)
\gamma $. The map $\rho : H @>>> H$ defined this way is an
automorphism of $H$.

\definition {Definition}$\rho $ is called the \it Nakayama
automorphism \rm of $H$ with respect to the
Frobenius homomorphism $\gamma $.\enddefinition 

\bigpagebreak
The following two propositions will help us to
give a conceptual proof of Radford's formula for ${\Cal S}^4$,
which is simpler than the original proof in \cite{10}.

Denote by $\overline {\Cal S}$ the (composition) inverse of the
antipode of $H$.

\proclaim {Proposition 3.6} Let $\gamma \in H^*$ be a nonzero right
integral, and let $\Gamma \in H$ be a left integral such that
$<\gamma , \Gamma > = 1$. If $t = \Cal S(\Gamma )$, and
$\alpha \in \Alg (H,k)$ is the modular function of $H$, we have:

1) $(\overline {\Cal S} (t_{(2)}) , t_{(1)})$ are dual bases for
$\gamma $.
  
2) $\forall h \in H$, 
$\rho (h) = <\alpha , h_{(1)}> {\overline {\Cal S}}^2 (h_{(2)} )$.
\endproclaim

\demo {Proof}

1) Follows from the fact that $( {\Gamma }_{(1)} , \Cal S ({\Gamma
}_{(2)} ))$ are dual bases for $\gamma $. (Or applying theorem
(3.1) to the Hopf algebra $H^{cop}$, which is obtained from
$H$ by taking the opposite comultiplication $ {\Delta }^{op} = \tau
\circ \Delta $.) In particular, $\langle \gamma, t \rangle = 1$, by
applying $\epsilon$ to the equation a) in 2.4. 

2) Let $h \in H$, then by 1) $$\rho (h) = \overline {\Cal S} (t_{(2)}
) <\gamma , t_{(1)} \rho (h) > =  \overline {\Cal S} (t_{(2)}
) <\gamma , h t_{(1)} >,$$ the last equality following from the
definition of $\rho$. Applying ${\Cal S}^2$,  and recalling that
$\gamma $ is a right integral in $H^*$, we get $$\multline {\Cal S}^2
\circ \rho (h) =<\gamma , h t_{(1)} > \Cal S (t_{(2)}) = <\gamma ,
h_{(1)} t_{(1)} > h_{(2)} t_{(2)} \Cal S (t_{(3)}) = \\ <\gamma ,
h_{(1)} t > h_{(2)} = <\gamma , <\alpha h_{(1)}> t > h_{(2)} = <\alpha
, h_{(1)}> h_{(2)}.\endmultline $$ Then
$$\rho (h) = <\alpha , h_{(1)}> {\overline {\Cal S}}^2 (h_{(2)} ),$$ 
as claimed. \qed \enddemo

\bigpagebreak

\proclaim {Proposition 3.7} If $a \in G(H)$ is the modular element
of $H$ and $t$ is as in (3.6), we have:

1) $(\Cal S (t_{(1)}) a, t_{(2)} )$ are dual bases for $\gamma $. 

2) $\forall h \in H$, $\rho (h) = 
a^{-1} ( {\Cal S }^2 (h_{(1)}) <\alpha ,h_{(2)}> ) a$.
\endproclaim 

\demo {Proof}

1) Using the definition of $a$, we have that $\forall h \in H$,
$$\multline \Cal S(t_{(1)})a <\gamma , t_{(2)} h> = \Cal S(t_{(1)})
t_{(2)} h_{(1)} <\gamma , t_{(3)} h_{(2)}> =\\  h_{(1)} < \gamma ,
th_{(2)}> = h_{(1)} <\epsilon , h_{(2)}> = h.\endmultline $$

2) By 1), $\forall h \in H$, we may write $$\rho (h) = \Cal
S(t_{(1)})a < \gamma , t_{(2)} \rho (h) >.$$ Then $$\multline a {
\overline {\Cal S} }^2 (\rho (h) )a^{-1} = a < \gamma , ht_{(2)}>
\overline {\Cal S} (t_{(1)}) = \\ h_{(1)} t_{(2)} <\gamma , h_{(2)}
t_{(3)}> \overline {\Cal S}(t_{(1)}) = h_{(1)} <\gamma , h_{(2)} t> =
h_{(1)} <\alpha , h_{(2)}>.\endmultline $$
 Hence, conjugating by $a^{-1}$ and applying ${\Cal S}^2$, we get
the result. 
\qed
\enddemo

\bigpagebreak
Consider the left and right $H^*$-module structures on $H$ given
by: $$\gather h^* \rightharpoonup h = h_{(1)} <h^*, h_{(2)}>, \\ h
\leftharpoonup h^* = <h^*, h_{(1)}> h_{(2)}, \endgather$$
$\forall h \in H , h^* \in H^*$. We then have:

\proclaim {Theorem 3.8 (Radford, 1976)} Let $a \in G(H)$, $\alpha \in
\Alg (H,k)$ be the modular elements of $H$. Then the following
formula holds , $\forall h \in H$: $${\Cal S}^4 (h) = a ( {\alpha
}^{-1} \rightharpoonup h \leftharpoonup \alpha ) a^{-1} = 
{\alpha }^{-1}\rightharpoonup (a h a^{-1}) \leftharpoonup \alpha .$$
\endproclaim

\demo {Proof}
We show first that $$
a ( {\alpha }^{-1} \rightharpoonup h \leftharpoonup \alpha ) a^{-1}
=  {\alpha }^{-1} \rightharpoonup (a h a^{-1}) \leftharpoonup
\alpha ,$$ $\forall h \in H$. For this, we compute 
$$a ( {\alpha }^{-1} \rightharpoonup h \leftharpoonup \alpha )
a^{-1} = <\alpha , h_{(1)}> a h_{(2)} a^{-1} <{\alpha }^{-1}, h_{(3)}>.
$$ On the other hand, as $a$ is a grouplike element of $H$, 
$$\multline {\alpha }^{-1}\rightharpoonup (a h a^{-1})
\leftharpoonup \alpha = <\alpha , ah_{(1)}
a^{-1}>ah_{(2)} a^{-1} <{\alpha }^{-1}, a h_{(3)} a^{-1}> =\\  
<\alpha , h_{(1)}> a h_{(2)} a^{-1} <{\alpha }^{-1},
h_{(3)}>,\endmultline $$ the last equality because $\alpha \in \Alg
(H,k)$. 

Now, by propositions (3.6) and (3.7), we have, $\forall h \in
H$, $$<\alpha , h_{(1)}> {\overline {\Cal S}}^2 (h_{(2)} )  
= \rho (h) = a^{-1} ( {\Cal S }^2 (h_{(1)}) <\alpha ,h_{(2)}> ) a.$$
Applying ${\Cal S}^2$ and conjugating by $a$, we get $$a <\alpha ,
h_{(1)}> h_{(2)} a^{-1} = {\Cal S }^4 (h_{(1)}) <\alpha , h_{(2)} >.$$
Multiplying with $<{\alpha }^{-1} , h_{(3)}>$, we find 
$$a ( {\alpha }^{-1} \rightharpoonup h \leftharpoonup \alpha )
a^{-1} = a (<\alpha , h_{(1)} > h_{(2)} <{\alpha }^{-1} , h_{(3)}> )
a^{-1} =  {\Cal S }^4 (h),$$ which finishes the proof. \qed \enddemo

\bigpagebreak
Theorem (3.8) has important consequences:

\bigpagebreak
\proclaim {Corollary 3.9} The order of the antipode is
finite.\endproclaim
\demo {Proof} Since $H$ is finite, and distinct powers of a
grouplike element are linearly independent (being themselves
grouplikes), every grouplike element in $H$ and also in $H^*$ has
finite order. Then, by Radford\rq s formula, ${\Cal S}^4$ and then
$\Cal S$ have finite order. \qed  \enddemo

\bigpagebreak
\proclaim {Corollary 3.10 (Larson, 1971)} 

a) If $H$ is unimodular, then ${\Cal S}^4$ coincides with the inner
automorphism of $H$ induced by a grouplike element. In
particular the order of the antipode is at most $4 \dim H$.

b) If $H$ and $H^*$ are unimodular, then ${\Cal S}^4 =
\id$.\qed \endproclaim 

\bigpagebreak
We want to show now an important trace formula involving the
antipode.  For this, we need  the following lemmas. 

\proclaim {Lemma 3.11} Let $A$ be a Frobenius algebra with
Frobenius homomorphism $\phi$ and dual bases $(r_i,l_i)$.

Let $\alpha \in k$, $e \in A$, such that $e^2 = \alpha e$. If $f \in
\End (eA)$ is a $k$-linear endomorphism of $eA$, then:

i) $\alpha \Tr (f) = <\phi , \sum_i {f(e l_i) r_i }>$.

ii) $\alpha \Tr (f) = <\phi , \sum_i {l_i f(e r_i)}>$. \endproclaim

\demo {Proof} 

 $\forall x \in A$, we have that $ex = \sum_i { <\phi, ex r_i> l_i }$,
then $\alpha e x = \sum_i { < \phi , ex r_i> e l_i }$. Thus, $\alpha
f(ex) = \sum_i {<\phi , exr_i> f( e l_i)}$. 

Hence under the canonical isomorphism ${(eA)}^* \otimes eA @>>> \End
(eA)$,  $$\sum_i { <\phi , \_\_  r_i >\otimes f(e l_i) }
\longmapsto \alpha f.$$ This proves i).

ii) is shown similarly.\qed \enddemo

\bigpagebreak                                                                                                                                                                                                                                                                                                                                                                                                                         
\definition {Definition} Let $V$ be a finite dimensional left
$H$-module, and let $\rho : H @>>> \End (V)$ be the algebra map
affording the module structure on $V$.
Then the element ${\Cal X}_V$ of $H^*$, defined by 
$<{\Cal X}_V,h> = \Tr (\rho (h))$ is called the \it character \rm of
$V$. \enddefinition

\remark {Basic Properties}
If $V$ and $W$ are finite dimensional $H$-modules then we
have the following relations, whose verifications are left to the
reader:

1) $< {\Cal X}_V , 1> = \dim V$.

