Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $a_{1},a_{2},\dots,a_{n}$ 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 $a_{1},a_{2}, \dots,a_{r}$ 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.

share|cite|improve this question
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
up vote 10 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.

share|cite|improve this answer

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 $(a_n)^{m+1}=0$.

Hence, we proved that $a_n$ is nilpotent, but this is enough. Indeed, since $f$ is invertible, $a_nx^n$ being nilpotent implies that $f-a_nX^n$ is unit and we can repeat (or more rigorously, perform induction on $\deg(f)$).

share|cite|improve this answer
Why does $a_{n-1}b_m+a_nb_{m-1}=0 \Rightarrow (a_n)^2b_{m-1}=0$ ? – Nethesis Oct 3 '14 at 0:45
Wait, got it, though surely that is only true if $a_n \in (Z)R$? – Nethesis Oct 3 '14 at 0:46
@Nethesis If by $Z$ you mean the center, it is probably because the ring is given commutative ;) – N. S. Oct 3 '14 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 '14 at 17:40
@zed111 In a commutative ring you have unit-nilpotent=unit. The proof is easy: if u is unit and v is nilpotent, then $v^n=0$ which means $$u^n=u^n-v^n=(u-v)(u^{n-1}+u^{n-2}v+...+v^{n-1})$$ This shows that $$(u-v)(u^{n-1}+u^{n-2}v+...+v^{n-1})(u^{-1})^n=1$$ – N. S. Apr 20 '15 at 15:47

protected by user26857 Nov 21 '15 at 23:01

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?