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.

  • 2
    $\begingroup$ not clear on the notation. $H$ is the Hopf algebra and $H^\ast$ is the algebra structure given by the coalgebra data on $H$? Does $H$ being cocommutative and semisimple mean $H^\ast$ is commutative and semisimple? because that answers a lot: it woudl definitely be a product of fields, but not obviously over $k$. They could be finite extensions of $k$. Unless $k$ were algebraically closed, then it would be all true. $\endgroup$
    – rschwieb
    Oct 29, 2020 at 14:34
  • $\begingroup$ I understand the A-W theorem, but I don't understand Hopf algebras. $\endgroup$
    – rschwieb
    Oct 29, 2020 at 14:35
  • 1
    $\begingroup$ 1st Question: Yes, by taking the dual map of the coproduct (with suitable restriction) as the product. The unit is the dual map of the counit. 2nd: Ah, of course, thanks! Yes, then $H^*$ is indeed commutative. That solves the problem. Cosemisimple in the finite-dimensional case just means $H^*$ is semisimple (see above). Assuming that the lecturer forgot to mention that $k$ is algebraically closed, then all falls into place neatly (c.f. Corollary 2.4.2. in "Finite dimensional algebras“ by Drozd and Kirichenko). $\endgroup$ Oct 29, 2020 at 15:20

1 Answer 1


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 $k$.


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