# invertible formal power series

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

 $\displaystyle f(X)\;:=\;\sum_{i=0}^{\infty}a_{i}X^{i}$ (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  $g(X):=\sum_{i=0}^{\infty}b_{i}X^{i}$.  Since

 $f(X)g(X)\;=\;\sum_{i=0}^{\infty}\sum_{j=0}^{i}a_{j}b_{i-j}X^{i}\;=\;1,$

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

 $\displaystyle g(X):=\sum_{i=0}^{\infty}b_{i}X^{i}$ (2)

the inverse        of $f(X)=\sum_{i=0}^{\infty}a_{i}X^{i}$, we first choose  $b_{0}:=a_{0}^{-1}$.  For all already defined coefficients $b_{0},\,b_{1},\,\ldots,\,b_{i-1}$ let the next coefficient be defined as

 $b_{i}\;:=\;-a_{0}^{-1}(a_{1}b_{i-1}\!+\!a_{2}b_{i-2}\!+\ldots+\!a_{i}b_{0}).$

This equation means that

 $\sum_{j=0}^{i}a_{j}b_{i-j}\;=\;a_{0}b_{i}\!+\!a_{1}b_{i-1}\!+\!a_{2}b_{i-2}\!+% \ldots+\!a_{i}b_{0}$

vanishes for all  $i=1,\,2,\,\ldots$;  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.

Title invertible formal power series InvertibleFormalPowerSeries 2016-04-27 10:47:14 2016-04-27 10:47:14 pahio (2872) pahio (2872) 8 pahio (2872) Theorem msc 13H05 msc 13F25 msc 13J05 RulesOfCalculusForDerivativeOfFormalPowerSeries