I have a question about a specific proof that all finite group schemes in characteristic 0 are etale. The proof is [here](http://www-personal.umich.edu/%7Easnowden/teaching/2013/679/L05.html), Proposition 8 in lecture notes by Andrew Snowden.


In his notation, let $A = k\oplus I$ be a (finite) local group scheme over a field $k$. For now, we are agnostic about the characteristic of the field. Let $x\_1,\dots,x\_n$ be elements in $I$ that form a basis for $\Omega\_{A/k} = I/I^2$. He defines derivations $D\_i: A \to A$ by $A \to A\otimes A \to A \otimes I/I^2 \to A$ where the first map is the comultiplication map, the second map is projection on the second factor and the third map sends $x\_i \to 1$ and the other $x\_j \to 0$ in the second factor.


He claims in the proof of the proposition that $D\_i(x\_i) = 1$ but I don't see why this has to be true. Let us do an example: Take $A = k[t]/(t-1)^p = \operatorname{Spec} \mu\_p$ in characteristic $p$ and we can take $n=1,x\_1 = t-1$. Then the comultiplication map sends $x\_1 \to x\_1\otimes x\_1 + x\_1\otimes 1 + 1\otimes x\_1$ and so $D\_1(x\_1) = x\_1+1$ if I did the computation right.


**Question 1:** Where did I go wrong?


**Question 2:** How do I geometrically interpret this proof? What is the $D\_i$ doing and why do they only exist for group schemes? I have seen other geometrically motivated proofs of this fact but I don't understand this one.



Proving that finite, connected group schemes in characteristic 0 are trivial
I don't understand that claim either.


It seems to me that if you follow the chain rule you get $$D\_i (\varphi(f)) = \sum\_{j=1}^n D\_i(x\_j) \varphi \left( \frac{\partial f}{\partial x\_j } \right). $$ You have $D\_i(x\_j) \equiv \delta\_{ij} \mod I$ so the matrix with entries $D\_i(x\_j)$ is invertible which means if $D\_i (\varphi(f))=0$ for all $i$ then $\frac{\partial f}{\partial x\_j}=0$ for all $j$ which again means that the kernel of $\varphi$ is invariant under $\frac{\partial}{\partial x\_j}$ and you can continue from there.


