# Artin-Wedderburn: Decomposition of a semisimple Dual Hopf algebra

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.

• 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. – rschwieb Oct 29 '20 at 14:34
• I understand the A-W theorem, but I don't understand Hopf algebras. – rschwieb Oct 29 '20 at 14:35
• 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). – M.C. Oct 29 '20 at 15:20

## 1 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 $$k$$.