Let $\Bbbk$ be a field. A well-known result (see Larson, Sweedler, An Associative Orthogonal Bilinear Form for Hopf Algebras and Pareigis, When Hopf algebras are Frobenius algebras) states that a bialgebra $B$ over $\Bbbk$ is a finite-dimensional Hopf algebra if and only if it is Frobenius as an algebra and the Frobenius morphism $\psi:B\to\Bbbk$ is a (left) integral on $B$ (i.e. $\sum b_1\psi(b_2)=\psi(b)1_B$ for all $b\in B$).
For didactical reasons, I would like to have an example of a (necessarily finite-dimensional) bialgebra $B$ which is Frobenius as an algebra but that is not a Hopf algebra. However, I didn't manage to construct one. Does anybody know one? Even a non-elementary one would be fine.
Alternatively, if somebody knows a proof of the fact that a Frobenius bialgebra is automatically a Hopf algebra, without further assumptions on the Frobenius homomorphism, then I would be very glad to see it.
Best Answer
Here is an example. Take k a field and S the multiplicative semigroup {0,1}. For notational convenience denote it S={1,x}. Consider B=k[S] the semigroup algebra.
B is bialgebra and not Hopf because S is a semigroup and nota a group.
$B\cong k[x]/(x^2-x)\cong k\times k$ is semisimple. In particular is a Frobenius algebra.