2) $V \backsimeq W$ $\Rightarrow$ ${\Cal X}_V = {\Cal X}_W$.

3) ${\Cal X}_{V \oplus W} = {\Cal X}_V + {\Cal X}_W$.

4) ${\Cal X}_{V \otimes W} = {\Cal X}_V {\Cal X}_W$.

5) ${\Cal X}_{V^*} = {\Cal X}_V \circ \Cal S$.\endremark

\bigpagebreak
Later we will develop a more detailed study of the characters of
the modules of a Hopf algebra. In the moment, we have the
following lemma. 

\proclaim {Lemma 3.12} 

1) ${{\Cal X}_H}^2 = (\dim H ) {\Cal X}_H$.

2) ${\Cal S}^2 ({\Cal X}_H) = {\Cal X}_H$ in $H^*$.    \endproclaim

\demo {Proof}

1) Let $V$ be any finite dimensional $H$-module. Denote by
$V_{\epsilon }$ the trivial $H$-module structure in the
underlying vector space $V$, i.e., the module structure given by $h
. v = <\epsilon , h> v$, $\forall v \in V$, $h \in H$.

Then the map $H \otimes V_{\epsilon } @>>> H \otimes V$,
such that $h \otimes v \mapsto h_{(1)} \otimes h_{(2)} v$, defines
an $H$-linear isomorphism. 

This implies that ${\Cal X}_H {\Cal X}_{V_{\epsilon } } = {\Cal X}_H
{\Cal X}_V$.

But ${\Cal X}_{V_{\epsilon }} = (\dim V ) \epsilon $.
So we proved that ${\Cal X}_H (\dim V) = {\Cal X}_H {\Cal X}_V$. 

Specializing $V$ in $H$, we get 1).

\bigpagebreak
2) Let $h \in H$, then $$<{\Cal S}^2 ({\Cal X}_H), h > = < {\Cal X}_H ,
{\Cal S}^2 (h) > = \Tr( L_{ {\Cal S}^2(h) } ).$$ Now, as ${\Cal S}^2$ is
an algebra automorphism of $H$, we have that $\forall h \in H$, 
$$\Tr( L_{ {\Cal S}^2(h) } ) = \Tr (L_h) = <{\Cal X}_H , h>, $$ and this
proves 2). \qed \enddemo

\bigpagebreak
We are now in a position to give a short proof of the following
important trace formula by Larson and Radford.

\proclaim {Theorem 3.13} Let $\gamma \in H^*$ be a nonzero right
integral, and let $\Gamma \in H$ be a left integral such that
$<\gamma , \Gamma > = 1$. Then $${\Tr }_{H^*} ({\Cal S}^2) =
<\epsilon , \Gamma > <\gamma , 1> =
 (\dim H) \Tr ({\Cal S}^2|_{{\Cal X}_H H^*}).$$ \endproclaim

\demo {Proof}

Let $\widetilde {\Gamma } \in H^{**}$, be defined by $<\widetilde
{\Gamma } ,h^*> = <h^*, \Gamma >$, $\forall h^* \in H^*$. Using the
fact that $\gamma $ is a Frobenius homomorphism in $H$ with
dual bases $({\Gamma }_{(1)} , \Cal S ({\Gamma }_{(2)} )$, it is easy
to see that $\widetilde {\Gamma }$ is a Frobenius homomorphism
in $H^*$ with dual bases $(\Cal S ({\gamma }_{(1)}), {\gamma
}_{(2)})$.
Applying lemma (3.11) to ${\Cal S}^2 \in \End (H^*)$, taking $\alpha
= 1$, $e = 1$, we get $$\Tr ({\Cal S}^2) =<\widetilde {\Gamma },
{\Cal S}^2 ({\gamma }_{(2)} ) \Cal S ({\gamma }_{(1)})> =
 <\widetilde {\Gamma } , <\gamma , 1> \epsilon > = <\gamma ,
1><\epsilon , \Gamma >.$$
Observe that this equality also follows from proposition (3.4)
applied to ${(H^{op})}^{cop}$.

Now,  by lemma (3.12), ${\Cal S}^2|_{{\Cal X}_H H^*}$ is an
endomorphism of ${\Cal X}_H H^*$ to which lemma 3.11 applies (with
$\alpha = \dim H$, $e = {\Cal X}_H$), giving that 
$$\multline (\dim H) \Tr ({\Cal S}^2|_{{\Cal X}_H H^*}) =
 <\widetilde {\Gamma }, {\Cal S}^2 ({\Cal X}_H {\gamma }_{(2)} ) \Cal
S ({\gamma }_{(1)})> = \\ 
<\widetilde {\Gamma },{\Cal S}^2 ({\Cal X }_H ){\Cal S}^2
({\gamma }_{(2)} ) \Cal S ({\gamma }_{(1)})> = 
<\widetilde {\Gamma } , {\Cal X }_H {\Cal S}^2
({\gamma }_{(2)} ) \Cal S ({\gamma }_{(1)})> = 
<\gamma , 1><{\Cal X }_H , \Gamma >.\endmultline \tag *$$
Again by lemma 3.11, now taking $f$ to be the left multiplication by
$\Gamma $ in $H$ ($\alpha = 1$, $e = 1$), we have 
$$<{\Cal X }_H , \Gamma > = <\gamma , \Cal S ({\Gamma }_{(2)})
\Gamma {\Gamma }_{(1)}> = <\gamma , <\epsilon , {\Gamma
}_{(2)}> \Gamma  {\Gamma }_{(1)} > = <\epsilon , \Gamma
><\gamma , \Gamma > = <\epsilon , \Gamma >.$$ (The second
equality because $\Gamma $ is a left integral in $H$). Then, by
($*$),  $$ (\dim H) \Tr ({\Cal S}^2|_{{\Cal X}_H H^*}) = <\epsilon ,
\Gamma ><\gamma , 1>, $$ which
finishes the proof of the theorem.  \qed \enddemo



Another consequence of the trace formula is the following
theorem, due to Larson and Radford (1988), on semisimple Hopf
algebras over fields of characteristic zero.

\proclaim {Theorem 3.14 (Larson - Radford, 1988)} Let $k$ be a field
of characteristic zero,
 and let $H$ be a finite Hopf algebra over $k$, then the following
statements are equivalent:

1) $H$ is semisimple.

2) $H^*$ is semisimple.

3) ${\Cal S}^2 = \id$. \endproclaim

\demo {Proof}

We have proved in (3.5) that 3) implies 1) and 2). For a
proof  of the part 2) $\Rightarrow $ 1), see \cite{7, Th. (3.3), p.
276}.

We show now that 1) and 2) together imply 3). 

Suppose that $H$ and $H^*$ are both semisimple. Then they are
unimodular, and so ${\Cal S}^4 = {({\Cal S}^2)}^2 = \id$ by (3.10).
Hence, the eigenvalues of ${\Cal S}^2$ and ${\Cal S}^2|_{{\Cal X}_H
H^*}$  are all $1$ or $-1$. 
 Call them, respectively,  ${\mu }_j$, ${\eta }_i$,
$1 \leq  j \leq n = \dim H$, $1 \leq i \leq m = \dim {\Cal X}_H H^*$.

Then $${\Tr }_{H^*} ({\Cal S}^2) = \sum_{j=1}^n {{\mu}_j}, \quad
\text {and} \quad {\Tr }_{H^*} ({\Cal S}^2|_{{\Cal X}_H H^*}) =
\sum_{i=1}^m {{\eta}_i}.$$ So, by theorem (3.13),
 $$\sum_{j=1}^n {{\mu}_j} = n \sum_{i=1}^m {{\eta}_i},$$ implying
that $$n |\ \sum_{i=1}^m {{\eta}_i} \ | \leq  \sum_{j=1}^n {
|{\mu}_j| } = n. $$ 
As $H$ and $H^*$ are semisimple,  $\sum_{i=1}^m {{\eta}_i} 
\neq 0$ by (3.5).  So, $\sum_{i=1}^m {{\eta}_i} = \pm 1$, or 
$\sum_{j=1}^n {{\mu}_j} = \pm n$.

But, as there is at least one ${\mu}_j$ which equals $1$ (since
${\Cal S}^2(1_{H^*}) = 1_{H^*}$ ), then all ${\mu}_j$ must equal
$1$, thus ${\Cal S}^2 = \id$.  
\qed \enddemo

\bigpagebreak
\example{Example}
Let $H$ be the Taft algebra of dimension $N^2$, $N \geq 2$.

Keeping the notations introduced in $\S1 $, we have $\Cal S(g)
= g^{-1}$,  $\Cal S(x) = -x g^{-1}$. Then $${\Cal S}^2 (g) = g,
\quad {\Cal S}^2(x) = g x g^{-1}.$$ So that ${\Cal S}^2$ is the inner
automorphism of $H$ induced by $g$. Then, as the order of $g$ in
the group of units of $H$ is $N$, the order of ${\Cal S}^2$ is also
$N$. In particular, ${\Cal S}^{2N} = \id$.

Now, suppose ${\Cal S}^r = \id$. Since the only powers of $\Cal S$
that are algebra maps are the even powers ($\Cal S$ being an
antialgebra map), it must be $r = 2q$, $q \in \Bbb N$. Then $N$
divides $q$, so $2N$ divides $r$, and this proves that the order of
$\Cal S$ is $2N$.

 
\endexample


\newpage 
\subhead The Nichols-Zoeller Theorem \endsubhead

We consider now (in the context of finite Hopf algebras) another
Kaplansky\rq s conjecture, namely, whether a Hopf algebra $H$ is free
over any Hopf subalgebra. If $H$ is a finite Hopf algebra, then the
Nichols-Zoeller theorem gives a positive answer to this question.

