Compute dual of an hopf algebra

I would like to compute the dual of the hopf algebra $k[\mathbb{Z}/n\mathbb Z]$, where comultiplication is $\Delta(g)=g \otimes g$, coinverse is $S(g)=g^{-1}$ and counit is $\epsilon(g)=_n$. I know the result is the same as the hopf algebra of $k[X]/(X^n-1)$ which corresponds to the group scheme $\mu_n$, but I am not sure how to show this, I have never computed explicitely the dual of anything.

Any hint? Thanks!

For clarity I'll write explicitly the whole group algebra. The Hopf algebra $H = k[\mathbb{Z}/n\mathbb{Z}]$ in question is the $n$-dimensional $k$-vector space with basis $e_0, \ldots, e_{n-1}$, the algebra structure \begin{aligned} m: H \otimes H \to H, \quad &e_i \otimes e_j \mapsto e_{i+j} \\ u: k \to H, \quad &1 \mapsto e_0 \end{aligned} the coalgebra structure \begin{aligned} \Delta: H \to H \otimes H, \quad &e_i \mapsto e_i \otimes e_i \\ \epsilon: H \to k, \quad &e_i \mapsto 1\end{aligned} and antipode $S: H \to H$, $e_i \mapsto e_{n - i}$ (where subscripts $e_i$ are understood to be modulo $n$).

The dual Hopf algebra $(H^\circ, m^\circ, u^\circ, \Delta^\circ, \epsilon^\circ, S^\circ)$ has underlying vector space $H^\circ = H^*$, so let $f_i: H \to k$ for $i = 1, \ldots, n$ be the basis of $H^*$ dual to the $e_i$. The new unit $u^\circ: k \to H^\circ$ is given by $1 \mapsto \epsilon$, making $\epsilon = f_0 + f_1 + \cdots + f_{n-1}$ the unit element in $H^\circ$.

Then we use the original comultiplication $\Delta$ to figure out the multiplication map $m^\circ$ in the dual:

$$m^\circ(f_i, f_j) = H \xrightarrow{\Delta} H \otimes H \xrightarrow{f_i \otimes f_j} k \otimes k \xrightarrow{\sim} k$$

Checking the action on the basis element $e_k$, $m^\circ(f_i, f_j)(e_k) = f_i(e_k) f_j(e_k)$, and so $m^\circ(f_i, f_i) = f_i$, while $m^\circ(f_i, f_j) = 0$ for $i \neq j$. Notice that $m^\circ(\epsilon, f_i) = f_i$ as expected.

You can go and find the coalgebra structures and antipode on $H^\circ$ yourself, but I'll also comment that the multiplication $m^\circ$ doesn't look obviously like the multiplication in $k[X]/(X^n - 1)$. If $k$ has a primitive $n$th root of unity $\omega$, and the characteristic of $k$ does not divide $n$, then writing

$$F_i = \frac{1}{n}\left(1 + \omega X + \omega^2 X^2 + \cdots + \omega^{n-1} X^{n-1}\right)$$

makes $F_0, \ldots, F_{n-1}$ a basis of $k[X]/(X^n - 1)$, satisfying $F_i^2 = 1$, $F_i F_j = 0$ for $i \neq j$, and $F_0 + \cdots + F_{n-1} = 1$. (The $F_i$ here are the decomposition of the group algebra ring into its primitive idempotents.) So the algebra structure on $H^\circ$ really does look like $k[X]/(X^n - 1)$, just in a different basis.

Let $H$ be your algebra and let $H^*$ be its dual. Let $g$ be a generator of $Z/nZ$, so that $\{1,g,g^2,\dots,g^{n-1}\}$ is a basis of $H$. Let $\{v_0,\dots,v_{n-1}\}$ be the dual basis of $H^*$, so that $v_i(g^j)=\delta_{i,j}$ for all $i,j\in\{0,\dots,n-1\}$.

The multiplication of $H^*$ is defined as follows: if $a$ and $b$ are two elements of $H^*$,then their product $ab$ in $H^*$ is the composition $a\otimes b\circ\Delta$. As $ab$ is an element of $H^*$, to compute it explicitely it is enough to find its values on the elements of our basis of $H$. If $i\in\{0,\dots,n-\}$ we have $$(ab)(g^i))=(a\otimes b)(\Delta(g^i))=(a\otimes b)(g^i\otimes g^i)=a(g^i)b(g^i).$$ Using this, we find immediately that if $i,j\in\{0,\dots,n-1\}$ we have $$v_iv_j=\delta_{i,j}v_i.$$ This decribes the algebra structure on $H^*$. Can you find its comultplication?

• I was computing the multiplication right now and I got that $a \otimes b \circ \Delta$ is the multiplication. Now I am computing the comultiplication. Is it enough to show that if the comultiplication and multiplication are the same then the Hopf algebra are the same? Or am I to compute also inverse, coinverse, unit and counit? – Ale Sep 28 '17 at 5:30
• Also, I really don't get what what I can infer from what you wrote, the multiplication in $\mu_n$ does not satisfy $e_ie_j=\delta_{i,j}e_i$ if I am not mistaken. It should just be: $m(X^i \otimes X_j)=X^{i+j}$. – Ale Sep 28 '17 at 5:51
• Well, my computation is correct. Joppy's answre, which is exactly the same as mine, gets the same multiplication. – Mariano Suárez-Álvarez Sep 28 '17 at 14:46