Each element of an algebraic group has an nth root

Question

I have a problem with this exercise form Shafarevich Basic Algebraic Geometry

Let G be an algebraic group and suppose it to be abelian. Define $\varphi_n:G\rightarrow G$ by $\varphi_n(g)=g^n$. Supposing that the ground field has characteristic 0, prove that d$_e\varphi_n$ is a nondegenerate linear map. Deduce from this that in a Abelian algebraic group the number of elements of order n is finite, and that every element has an nth root.

I cannot realize how to determine d$_e\varphi_n$, is it just $ng^{n-1}$?

Please help!

Answer 1

The map $\phi_n$ is the composition $$G \stackrel{\Delta}{\longrightarrow} G^n \stackrel{m}{\longrightarrow} G,$$ where $\Delta$ is the diagonal map and $m$ is the multiplication map. So $$d_e \phi_n = d_e m \circ d_e \Delta.$$ Now, $T_e G^n = (T_e G)^n$, and it is easy to see that $d_e \Delta$ is the diagonal inclusion and $d_e m$ is the sum map. So $d_e \phi_n$ is simply multiplication by $n$.