\bigpagebreak If $H$ is a Hopf algebra, then a subalgebra $R$ of $H$
is called a \it Hopf subalgebra, \rm if $\Delta (R) \subseteq R
\otimes R$ and $\Cal S (R) \subseteq R$.

\remark{Remark} A finite dimensional subbialgebra $R$ of a Hopf
algebra $H$ (i.e., a subalgebra $R$ such that $\Delta (R) \subseteq R
\otimes R$) is a Hopf subalgebra.

Indeed, we have to show that $\id_{R} \in \End(R)$ is invertible.
But $\id_{R}$ is invertible in $\Hom (H, R)$ (which contains
$\End(R)$ via the inclusion $R\to H$), since $H$ has an antipode.
Therefore convolution with $\id_{R}$ is injective in $\End(R)$, and
hence bijective since $\End(R)$ is finite dimensional. Thus $\id_{R}
\in \End(R)$ is invertible. \endremark

\bigpagebreak
\proclaim {Theorem 3.15. (Nichols-Zoeller, 1989)} Let $H$ be a finite
Hopf algebra, and let $R \subseteq H$ be a Hopf subalgebra. Then
$H$ is a free $R$-module. \endproclaim

\proclaim {Corollary 3.16. "Lagrange\rq s Theorem for Hopf algebras"
} If $R \subseteq H$ are finite Hopf algebras, then the order of $R$
divides the order of $H$.\qed \endproclaim

\bigpagebreak
In order to prove the theorem, we need a series of lemmas.

\bigpagebreak
\definition {Definition} A finite dimensional $k$-algebra $R$ is
called a \it quasi-Frobenius algebra, \rm if the $R$-modules
${}_RR$ and $R_R$ are injective. \enddefinition 

Recall that for an algebra $R$, a left $R$-module $M$ is called
\it faithful, \rm if its annihilator $\Ann (M) : = \{ r \in R : rM =
0 \}$ equals $0$. 

\bigpagebreak 
\proclaim {Lemma 3.17} Let $R$ be a finite dimensional
quasi-Frobenius algebra, and let $M$ be a finitely generated left
$R$-module. Then $M$ is faithful if and only if there exist a free
left $R$-module $F \neq 0$, and a non faithful left $R$-module
$N$, such that $$M^r \backsimeq F \oplus N, $$ for some integer $r
\geq 1$.\endproclaim
\demo {Proof} Follows from \cite{1, Th. (59.3), p. 404}. \qed \enddemo

\bigpagebreak
Let now $H$ be a Hopf algebra, and let $R \subseteq H$ be a
subalgebra such that $\Delta (R) \subseteq H \otimes R$.

If $M$ is a left $R$-module, and we consider $H$ as a trivial
$R$-module (i.e.,  $r . h = <\epsilon , r>h$,  $\forall r \in R, h \in
H$), then $H \otimes {}_.M$ is a left $R$-module via $$R @>\Delta >> H
\otimes R @>>> \End (H) \otimes \End (M) @>>> \End (H \otimes
M),$$ that is, $$r . (h \otimes m) = <\epsilon , r_{(1)}>h \otimes
r_{(2)} . m = h \otimes r . m, \tag 1$$ for $r \in R$, $h \in H$ and  $m
\in M$. In fact, $H \otimes {}_.M \backsimeq M^{\dim H}$ as left
$R$-modules.  

Also, if ${}_.H$ denotes the left $R$-module structure on $H$
given by left multiplication, then ${}_.H \otimes {}_.M$ is a left
$R$-module via $$r. (h \otimes m) = r_{(1)}h \otimes r_{(2)} . m.
\tag 2$$

\remark {Remark} Observe that the $R$-actions considered on
$H$ are both restrictions to $R$ of the analogous $H$-actions.
Thus the fact that $\Delta (R) \subseteq H \otimes R$ allowed us
to give $H \otimes {}_.M$ and ${}_.H \otimes {}_.M$ a left $R$-module
structure, as in (1) and (2). \endremark 

With this in mind, we have:

\bigpagebreak
\proclaim {Lemma 3.18} Let $R \subseteq H$ be a subalgebra such that
$\Delta (R) \subseteq H \otimes R$. Suppose that the antipode, $\Cal
S$, of $H$ is bijective. Then, $\forall n \in \Bbb N$,

1) $H \otimes {}_.R^n \backsimeq {}_.H \otimes {}_.R^n$, as left
$R$-modules.

2) $H \otimes {}_.H^n \backsimeq {}_.H \otimes {}_.H^n$, as left
$H$-modules. \endproclaim

\demo {Proof}

1) It is enough to show it in the case $n = 1$. 
 For this, define $\phi : H \otimes {}_.R @>>> {}_.H \otimes {}_.R$, by
$\phi (h \otimes r) = r_{(1)} h \otimes r_{(2)}$.

$\phi $ is well defined since $\Delta (R) \subseteq H \otimes R$,
and it is clear that it is $R$-linear.

Moreover, the map $\psi : {}_.H \otimes {}_.R @>>> H \otimes {}_.R$,
given by $\psi (h \otimes r) = \overline {\Cal S} (r_{(1)}) h
\otimes r_{(2)}$, is inverse to $\phi $ (where $ \overline {\Cal S}$
denotes the composition inverse of the antipode $\Cal S$). 

So $\phi $ is an $R$-isomorphism, and this proves 1).

 2) follows from 1) by taking $R = H$. \qed \enddemo   

\bigpagebreak
Denote, for any subalgebra $R$ of $H$, $R^+ : = R \bigcap \ker
(\epsilon )$. It is immediate that $R^+ H$ is an $R$-submodule of
${}_.H$. In particular, the quotient space $H /  R^+ H$ is a left
$R$-module, we have also that for $r \in R$, $\overline r =
\overline {\epsilon (r) 1}$, where $\overline r$ denotes the class
of $r$ in $H / {}R^+H$. 

\proclaim {Lemma 3.19} Let $R \subseteq H$ be a subalgebra of $H$ as
in  Lemma 3.18, then $$H {\otimes }_R H \backsimeq  {}_.H \otimes
(H /  R^+ H),$$ as left $H$-modules.\endproclaim

\demo {Proof}

Define $\phi : H {\otimes }_R H @>>> {}_.H \otimes (H / R^+H)$,
by $\phi (x \otimes y) = x y_{(1)} \otimes \overline {y_{(2)}}$.

To see that $\phi $ is well defined, let $r \in R$, then $$\phi (x
\otimes ry) = xr_{(1)} y_{(1)} \otimes \overline {r_{(2)}y_{(2)}}
= xr_{(1)} y_{(1)} \otimes <\epsilon , r_{(2)} >\overline {y_{(2)}}
= xr y_{(1)} \otimes \overline {y_{(2)}} = \phi (xr \otimes y). $$
Clearly $\phi $ is $H$-linear. Moreover, the
map $\psi :
 {}_.H \otimes (H / R^+H) @>>>  H {\otimes }_R H$  given by $\psi (x
\otimes \overline y) = x \Cal S(y_{(1)}) \otimes y_{(2)}$, is
inverse of $\phi $ ($\psi$ is well-defined because $\Delta (R)
\subseteq H \otimes R$). So $\phi $ is an isomorphism of
$H$-modules, as claimed.\qed \enddemo

\bigpagebreak
An \it augmented algebra \rm is a pair $(R,\epsilon )$, where $R$ is
a finite dimensional $k$-algebra and $\epsilon : R @>>> k$ is an
algebra map. In particular, if $H$ is a Hopf algebra
with counit $\epsilon $, $(R,\epsilon )$ is an augmented algebra for
every subalgebra $R$ of $H$. 

Observe that if $(R, \epsilon )$ is augmented, then we may consider
$k \in {}_R \Cal M$ via $\epsilon $.  

\proclaim {Lemma 3.20} Let $R$ be a finite dimensional augmented
$k$-algebra, and let $M$ be a finitely generated left $R$-module
such that $M^n$ is free over $R$ for some $n \geq 1$. Then $M$ is
free over $R$.\endproclaim

\demo {Proof}

By hypotheses, $M^n \backsimeq R^m$, for some $m \in \Bbb N$.
Then $${(M {\otimes }_R k)}^n \backsimeq {(R {\otimes }_R k)}^m
\backsimeq {k}^m.$$ Comparing dimensions, we get that $m = n
d$, where $d = \dim (M {\otimes }_R k)$. Hence, $$M^n \backsimeq
R^m \backsimeq {(R^d)}^n.$$ Now, by the Krull-Schmidt theorem
(see \cite{1, Th. (14.5), p. 83}), $M \backsimeq R^d$ is free over $R$.
\qed \enddemo 

We finally have:

\bigpagebreak
\demo {Proof of the Nichols-Zoeller\rq  Theorem}

If $R$ is a finite dimensional Frobenius algebra, then $R$ is a
quasi-Frobenius algebra (see for instance \cite{1, Th. (61.3), p. 414}).

Also, if $R$ is a Hopf subalgebra of $H$, as $R$ itself is a finite
Hopf algebra, it follows from theorem (2.3) that
$R$ is a Frobenius algebra. Then we may apply the preceding
lemmas.

Consider the left $R$-module ${}_.H$. As ${}_.H$ is a faithful
$R$-module, by lemma (3.17), ${}_.H^r \backsimeq F \oplus N$,
for some $r \geq 1$, where $F \backsimeq R^m$ is a free left
$R$-module, and $N$ is a non faithful left $R$-module.

