Invertible element can't be nilpotent?

Question

When I was reviewing cracking GRE subject mathematics 4-th edition, I was confused about the proof in Page 247.

Consider invertible element c in a Ring, $cc^{-1} = 1$. Then for any integer n, $(cc^{-1})^{n} = 1$, which means $c^{n}c^{-n} = 1$. Then $c^{n}$ can't be zero for any n, which means c can't be nilpotent.

I was wondering if $(cc^{-1})^{n} = c^{n}c^{-n}$ is satisfied for any Ring? I think it's satisfied only if it comes to a commutative ring (multiplication is commutative).

Any help would be greatly appreciated!

More generally nilpotent $\Rightarrow$ zero-divisor $\Rightarrow$ not invertible, since invertibles (units) are cancellable. — Bill Dubuque, Aug 29 '22 at 16:16
This is not accurate for another reason. In the zero ring, $0$ is both invertible and nilpotent. Of course, this is the only counterexample. — Mark Saving, Aug 29 '22 at 16:59
I guess the proof is avoiding the (far simpler?) proof by contradiction? If $c^n=0$ with $n>1$ minimal, then obviously $c^{n-1}=0$ after cancelling a copy of $c$, and you have a contradiction without any need to discuss commuting. — rschwieb, Aug 30 '22 at 17:12

score 7 · Answer 1 · answered Aug 29 '22 at 15:30

7

In any ring an invertible element commutes with its inverse.

answered Aug 29 '22 at 15:30

Ethan Bolker

103,433

score 3 · Answer 2 · answered Aug 29 '22 at 16:46

In any ring $R$, if any two elements $a,b\in R$ commutes then the identity $$(ab)^n=a^nb^n$$ holds true. It is not necessary that the whole ring should be commutative. Therefore in your case because $cc^{-1}=1=c^{-1}c$, it implies that $c$ and $c^{-1}$ commutes. So that $c^nc^{-n}=(cc^{-1})^n=1.$

score 2 · Answer 3 · answered Aug 29 '22 at 15:29

2

Both $(cc^{-1})^n = (c c^{-1}) (c c^{-1}) \cdots (c c^{-1}) = 1$ and $c^n c^{-n} = (c c \cdots c) (c^{-1} c^{-1} \cdots c^{-1}) = 1$, so naturally they are equal.

answered Aug 29 '22 at 15:29

JKL

2,189

2

Because $c$ and $c^{-1}$ commute. – anon Aug 29 '22 at 16:05
3

@runway44 This proof doesn’t use that (even though it’s true): In the first case, you simplify each pair to a $1$ and end up with a product of ones; in the second case, you simplify from the middle outwards, i.e. you place parentheses like this: $c(c(\cdots(c(cc^{-1})c^{-1})\cdots)c^{-1})c^{-1}$. – Eike Schulte Aug 30 '22 at 06:58

Invertible element can't be nilpotent?

3 Answers3