The unit group of a finite dimensional associative algebra is a Lie group?

Question

I am reading Serre's "Lie algebras and Lie groups" p.103. Let $k$ be a complete valued field(for example $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{Q}_p$) and $R$ be a finite dimensional associative $k$-algebra. Surely $R$ is an additive Lie group. The book asserts that the unit group $G_m(R)$ is a multiplicative Lie group and also contains the proof, but I cannot understand it. I copy the text here.

"We contend that $G_m(R)$ is an analytic group which is open as a subset of $R$. To show that $G_m(R)$ is open in $R$ it suffices to show that there is a neighborhood of $1$ contained in $G_m(R)$. Now, there exists an open neighborhood $U$ of $0$ in $R$ such that for $x \in U$ the series $\sum x^n$ converges. It follows $V=\{1-x:x \in U\} \subset G_m(R)$ and $V$ is a neighborhood of $1$. To show that $G_m(R)$ is an analytic group it remains to show that multiplication is a morphism. This follows since multiplication in $R$ is bilinear."

I cannot understand the first step and the final step:

1. Why does there exist an open set $U$ which satisfies $\sum x^n$ converges?
2. Why is multiplication a manifold morphism? (Also, It seems that we need $x\mapsto x^{-1}$ is a morphism.)

From googling, I've found (ex1) of http://www.math.cornell.edu/~sjamaar/classes/6520/problems/2016-10-26.pdf , but still I cannot solve it.

Answer 1

1. For $x$ of norm $<1$, $(\displaystyle\sum_{k=0}^n x^k)_n$ is a Cauchy sequence (by the triangle inequality) , thus by completeness of $k$ and finite dimensionality of $R$, it converges in $R$. The open neighbourhood is $||x||<1$.

2. Multiplication $R\times R\to R$ is bilinear so it must be smooth, thus so is its restriction to $G_m(R)$.

It remains the question of why $x\mapsto x^{-1}$ is smooth. First of all note that for $y=1-x\in V$, $y^{-1}= \sum_n x^n$, so for $y\in V$, $y^{-1}=\sum_n (1-y)^n$, thus the inversion is smooth on a neighbourhood of $1$.

If $x\in G_m(R)$, and $y\in xV$, then $y=xv$ for some $v\in V$ and $y^{-1}= v^{-1}x^{-1}= (\sum_n (1-v)^n)x^n$, so $y^{-1}= (\sum_n (1-x^{-1}y)^n)x^{-1}$ so it is smooth as well