Observe that, calling  $t : = \dim H$, $$\multline F^t \oplus N^t
\backsimeq {}_.H^{rt} \backsimeq H \otimes {}_.H^r  \underset
\text {by lemma 3.18}\to \backsimeq {}_.H \otimes {}_.H^r
\backsimeq  \\ ({}_.H \otimes {}_.F) \oplus ({}_.H \otimes {}_.N) 
\underset \text {by lemma 3.18} \to \backsimeq F^t \oplus ({}_.H
\otimes {}_.N). \endmultline $$ 
Then, by the Krull-Schmidt theorem, it follows that  $N^t
\backsimeq {}_.H  \otimes {}_.N$. Now, we have a surjection of
$R$-modules 
$$N^t \backsimeq {}_.H  \otimes {}_.N @>\id \otimes \text
{can} >> {}_.H \otimes {}_.(N / R^+ N).$$
But, as $R$ acts trivially on $(N / R^+ N)$ and ${}_.H$ is a
faithful $R$-module, ${}_.H \otimes (N / R^+ N)$ is faithful. So,
as $N$ is not faithful, $N / R^+ N = 0$. Hence, we find that $${(H /
R^+H)}^r \backsimeq {(R/R^+R)}^m \oplus (N/R^+N) \backsimeq
{(R/R^+R)}^m \backsimeq k^m.\tag *$$ 

We have also that $$H {\otimes }_R H^r \backsimeq (H
{\otimes }_R F) \oplus (H {\otimes }_R N) \backsimeq H^m
\oplus (H {\otimes }_R N), $$ and on the other hand,
$$\multline H {\otimes }_R H^r \backsimeq {(H {\otimes
}_R H )}^r \underset \text {by lemma (3.19)} \to \backsimeq
{(H \otimes \ H/R^+H)}^r \backsimeq \\ H  \otimes
{(H/R^+H)}^r \underset \text {by} (*) \to \backsimeq H  \otimes
k^m \backsimeq H^m. \endmultline $$ So $H {\otimes }_R N =
0$.

But the inclusion $R_R \subseteq H_R$ is $R$-direct, since $R_R$
is injective. Thus, $$N \backsimeq R {\otimes }_R N
\hookrightarrow H {\otimes }_R N = 0.$$ Then $N = 0$, implying
that $H^r \backsimeq F$ is free over $R$, and by lemma (3.20),
that $H $ is free over $R$. \qed \enddemo

\bigpagebreak
\remark {Remark} 
It is possible to show that if $R$ is a subalgebra of $R$ such that
$\Delta (R) \subseteq H \otimes R$, the following statements are
equivalent:

1) $R$ is a Frobenius algebra.

2) $R$ is a quasi-Frobenius algebra.

3) ${}_RH$ and $H_R$ are $R$-free. \endremark

An open problem is whether a subalgebra $R$
of $H$ satisfying $\Delta (R) \subseteq H \otimes R$ is \it always
\rm a Frobenius algebra.


\newpage
\head \S4 Character Theory for Finite Hopf algebras
\endhead

\bigpagebreak
We want to develop now a character theory for finite
semisimple Hopf algebras, similar to that when $H = kG$, the
group algebra of a finite group $G$ (see references \cite{1} or
\cite{3} on the subject).

From now on, $k$ will denote an algebraically closed field of
characteristic $0$, and $H$ will be a finite semisimple Hopf
algebra over $k$. By the theorem of Larson and Radford (3.14) we
know that $H^{*}$ is also semisimple and $\Cal S^{2} = \id$.

\bigpagebreak
As $H$ is semisimple, every finite dimensional irreducible
$H$-module is isomorphic to a direct summand of $H$. Then, there
exist only a \it finite \rm number of such modules up to
isomorphism (because $H$ is finite). We denote them by them $V_1,
\dots , V_n$, $V_1 = k_{\epsilon }$.

\definition {Definition} The  \it irreducible characters \rm of $H$
are by definition the characters of the irreducible $H$-modules 
$V_1, \dots , V_n$. 

We denote by ${\Cal X}_i : = {\Cal X}_{V_i}$, and by
${\Cal X}_{\overline i} : =  {\Cal X}_{{V_i}^*}$. 

The \it character rings \rm of $H$ are defined to be the following
subrings of $H^{*}$:  $$R_{\Bbb Q}(H) : = \sum_{i=1}^n {\Bbb Q {\Cal
X}_i}, \quad \text {and} \quad R_k(H) : = \sum_{i=1}^n {k {\Cal
X}_i}.$$ These are in fact subrings by the basic
properties of the characters, see the paragraph before
Lemma 3.12. \enddefinition

\bigpagebreak
We then have the inclusions $$R_{\Bbb Q}(H) \subseteq R_k(H)
\subseteq H^*.$$

\bigpagebreak
\remark {Remark} With the above definition, we get that the
character of the left regular representation of $H$ may be written
in the form $${\Cal X}_H = \sum_{i=1}^n {n_i {\Cal X}_i}, $$ where
$n_i$ are integers, $n_i \geq 1$, and $n_1 = 1$. Observe also that
${\Cal X}_1 = \epsilon$. \endremark

\bigpagebreak
Let $V, W$ be left $H$-modules. Then $H$ acts on $\Hom
(W,V)$, in the form $(h . \phi )(w) = h_{(1)} . \phi (\Cal S (h_{(2)}) . 
w$).

\proclaim {Lemma 4.1} Let $V$, $W$ as above, then ${\Hom }_H (W,V)
= {\Hom (W,V)}^H$, the isotypic component of trivial type in
$\Hom (W,V)$.\endproclaim 

\demo {Proof} Let $\phi :W @>>> V$ be an $H$-linear map, and let $h
\in H$, $w \in W$. Then $$(h . \phi )(w) = h_{(1)} . \phi (\Cal
S(h_{(2)}) . w) = h_{(1)} \Cal S(h_{(2)}) . \phi (w) = <\epsilon , h>
\phi (w).$$ So $\phi \in {\Hom (W,V)}^H$.

Conversely, suppose $\phi \in {\Hom (W,V)}^H$, and let $h \in H$,
$w \in W$. We have $$\multline \phi (h . w) = <\epsilon , h_{(1)}> \phi
(h_{(2)} . w) = (h_{(1)} . \phi ) ( h_{(2)} . w) = \\ h_{(1)}. \phi (\Cal
S(h_{(2)}) h_{(3)} . w) = h_{(1)} <\epsilon , h_{(2)}>. \phi (w) = h.
\phi (w). \endmultline $$ So $\phi \in {\Hom }_H(W,V)$. \qed
\enddemo

\bigpagebreak
If $V, W$ be finite dimensional $H$-modules. Then we have a
canonical $k$-isomorphism $V \otimes W^* \backsimeq \Hom
(W,V)$, given by $(v \otimes \phi )(w) = <\phi , w>v$.

This defines an $H$-linear isomorphism. So in particular, ${\Cal
X}_{\Hom (W,V)} = {\Cal X}_V {\Cal X}_{W^*}$.

\bigpagebreak
\proclaim {Lemma 4.2. (Schur's lemma)} Let $V, W$ be irreducible
$H$-modules, then ${\Hom }_H(W,V) = {\delta }_{V,W}
k$. (Here  ${\delta }_{V,W}$ equals $0$ if $V$ and $W$
are non isomorphic, and $1$ if they are isomorphic).\endproclaim

\demo {Proof} 

As $V$ and $W$ are irreducible, any $H$-linear map $\phi : W @>>>
V$ must be zero or an isomorphism. Thus, if $V$ and $W$ are not
isomorphic, ${\Hom }_H(W,V) = 0$.

Now, let $\phi : V @>>> V$ be an $H$-linear map. As $k$ is
algebraically closed, we may choose an eigenvalue of $\phi $,  $a
\in k$.  Hence, $\ker ( \phi - a \id) \neq 0$. And by the
irreducibility of $V$, $\ker ( \phi - a \id) = V$. So $\phi = a \id$.

This defines a map ${\Hom }_H(W,V) @>>> k$, which is clearly an
 isomorphism. \qed \enddemo

\bigpagebreak
(4.1) and (4.2) imply

\proclaim {Corollary 4.3} Let $1 \leq i,j \leq n$, then 
$${\Cal X}_i {\Cal
X}_{\overline j} = {\delta }_{ij} {\Cal X}_1+ \sum_{k \geq 2}{c_k
{\Cal X}_k}, $$ for some non negative integers $c_k$.
\qed \endproclaim

\bigpagebreak
As $H$ is a semisimple $k$-algebra, applying Wedderburn\rq s theorem
\cite{1, Th. (26.4), p.175}, $H$ is isomorphic to a product of full matrix
algebras over skew fields of finite degree over $k$. Now, as $k$
is algebraically closed, every such skew field must coincide with
$k$. Thus $$H \backsimeq  \prod_{i=1}^n {M_{n_i}(k)}. \tag * $$
Moreover, we can assume that  $V_i$ is the natural representation of
$M_{n_i}(k)$ of dimension  $n_i$, then $V_i$ is an irreducible
$H$-module, and we have an $H$-isomorphism: $M_{n_i}(k)
\backsimeq {V_i}^{n_i}$.

Call $E_i \in M_{n_i}(k)$ the $n_i \times n_i$ identity matrix, and
let $e_i \in H$ be the element which corresponds to $(0, \dots ,
E_i, \dots , 0)$ via the isomorphism $(*)$.

Then, the $e_i$ form a set of orthogonal idempotents in $H$, that
is,  $$\gather e_i e_j = 0, \quad \text {if} \ i \neq j, \\ {e_i}^2 =
e_i, \\ e_1+\dots +e_m = 1. \endgather $$
In particular, $H = \bigoplus {He_i}$.

We have, for $1 \leq i \leq n$, that $H e_i \backsimeq M_{n_i}(k)$,
hence $He_i \backsimeq {V_i}^{n_i}$.

