# nichols-zoeller theorem

Let $H$ be a Hopf algebra over a field $k$ with an antipode $S$. We will say that $K\subseteq H$ is a Hopf subalgebra^{} if $K$ is both subalgebra and subcoalgebra of underlaying algebra and coalgebra structures^{} of $H$, and additionaly $S(K)\subseteq K$. In particular a Hopf subalgebra $K\subseteq H$ is an algebra over $k$, so $H$ may be regarded as a $K$-module.

The Nichols-Zoeller Theorem. If $K\subseteq H$ is a Hopf subalgebra of a Hopf algebra $H$, then $H$ is free as a $K$-module. In particular, if $H$ is finite dimensional, then ${\mathrm{dim}}_{k}K$ divides ${\mathrm{dim}}_{k}H$.

Remark 1. This theorem shows that Hopf algebras are very similar^{} to groups, because this is a Hopf analogue of the Lagrange Theorem.

Remark 2. Generally this theorem does not need to hold if $H$ is only an algebra. For example, consider $H={\mathbb{M}}_{n}(k)$ the matrix algebra, where $n\ge 2$ and let $T\subseteq H$ be the upper triangular matrix^{} subalgebra. It is well known that ${\mathrm{dim}}_{k}H={n}^{2}$ and ${\mathrm{dim}}_{k}T=\frac{n(n+1)}{2}$. Of course $\frac{n(n+1)}{2}$ does not divide ${n}^{2}$ for $n\ge 2$. Thus the Nichols-Zoeller Theorem does not hold for algebras.

Title | nichols-zoeller theorem |
---|---|

Canonical name | NicholszoellerTheorem |

Date of creation | 2013-03-22 18:58:34 |

Last modified on | 2013-03-22 18:58:34 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 6 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 16W30 |