invertible formal power series

Theorem.  Let $R$ be a commutative ring with non-zero unity.  A formal power series

$f(X):={\displaystyle \sum _{i=0}^{\mathrm{\infty }}}{a}_i{X}^i$(1)

is invertible in the ring $R[[X]]$  iff  $a_0$ is invertible in the ring $R$.

Proof.  $1^{\circ }$.  Let $f(X)$ have the multiplicative inverse  $g(X):={\sum }_{i=0}^{\mathrm{\infty }}{b}_i{X}^i$.  Since

$$f(X)g(X)=\sum _{i=0}^{\mathrm{\infty }}\sum _{j=0}^i{a}_j{b}_{i-j}{X}^i=\mathrm{\hspace{0.33em}1},$$

we see that  $a_0{b}_0=1$, i.e. $a_0$ is an invertible element (unit) of $R$.

$2^{\circ }$.  Assume conversely that $a_0$ is invertible in $R$.  For making from a formal power series

$g(X):={\displaystyle \sum _{i=0}^{\mathrm{\infty }}}{b}_i{X}^i$(2)

the inverse of $f(X)={\sum }_{i=0}^{\mathrm{\infty }}{a}_i{X}^i$, we first choose  $b_0:=a_0^{-1}$.  For all already defined coefficients $b_0,b_1,\mathrm{\dots },b_{i-1}$ let the next coefficient be defined as

$$b_i:=-a_0^{-1}(a_1{b}_{i-1}+a_2{b}_{i-2}+\mathrm{\dots }+a_i{b}_0).$$

This equation means that

$$\sum _{j=0}^i{a}_j{b}_{i-j}=a_0{b}_i+a_1{b}_{i-1}+a_2{b}_{i-2}+\mathrm{\dots }+a_i{b}_0$$

vanishes for all  $i=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots }$;  since  $a_0{b}_0=1$,  the product of the formal power series (1) and (2) becomes simply equal to 1.  Accordingly, $f(x)$ is invertible.