Kahler differentials of a Hopf Algebra

Question

Let $A$ be a $k$-Hopf Algebra over some ring $k$, with augmentation ideal $J_A=$ ker $(\epsilon:A\rightarrow k)$

I would like to prove that the module of Khaler Differentials $\Omega_A$ of $A$ over $k$ is isomorphic to the tensor product $A\otimes_k J/J^2$.

I found this theorem in W.C. Waterhouse Intro to Aff. Group Schemes (11.3 pag 85), but I don't get his proof.

Related question: Kernel of an algebra map and module of Kahler Differentials

Answer 1

By this result we get an isomorphism $\Omega_A \otimes_\epsilon k = J/J^2 $, and from this you get by tensoring for $A$ the isomorphism $\Omega_A = A \otimes_k J/J^2$.

Answer 2

If $H$ is a Hopf algebra and $M$ an $H$-module, you can view $H\otimes M$ with diagonal action $(h\cdot(k\otimes m)=h_1k\otimes h_2m$) or as free left $H$-module (that is, $h\cdot (k\otimes m)=hk\otimes m$). In the second version, the module structure of $M$ doesn't have any role... nevertheless, both structure are isomorphic. For that, let us denote $V_M$ the underlying vector space of $M$ and denote $H\otimes V_M$ the second $H$-module structure on $H\otimes M$. The map $$H\otimes M\to H\otimes V_M$$ $$k\otimes m\mapsto k_1\otimes S(k_2)m$$ realizes the desired isomorphism. One can check that the inverse is $h\otimes m\mapsto h_1\otimes h_2m$.

With this in mind, you can consider $m:H\otimes H\to H$ given by $m(h\otimes h')=hh'$, and you can view it under the above isomorphism and wonder who -in $H\otimes V_H$- corresponds to the Kernel of the multiplication map $\subset H\otimes H$, and you will see that it is $Ker(\epsilon\otimes H)$.

If your Hopf algebra is commutative, then the Kernel of the multiplication is an ideal, look at its square under this isomorphism and you are done. Or, maybe I did the right-sided version of what you wanted..