# If $a_0\in R$ is a unit, then $\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$

Let $R$ a ring, and let $$\displaystyle R[[x]]=\left\{\sum_{k=0}^{\infty}a_k x^k\;\middle\vert\; a_k\in R\right\}$$ with addition and multiplication as defined for polynomials. We have that $R[[x]]$ is a ring containing $R[x]$ as a subring.

How to prove that if $a_0\in R$ is a unit, then $\displaystyle\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$?

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Let $a=\sum_{k=0}^\infty a_kx^k\in R[[x]]$, where $a_0$ is a unit. We want to construct some $b=\sum_{k=0}^\infty b_kx^k\in R[[x]]$ such that $ab=1$, or after expanding, $$ab=a_0b_0+(a_1b_0+a_0b_1)x+\cdots=1+0x+0x^2+\cdots$$ We therefore need $b_0=a_0^{-1}$ (recall that $a_0$ is a unit). We want to have $a_1b_0+a_0b_1=0$, so our only choice for $b_1$ is $$b_1=\frac{-a_1b_0}{a_0}=-a_1a_0^{-2}.$$ We want $a_2b_0+a_1b_1+a_0b_2=0$, so we must have $$b_2=\frac{-a_2b_0-a_1b_1}{a_0}=-a_2a_0^{-2}+a_1^2a_0^{-3}.$$ Prove that, by continuing this process, you get a $b$ such that $ab=1$.

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Zev when write $b_1=-a_1a_{0}^{-1}$ you use that $R$ is conmutative to find $b_1$? –  Andres Jun 19 '12 at 21:12
That's true (I'm just accustomed to working with commutative rings) but it isn't too hard to modify what I've written to be correct for non-commutative rings too. –  Zev Chonoles Jun 20 '12 at 12:11

Hint $\rm\displaystyle\quad 1\: =\: (a-xf)(b-xg)\ \Rightarrow\ ab=1$ $$\Rightarrow\ \ \displaystyle\rm\frac{1}{a-xf}\ =\ \frac{b}{1-bxf}\ =\ b\:(1+bxf+(bxf)^2+(bxf)^3+\:\cdots\:)$$

By the way, such rings are called rings of (formal) power series.

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Bill as I use this for my exercise? –  Andres Jun 19 '12 at 20:50
@Andres I'm not sure what your question means. If you're asking if the above is rigorous, then, yes, it is, as long as you understand that the RHS converges since the order of $\rm\:(xf)^n\to\infty$ as $\rm\:n\to\infty.\:$ See here for more on convergence of formal power series. –  Bill Dubuque Jun 19 '12 at 21:11
@Andres: To use this answer for the exercise, let $a-xf$ be your power series, where $a=a_0$, $b$ is its inverse, and $f=-\sum_{k=1}^\infty a_kx^{k-1}$ (I use more or less the same idea in my answer). –  anon Jun 19 '12 at 21:48

One trick is to get the geometric series involved. We can write

$$\left(\sum_{k=0}^\infty a_k x^k\right)^{-1}=\frac{1}{a_0}\frac{1}{1+\underbrace{\left(\sum\limits_{k=1}^\infty a_0^{-1}a_kx^k\right)}_{u(x)}}. \tag{\circ}$$

Now $x|u(x)$, so in the $l$-adic topology the sequence $u(x),u(x)^2,u(x)^3,\cdots$ converges to $0$, or equivalently the terms contributed to the coefficient of a $x^n$ from these powers is gauranteed to be finite for all $n$. Thus if you expand $(\circ)$ in a geometric series, the element of $R[[x]]$ it converges to will be a legitimate multiplicative inverse (by the usual algebra establishing the geo sum formula).

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