As a consequence, we get 

\proclaim {Proposition 4.4} Each $V_i$ occurs in the left regular
representation of $H$ with multiplicity equals to its $k$-dimension:
$n_i = {\Cal X}_i(1)$. In terms of characters, we have 
$${\Cal X}_H = \sum_{i=1}^n {n_i {\Cal X}_i}$$ In
particular, $\dim H = \sum_{i=1}^n {{n_i}^2}$. \qed \endproclaim

It is clear also, from
$(*)$, that if $Z(H)$ denotes the center of $H$, then $$Z(H) =
\sum_{i=1}^n {ke_i}.$$

 Observe that $e_i$ acts as the identity on
$V_i$ and anihilates $V_j$, $\forall j \neq i$. That is, $${\Cal
X}_i(e_i) = n_i, \quad \text {and} \ {\Cal X}_j (e_i) = 0 , {} \text
{if} \ j\neq i.$$

\proclaim {Proposition 4.5} Let $\lambda \in H^*$ be a nonzero
integral such that $<\lambda , 1> =1$. Then $Z(H)
\lambda = R_k(H)$.\endproclaim

\demo {Proof} As $Z(H) = \sum_{i=1}^n {ke_i}$, the claim will be
proved if we show that $$e_i \lambda = {(\dim H)}^{-1} n_i
{\Cal X}_i.$$
Observe that as $<\lambda , 1> = 1$, then 
${\Cal X}_H = (\dim H) \lambda$, by (3.4).

Now,
if $h \in H$, then  $$\multline <e_i \lambda , h> = <\lambda , h
e_i> = {(\dim H)}^{-1} <{\Cal X}_H , he_i> = \\ {(\dim H)}^{-1} <{\Tr
}_{He_i} , h> = {(\dim H)}^{-1} <n_i {\Cal X}_i , h > = {(\dim
H)}^{-1} n_i < {\Cal X}_i , h>. \endmultline $$ \qed \enddemo 

\bigpagebreak
Let $\Lambda \in H$ be an integral such that $<\lambda , \Lambda > =
1$. Then we have, as $H$ and $H^*$ are unimodular, that $\lambda $
is a Frobenius homomorphism of $H$ with dual bases $({\Lambda
}_{(1)} , \Cal S( {\Lambda }_{(2)}))$. 

This gives the following 

\proclaim {Corollary 4.6} Let $1 \leq i \leq n$, then $$(\dim H / n_i
)\ e_i = {\Lambda }_{(1)} <{\Cal X}_i , \Cal S({\Lambda }_{(2)})>.$$
\endproclaim

\demo {Proof} Using the identity $e_i \lambda = {(\dim H)}^{-1} n_i
{\Cal X}_i, $ we compute $${(\dim H)}^{-1} n_i  {\Lambda }_{(1)} 
<{\Cal X}_i , \Cal S({\Lambda }_{(2)})> =  {\Lambda }_{(1)} 
<e_i \lambda , \Cal S({\Lambda }_{(2)})> =  {\Lambda }_{(1)}
<\lambda  , \Cal S({\Lambda }_{(2)}) e_i > = e_i. \qed $$ \enddemo

\bigpagebreak
\remark {Remark} When $H = kG$, the group algebra of a finite
group $G$, then for $\Lambda = \sum_{g \in G} g$, 
corollary (4.6) says that $$(|G| / n_i )\ e_i = \sum_{g \in G} {g
<{\Cal X}_i , g^{-1}>}.$$ 
 Now, as $G$ is a finite group, ${\rho }_i (g)$
has finite order as an element of $GL (V_i)$, $\forall g \in G$
(where, as usual, ${\rho }_i$ denotes the representation of $H = kG$
in $V_i$).

Hence the eigenvalues of ${\rho }_i (g)$ are roots of unity in $k$.
In particular, they are all contained in a subring $R$ of algebraic
integers such that $R$ is finitely generated as $\Bbb Z$-module. Any
element of the group algebra $RG$ is integral over $\Bbb Z$ (since
it generates a subring which is  finitely generated as $\Bbb
Z$-module). Hence $(|G| / n_i )\ e_i$ is integral, and $|G| /n_i$
is an algebraic integer.

But $|G| / n_i $ is a rational number, so it must be an integer. Then 
$n_i$ divides $|G|$.

We have seen then that, for a finite group $G$,  the dimensions
of the irreducible representations of $G$ \it divide \rm the order
of $G$. \endremark

\bigpagebreak
 We want to prove now a theorem on orthogonality of characters due to
Larson (1971). 

Recall that for $f \in H^*$, $<{\Tr }_{H^*},f> = \Tr (L_f)$. Define
for \ $1 \leq i, j \leq n$, $({\Cal X}_i , {\Cal X}_j) : = <{\Tr
}_{H^*} , {\Cal X}_i{\Cal X}_{\overline j}>$.

\proclaim {Theorem 4.7 (Orthogonality relations)} Let $1 \leq i, j
\leq n$, then $$({\Cal X}_i , {\Cal X}_j) = {\delta }_{ij} \dim H.$$
\endproclaim

\demo {Proof}

By corollary (4.6), we have that for $ 1 \leq i,j \leq n$,
${\Cal X}_i {\Cal X}_{\overline j} = {\delta }_{ij} {\Cal X}_1+ \sum_{k
\geq 2}{c_k {\Cal X}_k}$,  with $c_k \geq 0$. Then,
$$( {\Cal X}_i ,  {\Cal X}_j ) = <{\Tr}_{H^*} , {\Cal X}_i {\Cal
X}_{\overline j}> = {\delta }_{ij} \dim H + \sum_{k \geq 2} { c_k
<{\Tr}_{H^*} , {\Cal X}_k >}.$$ (Observe that $<{\Tr}_{H^*} , {\Cal
X}_1 > = <{\Tr}_{H^*} , \epsilon > = \dim H^* = \dim H$). 

\bigpagebreak
We claim that $<{\Tr}_{H^*} , {\Cal X}_k > = 0$, for all $k \geq 2$.

\bigpagebreak
From (3.4), we know that if the integrals $\Lambda \in H$, 
$\lambda \in H^{*}$ are chosen as above, then ${\Tr}_{H^*} = (\dim H)
\  \widetilde {\Lambda }$,  (where $<\widetilde {\Lambda }, h^*> =
<h^*, \Lambda >$, $\forall h^* \in H^*$).

Now, as $H$ is semisimple, and thus unimodular, the space of
integrals is contained in the center of $H$.

Let then $k \geq 2$, and let ${\rho }_k : H @>>> \End (V_k)$ denote
the algebra map affording the $H$-module structure in $V_k$. Then
the image of the integral is an $H$-linear map since the integral is
central. By Schur's Lemma (4.2) there exists an element $a \in k$
such that  ${\rho }_k (\Lambda ) = a \id$.

Let $h \in H^+ = \ker (\epsilon )$. We have 
$$a {\rho }_k (h) = {\rho }_k (\Lambda ) {\rho }_k (h) =
 {\rho }_k (\Lambda h ) = <\epsilon , h> {\rho }_k (\Lambda ) = 0.$$ 

Now, as $k \geq 2$, we may find $h \in H^+$ such that ${\rho }_k (h)
\neq 0$, and this implies that $a = 0$. But then $\Lambda . V_k =
0$, so $<{\Tr}_{H^*} , {\Cal X}_k > =  < {\Cal X}_k ,  \Lambda > =
0$, as we claimed. \qed \enddemo

\bigpagebreak
\proclaim {Corollary 4.8} The irreducible characters of $H$, ${\Cal
X}_1, \dots , {\Cal X}_n$, are $k$-linearly independent in $H^*$,
and the map $R_{\Bbb Q}(H) {\otimes }_{\Bbb Q} k @>>> 
R_k(H)$, $x \otimes 1 \mapsto x$, is an isomorphism of
$k$-algebras. \qed \endproclaim 

\bigpagebreak
Theorem (4.7) implies, moreover, that ${(\dim H)}^{-1/2} {\Cal X}_i$,
$1 \leq i \leq n$, is an orthonormal basis of $R_k(H)$ with respect
to the form $(\ , \ )$. 

In particular,  if $V$ is a finite
dimensional module for $H$, then its character ${\Cal X}_V$ is
expressible in a unique way as a linear combination $${\Cal X}_V =
\sum_i { m_i{\Cal X}_i }, \tag 1 $$ where $m_i = {(\dim H)}^{-1}
({\Cal X}_V, {\Cal X}_i)$. So that if $V, W$ are finite dimensional
$H$-modules, $V \backsimeq  W$ iff ${\Cal X}_V = {\Cal X}_W$.

\bigpagebreak
\remark {Remark} As $H$ is semisimple (and thus ${\Cal S}^2 =
\id$), we have that for every finite dimensional
$H$-module $V$, $V^{**} \backsimeq V$ as $H$-modules.

Also, if $W$ is a finite dimensional $H$-module, the canonical
isomorphism ${(V \otimes W)}^* \backsimeq W^* \otimes V^*$ is
an $H$-isomorphism.

Putting these together, we find that the map $x \mapsto
\overline x$, defined by ${\Cal X}_i \mapsto {\Cal X}_{\overline
i}$, gives an involutory antialgebra map in $R_k(H)$ (and also in
$R_{\Bbb Q}(H)$).\endremark 

\bigpagebreak
Corollary (4.8) says that $R_k(H)$ is obtained
by $R_{\Bbb Q} (H)$ by extension of scalars. Using this fact, we will
prove the following result, which is due to Zhu (1994).

\proclaim {Theorem 4.9} $R_k(H)$ is a semisimple
$k$-algebra.\endproclaim   

\demo {Proof} We first show that $R_{\Bbb Q}(H)$ is semisimple. 

For this, observe that if $0 \neq x \in R_{\Bbb Q}(H)$, say $x =
\sum_i x_i {\Cal X}_i$, $x_i \in \Bbb Q$, then by (4.3) 
$$x \overline
x = \sum_{i,j} {x_i x_j {\Cal X}_i {\Cal X}_{\overline j} } =
\sum_{i,j} { x_i x_j \ ( {\delta }_{ij} {\Cal X}_1 + \sum_{k\geq 2}
{ c_{ijk} {\Cal X}_k } ) } = (\sum_i { {x_i}^2 }) \ {\Cal X}_1  +
\sum_{k\geq 2} { b_k {\Cal X}_k },$$ 
for some coefficients $c_{ijk}$ and some rational numbers $b_{k}$.

