Comparing proof of $R[[x]]^*$ consists of $\sum_{n \geq 0}a_{n}x^n (a_{n} \in R) ,a_{0} \in R^* ,R$ an integral domain to $R$ a ring.

Question

I want to answer letter $(b)$ in this question:

Let $R$ be an integral domain and let $R[[x]]$ be the corresponding ring of formal power series.

$(a)$ Show that $R[[x]]$ is an integral domain.

$(b)$Show that $R[[x]]^*$ consists of the series $\sum_{n \geq 0}a_{n}x^n (a_{n} \in R)$ such that $a_{0} \in R^*.$

My question is:

I found a solution here If $a_0\in R$ is a unit, then $\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$ in case of $R$ a ring, I am wondering how will my solution differ as my $R$ is an integral domain?

Answer 1

Is it clear to you that for $R$ any unital ring (it may be non-commutative in which case $x$ commutes with the elements of $R$) and $f=\sum_{n\ge 0}a_n x^n\in R[[x]]$,

If $a_0\in R^\times$ then $$g=\sum_{k\ge 0} (1-a_0^{-1}f)^k\in R[[x]]$$ and (using that $a_0^{-1}f$ commutes with $1-a_0^{-1}f$) $$a_0^{-1}fg= 1$$ thus $ga_0^{-1} = f^{-1}$ ?

Conversely if $(\sum_{n\ge 0}a_n x^n)(\sum_{n\ge 0}b_n x^n)=1$ then $a_0b_0=1$ ie. $a_0\in R^\times$.