Prove $\alpha \in R[[x]]$ is a unit iff $a_0 \in R$ is a unit

Prove $\alpha \in R[[x]]$ is a unit iff $a_0 \in R$ is a unit.

$"\Rightarrow"$ suppose $\alpha \in R[[x]]$ is a unit, where $\alpha = \sum_{n=0}^{\infty} a_nx^n$

then there exists an inverse $\beta = \sum_{n=0}^{\infty} b_nx^n$

stuck at this spot,

$(\sum_{n=0}^{\infty} a_nx^n)(\sum_{n=0}^{\infty} b_nx^n) = \sum_{n=0}^{\infty}(\sum_{k=0}^{n}(a_kb_{n-k}))x^n$

A little mixed up on how this becomes $1 = \sum_{n=0}^{\infty}e_nx^n$ where $e_0 =1, e_{n>0} = 0$ and why that would make $a_0b_0 = 1$ which would make $a_0 \in R$ a unit. I haven't tried the other direction yet.

EDIT

Maybe I get it, because we assumed this sum is a unit then its inverse exists and so the $x^0=1$ and all $a_ib_i=1$ thus $a_0b_0=1$ which gives $a_0$ is a unit in $R$

Solution 1:

We know (if $\beta = \alpha^{-1}$) that $$ 1 = \sum_n e_n x^n = \sum_n \sum_{k=0}^n a_k b_{n-k} x^n $$ That is $$ \forall n \ge 0 : e_n = \sum_{k=0}^n a_k b_{n-k} $$ especially (let $n = 0$): $$ 1 = e_0 = a_0 b_0 $$ (hence $a_0$ is a unit).

For the other direction, suppose $a_0$ is a unit with inverse $b_0 := a_0^{-1}$. We want to define $\beta = \sum_n b_n x^n$ such that $$ 0 = \sum_{k=0}^n a_k b_{n-k}, \quad n \ge 1 $$ This is achieved by inductively defining $$ b_n := -a_0^{-1} \sum_{k=1}^n a_k b_{n-k} $$ Then $\beta := \sum_n b_n x^n$ is an inverse to $\alpha$.

Solution 2:

$\alpha \beta =1$ means $(a_0 + a_1 x+\cdots)(b_0 + b_1 x+ \cdots)=1+0x+\cdots$ so $a_0b_0+ (a_1b_0+a_0 b_1)x+\cdots=1+0x+\cdots$ therefore $a_0b_0=1$