And, as $x_i \i \Bbb Q$ are not
all zero, $\sum_i {{x_i}^2} \neq 0$, so $x \overline x \neq 0$.

Suppose now that there exists $0 \neq x \in \rad (R_{\Bbb Q}(H)
)$. Call $y : = x \overline x$. Then $y \neq 0$ and 
$y \in \rad (R_{\Bbb Q}(H))$. Moreover, $\overline y = y$.

So, $y^2 = y \overline y \neq 0$, $y^4 = y^2 \overline {y^2} \neq
0$, and in general $y^{2^t} = y^{2^{t-1}} \overline { y^{2^{t-1}} }
\neq 0$, for all $t \geq 1$. 

But this is absurd, since all the
elements in $\rad (R_{\Bbb Q}(H))$ are nilpotent.

Then $x$ must equal $0$, so  $\rad (R_{\Bbb Q}(H)) = 0$ and 
$R_{\Bbb Q}(H)$ is semisimple. 

By the results in \cite{1, $\S69 $, p. 459}, as $R_{\Bbb Q}(H)$ is
semisimple, then $R_{\overline {\Bbb Q}}$ also is. Then
$R_{\overline {\Bbb Q}}(H)$ is a direct product $$R_{\overline
{\Bbb Q}} = \prod M_{n_i}(\overline {\Bbb Q}),$$of full matrix
algebras over $\overline {\Bbb Q}$. But as $k$ is algebraically
closed and $\char k = 0$, $\overline {\Bbb Q} \subseteq k$, and we
also have $$M_{n_i}(\overline {\Bbb Q}) {\otimes }_{\overline
{\Bbb Q}} k \backsimeq M_{n_i}(k).$$

Now, as  $R_k(H)$ is
the extension by scalars of $R_{\overline {\Bbb Q}}(H)$,  
$R_k(H)$ is semisimple,
as we claimed. 
\qed \enddemo  

\bigpagebreak
\subhead The Class Equation for Hopf algebras \endsubhead

\bigpagebreak
We state now some results concerning algebraic number theory,
that will be needed in the sequel. The reader can find in 
\cite{1, Ch. III}
the proofs of these results.

\bigpagebreak
\definition {Definition} An integral domain $R$ is called a \it
Dedekind domain, \rm if it satisfies the following conditions:

i) $R$ is a noetherian ring.

ii) Every non-zero prime ideal in $R$ is maximal.

iii) $R$ is integrally closed in its field of quotients, $K$. (That is, if
$a \in K$ is integral over $R$, then $a \in R$).\enddefinition

If $M$ is a torsion free module over a Dedekind domain $R$,
then the \it rank \rm of $M$ is the dimension of $M {\otimes }_R
K$ over $K$. If $M$ is finitely generated, then it has finite rank.

\bigpagebreak
Let $K$ be an algebraic number field (i.e., a finite extension of the
field $\Bbb Q$ of rational numbers), and let $R$ denote the ring of
integers of $K$, $R = \{a \in K : a \ \text {is an algebraic integer} \}
$. Then $R$ is a Dedekind domain (whose field of quotients is $K$). 
 
If $R$ is an integral domain, and $K$ is its field of quotients, 
a finitely generated $R$-submodule $0 \neq I$ of $K$, is called a \it
fractional ideal \rm of $R$.

\proclaim {Proposition 4.10}
Let $M$ be a  finitely generated, projective module of rank $t$
over a  Dedekind domain $R$. Then there exist $m_1, \dots ,
m_t \in M$, and fractional ideals $I_1, \dots , I_t$ of $R$ such
that $$M = I_1m_1 + \dots + I_t m_t, $$ where the indicated sum
is direct. \qed \endproclaim   

\bigpagebreak
\proclaim {Lemma 4.11} Let $K$ be an algebraic number field, $R$ its
ring of integers, and call $\overline K$ the algebraic closure of
$K$.

Suppose $A$ is an $R$-algebra finitely generated and projective
as an $R$-module, such that $A {\otimes }_R K$ is
semisimple.

Then, there exists an isomorphism of $\overline K$-algebras
$$A {\otimes }_R \overline K @>>> \prod_{\alpha = 1}^s {
M_{n_{\alpha }} (\overline K) }, \quad e_{ {\alpha }_{ij}} 
\mapsto E_{ {\alpha }_{ij}}, $$
for some $s \geq 1$, and natural numbers $n_{\alpha }$ (where 
$E_{ {\alpha }_{ij}}$ denotes the $s$-tuple consisting of all zeros,
except for the $\alpha $-place, where it equals the $n_{\alpha }
\times n_{\alpha }$ matrix whose entries are $a_{kl} = {\delta
}_{ik} {\delta }_{jl}$), such that every $a \in A$ may be written in
the form $$a = 
\sum_{\alpha , i, j} { r_{ {\alpha }_{ij} } e_{ {\alpha }_{ij} } }, $$ with 
$ r_{ {\alpha }_{ij} }$ algebraic integers.\endproclaim

\demo {Proof}

(Observe that we are identifying $a \in A$ with $a \otimes 1$ in
$A {\otimes }_R \overline K$).

As $A {\otimes }_R \overline K \backsimeq (A {\otimes }_R K)
{\otimes }_K \overline K$, then $A {\otimes }_R \overline K $ is
semisimple, and we may assume that 
$$A {\otimes }_R \overline K \simeq \prod_{\alpha =1}^s { {\End
}_{\overline K} (V_{\alpha }) }, $$ where $V_{\alpha }$ are
irreducible $A {\otimes }_R \overline K$-modules (say, of
dimension $n_{\alpha }$).

The claim will be proved if we can show that for each irreducible
$A {\otimes }_R \overline K$-module, there exists a basis over
$\overline K$, such that the matrices representing elements of
$A$ in that basis, have all entries algebraic integers.

Now, let $V$ be an irreducible $A {\otimes }_R \overline
K$-module, with $\overline K$-basis $v_1, \dots , v_t$. The
matrices representing elements of $A$ in this basis have all
entries in a finite field extension of $K$ (since $A$ is finitely
generated over $R$). Instead of extending $K$ we may assume
that all these entries are in $K$.



Call $U : = \sum_i{ Av_i } \subseteq V$. Then $U$ is a
finitely generated, projective $R$-module of rank $t$, and thus
there exist $u_1, \dots , u_t \in U$, and fractional ideals $I_1,
\dots , I_t$ of $R$ such that $$U = \sum_i{ I_iu_i }.$$

Also, we may find $K \subseteq K'$, an algebraic number field
extension ($R \subseteq R'= $ \ the ring of integers of $K'$),
such that all fractional ideals in $K$ become principal in $K'$. (See
\cite{1, Th. (20.14), p.130}).

In particular, there exist $s_i \in K'$, such that $R' I_i = R' s_i$, for
all $i = 1, \dots , t$.
Then $$R' U = \sum_i{ R' I_iu_i } = \sum_i{  R' s_iu_i } = 
\sum_i{  R' {v_i}' }.$$ where ${v_i}' = s_i u_i$. Now, in the basis
$({v_i}')$, the matrices representing elements of
$A{\otimes }_RR' \subseteq A{\otimes }_RK'$ have all entries in
$R'$, and thus are algebraic integers. It remains to observe the
inclusion: $A = A{\otimes }_R R \subseteq A{\otimes }_RR'$. \qed
\enddemo

\bigpagebreak
\definition {Definition} Let $A$ be a finite dimensional
$k$-algebra. A $k$-linear map $t:A @>>> k$, is called a \it trace
like \rm function if it satisfies the following conditions:

a) $\forall x, y \in A$, $t(xy) = t(yx)$.

b) If $0 \neq e$ is an idempotent in $A$, then $0 \neq t(e)$ is a
natural number.

c) If $e, f$ are nonzero idempotents in $A$, such that $eA
\underset A \to \backsimeq fA$, then $t(e) = t(f)$. \enddefinition
\remark{Remark}Note that the usual trace map $\Tr_{A}$ is trace
like. But also $\Tr_{A}$ restricted to any subalgebra $B$ of $A$ is
trace like. Indeed, a) is clear, and b) follows since $\Tr_{A}(e) =
\dim(eA)$. To prove c), note that $eB\otimes _{B}A \simeq eA$.
\endremark

\bigpagebreak
The following lemma is due, independently, to Kac (1971) and
Zhu (1994).

\bigpagebreak
A non-zero idempotent $e$ in an algebra $R$ is called \it primitive,
\rm if it can not be expressed as the sum of two non-zero orthogonal
idempotents.

\proclaim {Lemma 4.12} In the situation of lemma (4.11), let 
$t: A {\otimes }_R \overline K @>>> \overline K$ be a trace like
function. 

Assume that there exist $0 \neq d \in \Bbb N$, and $a_i, b_i \in
A$, $1 \leq i \leq n : = \dim (A {\otimes }_R \overline K)$, such
that $t(a_k b_l) = {\delta }_{kl}d$, $\forall k, l$.

Then, if $0 \neq e \in A {\otimes }_R \overline K$ is a primitive
idempotent, $t(e)$ divides $d$.\endproclaim 

\demo {Proof} Keeping the notations in lemma (4.11), we
may write $$a_k = \sum_{ \alpha , i, j } { r_{ k, {\alpha }_{ij} }
e_{ {\alpha }_{ij} } }, \quad b_l = \sum_{ \beta , \mu , \nu }
{ s_{ l,{\beta }_{\mu \nu } } e_{ {\beta }_{\mu \nu } } }, $$ where  
$r_{ k, {\alpha }_{ij} }$ and $s_{ l,{\beta }_{\mu \nu } }$ are algebraic
integers. 

