If a unit and nilpotent element commute in a ring, their sum is also a unit.

Question

Let $R$ be a ring (not necessarily commutative). Let $u\in R^\times$, (so $u$ is a unit), and let $a\in R$ be a nilpotent element. Show that if $ua=au$ then $u+a\in R^\times$.

I have that if $a$ is nilpotent, then $1-a$ is a unit. Then, I know that this means $\exists b\in R$ such that $ (1-a)b=1$. Additionally I have that $\exists c\in R$ such that $uc=1$. Then I was thinking maybe I could do some kind of algebraic manipulation like $uc=(1-a)b$ but that seems to only lead to dead ends.

Any hints or pointers in the right direction appreciated.

The title seems to have nothing to do wih the actual question... — Mariano Suárez-Álvarez, Jan 20 '17 at 01:38
Again, the title does not have anything to do with the question. Moreover, the title claims something false. In $\mathbb Z$ the elements $2$ and $3$ commute and their sum is not a unit. — Mariano Suárez-Álvarez, Jan 20 '17 at 01:42
$u + a = u (1+u^{-1} a)$ so I would try to show that $u^{-1} a$ is nilpotent. — Torsten Schoeneberg, Jan 20 '17 at 01:46
Pointer : use the site's search feature first. There are at least two duplicates of this, and they show up at the very top of the most naive search. — rschwieb, Jan 20 '17 at 11:35

score 1 · Answer 1 · answered Jan 20 '17 at 01:45

1

It suffices to show that $1+\frac{a}{u}$ is a unit. Since $\frac{a}{u}$ is still nilpotent, it suffices to prove that $1+a$ is a unit for nilpotent $a$, and this follows from $$\frac{1}{1+a}=1+a+a^2+\cdots +a^n$$

answered Jan 20 '17 at 01:45

Rene Schipperus

39,526

If a unit and nilpotent element commute in a ring, their sum is also a unit.

1 Answers1