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Prove $\alpha \in R[[x]]$ is a unit iff $a_0 \in R$ is a unit.

$"\Rightarrow"$ suppose $\alpha \in R[[x]]$ is a unit, where $\alpha = \sum_{n=0}^{\infty} a_nx^n$

then there exists an inverse $\beta = \sum_{n=0}^{\infty} b_nx^n$

stuck at this spot,

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

A little mixed up on how this becomes $1 = \sum_{n=0}^{\infty}e_nx^n$ where $e_0 =1, e_{n>0} = 0$ and why that would make $a_0b_0 = 1$ which would make $a_0 \in R$ a unit. I haven't tried the other direction yet.

EDIT

Maybe I get it, because we assumed this sum is a unit then its inverse exists and so the $x^0=1$ and all $a_ib_i=1$ thus $a_0b_0=1$ which gives $a_0$ is a unit in $R$

user26857
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oliverjones
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  • This looks like a duplicate of this older question. Oliver, do remember to search the site before and also immediately after posting! The site SW will compile a list of Related questions to the right margin. I picked this from there (consequently promoting it to the list of Linked questions). Leaving it to the others to decide, whether this needs to be closed as a duplicate. – Jyrki Lahtonen Mar 27 '15 at 07:11
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    Not quite a duplicate, I'd say, since this question asks for "iff" whereas the other only goes in one direction. If anything, that question should logically be subsumed under this one. But Jyrki''s advice about searching is most excellent. Cheers to all! – Robert Lewis Mar 27 '15 at 07:35

2 Answers2

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We know (if $\beta = \alpha^{-1}$) that $$ 1 = \sum_n e_n x^n = \sum_n \sum_{k=0}^n a_k b_{n-k} x^n $$ That is $$ \forall n \ge 0 : e_n = \sum_{k=0}^n a_k b_{n-k} $$ especially (let $n = 0$): $$ 1 = e_0 = a_0 b_0 $$ (hence $a_0$ is a unit).

For the other direction, suppose $a_0$ is a unit with inverse $b_0 := a_0^{-1}$. We want to define $\beta = \sum_n b_n x^n$ such that $$ 0 = \sum_{k=0}^n a_k b_{n-k}, \quad n \ge 1 $$ This is achieved by inductively defining $$ b_n := -a_0^{-1} \sum_{k=1}^n a_k b_{n-k} $$ Then $\beta := \sum_n b_n x^n$ is an inverse to $\alpha$.

martini
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    I know that for a polynomial to be a unit it is necessary that the constant term is unit and the other coefficients are nilpotent. But one can replicate the same inductive steps and show that "if the constant term is unit then the polynomial is a unit". What am I missing? – Babai Sep 22 '19 at 11:08
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    @Babai defining $b_n$ this way makes the $n$-th coefficient zero but creates higher order terms in $\alpha\beta$ which are not zero, this gets resolved by choosing an appropriate coefficient of $\beta$ for that order term. With finite degree polynomials there is no choice for that higher order term. Try defining the "inverse" this way for an easy polynomial e.g. of degree 2. It will become clear. – Jarne Renders Oct 31 '19 at 13:33
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$\alpha \beta =1$ means $(a_0 + a_1 x+\cdots)(b_0 + b_1 x+ \cdots)=1+0x+\cdots$ so $a_0b_0+ (a_1b_0+a_0 b_1)x+\cdots=1+0x+\cdots$ therefore $a_0b_0=1$

user 1
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