The $t(e_{{\alpha }_{ii}})$ are natural numbers. Since the
$e_{{\alpha }_{ii}}$ form a complete set of orthogonal primitive
idempotents, there is an ${\alpha}_{ii}$ such that $e$ and
$e_{{\alpha}_{ii}}$ generate isomorphic right ideals. Hence, by
property c), $t(e)= t(e_{{\alpha }_{ii}})$.

In view of the relations 
$$e_{ {\alpha }_{ij} } e_{ {\beta }_{\mu \nu } }  = {\delta
}_{\alpha , \beta } {\delta }_{\mu , j}e_{ {\alpha }_{i\nu} }, \quad
\forall \alpha , \beta , \mu ,\nu ,i,j \ ,$$ which, in particular,
imply that for $\nu \neq i$  $$t(e_{ {\alpha }_{i\nu } }  ) =  t(
e_{ {\alpha }_{ii} }  e_{ {\alpha }_{i\nu } }  - 
e_{ {\alpha }_{i\nu } }  e_{ {\alpha }_{ii} } ) = 0, $$
we get, 
$$a_k b_l = \sum{ r_{ k, {\alpha }_{ij} } s_{ l,{\beta }_{\mu \nu } }
e_{ {\alpha }_{ij} } e_{ {\beta }_{\mu \nu } } } = 
\sum_{\alpha ,i,j,\nu } { r_{ k, {\alpha }_{ij} } s_{ l,{\alpha }_{j \nu }
} e_{ {\alpha }_{i\nu } }  }, $$ and then $$d {\delta }_{kl} = t(a_k
b_l) = 
\sum_{\alpha ,i,j,\nu } { r_{ k, {\alpha }_{ij} } s_{ l,{\alpha }_{j \nu }
} t( e_{ {\alpha }_{i\nu } } ) } = 
\sum_{\alpha ,i,j} { r_{ k, {\alpha }_{ij} } s_{ l,{\alpha }_{j i }
} t( e_{ {\alpha }_{ii } } ) }.$$
Since $n = \text {number of all} \ e_{ {\alpha }_{ij} }$: $1 \leq
\alpha \leq s$, $1 \leq i,j \leq n_{\alpha }$, there is a bijection 
$$\{ 1, \dots , n \} @>>> \{ {\alpha }_{ij} : 1 \leq \alpha \leq s\ , \
1\leq i,j \leq n_{\alpha } \} .$$ Define $$ a_{ k, {\alpha }_{ij}} = 
 r_{ k, {\alpha }_{ij}} {(t( e_{ {\alpha }_{ii } } ) /d)}^{1/2}, \quad 
 b_{ {\alpha }_{ij},l }=s_{ l, {\alpha }_{ji}} {(t( e_{ {\alpha }_{ii } } )
/d)}^{1/2}.$$ 

By the calculation above, the $n \times n$ matrices 
$A = ( a_{ k, {\alpha }_{ij}})$ and $B = ( b_{ {\alpha }_{ji}, l})$ verify
that $AB = \id$, but this implies that $BA = \id$.

Then $\forall \alpha , i$, $$\sum_l { b_{ {\alpha }_{ii},l }
 a_{ l, {\alpha }_{ii} } } = 1, \quad \text {or} \ \sum_l { 
s_{ l, {\alpha }_{ii} } r_{ l, {\alpha }_{ii} }} = d {t( e_{ {\alpha }_{ii } }
)}^{-1} ,$$ which is an algebraic integer. But as it is a rational
number, it must be an integer. So $t( e_{ {\alpha }_{ii } } )$
divides $d$. \qed  \enddemo

\bigpagebreak
The following theorem is due to Kac and Zhu.
\footnote{Added in 1997: In a recent paper by M. Lorenz \cite{M.
Lorenz, On the class equation for Hopf algebras, Proc. Amer. Math.
Soc. (1997)} a shorter proof of 4.13 is giving using less algebraic
number theory.}

\proclaim {Theorem 4.13. "Class Equation" }
Let $\lambda \in H^*$ be an integral, such that $<\lambda , 1> =
1$, and let  $\  \lambda = e_1 , e_2, \dots , e_n$, be a complete
set of orthogonal primitive idempotents in $R_k(H)$. Then $$\dim H =
\sum_{i=1}^n { \dim (e_i H^*) },$$ with $\ \dim (e_1 H^*) = 1$, and
$\dim (e_i H^*)\  /  \dim H$, $\forall 1 \leq i \leq
n$.\endproclaim  

\demo {Proof} 
Since $H^* = \bigoplus_i { e_i \ H^*}$, 
$\dim H = \dim H^* = \sum_{i=1}^n { \dim (e_i H^*) }$.
It is clear also, as $\lambda $ is an integral, that $\dim (e_1 H^*) =
1$.

\bigpagebreak
Call $A = R_{\Bbb Z}(H) = \sum_i { \Bbb Z \ {\Cal X}_i}$, which is a
free $\Bbb Z$-module of finite rank $n$, and such that $A{\otimes
}_{\Bbb Z}\ \Bbb Q = R_{\Bbb Q}(H)$ is semisimple.

 Let $t = {\Tr }_{H^*}: R_k(H) @>>> k$; it is a trace like function
by the remark after the definition of such functions.



Now, if we consider $a_i =  {\Cal X}_i$, $b_i =  {\Cal
X}_{\overline i}$, $1 \leq i \leq n$, ($a_i, b_i \in A$), then
lemma (4.12) applies with $d = \dim H$ by the orthogonality
relations (4.7).  (Observe that, as $k$ is algebraically closed and
$\char k = 0$, $A{\otimes }_{\Bbb Z}\ \overline {\Bbb Q} \subseteq
R_k(H)$). 

On the other hand, as $\ e_1, \dots , e_m$ are orthogonal
idempotents, $t(e_i) = \dim (e_i H^*)$.

Thus, $t(e_i) = \dim (e_i H^*) \ / d = \dim H$, which finishes the
proof of the theorem. \qed \enddemo

\bigpagebreak
As a consequence of theorem (4.13), we  show a theorem of
 Masuoka \cite{9} on semisimple Hopf algebras of dimension $p^m$,
$p$ a prime generalizing previous results of Kac \cite{3} and Zhu
\cite{21}. We need the following lemma.

\proclaim {Lemma 4.14} Let $\lambda \in H^*$ an integral, such that
$<\lambda , 1> = 1$. Then

1) $\dim (h\lambda \ H^*) = 1$, $h \in H$, if and only if,
there exist $0 \neq \alpha \in k$ and $g \in G(H)$ such that $h =
\alpha g$.

2) If $a \in R_k(H)$ is such that $\dim (a H^*) = 1$, and $aH^*$ is
not the trivial $H^*$-module, then there exist $0 \neq \alpha \in
k$, and $1 \neq g \in G(H) \bigcap Z(H)$, such that $a = \alpha g
\lambda $.\endproclaim

\demo {Proof}

1) Let $h \in H$. Then $\dim (h \lambda H^*) = 1$, if and only if
there exists $g \in G(H)$, such that $\forall p \in H^*$, $(h
\lambda ) p = (h \lambda )<p,g>$.

But that is equivalent to the existence of a $g \in G(H)$ such that
$\forall x \in H$, $$<h\lambda , x_{(1)}>x_{(2)} = <h\lambda , x>g,
\quad \text {or} \quad <gh\lambda , x_{(1)}g^{-1}> x_{(2)} g^{-1} =
<gh\lambda , x g^{-1}>\ 1,$$ if and only if $gh\lambda $ is an
integral in $H^*$. Which is equivalent to the existence of a nonzero
$\alpha \in k$ such that $gh\lambda = \alpha \lambda$, i.e., $h =
\alpha g^{-1}$.

\bigpagebreak
2) As $a \in R_k(H)$ (and $a \neq 0)$, there exists $0 \neq x \in
Z(H)$ such that $a = x \lambda$. Then, by 1), there exists $0 \neq
\alpha \in k$ and $g \in G(H)$, such that $x = \alpha g$. In
particular, $g \in Z(H)$ by (4.6).

Suppose $g = 1$. Then $x \lambda = \alpha \lambda $, and so $a
H^* = \lambda H^*$, which is the trivial $H^*$-module. But this
contradicts our assumption on $a$, so $g \neq 1$.\qed \enddemo 

\bigpagebreak
\proclaim {Theorem 4.15} Let $H$ be a semisimple Hopf algebra such
that $\dim H = p^m$, with $p$ a prime, and $m \geq 1$. Then $H$
contains a central grouplike $g \neq 1$.\endproclaim

\demo {Proof} By theorem (4.13), we may write $$p^m = 1 +
\sum_i{ \dim (e_iH^*) },$$ with $e_i \neq \lambda $ a primitive
idempotent in $H^*$, and $\dim (e_iH^*) \ / \dim H = p^m$, $\forall
i$.

Then, there must be a primitive idempotent $e_i \neq \lambda $,
such that $\dim (e_i H^*) = 1$.

Now, as $e_i \neq \lambda $, $e_i H^*$ is not the trivial
$H^*$-module, so the previous lemma implies, in particular, that
$H$ possesses a central grouplike $g \neq 1$.\qed \enddemo 

\bigpagebreak

\subhead Hopf algebras of prime order \endsubhead
We now apply  the previous results to arbitrary, not necessarily
semisimple, Hopf algebras of prime order. 
$k$ will denote an algebraically closed field of characteristic
zero.

\bigpagebreak
Let $p$ be a prime, and let ${\Bbb Z}_p$ be the cyclic group of
order $p$.  Consider the Hopf algebra $H = k {\Bbb Z}_p $.

Observe that $H^* =  k^{{\Bbb Z}_p}$ contains a grouplike $G$ of
order $p$ (defined by $G(a) = {\omega }_p$, where $a$ is a
generator of ${\Bbb Z}_p$, and ${\omega }_p$ is a primitive $p$-th
root of unity in $k$). Then $ k^{{\Bbb Z}_p}$ contains the $k$-linear
span of the powers of $G$, which is isomorphic (as a Hopf
algebra) to $k  {\Bbb Z}_p$. So, because they have the same
finite order, $H^* = k^{{\Bbb Z}_p} \backsimeq k{\Bbb Z}_p = H$.  

