1. Context
My lecture notes prove that any cocommutative finite-dimensional Hopf algebra over a field $k$ of characteristic zero is semisimple and cosemisimple.
They try to argue from there that any finite-dimensional, cocommutative Hopf algebra over a field of characteristic zero is isomorphic to a group algebra:
Since $H^*$ is semisimple, it is, as an algebra, isomorphic to $H^* \cong k \times. . . \times k$ by the Artin-Wedderburn theorem. The projection $p_i$ to the $i$-th factor is a morphism of algebras or, put differently, a grouplike element in $H^{**} \cong H$. All projections give a basis of $H$ consisting of grouplike elements. Thus $H$ is a group algebra of a finite group.
2. Question
- Why does the isomorphism $H^* \cong k \times. . . \times k$ exist? Where is the Artin-Wedderburn theorem used?
The Artin-Wedderburn theorem gives an isomorphism $H^* \cong \prod M_{n_i}(D_i)$ where the $n_{i}$ are natural numbers, the $D_i$ are finite dimensional division algebras over $k$ and $M_{n_i}(D_i) $ is the algebra of $n_i \times n_i $matrices over $D_i$. If $k$ were algebraically closed we would even know that $H^* \cong \prod M_{n_i}(k)$ holds. How to proceed? I am not familiar with Artin-Wedderburn, I guess. So any hint would be appreciated.
Best Answer
Since you confirmed it in the comment above:
It sounds like the intended context was an algebraically closed field, and that $H^\ast$ is a commutative semisimple ring.
Using the Artin-wedderburn theorem it is easy to see why all the matrices must have side length $1$ in order to be commutative. That makes a product of fields, each one a finite field extension of $k$. But adding algebraiclly closed means that these fields are in fact just $k$.