Taft algebra

Example

Let $n \ge 2$ and let $k$ be a field that contains a primitive $n$-th root of unity $\omega$. The Taft algebra $T_n(\omega)$ is the $k$-algebra generated by $g$ and $x$ with the relations $$g^n = 1, \qquad x^n = 0, \qquad x g = \omega\, g x.$$ It has the $k$-basis $g^i x^j$ with $0 \le i, j \le n - 1$, so its dimension is $n^2$. It is a Hopf algebra with $$\Delta(g) = g \otimes g, \quad \Delta(x) = x \otimes 1 + g \otimes x, \quad \varepsilon(g) = 1, \quad \varepsilon(x) = 0, \quad S(g) = g^{-1}, \quad S(x) = -g^{-1} x.$$ The check that $\Delta$ respects the relations: $\Delta(g)^n = g^n \otimes g^n = 1 \otimes 1$. For $\Delta(x) \Delta(g) = \omega\, \Delta(g) \Delta(x)$, both sides equal $\omega(gx \otimes g + g^2 \otimes gx)$ after using $xg = \omega gx$. For $\Delta(x)^n = 0$, put $A = g \otimes x$ and $B = x \otimes 1$; then $BA = \omega AB$, so the $q$-binomial formula with $q = \omega$ gives $(A + B)^n = A^n + B^n$, and $A^n = g^n \otimes x^n = 0$, $B^n = x^n \otimes 1 = 0$. The antipode check on $x$: $S(x) \cdot 1 + S(g) x = -g^{-1}x + g^{-1}x = 0 = \varepsilon(x)$, and $x S(1) + g S(x) = x - x = 0$.