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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?

Wang Kah Lun
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    An integral domain is a ring with some ~ bonus property ~ that says zero divisors don't exist. Then any proof that works for all rings will also work for integral domains. This is akin to proving something for all primes, and then asking if the same proof works for odd primes. Of course it does! And the same is true of your question ^_^ – HallaSurvivor Nov 04 '20 at 08:01

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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. $a_0\in R^\times$.

reuns
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