$1+a$ and $1-a$ in a ring are invertible if $a$ is nilpotent

Question

Let $(A, +, \cdot)$ be a ring with $1$. An element $a\in A$ is nilpotent if there exists $n\in \mathbb{N}$ so that $a^n=0$.

Show that if $a$ is nilpotent then $1+a$ and $1-a$ are invertible.

score 4 · Answer 1 · edited Jul 25 '13 at 07:41

4

$1 = 1 - {a^n} = (1 - a)(1 + a + \cdots + {a^{n - 1}})$

If $n$ is an odd number it's the same for $1+a$: $1 = 1 + {a^n} = (1 + a)\sum\limits_{k = 0}^{n - 1} {{{( - 1)}^k}{a^k}} $.

edited Jul 25 '13 at 07:41

answered Jul 24 '13 at 23:59

Jean G

6

Perhaps you could note that you choose $n$ large enough so that $a^n=0$, and take $n$ odd so that you can apply the second formula. – paul garrett Jul 25 '13 at 00:14
2

It's not necessary to treat $1+a$ separately: just replace $a$ by $-a$. (Why can we do it?) – Jul 25 '13 at 07:43

1 Answers1