1
$\begingroup$

I'm having trouble solving the part C of the problem on Dummit and Foote 7.2.3

Prove that $$\alpha = \sum_{n=0}^\infty a_nx^n$$ is a unit in $R[x]$ if and only if $a_0$ is a unit in $R$.

I had no trouble proving the direction where we assume $\sum_{n=0}^\infty = a_nx^n$ is a unit in $R$, but I have no clue about how to go about the other direction.

My approach was to construct an element $\beta$ in $R[[x]]$ such that $\alpha\beta = 1$, but not sure how to proceed to do so.

Any help would be greatly appreciated!

  • 0
    Use the definition of multiplication for power series. – 2017-01-18
  • 0
    @Masacroso I understand I need to do that but I'm not sure how to explicitly define the coefficients of $\beta$ – 2017-01-18
  • 1
    I think you can use the definition of multiplication for power series, and you can explicitly define the coefficients of $\beta$, albeit recursively. Should get something like $b_{0}=a_{0}^{-1}$, $b_{n}=\sum_{i=1}^{n}-b_{0}a_{i}b_{n-i}$. – 2017-01-18
  • 0
    Oh, ok... I think that what you call unit is the same as unity... I dont know this concept, sorry. – 2017-01-18
  • 1
    https://en.wikipedia.org/wiki/Formal_power_series#Inverting_series – 2017-01-18

1 Answers 1

2

For $f(x) \in R[[x]]$ be a unit in $R[[x]]$, it must exists $g(x) \in R[[x]]$ such that

$$f(x)g(x)=(\sum_{n=0}^{\infty}a_nx^n)(\sum_{n=0}^{\infty}b_nx^n)=\sum_{n=0}^{\infty}(\sum_{k=0}^{n}a_kb_{n-k})x^n=1.$$

We see that this is possible in and only if $a_0b_0=1$, $a_0b_1+a_1b_0=0$, ..., $\sum_{k=0}^{n}a_{k}b_{n-k}=0$, for all $n\in\mathbb{N}$. So, this is possible if and only if $a_0$ is a unit in $R$, with inverse $b_0$. In that caseWe solve $a_0b_0=1$, then we solve $a_0b_1+a_1b_0=0$, and go on...