\bigpagebreak
\proclaim {Theorem 4.16. (Zhu, 1994)} Let $H$ be a Hopf algebra over
$k$ of prime order $p$. Then $H$ is isomorphic to the algebra of
functions on the cyclic group ${\Bbb Z}_p$.\endproclaim  

\demo {Proof} 
If $p = 2$, it is easily seen that $H$ must be commutative and
cocommutative, and the result follows from the theorem which asserts
that every finite dimensional cocommutative 
Hopf algebra over an algebraically
closed field of characteristic zero is isomorphic to a group
algebra $kG$ for some finite group $G$, and it is clear that $G$
must be isomorphic to ${\Bbb Z}_2$. 

Assume now $p>2$. We consider separetaly three
cases.

\bigpagebreak
\it First Case:\rm \ $H$ or $H^*$ not unimodular.

Suppose $H$ is not unimodular (the proof is analogous for $H^*$).
Then the modular function $\alpha \in H^*$ (see $\S3$) is a
nontrivial grouplike element in $H^*$. By  
theorem (3.15), the linear span of the powers of $\alpha $, $k<\alpha
>$
 (which is a Hopf subalgebra of $H$), must coincide with $H$. So in
particular it is isomorphic to $k {\Bbb Z}_p$, and we have $$H
\backsimeq H^{**} = {(k<\alpha >)}^* \backsimeq {k {\Bbb
Z}_p}^* \backsimeq k {\Bbb Z}_p.$$ 

\bigpagebreak
\it Second Case: \rm \ $H$ and $H^*$ unimodular, but $H$ or $H^*$
not semisimple.

By (3.4), we must have $\Tr ({\Cal S}^2) = 0$. And
as $H$ and $H^*$ are unimodular, ${\Cal S}^4 = \id$. But then, the
eigenvalues of ${\Cal S}^2$ are all $1$ or $-1$, and their sum is
$0$. This implies that the order of $H$ must be even, which is an
absurd. So this case is not possible.

\bigpagebreak
\it Third Case : \rm \ $H$ and $H^*$ both semisimple.

By theorem (4.15), $H$ contains a central grouplike $g \neq 1$.
Hence, by the Nichols-Zoeller\rq s theorem, $H = k<g> \backsimeq k
{\Bbb Z}p$. Note that we did not use "2) $\implies$ 1)" of Theorem
(3.13). \qed \enddemo

\newpage

\head Appendix \endhead

Here is a complete list of the conjectures made by Kaplansky
(except one which is easy).

\proclaim {1} If $C$ is a Hopf subalgebra of the Hopf algebra $B$
then $B$ is a free left $C$-module.
\endproclaim

\remark {Remark} 
In general this is false. A counterexample is given in \cite{12}.
Some particular cases are known, most notably if $B$ is finite
dimensional \cite{11}, as seen in the text, but also if $C$ is finite
dimensional and either semisimple or normal. The following  form of
the conjecture is still open: is $B$ is faithfully flat over
$C$? And dually, is a Hopf algebra faithfully coflat over its
quotient Hopf algebras?
\endremark 

\bigpagebreak
\proclaim {2} Call a coalgebra $C$  admissible  if it admits
an algebra structure making it a Hopf algebra. The conjecture
states that $C$ is admissible if and only if every finite subset of
$C$ lies in a finite dimensional admissible subcoalgebra.
\endproclaim

\bigpagebreak
\proclaim {3} A Hopf algebra of characteristic zero has no
non-zero central nilpotents.\endproclaim

\bigpagebreak
\proclaim {4} If $H$ is a finite dimensional Hopf algebra, such that 
$H$ or $H^*$ are semisimple, the square of the antipode is the
identity.\endproclaim

\remark {Remark} We have seen this is true if the characteristic
of the ground field $k$ equals zero.\endremark  

\bigpagebreak
\proclaim {5} If $H$ is a (semisimple) Hopf algebra over  the
algebraically closed field $k$, then the sizes of the matrices
occuring in any full matrix constituent of $H$ divides the
dimension of $H$.\endproclaim

\remark{Remark} This is seen in the text if $H$ is a group algebra.
\endremark

\bigpagebreak
\proclaim {6} If $H$ and $H^*$ are semisimple the characteristic
of $k$ does not divide the dimension of $H$.\endproclaim

\remark {Remark} This is a consequence of theorem (3.13) first
proved by Larson and Radford.\endremark

\bigpagebreak
\proclaim {7} If the dimension of $H$ is prime ($k$ algebraically
closed), then $H$ is commutative and
cocommutative.\endproclaim

\remark {Remark} This is the result of Zhu shown in the text (th.
4.16) if the characteristic of $k$ is zero, and is still open if the
characteristic is positive.\endremark

\bigpagebreak\proclaim {8} If the characteristic of $k$ does not
divide the dimension of $H$, the dimension of the radical is the
same in $H$ and in $H^*$.\endproclaim

\remark {Remark} If the $\char k = p$  divides the dimension of
$H$ one has the following counterexample: $H = {\Bbb Z}_p$ is not
semisimple, but $H^*$ is.

We have seen that in characteristic zero $H$ is semisimple iff
$H^*$ is. But the conjecture is false even in characteristic zero
(take the Frobenius- Luztig kernel of $sl (2)$)\footnote{This 
counterexample was suggested by N. Andruskiewitsch.}.\endremark

\bigpagebreak \proclaim {9} Again under the hipothesis that the
characteristic of $k$ does not divide the dimension of $H$, the
conjecture states that there are only a finite number (up to
isomorphism) of Hopf algebras of a given dimension.\endproclaim

\remark {Remark} The following result is known (D. Stefan, 1995): 
The set of types of semisimple and cosemisimple Hopf algebras of a
given dimension is finite (in any characteristic).\endremark


\newpage
\head REFERENCES \endhead 

\bigpagebreak
[1]. C.W. Curtis and I. Reiner. \it Representation Theory of finite
groups and associative algebras, \rm Interscience, New York,
1962.

\bigpagebreak
[2]. N. Jacobson. \it Lie Algebras, \rm Interscience, New York,
1962.

\bigpagebreak
[3].  G. I. Kac. \it Certain arithmetic properties of ring groups,
\rm Funct. anal. appl. \bf 6 \rm , (158-160), 1972.

\bigpagebreak
[4]. J. C. Jantzen. \it Representations of algebraic groups,
\rm Academic Press, 1987.

\bigpagebreak
[5]. R. G. Larson. \it Characters of Hopf algebras, \rm J. of Algebra
\bf 17 \rm , (352-368), 1971.

\bigpagebreak
[6]. R. G. Larson and D. E. Radford. \it Semisimple cosemisimple
Hopf algebras, \rm Amer. J. of Math. \bf 110\rm , (187-195), 1988.

\bigpagebreak
[7]. R. G. Larson and D. E. Radford. \it Finite dimensional
cosemisimple Hopf algebras in characteristic 0 are semisimple,
\rm J. of Algebra \bf 117 \rm , (267-289), 1988.

\bigpagebreak
[8]. R. G. Larson and M. Sweedler. \it An associative orthogonal
bilinear form for Hopf algebras,
\rm Amer. J.  Math. \bf 91 \rm , (75-93), 1969.

\bigpagebreak
[9]. A. Masuoka. \it The $p^{n}$-theorem for semisimple hopf
algebras, \rm Proc. Amer. Math. Soc.,  to appear.

\bigpagebreak
[10]. S. Montgomery. \it Hopf algebras and their actions on rings,
\rm CBMS, 1993.

\bigpagebreak
[11]. W. D. Nichols and M. B. Zoeller. \it A Hopf algebra freeness
theorem, \rm Amer. J.  Math. \bf 111\rm , (381-385), 1989.

\bigpagebreak
[12]. U. Oberst and H.J. Schneider. \it Untergruppen formeller
Gruppen von endlichen Index, \rm J. of Algebra \bf 31\rm , (10-44),
1974.

\bigpagebreak
[13]. U. Oberst and H.J. Schneider. \it Uber untergruppen
endlicher algebraischer Gruppen, \rm Manuscripta Math. \bf 8\rm ,
(217-241), 1973.

\bigpagebreak
[14]. D. E. Radford. \it The order of the antipode of a finite
dimensional Hopf algebra is finite,\rm \  Amer. J.  Math. \bf 98\rm ,
(333-355), 1976.   

\bigpagebreak
[15]. J. P. Serre. \it Linear Representations of Finite Groups, \rm 
Springer-Verlag, New York, 1977.

\bigpagebreak
[16]. M. E. Sweedler. \it Hopf Algebras,\rm \  Benjamin, New York,
1969.

\bigpagebreak
[17]. M. E. Sweedler, \it Integrals for Hopf algebras, \rm Ann. Math.
\bf 89\rm , (323-335), 1969. 

\bigpagebreak
[18]. E. J. Taft. \it The order of the antipode of finite dimensional
Hopf algebras, \rm Proc. Nat. Acad. Sci. USA \bf 68\rm ,
(2631-2633), 1971. 


\bigpagebreak
[19]. E. J. Taft and R. L. Wilson. \it There exist finite dimensional
Hopf algebras with antipode of arbitrary even order, \rm J. of
Algebra \bf 62\rm , (283-291), 1980. 

\bigpagebreak
[20]. D. Wigner. \it An identity in the free Lie algebras, \rm Proc.
Amer. Math. Soc. \bf 106\rm , (639-640), 1989. 

\bigpagebreak
[21]. Y. Zhu. \it Hopf algebras of prime dimension, \rm Inter. Math.
Research Not. \bf 1\rm , (53-59), 1994. 

\bigpagebreak
[22]. M. Lorentz. \it On the class equation for Hopf algebras, \rm
Inter. Math. Research Not. \bf 1\rm , (53-59), 1997.  \enddocument






































































