Taft-Hopf Algebra has dimension $N^2$? 
*

*Definition of the Taft-Hopf Algebra
Let $k$ be a field. Let be $N$ a positive integer such that there exists a primitive $N$-th root of unitiy $\zeta$ over $k$.
Denote by $(H, \mu, 1_H)$ the unital, associative algebra over $k$ that is generated by elements $g$ and $x$, subject to the three relations:
$g^N=1$, $x^N=0$, $xg=\zeta gx$.
One can show that there are algebra maps:


*

*$\eta: k \rightarrow H; 1 \mapsto 1_H$.

*$\Delta: H \rightarrow H \otimes H; g \mapsto g\otimes g, x \mapsto 1 \otimes x + x \otimes g$.

*$\epsilon: H \rightarrow k; g \mapsto 1, x \mapsto 0$.

*$S: H \rightarrow H^{opp}; g \mapsto g^{-1}, x \mapsto -xg^{-1}$.

Then (one can show) that $(H, \mu, \eta, \Delta, \epsilon, S)$ is a Hopf algebra. We call it the Taft-Hopf Algebra.


*Question(s)


*

*In my lecture notes it says, for a given $N$ the Taft-Hopf algebra has dimension $N^2$. Why?

*What would be a basis?

 A: This has nothing to do with the Hopf algebra structure.
It follows from the relations $g^N = 1$ that we can define for every element $[n]$ of $\mathbb{Z}/N$ the power
$$
  g^{[n]} := g^n \,.
$$
We show in the following that the elements $g^{[n]} x^m$ with $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$ are a basis of $H$.

Vector space generating set
We first show that these elements span the algebra $H$ as a vector space.
We know from the construction of $H$ that it is spanned as a vector space by the monomials
$$
  p_1 \dotsm p_r
$$
with $r \geq 0$ and $p_i \in \{g, x\}$.
It follows from the relation $x g = \zeta g x$ that we can reorder the factors $z_i$ in any such monomial up to some nonzero factor (namely a power of $\zeta$).
It follows that $H$ is already spanned by those monomials of the form
$$
  g^n x^m
$$
with $n, m \geq 0$.
It follows from the relations $g^N = 1$ and $x^N = 0$ the vector space generators can equivalently be written as
$$
  g^{[n]} x^m
$$
with $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.

Linearly independent
To show that these monomials are linearly independent we use a standard trick from representation theory.
We first compute the action of the two algebra generators $g$ and $x$ on our vector space generating set.
We find from the relations $xg = \zeta gx$ and $x^N = 0$ that
\begin{align*}
  g \cdot g^{[n]} x^m
  &=
  g^{[n+1]} x^m \,,
  \\
  x \cdot g^{[n]} x^m
  &=
  \begin{cases}
    \zeta^n g^{[n]} x^{m+1} & \text{if $0 \leq m \leq N-2$,} \\
    0                       & \text{if $m = N-1$,}
  \end{cases}
\end{align*}
for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.
Let now $V$ be the free vector space with basis
$$
  G^{[n]} X^m
  \qquad
  \text{with $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.}
$$
We denote the free algebra on the generators $g$ and $x$ by $F$.
We can define an $F$-module structure on the vector space $V$ via
\begin{align*}
  g \cdot G^{[n]} X^m
  &=
  G^{[n+1]} X^m \,,
  \\
  x \cdot G^{[n]} X^m
  &=
  \begin{cases}
    \zeta^n G^{[n]} X^{m+1} & \text{if $0 \leq m \leq N-2$,} \\
    0                       & \text{if $m = N-1$,}
  \end{cases}
\end{align*}
for all $[n] \in \mathbb{Z}$ and $m = 0, \dotsc, N-1$.
This module structure is compatible with the relations of $H$ because
\begin{align*}
  xg \cdot G^{[n]} X^m
  &=
  x \cdot G^{[n+1]} X^m
  \\
  &=
  \zeta^{n+1} G^{[n+1]} X^{m+1}
  \\
  &=
  \zeta^{n+1} g \cdot G^{[n]} X^{m+1}
  \\
  &=
  \zeta g x \cdot G^{[n]} X^m
\end{align*}
for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-2$,
and similarly
\begin{align*}
  xg \cdot G^{[n]} X^{N-1}
  &=
  x \cdot G^{[n+1]} X^{N-1}
  \\
  &=
  0
  \\
  &=
  \zeta g \cdot 0
  \\
  &=
  \zeta g x \cdot G^{[n]} X^{N-1} \,,
\end{align*}
for all $[n] \in \mathbb{Z}/N$, as well as
$$
  g^N \cdot G^{[n]} X^m
  =
  G^{[n+N]} X^m
  =
  G^{[n]} X^m
$$
for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$, and
$$
  x^N \cdot G^{[n]} X^m
  =
  0
  =
  0 \cdot G^{[n]} X^m
$$
for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.
It follows that the $F$-module struture on $V$ descends to an $H$-module structure on $V$.
For this $H$-module structure on $V$ we now have
$$
  g^{[n]} x^m
  \cdot
  G^{[0]} X^{0}
  =
  \zeta^{0 \cdot m} g^{[n]} \cdot G^{[0]} X^m
  =
  g^{[n]} \cdot G^{[0]} X^m
  =
  G^{[n]} X^m
$$
for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.
The elements $G^{[n]} X^m$ are linearly independent in $V$ (since they form a basis).
It thus follows that the elements $g^{[n]} x^m$ are linearly independent in $H$.
Indeed, if we have some linear combination
$$
  0
  =
  \sum_{[n], m}
  \lambda_{[n], m}
  g^{[n]} x^m
$$
in $H$ then it follows that
$$
  0
  =
  \sum_{[n], m}
  \lambda_{[n], m}
  g^{[n]} x^m
  \cdot
  G^{[0]} X^{0}
  =
  \sum_{[n], m}
  \lambda_{[n], m}
  G^{[n]} X^m
$$
and therefore $\lambda_{[n], m} = 0$ for all $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$.
We have altogether shows that the elements $g^{[n]} x^m$ of $H$ with $[n] \in \mathbb{Z}/N$ and $m = 0, \dotsc, N-1$ are vector space generators for $H$ as well as linearly independent.
These elements are hence a basis of $H$, as claimed in the beginning.
