$1-a$ is a unit in the ring $R$

Question

I have $R$ a commutative ring with unity and $a\in R$ such that $a^3=0$. How do I show that $1-a$ is a unit?

If $a=0$, we are done, so I assume that $a\neq 0$, then can I say that $a$ and $a^2$ are zero divisors?

Also, if there exists $b\in R$ such that $(1-a)b=1$, what should I do with this?

I don't really have much light on this one; can someone help me out, please?

We don't need that $R$ is commutative. In fact, everything takes place in $\mathbb{Z}[a]$ anyway. — Martin Brandenburg, Apr 22 '13 at 08:02
@IvordesGreenleaf have you ever seen the formula for the sum of a geometric series? $\frac{1-r^{n-1}}{1-r}=1+r+r^2+\dots+r^n$ — obataku, Apr 22 '13 at 15:39
Yes, I've seen this, are you saying that I can use this as another approach? — Akaichan, Apr 22 '13 at 19:39

score 9 · Accepted Answer · answered Apr 22 '13 at 03:10

9

We have$$(1-a)(a^2+a+1)=-a^3+1=1.$$

answered Apr 22 '13 at 03:10

Clayton

24,751

1

wooow... >_< I kinda feel ashamed now :( – Akaichan Apr 22 '13 at 03:13
@IvordesGreenleaf: Don't worry about it! It happens :) – Clayton Apr 22 '13 at 03:17

score 1 · Answer 2 · edited Apr 22 '13 at 15:36

Also, if there exists a $b$ such that $(1-a)b=1$, what should I do with this?

That's a nice idea how to come up with the inverse.

If $(1-a)b=1$, then $b=1+ab$, hence $b=1+a(1+ab)$, hence $b=1+a(1+a(1+ab))=1+a+a^2$ since $a^3=0$. Now you have to check that, in fact, $1+a+a^2$ is the inverse.

More generally, if $a^n=0$, then $1+a+\dotsc+a^{n-1}$ is inverse to $1-a$.

$1-a$ is a unit in the ring $R$

2 Answers2