I'm reading ''COHOMOLOGY OPERATIONS AND APPLICATIONS HOMOTOPY THEORY'' by R. E. Mosher and M.C. Tangora. At the beginning of the six chapter the following is explained:
Assume $R$ is afield and let $A = (A_i)_i$ be a Hopf algebr over $R$ of finite type (that is each $A_i$ is finite-dimensional over $R$). We define the dual Hopf algebra to $A,$ $A^*$ by setting $$A^* = (A_i^*)_i,\quad A_i^* = \mbox{Hom}_R(A_i, R)$$ that is the dual to $A_i$ as a vector space over $R.$ So $A^*$ is a Hopf algebra over $R$ of finite type.
My question is
Show that $A$ and $A^*$ are isomorphic as $R$-module but certainly not as $R$-algebra in general?
I know that $A$ and $A^*$ are isomorphic as $R$-module but I don't understand why $A$ and $A^*$ not as $R$-algebra in general.
I hope that someone, can help. Thank you in advance!
Proposition: Let $(H, \mu, \eta, \Delta, \varepsilon,S)$ a finite dimensional $k$-Hopf algebra. Then the dual v.s. $Η^{\ast}$, equipped with the $k$-algebra structure the dual of the $k$-coalgebra $(H, \Delta, \varepsilon)$ and $k$-coalgebra structure the dual of the $k$-algebra $(H, \mu, \eta)$ form a $k$-Hopf algebra: $$ (H^{\ast} \; , \; \Delta^{\ast}\!\!\!\circ\!\!\lambda \; , \; \varepsilon^{\ast}\!\!\!\circ\!\!\psi \; , \; \lambda^{-1}\!\!\circ\!\mu^{\ast} \; , \; \psi^{-1}\!\!\circ\!\eta^{\ast},S^*) $$ which will be called the dual Hopf algebra of $Η$.
(In the above, we denote with $\psi$, the natural v.s. isomorphism $k\cong k^*$ defined by $\psi(\kappa)=f$ with $f(1)=\kappa$ and $\psi^{-1}(f)=f(1)$ and with $\lambda:A^*\otimes A^*\rightarrow(A\otimes A)^*$ the linear v.s. injection, defined by $\lambda(f\otimes g)(u\otimes v)=f(u)g(v)\in k$. In case $A$ is of finite $k$-dimension then $\lambda$ is a vector space isomorphism).
Thus, the answer to your question is that the algebraic structure of the dual Hopf algebra is determined by the coalgebraic structure of the initial Hopf algebra. So, it does not have to be necessarily isomorphic to the initial algebra.
Example: As an example of the above, you can see the construction of the dual group Hopf algebra $(kG)^*$ of the group Hopf algebra $kG$ of a finite group $G$:
Let $G$ a finite group and $k$ the field. For simplicity consider $k$ algebraically closed and of characteristic zero. Let the dual space $(kG)^{*}$: $$ k^{G}=Map(G,k) \cong Hom(kG,k)=(kG)^{*} $$ $k^{G}$ is the v.s. of set maps from $G$ to $k$. Since $dim_{k}(kG)^{*}=|G|< \infty$ we can pick as a $k$-basis of the dual space the dual of the $k\mathbb{G}$ basis $\{ g /g \in G \}$, i.e.: $\{ p_{g} /g \in G \}$ where $p_{g}(h) = \delta_{g,h}$ for all $g \in G$ and for all $h \in G$.
Applying the above proposition, we get the following maps for the dual $k$-Hopf algebra $(kG)^{*}$ of the group Hopf algebra $kG$:
comultiplication: $$ \Delta(p_{g})=\sum_{\begin{array}{c} g_{1},g_{2} \in G \\ g_{1}g_{2}=g \\ \end{array}}p_{g_{1}} \otimes p_{g_{2}}=\sum_{x \in G}p_{x} \otimes p_{x^{-1}g} $$
counity: $$ \varepsilon(p_{g}) = p_{g}(1_{G}) = \delta_{g,1_{G}} $$
multiplication: $$ \big( p_{g_{1}} \star p_{g_{2}} \big)(h)=p_{g_{1}}(h)p_{g_{2}}(h)=\delta_{g_{1},h} \delta_{g_{2},h}= \left\{ \begin{array}{l} 0, \;\;\;\;\;\; g_{1} \neq g_{2} \\ \delta_{g,h}, \;\; g_{1} = g_{2} = g \\ \end{array} \right. $$ thus: $p_{g}^{2}=p_{g}$ and $p_{g} \star p_{h}=0$ if $g \neq h$. In other words, the basis $\{ p_{g} /g \in G \}$ of $(kG)^{*}$ consists of orthogonal idempotents, implying that:
the unity map is determined by $$ \sum_{g \in G}p_{g}=1_{(kG)^{*}} $$ The above imply the $k$-algebra isomorphism $$ (kG)^{*} \cong k \times k \times \cdots \times k \equiv k^{(n)} $$ given explicitly by: $p_{g} \leftrightarrow (0,\ldots,1,0,\ldots,0) $ ($1$ lies at the $g$-th entry and $n=|G|$).
We thus conclude that $(kG)^{*}$ is commutative and semisimple (remember that $kG$ is cocommutative and cosemisimple but in general not commutative since the group $G$ was not assumed to be abelian; this an example of duality). Thus $kG$ and $(kG)^*$ are certainly not isomorphic as algebras (in general).
Finally, the antipode of the dual is given by: $S^{*}(p_{g}) = p_{g^{-1}}$