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I am trying to prove a result, for which I have got one part, but I am not able to get the converse part.

Theorem. Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\cdots,a_{n}$ are all nilpotent in $R$.

Proof. Suppose $f(X)=a_{0}+a_{1}X+\cdots +a_{n}X^{n}$ is such that $a_{0}$ is a unit in $R$ and $a_{1},a_{2}, \cdots,a_{r}$ are all nilpotent in $R$. Since $R$ is commutative, we get that $a_{1}X,a_{2}X^{2},\cdots,a_{n}X^{n}$ are all nilpotent and hence also their sum is nilpotent. Let $z = \sum a_{i}X^{i}$ then $a_{0}^{-1}z$ is nilpotent and so $1+a_{0}^{-1}z$ is a unit. Thus $f(X)=a_{0}+z=a_{0} \cdot (1+a_{0}^{-1}z)$ is a unit since product of two units in $R[X]$ is a unit.

I have not been able to get the converse part and would like to see the proof for the converse part.

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If $x$ is nilpotent then $1-x$ is a unit. –  PEV Jan 26 '11 at 20:47
    
@Arturo: May i know, why you edited my question? –  anonymous Jan 26 '11 at 20:48
    
@Chandru1: Why not check the edit history? You had is a unit in $R[X]$ is a unit in $R[X]$. I removed the unnecessary second "is a unit in $R[X]$". –  Arturo Magidin Jan 26 '11 at 20:49
    
@Arturo: Oh, sorry i didn't know that. –  anonymous Jan 26 '11 at 20:50
    
@Arturo: Anyhow, thanks for making the correction. –  anonymous Jan 26 '11 at 20:50

3 Answers 3

up vote 8 down vote accepted

If $R$ is a domain then easily $f(X)$ a unit implies that $a_i = 0$ for $i>0$. Now $R\to R/\mathfrak p$, for $\mathfrak p$ prime, reduces to the domain case, yielding that the $a_i$, $i>0$ are in every prime ideal. But the intersection of all prime ideals is the nilradical, the set of all nilpotent elements - as you proved a few days ago.

See also my post here on reduction to domains by factoring out prime ideals.

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Let $f=\sum_{k=1}^n a_kX^k, g= \sum_{k=1}^m b_kX^k$. If $f g=1$ then clearly $a_0,b_0$ are units and:

$$a_nb_m=0$$ $$a_{n-1}b_m+a_nb_{m-1}=0 \Rightarrow (a_n)^2b_{m-1}=0 $$ $$a_{n-2}b_m+a_{n-1}b_{m-1}+a_nb_{m-2}=0 \Rightarrow (a_n)^3b_{m-2}=0 $$ $$.....$$ $$.....+a_{n-2}b_2+a_{n-1}b_1+a_nb_0=0 \Rightarrow (a_n)^{m+1}b_{0}=0 $$

Since $b_0$ is an unit it follows that $(a_n)^{m+1}=0$.

We proved that $a_n$ is nilpotent. But is enough. Indeed, $f$ invertible, $a_nx^n$ is nilpotent implies $f-a_nX^n$ is unit and we can repeat (or better induction by $\deg(f)$.

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Why does $a_{n-1}b_m+a_nb_{m-1}=0 \Rightarrow (a_n)^2b_{m-1}=0$ ? –  Nethesis Oct 3 at 0:45
    
Wait, got it, though surely that is only true if $a_n \in (Z)R$? –  Nethesis Oct 3 at 0:46
    
@Nethesis If by $Z$ you mean the center, it is probably because the ring is given commutative ;) –  N. S. Oct 3 at 1:15
    
Ah right, sorry, I was looking for an answer about a non commutative ring and din't see that in the question, my bad @N.S. –  Nethesis Oct 5 at 17:40
    
@Nethesis No problem. I am not to familiar with polynomials in non-commutative rings, how do you define multiplication? What is $(ax) \cdot (bx^2)$? –  N. S. Oct 5 at 18:05

Let $f(x) = x^n a_n + x^{n-1}a_{n-1} + \dots + a_0$ be a unit, i.e. $$\exists g(x) \in R[x] : f(x) g(x) = 1 \forall x \in R$$

Let $g(x) = x^m b_m + x^{m-1}b_{m-1} + \dots + b_0$ then $$f(x)g(x) = x^{m + n}a_n b_m + \dots + a_0 b_0 = 1 \forall x \in R$$

In particular, for $x = 0$ it follows that $a_0 b_0 = 1$, so it follows that $a_0$ is a unit.

Will try and add the rest of the proof later.

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Yes; but why are $a_1,\ldots,a_n$ nilpotent? –  Arturo Magidin Jan 26 '11 at 21:05
    
One can use a backwards induction. The element $a_n$ is nilpotent because $a_n b_m = 0$. What about $a_{n-1}$? Well, the coefficient of $x^{n-1}$ is $0 = a_{n-1}b_1 + a_n b_0$. Therefore $a_{n-1}b_1 = -a_n b_0$. So $a_{n-1}b_1b_m = -a_nb_0 b_m = 0$, so the element $b_1b_m$ kills $a_{n-1}$. The inductive step works very similarly. –  Jason DeVito Jan 26 '11 at 21:26
    
@JasonDeVito , why $a_n b_m=0$ implies that $a_n$ is nilpotent ? can you explain this more plz ? –  Maths Lover May 15 '13 at 15:30
    
@MathsLover Yeah, you got a crucial point here. –  Hagen von Eitzen May 15 '13 at 16:03
    
@HagenvonEitzen what do u mean ? –  Maths Lover May 15 '13 at 16:04

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